PHY2403H Quantum Field Theory. Lecture 10: Lorentz boosts, generator of spacetime translation, Lorentz invariant field representation. Taught by Prof. Erich Poppitz

DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

Lorentz transform symmetries.

From last time, recall that an infinitesimal Lorentz transform has the form
\label{eqn:qftLecture10:20}
x^\mu \rightarrow x^\mu + \omega^{\mu\nu} x_\nu,

where
\label{eqn:qftLecture10:40}
\omega^{\mu\nu} = -\omega^{\nu\mu}

We showed last time that $$\omega^{ij}$$ induces a rotation, and will show today that $$\omega^{0i}$$ is a boost.

We introduced a three index current, factoring out explicit dependence on the incremental Lorentz transform tensor $$\omega^{\mu\nu}$$ as follows
\label{eqn:qftLecture10:80}
J^{\nu \mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^\mu T^{\nu\rho} },

and can easily show that this current has the desired zero four-divergence property
\label{eqn:qftLecture10:100}
\begin{aligned}
\partial_\nu J^{\nu \mu\rho}
&= \inv{2} \lr{
(\partial_\nu x^\rho) T^{\nu\mu}
+
x^\rho {\partial_\nu T^{\nu\mu} }
– (\partial_\nu x^\mu) T^{\nu\rho}
– x^\mu {\partial_\nu T^{\nu\rho} }
} \\
&= \inv{2} \lr{
(\partial_\nu x^\rho) T^{\nu\mu}
– (\partial_\nu x^\mu) T^{\nu\rho}
} \\
&= \inv{2} \lr{
T^{\rho\mu}
+
– T^{\mu\rho}
} \\
&= 0,
\end{aligned}

since the energy-momentum tensor is symmetric.

Defining charge in the usual fashion $$Q = \int d^3 x j^0$$, so we can define a charge for each pair of indexes $$\mu\nu$$, and in particular
\label{eqn:qftLecture10:120}
Q^{0k} = \int d^3 x J^{0 0 k} = \inv{2} \int d^3 x \lr{ x^k T^{00} – x^0 T^{0k} }

\label{eqn:qftLecture10:540}
\begin{aligned}
\dot{Q}^{0k}
&= \int d^3 x \dot{J}^{0 0k} \\
&= \inv{2} \int d^3 x \lr{ x^k \dot{T}^{00} – x^0 \dot{T}^{0k} }
\end{aligned}

However, since $$0 = \partial_\mu T^{\mu \nu} = \dot{T}^{0 \nu} + \partial_j T^{j \nu}$$, or $$\dot{T}^{0 \nu} = -\partial_j T^{j \nu}$$,
\label{eqn:qftLecture10:560}
\begin{aligned}
\dot{Q}^{0k}
&= \inv{2} \int d^3 x \lr{ x^k (-\partial_j T^{j0}) – T^{0k} – x^0 (-\partial_j T^{jk}) } \\
&= \inv{2} \int d^3 x \lr{
\partial_j (-x^k T^{j0}) + (\partial_j x^k) T^{j0}
– T^{0k} + x^0 \partial_j T^{jk}
} \\
&= \inv{2} \int d^3 x \lr{
\partial_j (-x^k T^{j0}) + {T^{k0}}
– {T^{0k}} + x^0 \partial_j T^{jk}
} \\
&= \inv{2} \int d^3 x \lr{
\partial_j (-x^k T^{j0})
+ x^0 \partial_j T^{jk}
} \\
&= \inv{2} \int d^3 x
\partial_j \lr{
-x^k T^{j0}
+ x^0 T^{jk}
},
\end{aligned}

which leaves just surface terms, so $$\dot{Q}^{0k} = 0$$.

Quantizing:

From our previous identification
\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,

we have
\label{eqn:qftLecture10:580}
T^{\nu\mu} = \partial^\nu \phi \partial^\mu \phi – g^{\nu\mu} \LL,

or
\label{eqn:qftLecture10:600}
\begin{aligned}
T^{00}
&= \partial^0 \phi \partial^0 \phi – \inv{2} \lr{ \partial_0 \phi \partial^0 \phi + \partial_k \phi \partial^k \phi } \\
&= \inv{2} \partial^0 \phi \partial^0 \phi – \inv{2} (\spacegrad \phi)^2,
\end{aligned}

and
\label{eqn:qftLecture10:620}
T^{0k} = \partial^0 \phi \partial^k \phi,

so we may quantize these energy momentum tensor components as
\label{eqn:qftLecture10:640}
\begin{aligned}
\hatT^{00} &= \inv{2} \hat{\pi}^2 + \inv{2} (\spacegrad \phihat)^2 \\
\hatT^{0k} &= \inv{2} \hat{\pi} \partial^k \phihat.
\end{aligned}

We can now start computing the commutators associated with the charge operator. The first of those commutators is
\label{eqn:qftLecture10:140}
\antisymmetric{\hatT^{00}(\Bx)}{\phihat(\By)}
=
\inv{2}
\antisymmetric{\hat{\pi}^2(\Bx)}{\phihat(\By)},

which can be evaluated using the field commutator analogue of $$\antisymmetric{F(p)}{q} = i F’$$ which is
\label{eqn:qftLecture10:660}
\antisymmetric{F(\hat{\pi}(\Bx))}{\phihat(\By)} = -i \frac{dF}{d \hat{\pi}} \delta(\Bx – \By),

to give
\label{eqn:qftLecture10:680}
\antisymmetric{\hatT^{00}(\Bx)}{\phihat(\By)}
= -i \delta^3(\Bx – \By) \hat{\pi}(\Bx)

The other required commutator is
\label{eqn:qftLecture10:160}
\begin{aligned}
\antisymmetric{\hatT^{0i}(\Bx)}{\phihat(\By)}
&=
\antisymmetric{\hat{\pi}(\Bx)\partial^i \phihat(\Bx)}{\phihat(\By)} \\
&=
\partial^i \phihat(\Bx)
\antisymmetric{\hat{\pi}(\Bx)
}{\phihat(\By)} \\
&= -i \delta^3(\Bx – \By) \partial^i \phihat(\Bx),
\end{aligned}

The charge commutator with the field can now be computed
\label{eqn:qftLecture10:180}
\begin{aligned}
i \epsilon \antisymmetric{\hatQ^{0k}}{\phihat(\By)}
&=
i
\frac{\epsilon}{2} \int d^3 x
\lr{
x^k
\antisymmetric{\hatT^{00}}{\phihat(\By)}

x^0
\antisymmetric{\hatT^{0k}}{\phihat(\By)}
} \\
&=
\frac{\epsilon}{2} \lr{ y^k \hat{\pi}(\By) – y^0 \partial^k \phihat(\By) } \\
&=
\frac{\epsilon}{2} \lr{ y^k \dot{\phihat}(\By) – y^0 \partial^k \phihat(\By) },
\end{aligned}

so to first order in $$\epsilon$$
\label{eqn:qftLecture10:200}
e^{i \epsilon \hatQ^{0k} } \phihat(\By)
e^{-i \epsilon \hatQ^{0k} }
=
\phihat(\By)
+ \frac{\epsilon}{2} y^k \dot{\phihat}(\By)
+ \frac{\epsilon}{2} y^0 \partial_k \phihat(\By)

For example, with $$k = 1$$
\label{eqn:qftLecture10:700}
\begin{aligned}
e^{i \epsilon \hatQ^{0k} } \phihat(\By)
e^{-i \epsilon \hatQ^{0k} }
&=
\phihat(\By)
+ \frac{\epsilon}{2} \lr{
y^1 \dot{\phihat}(\By)
+
y^0 \PD{y^1}{\phihat}(\By)
} \\
&=
\phihat(y^0 + \frac{\epsilon}{2} y^1,
y^1 + \frac{\epsilon}{2} y^2, y^3).
\end{aligned}

This is a boost. If we compare explicitly to an infinitesimal Lorentz transformation of the coordinates
\label{eqn:qftLecture10:220}
\begin{aligned}
x^0 \rightarrow x^0 + \omega^{01} x_1 &= x^0 – \omega^{01} x^1 \\
x^1 \rightarrow x^1 + \omega^{10} x_0 &= x^1 – \omega^{01} x_0 = x^1 – \omega^{01} x^0
\end{aligned}

we can make the identification
\label{eqn:qftLecture10:240}
\frac{\epsilon}{2} = – \omega^{01}.

We now have the explicit form of the generator of a spacetime translation

\label{eqn:qftLecture10:260}
\boxed{
\hatU(\Lambda) = \exp\lr{-i \omega^{0k} \int d^3 x \lr{ \hatT^{00} x^k – \hatT^{0k} x^0 }}
}

An explicit boost along the x-axis has the form
\label{eqn:qftLecture10:300}
\hatU(\Lambda) \phihat(t, \Bx)
\hatU^\dagger(\Lambda)
=
\phihat\lr{ \frac{t – vx}{\sqrt{1 – v^2}}, \frac{x – vt}{\sqrt{1 – v^2}}, y, z },

and more generally
\label{eqn:qftLecture10:320}
\hatU(\Lambda) \phihat(x) \hatU^\dagger(\Lambda) =
\phihat(\Lambda x)

where $$x$$ is a four vector, $$(\Lambda x)^\mu = {{\Lambda}^\mu}_\nu x^\nu$$, and $${{\Lambda}^\mu}_\nu \approx {{\delta}^\mu}_\nu + {{\omega}^\mu}_\nu$$.

Transformation of momentum states

In the momentum space representation

\label{eqn:qftLecture10:340}
\begin{aligned}
\phihat(x)
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}} \lr{
e^{i (\omega_\Bp t – \Bp \cdot \Bx)} \hat{a}_\Bp
+
e^{-i (\omega_\Bp t – \Bp \cdot \Bx)} \hat{a}^\dagger_\Bp
} \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}} \evalbar{
\lr{
e^{i p^\mu x^\mu } \hat{a}_\Bp
+
e^{-i p^\mu x^\mu } \hat{a}^\dagger_\Bp
}
}{p_0 = \omega_\Bp}
\end{aligned}

\label{eqn:qftLecture10:720}
\begin{aligned}
\hatU(\Lambda) \phihat(x) \hatU^\dagger(\Lambda)
&=
\phihat(\Lambda x) \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}} \evalbar{
\lr{
e^{i p^\mu {{\Lambda}^\mu}_\nu x^\nu }
\hat{a}_\Bp
+
e^{-i p^\mu {{\Lambda}^\mu}_\nu x^\nu } \hat{a}^\dagger_\Bp
}
}{p_0 = \omega_\Bp}
\end{aligned}

This can be put into an explicitly Lorentz invariant form
\label{eqn:qftLecture10:n}
\begin{aligned}
\phihat(\Lambda x)
&=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta(p_0^2 – \Bp^2 – m^2) \Theta(p^0) \sqrt{2 \omega_\Bp}
e^{i p^\mu {{\Lambda}^\mu}_\nu x^\nu }
\hat{a}_\Bp + \text{h.c.} \\
&=
\int \frac{dp^0 d^3 p}{(2\pi)^3}
\lr{
\frac{\delta(p_0 – \omega_\Bp)}{2 \omega_\Bp}
+
\frac{\delta(p_0 + \omega_\Bp)}{2 \omega_\Bp}
}
\Theta(p^0) \sqrt{2 \omega_\Bp} \hat{a}_\Bp + \text{h.c.},
\end{aligned}

which recovers \ref{eqn:qftLecture10:720} by making use of the delta function identity $$\delta(f(x)) = \sum_{f(x_\conj) = 0} \frac{\delta(x – x_\conj)}{f'(x_\conj)}$$, since the $$\Theta(p^0)$$ kills the second delta function.

We now have a more explicit Lorentz invariant structure
\label{eqn:qftLecture10:380}
\phihat(\Lambda x)
=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta(p_0^2 – \Bp^2 – m^2) \Theta(p^0) \sqrt{2 \omega_\Bp}
e^{i p^\mu {{\Lambda}^\mu}_\nu x^\nu }
\hat{a}_\Bp + \text{h.c.}

Recall that a boost moves a spacetime point along a parabola, such as that of fig. 1, whereas a rotation moves along a constant “circular” trajectory of a hyper-paraboloid. In general, a Lorentz transformation may move a spacetime point along any path on a hyper-paraboloid such as the one depicted (in two spatial dimensions) in fig. 2. This paraboloid depict the surfaces of constant energy-momentum $$p^0 = \sqrt{ \Bp^2 + m^2 }$$. Because a Lorentz transformation only shift points along that energy-momentum surface, but cannot change the sign of the energy coordinate $$p^0$$, this means that $$\Theta(p^0)$$ is also a Lorentz invariant.

fig. 1. One dimensional spacetime surface for constant (p^0)^2 – p^2 = m^2.

fig. 2. Surface of constant squared four-momentum.

Let’s change variables
\label{eqn:qftLecture10:400}
p^\lambda = {{\Lambda}^\lambda}_\rho {p’}^{\rho}

so that
\label{eqn:qftLecture10:420}
\begin{aligned}
p_\mu
{{\Lambda}^\mu}_\nu x^\nu
&=
{{\Lambda}^\lambda}_\rho {p’}^\rho g_{\lambda\nu} {{\Lambda}^\nu}_\sigma x^{\sigma} \\
&=
{p’}^\rho
\lr{ {{\Lambda}^\lambda}_\rho
g_{\lambda\nu} {{\Lambda}^\nu}_\sigma } x^{\sigma} \\
&=
{p’}^\rho g_{\rho\sigma} x^\sigma
\end{aligned}

which gives
\label{eqn:qftLecture10:440}
\begin{aligned}
\phihat(\Lambda x)
&=
\int \frac{d{p’}^0 d^3 p’}{(2\pi)^3} \delta({p’}_0^2 – {\Bp’}^2 – m^2) \Theta(p^0) \sqrt{2 \omega_{\Lambda \Bp’}} e^{i p’ \cdot x} \hat{a}_{\Lambda \Bp’} + \text{h.c.} \\
&=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta({p}_0^2 – {\Bp}^2 – m^2) \Theta(p^0) \sqrt{2 \omega_{\Lambda \Bp}} e^{i p \cdot x} \hat{a}_{\Lambda \Bp} + \text{h.c.}
\end{aligned}

Since
\label{eqn:qftLecture10:460}
\phihat(x)
=
\int \frac{dp^0 d^3 p}{(2\pi)^3} \delta({p}_0^2 – {\Bp}^2 – m^2) \Theta(p^0) \sqrt{2 \omega_{\Bp}} e^{i p \cdot x} \hat{a}_{\Bp} + \text{h.c.}

we can now conclude that the creation and annihilation operators transform as

\label{eqn:qftLecture10:480}
\boxed{
\sqrt{2 \omega_{\Lambda \Bp}} \hat{a}_{\Lambda \Bp}
=
\hatU(\Lambda)
\sqrt{2 \omega_{ \Bp}} \hat{a}_{ \Bp}
\hatU^\dagger(\Lambda)
}

In particular
\label{eqn:qftLecture10:500}
\sqrt{2 \omega_{ \Bp}} \hat{a}^\dagger_{ \Bp} \ket{0} = \ket{\Bp}

and noting that $$\hatU(\Lambda) \ket{0} = \ket{0}$$ (i.e. the ground state is Lorentz invariant), we have
\label{eqn:qftLecture10:520}
\begin{aligned}
\sqrt{2 \omega_{\Lambda \Bp}} \hat{a}^\dagger_{\Lambda \Bp} \ket{0}
&=
\hatU(\Lambda) \sqrt{ 2\omega_\Bp} \hat{a}^\dagger_\Bp \hatU^\dagger(\Lambda) \hatU(\Lambda) \ket{0} \\
&=
\hatU(\Lambda) \sqrt{ 2\omega_\Bp} \hat{a}^\dagger_\Bp \ket{0} \\
&=
\hatU(\Lambda) \ket{\Bp}.
\end{aligned}

PHY2403H Quantum Field Theory. Lecture 8: 1st Noether theorem, spacetime translation current, energy momentum tensor, dilatation current. Taught by Prof. Erich Poppitz

DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

1st Noether theorem.

Recall that, given a transformation
\label{eqn:qftLecture8:20}
\phi(x) \rightarrow \phi(x) + \delta \phi(x),

such that the transformation of the Lagrangian is only changed by a total derivative
\label{eqn:qftLecture8:40}
\LL(\phi, \partial_\mu \phi) \rightarrow
\LL(\phi, \partial_\mu \phi)
+ \partial_\mu J_\epsilon^\mu,

then there is a conserved current
\label{eqn:qftLecture8:60}
j^\mu = \PD{(\partial_\mu \phi)}{\LL} \delta_\epsilon \phi – J_\epsilon^\mu.

Here $$\epsilon$$ is an x-independent quantity (i.e. a \underline{global symmetry}).
This is in contrast to “gauge symmetries”, which can be more accurately be categorized as a redundancy in the description.

As an example, for $$\LL = (\partial_\mu \phi \partial^\mu \phi – m^2 \phi^2)/2$$, let
\label{eqn:qftLecture8:80}
\phi(x) \rightarrow \phi(x) – a^\mu \partial_\mu \phi

\label{eqn:qftLecture8:100}
\LL(\phi, \partial_\mu \phi) \rightarrow
\LL(\phi, \partial_\mu \phi)
– a^\mu \partial_\mu \LL
=
\LL(\phi, \partial_\mu \phi)
+ \partial_\mu \lr{ -{\delta^\mu}_\nu a^\nu \LL }

Here $$J^\mu_\epsilon = \evalbar{J^\mu_\epsilon}{\epsilon = a^\nu}$$, and the current is
\label{eqn:qftLecture8:120}
J^\mu = (\partial^\mu \phi)(-a^\nu \partial_\nu \phi) + {\delta^{\mu}}_\nu a^\nu \LL.

In particular, we have one such current for each $$\nu$$, and we write
\label{eqn:qftLecture8:140}
{T^\mu}_\nu =
-(\partial^\mu \phi)(\partial_\nu \phi) + {\delta^{\mu}}_\nu \LL.

By Noether’s theorem, we must have
\label{eqn:qftLecture8:160}
\partial_\mu
{T^\mu}_\nu = 0, \quad \forall \nu.

Check:

\label{eqn:qftLecture8:1380}
\begin{aligned}
\partial_\mu {T^\mu}_\nu
&=
-(\partial_\mu \partial^\mu \phi)(\partial_\nu \phi)
-(\partial^\mu \phi)(\partial_\mu \partial_\nu \phi)
+ {\delta^{\mu}}_\nu
\partial_\mu \lr{
\inv{2} \partial_\alpha \phi \partial^\alpha \phi – \frac{m^2}{2} \phi^2
} \\
&=
-(\partial_\mu \partial^\mu \phi)(\partial_\nu \phi)
-(\partial^\mu \phi)(\partial_\mu \partial_\nu \phi)
+
\inv{2} (\partial_\nu \partial_\mu \phi) (\partial^\mu \phi )
+
\inv{2} (\partial_\mu \phi) (\partial_\nu \partial^\mu \phi )
– m^2 (\partial_\nu \phi) \phi \\
&=
-\lr{ \partial_\mu \partial^\mu \phi + m^2 \phi }(\partial_\nu \phi)
-(\partial_\mu \phi)(\partial^\mu \partial_\nu \phi)
+
\inv{2} (\partial_\nu \partial^\mu \phi) (\partial_\mu \phi )
+
\inv{2} (\partial_\mu \phi) (\partial_\nu \partial^\mu \phi )
&= 0.
\end{aligned}

Example: our potential Lagrangian

\label{eqn:qftLecture8:180}
\LL = \inv{2} \partial^\mu \phi \partial_\nu \phi – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4

Written with upper indexes
\label{eqn:qftLecture8:200}
\begin{aligned}
T^{\mu\nu}
&= -(\partial^\mu \phi)(\partial^\nu \phi) + g^{\mu\nu} \LL \\
&= -(\partial^\mu \phi)(\partial^\nu \phi) + g^{\mu\nu} \lr{
\inv{2} \partial^\alpha \phi \partial_\alpha \phi – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4
}
\end{aligned}

There are 4 conserved currents $$J^{\mu(\nu)} = T^{\mu\nu}$$. Observe that this is symmetric ($$T^{\mu\nu} = T^{\nu\mu}$$).

We have four associated charges
\label{eqn:qftLecture8:220}
Q^\nu = \int d^3 x T^{0 \nu}.

We call
\label{eqn:qftLecture8:240}
Q^0 = \int d^3 x T^{0 0},

the energy density, and call
\label{eqn:qftLecture8:260}
P^i = \int d^3 x T^{0 i},

(i = 1,2,3) the momentum density.

writing this out explicitly the energy density is
\label{eqn:qftLecture8:280}
\begin{aligned}
T^{00}
&= – \dot{\phi}^2 + \inv{2} \lr{ \dot{\phi}^2 – (\spacegrad \phi)^2 – \frac{m^2}{2}\phi^2 – \frac{\lambda}{4} \phi^4} \\
&= -\lr{
\inv{2} \dot{\phi}^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m^2}{2}\phi^2 + \frac{\lambda}{4} \phi^4
},
\end{aligned}

and
\label{eqn:qftLecture8:300}
T^{0i} = \partial^0 \phi \partial^i \phi,

\label{eqn:qftLecture8:320}
P^{i} = -\int d^3 x\partial^0 \phi \partial^i \phi

Since the energy density is negative definite (due to an arbitrary choice of translation sign), let’s redefine $$T^{\mu\nu}$$ to have a positive sign
\label{eqn:qftLecture8:340}
T^{00}
\equiv
\inv{2} \dot{\phi}^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m^2}{2} \phi^2 + \frac{\lambda}{4} \phi^4,

and
\label{eqn:qftLecture8:360}
P^{i} = \int d^3 x\partial^0 \phi \partial^i \phi

As an operator we have
\label{eqn:qftLecture8:380}
\hatQ = \int d^3 x \hatT^{00} =
\int d^3 x
\lr{
\inv{2} \hat{\pi}^2 + \inv{2} (\spacegrad \phihat)^2 + \frac{m^2}{2} \phihat^2 + \frac{\lambda}{4} \phihat^4
}.

\label{eqn:qftLecture8:400}
\hatP^{i} = \int d^3 x \hat{\pi} \partial^i \phi

We showed that
\label{eqn:qftLecture8:420}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}

This implied that $$\phihat, \hat{\pi}$$ obey the classical EOMs
\label{eqn:qftLecture8:440}
\ddt{\phihat} = i \antisymmetric{\hat{H}}{\phihat} = \ddt{\hat{\pi}}

\label{eqn:qftLecture8:460}
\ddt{\hat{\pi}} = i \antisymmetric{\hatH}{\hat{\pi}} = …

In terms of creation and annihilation operators (for the $$\lambda = 0$$ free field), up to a constant
\label{eqn:qftLecture8:480}
\begin{aligned}
\hatH
&= \int d^3 x \hatT^{00} \\
&= \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \hat{a}_\Bp^\dagger \hat{a}_\Bp
\end{aligned}

Can show that:

\label{eqn:qftLecture8:500}
\begin{aligned}
\hatP^i
&= \int d^3 x \hat{\pi} \partial^i \phihat \\
&= \cdots \\
&= \int \frac{d^3 p}{(2 \pi)^3} p^i \hat{a}_\Bp^\dagger \hat{a}_\Bp
\end{aligned}

Now we see the energy and momentum as conserved quantities associated with spacetime translation.

Unitary operators

In QM we say that $$\hat{\Bp}$$ “generates translations”.

With $$\hat{\Bp} \equiv -i \Hbar \spacegrad$$ that translation is
\label{eqn:qftLecture8:520}
\hatU = e^{i \Ba \cdot \hat{\Bp}} = e^{\Ba \cdot \spacegrad}

In particular
\label{eqn:qftLecture8:540}
\bra{\Bx} \hatU \ket{\psi} = e^{\Ba \cdot \hat{\Bp} } \psi(\Bx) = \psi(\Bx + \Ba).

In one dimension
\label{eqn:qftLecture8:560}
\begin{aligned}
\hatU \hat{x} \hatU^\dagger
&=
e^{\Ba \cdot \hat{p} } \psi(\Bx)
e^{-\Ba \cdot \hat{p} } \\
&= \hat{\Bx} + a \hat{\mathbf{1}}.
\end{aligned}

This uses the Baker-Campbell-Hausdorff formula.

Theorem: Baker-Campbell-Hausdorff

\label{eqn:qftLecture8:600}
e^{B} A e^{-B} = \sum_{n = 0}^\infty \inv{n!} \antisymmetric{B \cdots}{\antisymmetric{B}{A}},

where the n-th commutator is denoted above

• $$n = 1$$ : $$\antisymmetric{B}{A}$$
• $$n = 2$$ : $$\antisymmetric{B}{\antisymmetric{B}{A}}$$
• $$n = 3$$ : $$\antisymmetric{B}{\antisymmetric{B}{\antisymmetric{B}{A}}}$$

Proof:

\label{eqn:qftLecture8:620}
\begin{aligned}
f(t)
&= e^{tB} A e^{-tB} \\
&= f(0) + t f'(0) + \frac{t^2}{2} f”(0) + \cdots \frac{t^n}{n!} f^{(n)}(0)
\end{aligned}

\label{eqn:qftLecture8:640}
f(0) = A

\label{eqn:qftLecture8:660}
\begin{aligned}
f'(t)
&=
e^{tB} B A e^{-tB}
+
e^{tB} A (-B) e^{-tB} \\
&=
e^{tB} \antisymmetric{B}{A} e^{-tB}
\end{aligned}

\label{eqn:qftLecture8:680}
\begin{aligned}
f”(t)
&=
e^{tB} B \antisymmetric{B}{A} e^{-tB}
+
e^{tB} \antisymmetric{B}{A} (-B) e^{-tB} \\
&=
e^{tB} \antisymmetric{B}{\antisymmetric{B}{A}} e^{-tB}.
\end{aligned}

From
\label{eqn:qftLecture8:700}
f(1)
= f(0) + f'(0) + \inv{2} f”(0) + \cdots \inv{n!} f^{(n)}(0)

we have
\label{eqn:qftLecture8:720}
e^{B} A e^{-B} = A +
\antisymmetric{B}{A} + \inv{2} \antisymmetric{B}{\antisymmetric{B}{A}} + \cdots

Example:
\label{eqn:qftLecture8:740}
\begin{aligned}
e^{a \partial_x} x e^{-a \partial_x }
&= x + a \antisymmetric{\partial_x}{x} + \cdots \\
&= x + a.
\end{aligned}

Application:

\label{eqn:qftLecture8:760}
e^{i \text{Hermitian} } = \text{unitary}

\label{eqn:qftLecture8:860}
e^{i \text{Hermitian} } \times
e^{-i \text{Hermitian} }
= 1

So
\label{eqn:qftLecture8:780}
\hatU(\Ba) =
e^{i a^j \hat{p}^j }

is a unitary operator representing finite translations in a Hilbert space.

\label{eqn:qftLecture8:800}
\begin{aligned}
\hatU(\Ba) \phihat(\Bx) \hatU^\dagger(\Ba)
&=
e^{i a^j \hat{p}^j }
\phihat(\Bx)
e^{-i a^k \hat{p}^k } \\
&=
\phihat(\Bx)
+ i a^j \antisymmetric{\hatP^j}{\phihat(\Bx)} + \frac{-a^{j_1} a^{j_2}}{2} \antisymmetric{\hatP^{j_1}}{\antisymmetric{\hatP^{j_2}}{\phihat(\Bx)}}
\end{aligned}

\label{eqn:qftLecture8:820}
\begin{aligned}
\antisymmetric{\hatP^j}{\phihat(\Bx)}
&=
\int d^3 y \antisymmetric{\hat{\pi}(\By) \partial^j \phihat(\By)}{\phihat(\Bx)} \\
&=
\int d^3 y \antisymmetric{\hat{\pi}(\By)}{\phihat(\Bx} \partial^j \phihat(\By) \\
&=
\int d^3 y (-i ) \delta^3(\By – \Bx) \partial^j \phihat(\By) \\
&=
-i \partial^j \phihat(\Bx).
\end{aligned}

\label{eqn:qftLecture8:840}
\begin{aligned}
\hatU(\Ba) \phihat(\Bx) \hatU^\dagger(\Ba)
&= \phihat(\Bx) + i a^j (-i) \partial^j \phihat(\Bx) + \cdots \\
&= \phihat(\Bx) + a^j \partial^j \phihat(\Bx) + \cdots \\
&= \phihat(\Bx + \Ba)
\end{aligned}

Continuous symmetries

For all infinitesimal transformations, continuous symmetries lead to conserved charges $$Q$$. In QFT we map these charges to Hermitian operators $$Q \rightarrow \hatQ$$. We say that these charges are “generators of the corresponding symmetry” through unitary operators
\label{eqn:qftLecture8:880}
\hatU = e^{i \text{parameter} \hatQ}.

These represent the action of the symmetry in the Hilbert space.

Example: spatial translation

\label{eqn:qftLecture8:900}
\hatU(\Ba) = e^{i \Ba \cdot \hat{\BP}}

Example: time translation

\label{eqn:qftLecture8:920}
\hatU(t) = e^{i t \hat{H}}.

Classical scalar theory

For $$d > 2$$ let’s look at
\label{eqn:qftLecture8:940}
S =
\int d^d x \lr{
\inv{2} \partial^\mu \phi \partial_\mu \phi – \frac{m^2}{2} \phi^2 – \lambda \phi^{d-2}
}

Take $$m^2, \lambda \rightarrow 0$$, the free massless scalar field.

We have a shift symmetry in this case since $$\phi(x) \rightarrow \phi(x) + \text{constant}$$.
The current is just
\label{eqn:qftLecture8:960}
\begin{aligned}
j^\mu
&= \PD{(\partial_\mu \phi)}{\phi} \delta \phi – J^\mu \\
&= \PD{(\partial_\mu \phi)}{\phi} \delta \phi \\
&= \text{constant} \times \partial^\mu \phi \\
&= \partial^\mu \phi,
\end{aligned}

where the constant factor has been set to one.
This current is clearly conserved since $$\partial_\mu J^\mu = \partial_\mu \partial^\mu \phi = 0$$ (the equation of motion).

These are called “Goldstein Bosons”.

With $$m = \lambda = 0, d = 4$$ we have

NOTE: We did this in class differently with $$d \ne 4, m, \lambda \ne 0$$, and then switched to $$m = \lambda = 0, d = 4$$, which was confusing. I’ve reworked my notes to $$d = 4$$ like the supplemental handout that did the same.

\label{eqn:qftLecture8:980}
S =
\int d^4 x \lr{
\inv{2} \partial^\mu \phi \partial_\mu \phi
}

Here we have a scale or dilatation invariance
\label{eqn:qftLecture8:1000}
x \rightarrow x’ = e^{\lambda} x,

\label{eqn:qftLecture8:1020}
\phi(x) \rightarrow \phi'(x’) = e^{-\lambda} \phi,

\label{eqn:qftLecture8:1040}
d^4 x \rightarrow d^4 x’ = e^{4\lambda} d^4 x,

The partials transform as
\label{eqn:qftLecture8:1400}
\partial^\mu \rightarrow
\PD{x’_\mu}{}
=
\PD{x’_\mu}{x_\mu}
\PD{x_\mu}{}
=
e^{-\lambda}
\PD{x_\mu}{}

so the partial of the field transforms as
\label{eqn:qftLecture8:1420}
\partial^\mu \phi(x) \rightarrow \PD{x’_\mu}{\phi'(x’)} = e^{-2\lambda} \partial^\mu \phi(x),

and finally
\label{eqn:qftLecture8:1060}
(\partial_\mu \phi)^2 \rightarrow e^{-4\lambda} \lr{ \partial_\mu \phi(x) }^2.

With a $$-4 \lambda$$ power in the transformed quadratic term, and $$4 \lambda$$ in the volume element, we see that the action is invariant.

To find Noether current, we need to vary the field and it’s derivatives
\label{eqn:qftLecture8:1100}
\begin{aligned}
\delta_\lambda \phi
&= \phi'(x) – \phi(x) \\
&= \phi'(e^{-\lambda} x’) – \phi(x) \\
&\approx \phi'(x’ -\lambda x’) – \phi(x) \\
&\approx \phi'(x’) – \lambda {x’}^\alpha \partial_\alpha \phi'(x’) – \phi(x) \\
&\approx (1 – \lambda) \phi(x) – \lambda {x’}^\alpha \partial_\alpha \phi'(x’) – \phi(x) \\
&= – \lambda(1 + x^\alpha \partial_\alpha ) \phi,
\end{aligned}

where the last step assumes that $$x’ \rightarrow x, \phi’ \rightarrow \phi$$, effectively weeding out any terms that are quadratic or higher in $$\lambda$$.

Now we need the variation of the derivatives of $$\phi$$
\label{eqn:qftLecture8:1440}
\delta \partial_\mu \phi(x)
=
\partial_\mu’ \phi'(x) – \partial_\mu \phi(x),

By \ref{eqn:qftLecture8:1420}
\label{eqn:qftLecture8:1460}
\begin{aligned}
\partial_\mu’ \phi'(x’)
&=
e^{-2\lambda} \partial_\mu \phi(x) \\
&=
e^{-2\lambda} \partial_\mu \phi(e^{-\lambda} x’) \\
&\approx
e^{-2\lambda} \partial_\mu
\lr{
\phi(x’) – \lambda {x’}^\alpha \partial_\alpha \phi(x’)
} \\
&\approx
\lr{
1 – 2 \lambda
}
\partial_\mu
\lr{
\phi(x’) – \lambda {x’}^\alpha \partial_\alpha \phi(x’)
},
\end{aligned}

so
\label{eqn:qftLecture8:1480}
\begin{aligned}
\delta \partial_\mu \phi
&=
– \lambda {x}^\alpha \partial_\alpha \partial_\mu \phi(x)
– 2 \lambda \partial_\mu \phi(x) + O(\lambda^2) \\
&=
– \lambda \lr{
{x}^\alpha \partial_\alpha + 2
}
\partial_\mu \phi(x).
\end{aligned}

\label{eqn:qftLecture8:1200}
\begin{aligned}
\delta \LL
&=
(\partial^\mu \phi) \delta (\partial_\mu \phi) \\
&= – \lambda \lr{ 2
\partial_\mu \phi
+ x^\alpha \partial_\alpha
\partial_\mu \phi
}
\partial^\mu \phi,
\end{aligned}

or
\label{eqn:qftLecture8:1500}
\begin{aligned}
\frac{\delta \LL }{-\lambda}
&=
4 \LL + x^\alpha \lr{ \partial_\alpha \partial_\mu \phi } \partial^\mu \phi \\
&=
4 \LL + x^\alpha \partial_\alpha \lr{ \LL } \\
&=
{4 \LL} + \partial_\alpha \lr{ x^\alpha \LL } – {\LL \partial_\alpha x^\alpha} \\
&=
\partial_\alpha \lr{ x^\alpha \LL }.
\end{aligned}

The variation in the Lagrangian density is thus
\label{eqn:qftLecture8:1520}
\delta \LL = \partial_\mu J^\mu_\lambda = \partial_\mu \lr{ -\lambda x^\mu \LL },

and the current is
\label{eqn:qftLecture8:1540}
J^\mu_\lambda = -\lambda x^\mu \LL.

The Noether current is
\label{eqn:qftLecture8:1240}
\begin{aligned}
j^\mu
&= \PD{(\partial_\mu \phi)}{\LL} \delta \phi – J^\mu \\
&= -\partial^\mu \phi \lr{ 1 + x^\nu \partial_\nu } \phi + \inv{2} x^\mu \partial_\nu \phi \partial^\nu \phi,
\end{aligned}

or after flipping signs
\label{eqn:qftLecture8:1280}
\begin{aligned}
j^\mu_{\text{dil}}
&= \partial^\mu \phi \lr{ 1 + x^\nu \partial_\nu } \phi – \inv{2} x^\mu
\partial_\nu \phi \partial^\nu \phi \\
&= x_\nu \lr{ \partial^\mu \phi \partial^\nu \phi – \inv{2} {\delta^{\nu}}_\mu \partial_\lambda \phi \partial^\lambda \phi }
+ \inv{2} \partial^\mu (\phi^2),
\end{aligned}

\label{eqn:qftLecture8:1300}
j^\mu_{\text{dil}} = -x_\nu T^{\nu \mu} + \inv{2} \partial^\mu (\phi^2),

The current and $$T^{\mu\nu}$$ can both be redefined $$j^{\mu’} = j^\mu + \partial_\nu C^{\nu\mu}$$ adding an antisymmetric $$C^{\mu\nu} = -C^{\nu\mu}$$

\label{eqn:qftLecture8:1320}
j^\mu_{\text{dil conformal}} = – x_\nu T^{\nu\mu}_{\text{conformal}}

\label{eqn:qftLecture8:1340}
\partial_\mu
j^\mu_{\text{dil conformal}} = – {{T_{\text{conformal}}}^\mu}_\mu

consequence: $$0 = T^{00} – T^{11} – T^{22} – T^{33}$$, which is essentially
\label{eqn:qftLecture8:1360}
0 = \rho – 3 p = 0.

PHY2403H Quantum Field Theory. Lecture 9: Unbroken and spontaneously broken symmetries, Higgs Lagrangian, scale invariance, Lorentz invariance, angular momentum quantization. Taught by Prof. Erich Poppitz

[Click here for a PDF of this post with nicer formatting (and a Mathematica listing that I didn’t include in this blog post’s latex export)]

DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

Last time

We followed a sequence of operations

1. Noether’s theorem
2. $$\rightarrow$$ conserved currents
3. $$\rightarrow$$ charges (classical)
4. $$\rightarrow$$ “correspondence principle”
5. $$\rightarrow \hatQ$$
• Hermitian operators
• “generators of symmetry”
\label{eqn:qftLecture9:20}
\hatU(\alpha) = e^{i \alpha \hatQ}

We found
\label{eqn:qftLecture9:40}
\hatU(\alpha) \phihat \hatU^\dagger(\alpha) = \phihat + i \alpha \antisymmetric{\hatQ}{\phihat} + \cdots

Example: internal symmetries:

(non-spacetime), such as $$O(N)$$ or $$U(1)$$.

In QFT internal symmetries can have different “\underline{modes of realization}”.

[I]

1. “Wigner mode”. These are also called “unbroken symmetries”.
\label{eqn:qftLecture9:60}
\hatQ \ket{0} = 0

i.e. $$\hatU(\alpha) \ket{0} = 0$$.
Ground state invariant. Formally $$:\hatQ:$$ annihilates $$\ket{0}$$.
$$\antisymmetric{\hatQ}{\hatH} = 0$$ implies that all eigenstates are eigenstates of $$\hatQ$$ in $$U(1)$$. Example from HW 1
\label{eqn:qftLecture9:80}
\hatQ = \text{“charge” under $$U(1)$$}.

All states have definite charge, just live in QU.
2. “Nambu-Goldstone mode” (Landau-ginsburg). This is also called a “spontaneously broken symmetry”\footnote{
First encounter example (HWII, $$SU(2) \times SU(2) \rightarrow SU(2)$$). Here a $$U(1)$$ spontaneous broken symmetry.}.
$$H$$ or $$L$$ is invariant under symmetry, but ground state is not.

fig. 1. Mexican hat potential.

fig. 2. Degenerate Mexican hat potential ( v = 0)

Example:
\label{eqn:qftLecture9:100}
\LL = \partial_\mu \phi^\conj \partial^\mu \phi – V(\Abs{\phi}),

where
\label{eqn:qftLecture9:120}
V(\Abs{\phi}) = m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi }^2.

When $$m^2 > 0$$ we have a Wigner mode, but when $$m^2 < 0$$ we have an issue: $$\phi = 0$$ is not a minimum of potential.
When $$m^2 < 0$$ we write
\label{eqn:qftLecture9:140}
\begin{aligned}
V(\phi)
&= – m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi}^2 \\
&= \frac{\lambda}{4} \lr{
\lr{ \phi^\conj \phi}^2 – \frac{4}{\lambda} m^2 } \\
&= \frac{\lambda}{4} \lr{
\phi^\conj \phi – \frac{2}{\lambda} m^2 }^2 – \frac{4 m^4}{\lambda^2},
\end{aligned}

or simply
\label{eqn:qftLecture9:780}
V(\phi)
=
\frac{\lambda}{4} \lr{ \phi^\conj \phi – v^2 }^2 + \text{const}.

The potential (called the Mexican hat potential) is illustrated in fig. 1 for non-zero $$v$$, and in
fig. 2 for $$v = 0$$.
We choose to expand around some point on the minimum ring (it doesn’t matter which one).
When there is no potential, we call the field massless (i.e. if we are in the minimum ring).
We expand as
\label{eqn:qftLecture9:160}
\phi(x) = v \lr{ 1 + \frac{\rho(x)}{v} } e^{i \alpha(x)/v },

so
\label{eqn:qftLecture9:180}
\begin{aligned}
\frac{\lambda}{4}
\lr{\phi^\conj \phi – v^2}^2
&=
\lr{
v^2 \lr{ 1 + \frac{\rho(x)}{v} }^2
– v^2
}^2 \\
&=
\frac{\lambda}{4}
v^4 \lr{ \lr{ 1 + \frac{\rho(x)}{v} }^2 – 1 } \\
&=
\frac{\lambda}{4}
v^4
\lr{
\frac{2 \rho}{v} + \frac{\rho^2}{v^2}
}^2.
\end{aligned}

\label{eqn:qftLecture9:200}
\partial_\mu \phi =
\lr{
v \lr{ 1 + \frac{\rho(x)}{v} } \frac{i}{v} \partial_\mu \alpha
+ \partial_\mu \rho
} e^{i \alpha}

so
\label{eqn:qftLecture9:220}
\begin{aligned}
\LL
&= \Abs{\partial \phi^\conj}^2 – \frac{\lambda}{4} \lr{ \Abs{\phi^\conj}^2 – v^2 }^2 \\
&=
\partial_\mu \rho \partial^\mu \rho + \partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }

\frac{\lambda v^4}{4} \frac{ 4\rho^2}{v^2} + O(\rho^3) \\
&=
\partial_\mu \rho \partial^\mu \rho
– \lambda v^2\rho^2
+
\partial_\mu \alpha \partial^\mu \alpha \lr{ 1 + \frac{\rho}{v} }.
\end{aligned}

We have two fields, $$\rho$$ : a massive scalar field, the “Higgs”, and a massless field $$\alpha$$ (the Goldstone Boson).

$$U(1)$$ symmetry acts on $$\phi(x) \rightarrow e^{i \omega } \phi(x)$$ i.t.o $$\alpha(x) \rightarrow \alpha(x) + v \omega$$.
$$U(1)$$ global symmetry (broken) acts on the Goldstone field $$\alpha(x)$$ by a constant shift. ($$U(1)$$ is still a symmetry of the Lagrangian.)

The current of the $$U(1)$$ symmetry is:
\label{eqn:qftLecture9:240}
j_\mu = \partial_\mu \alpha \lr{ 1 + \text{higher dimensional $$\rho$$ terms} }.

When we quantize
\label{eqn:qftLecture9:260}
\alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp

\label{eqn:qftLecture9:280}
j^\mu(x) = \partial^\mu \alpha(x) =
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ i \omega_\Bp – i \Bp } e^{i \omega_p t – i \Bp \cdot \Bx} \hat{a}_\Bp^\dagger +
\int \frac{d^3p}{(2\pi)^3 \sqrt{ 2 \omega_p }} \lr{ -i \omega_\Bp + i \Bp } e^{-i \omega_p t + i \Bp \cdot \Bx} \hat{a}_\Bp.

\label{eqn:qftLecture9:300}
j^\mu(x) \ket{0} \ne 0,

instead it creates a single particle state.

Examples of symmetries

In particle physics, examples of Wigner vs Nambu-Goldstone, ignoring gravity the only exact internal symmetry in the standard module is
$$(B\# – L\#)$$, believed to be a $$U(1)$$ symmetry in Wigner mode.

Here $$B\#$$ is the Baryon number, and $$L\#$$ is the Lepton number. Examples:

• $$B(p) = 1$$, proton.
• $$B(q) = 1/3$$, quark
• $$B(e) = 1$$, electron
• $$B(n) = 1$$, neutron.
• $$L(p) = 1$$, proton.
• $$L(q) = 0$$, quark.
• $$L(e) = 0$$, electron.

The major use of global internal symmetries in the standard model is as “approximate” ones. They become symmetries when one neglects some effect( “terms in $$\LL$$”).
There are other approximate symmetries (use of group theory to find the Balmer series).

Example from HW2:

QCD in limit
\label{eqn:qftLecture9:320}
m_u = m_d = 0.

$$m_u m_d \ll m_p$$ (the products of the up-quark mass and the down-quark mass are much less than a composite one (name?)).
$$SU(2)_L \times SU(2)_R \rightarrow SU(2)_V$$

EWSB (Electro-Weak-Symmetry-Breaking) sector

When the couplings $$g_2, g_1 = 0$$. ($$g_2 \in SU(2), g_1 \in U(1)$$).

Scale invariance

\label{eqn:qftLecture9:340}
\begin{aligned}
x &\rightarrow e^{\lambda} x \\
\phi &\rightarrow e^{-\lambda} \phi \\
A_\mu &\rightarrow e^{-\lambda} A_\mu
\end{aligned}

Any unitary theory which is scale invariant is also \underline{conformal} invariant. Conformal invariance means that angles are preserved.
The point here is that there is more than scale invariance.

We have classical internal global continuous symmetries.
These can be either

1. “unbroken” (Wigner mode)
\label{eqn:qftLecture9:360}
\hatQ\ket{0} = 0.
2. “spontaneously broken”
\label{eqn:qftLecture9:380}
j^\mu(x) \ket{0} \ne 0

(creates Goldstone modes).
3. “anomalous”. Classical symmetries are not a symmetry of QFT.
Examples:

• Scale symmetry (to be studied in QFT II), although this is not truly internal.
• In QCD again when $$\omega_\Bq = 0$$, a $$U(1$$ symmetry (chiral symmetry) becomes exact, and cannot be preserved in QFT.
• In the standard model (E.W sector), the Baryon number and Lepton numbers are not symmetries, but their difference $$B\# – L\#$$ is a symmetry.

Lorentz invariance.

We’d like to study the action of Lorentz symmetries on quantum states. We are going to “go by the book”, finding symmetries, currents, quantize, find generators, and so forth.

Under a Lorentz transformation
\label{eqn:qftLecture9:400}
x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu,

We are going to consider infinitesimal Lorentz transformations
\label{eqn:qftLecture9:420}
{\Lambda^\mu}_\nu \approx
{\delta^\mu}_\nu + {\omega^\mu}_\nu
,

where $${\omega^\mu}_\nu$$ is small.
A Lorentz transformation $$\Lambda$$ must satisfy $$\Lambda^\T G \Lambda = G$$, or
\label{eqn:qftLecture9:800}
g_{\mu\nu} = {{\Lambda}^\alpha}_\mu g_{\alpha \beta} {{\Lambda}^\beta}_\nu,

into which we insert the infinitesimal transformation representation
\label{eqn:qftLecture9:820}
\begin{aligned}
0
&=
– g_{\mu\nu} +
\lr{ {\delta^\alpha}_\mu + {\omega^\alpha}_\mu }
g_{\alpha \beta}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
\lr{
g_{\mu \beta}
+
\omega_{\beta\mu}
}
\lr{ {\delta^\beta}_\nu + {\omega^\beta}_\nu } \\
&=
– g_{\mu\nu} +
g_{\mu \nu}
+
\omega_{\nu\mu}
+
\omega_{\mu\nu}
+
\omega_{\beta\mu}
{\omega^\beta}_\nu.
\end{aligned}

The quadratic term can be ignored, leaving just
\label{eqn:qftLecture9:840}
0 =
\omega_{\nu\mu}
+
\omega_{\mu\nu},

or
\label{eqn:qftLecture9:860}
\omega_{\nu\mu} = – \omega_{\mu\nu}.

Note that $$\omega$$ is a completely antisymmetric tensor, and like $$F_{\mu\nu}$$ this has only 6 elements.
This means that the
infinitesimal transformation of the coordinates is
\label{eqn:qftLecture9:440}
x^\mu \rightarrow {x’}^\mu \approx x^\mu + \omega^{\mu\nu} x_\nu,

the field transforms as
\label{eqn:qftLecture9:460}
\phi(x) \rightarrow \phi'(x’) = \phi(x)

or
\label{eqn:qftLecture9:760}
\phi'(x^\mu + \omega^{\mu\nu} x_\nu) =
\phi'(x) + \omega^{\mu\nu} x_\nu \partial_\mu\phi(x) = \phi(x),

so
\label{eqn:qftLecture9:480}
\delta \phi = \phi'(x) – \phi(x) =
-\omega^{\mu\nu} x_\nu \partial_\mu \phi.

Since $$\LL$$ is a scalar
\label{eqn:qftLecture9:500}
\begin{aligned}
\delta \LL
&=
-\omega^{\mu\nu} x_\nu \partial_\mu \LL \\
&=

\partial_\mu \lr{
\omega^{\mu\nu} x_\nu \LL
}
+
(\partial_\mu x_\nu) \omega^{\mu\nu} \LL \\
&=
\partial_\mu \lr{

\omega^{\mu\nu} x_\nu \LL
},
\end{aligned}

since $$\partial_\nu x_\mu = g_{\nu\mu}$$ is symmetric, and $$\omega$$ is antisymmetric.
Our current is
\label{eqn:qftLecture9:520}
J^\mu_\omega
=

\omega^{\mu\nu} x_\mu \LL
.

Our Noether current is
\label{eqn:qftLecture9:540}
\begin{aligned}
j^\nu_{\omega^{\mu\rho}}
&= \PD{\phi_{,\nu}}{\LL} \delta \phi – J^\mu_\omega \\
&=
\partial^\nu \phi\lr{ – \omega^{\mu\rho} x_\rho \partial_\mu \phi } + \omega^{\nu \rho} x_\rho \LL \\
&=
\omega^{\mu\rho}
\lr{
\partial^\nu \phi\lr{ – x_\rho \partial_\mu \phi } + {\delta^{\nu}}_\mu x_\rho \LL
} \\
&=
\omega^{\mu\rho} x_\rho
\lr{
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL
}
\end{aligned}

We identify
\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,

so the current is
\label{eqn:qftLecture9:580}
\begin{aligned}
j^\nu_{\omega_{\mu\rho}}
&=
-\omega^{\mu\rho} x_\rho
{T^\nu}_\mu \\
&=
-\omega_{\mu\rho} x^\rho
T^{\nu\mu}
.
\end{aligned}

Define
\label{eqn:qftLecture9:600}
j^{\nu\mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} },

which retains the antisymmetry in $$\mu \rho$$ yet still drops the parameter $$\omega^{\mu\rho}$$.
To check that this makes sense, we can contract
$$j^{\nu\mu\rho}$$ with $$\omega_{\rho\mu}$$
\label{eqn:qftLecture9:880}
\begin{aligned}
j^{\nu\mu\rho} \omega_{\rho\mu}
&= -\inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} }
\omega_{\mu\rho} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\mu} T^{\nu\rho}
\omega_{\rho\mu} \\
&=
-\inv{2} x^\rho T^{\nu\mu}
\omega_{\mu\rho}
– \inv{2} x^{\rho} T^{\nu\mu}
\omega_{\mu\rho} \\
&=
– x^{\rho} T^{\nu\mu}
\omega_{\mu\rho},
\end{aligned}

which matches \ref{eqn:qftLecture9:580} as desired.

Example. Rotations $$\mu\rho = ij$$

\label{eqn:qftLecture9:620}
\begin{aligned}
J^{0 i j} \epsilon_{ijk}
&=
\inv{2} \lr{ x^i T^{0j} – x^{j} T^{0i} } \epsilon_{ijk} \\
&=
x^i T^{0j} \epsilon_{ijk}.
\end{aligned}

Observe that this has the structure of $$(\Bx \cross \Bp)_k$$, where $$\Bp$$ is the momentum density of the field.
Let
\label{eqn:qftLecture9:640}
L_k \equiv Q_k = \int d^3 x J^{0ij} \epsilon_{ijk}.

We can now quantize and build a generator
\label{eqn:qftLecture9:660}
\begin{aligned}
\hatU(\Balpha)
&= e^{i \Balpha \cdot \hat{\BL}} \\
&= \exp\lr{i \alpha_k
\int d^3 x x^i \hat{T}^{0j} \epsilon_{ijk}
}
\end{aligned}

From \ref{eqn:qftLecture9:560} we can quantize with $$T^{0j} = \partial^0 \phi \partial^j \phi \rightarrow \hat{\pi} \lr{\spacegrad \phihat}_j$$, or
\label{eqn:qftLecture9:900}
\begin{aligned}
\hatU(\Balpha)
&=
\exp\lr{i \alpha_k
\int d^3 x x^i \hat{\pi} (\spacegrad \phihat)_j \epsilon_{ijk}
} \\
&=
\exp\lr{i \Balpha \cdot
\int d^3 x \hat{\pi} \spacegrad \phihat \cross \Bx
}
\end{aligned}

\label{eqn:qftLecture9:680}
\begin{aligned}
\phihat(\By) \rightarrow \hatU(\alpha) \phihat(\By) \hatU^\dagger(\alpha)
&\approx
\phihat(\By) +
i \Balpha \cdot
\antisymmetric{
\int d^3 x \hat{\pi}(\Bx) \spacegrad \phihat(\Bx) \cross \Bx
}
{
\phihat(\By)
} \\
&=
\phihat(\By) +
i \Balpha \cdot
\int d^3 x
(-i) \delta^3(\Bx – \By)
&=
\phihat(\By) +
\Balpha \cdot \lr{ \spacegrad \phihat(\By ) \cross \By}
\end{aligned}

Explicitly, in coordinates, this is
\label{eqn:qftLecture9:700}
\begin{aligned}
\phihat(\By)
&\rightarrow
\phihat(\By) +
\alpha^i
\lr{
\partial^j \phihat(\By) y^k \epsilon_{jki}
} \\
&=
\phihat(\By) –
\epsilon_{ikj} \alpha^i y^k \partial^j \phihat \\
&=
\phihat( y^j – \epsilon^{ikj} \alpha^i y^k ).
\end{aligned}

This is a rotation. To illustrate, pick $$\Balpha = (0, 0, \alpha)$$, so $$y^j \rightarrow y^j – \epsilon^{ikj} \alpha y^k \delta_{i3} = y^j – \epsilon^{3kj} \alpha y^k$$, or
\label{eqn:qftLecture9:n}
\begin{aligned}
y^1 &\rightarrow y^1 – \epsilon^{3k1} \alpha y^k = y^1 + \alpha y^2 \\
y^2 &\rightarrow y^2 – \epsilon^{3k2} \alpha y^k = y^2 – \alpha y^1 \\
y^3 &\rightarrow y^3 – \epsilon^{3k3} \alpha y^k = y^3,
\end{aligned}

or in matrix form
\label{eqn:qftLecture9:720}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & \alpha & 0 \\
-\alpha & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
y^1 \\
y^2 \\
y^3 \\
\end{bmatrix}.

PHY2403H Quantum Field Theory. Lecture 7: Symmetries, translation currents, energy momentum tensor. Taught by Prof. Erich Poppitz

DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

Symmetries

Given the complexities of the non-linear systems we want to investigate, examination of symmetries gives us simpler problems that we can solve.

• “internal” symmetries. This means that the symmetries do not act on space time $$(\Bx, t)$$. An example is
\label{eqn:qftLecture7:20}
\phi^i =
\begin{bmatrix}
\psi_1 \\
\psi_2 \\
\vdots \\
\psi_N \\
\end{bmatrix}

If we map $$\phi^i \rightarrow O^i_j \phi^j$$ where $$O^\T O = 1$$, then we call this an internal symmetry.
The corresponding Lagrangian density might be something like
\label{eqn:qftLecture7:40}
\LL = \inv{2} \partial_\mu \Bphi \cdot \partial^\mu \Bphi – \frac{m^2}{2} \Bphi \cdot \Bphi – V(\Bphi \cdot \Bphi)

• spacetime symmetries: Translations, rotations, boosts, dilatations. We will consider continuous symmetries, which can be defined as a succession of infinitesimal transformations.
An example from $$O(2)$$ is a rotation
\label{eqn:qftLecture7:60}
\begin{bmatrix}
\phi^1 \\
\phi^2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
\cos\alpha & \sin\alpha \\
-\sin\alpha & \cos\alpha \\
\end{bmatrix}
\begin{bmatrix}
\phi^1 \\
\phi^2
\end{bmatrix},

or if $$\alpha \sim 0$$
\label{eqn:qftLecture7:80}
\begin{bmatrix}
\phi^1 \\
\phi^2 \\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & \alpha \\
-\alpha & 1\\
\end{bmatrix}
\begin{bmatrix}
\phi^1 \\
\phi^2
\end{bmatrix}
=
\begin{bmatrix}
\phi^1 \\
\phi^2
\end{bmatrix}
+
\alpha
\begin{bmatrix}
\phi^2 \\
-\phi^1
\end{bmatrix}

In index notation we write
\label{eqn:qftLecture7:100}
\phi^i \rightarrow \phi^i + \alpha e^{ij} \phi^j,

where $$\epsilon^{12} = +1, \epsilon^{21} = -1$$ is the completely antisymmetric tensor. This can be written in more general form as
\label{eqn:qftLecture7:120}
\phi^i \rightarrow \phi^i + \delta \phi^i,

where $$\delta \phi^i$$ is considered to be an infinitesimal transformation.

Definition: Symmetry

A symmetry means that there is some transformation
\begin{equation*}
\phi^i \rightarrow \phi^i + \delta \phi^i,
\end{equation*}
where $$\delta \phi^i$$ is an infinitesimal transformation, and the equations of motion are invariant under this transformation.

Theorem: Noether’s theorem (1st).

If the equations of motion re invariant under $$\phi^\mu \rightarrow \phi^\mu + \delta \phi^\mu$$, then there exists a conserved current $$j^\mu$$ such that $$\partial_\mu j^\mu = 0$$.

Noether’s first theorem applies to global symmetries, where the parameters are the same for all $$(\Bx, t)$$. Gauge symmetries are not examples of such global symmetries.

Given a Lagrangian density $$\LL(\phi(x), \phi_{,\mu}(x))$$, where $$\phi_{,\mu} \equiv \partial_\mu \phi$$. The action is
\label{eqn:qftLecture7:160}
S = \int d^d x \LL.

EOMs are invariant if under $$\phi(x) \rightarrow \phi'(x) = \phi(x) + \delta_\epsilon \phi(x)$$, we have
\label{eqn:qftLecture7:180}
\LL(\phi) \rightarrow \LL'(\phi’) = \LL(\phi) + \partial_\mu J_\epsilon^\mu(\phi) + O(\epsilon^2).

Then there exists a conserved current. In QFT we say that the E.O.M’s are “on shell”. Note that \ref{eqn:qftLecture7:180} is a symmetry since we have added a total derivative to the Lagrangian which leaves the equations of motion of unchanged.

In general, the change of action under arbitrary variation of $$\delta \phi$$ of the fields is
\label{eqn:qftLecture7:200}
\begin{aligned}
\delta S
&=
\int d^d x \delta \LL(\phi, \partial_\mu \phi) \\
&=
\int d^d x \lr{
\PD{\phi}{\LL} \delta \phi
+
\PD{(\partial_\mu \phi)}{\LL} \delta \partial_\mu \phi
} \\
&=
\int d^d x \lr{
\partial_\mu \lr{ \PD{(\partial_\mu \phi)}{\LL} } \delta \phi
+
\PD{(\partial_\mu \phi)}{\LL} \partial_\mu \delta \phi
} \\
&=
\int d^d x
\partial_\mu \lr{ \frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta \phi }
\end{aligned}

However from \ref{eqn:qftLecture7:180}
\label{eqn:qftLecture7:220}
\delta_\epsilon \LL = \partial_\mu J_\epsilon^\mu(\phi, \partial_\mu \phi),

so after equating these variations we fine that
\label{eqn:qftLecture7:240}
\delta S = \int d^d x \delta_\epsilon \LL = \int d^d x \partial_\mu J_\epsilon^\mu,

or
\label{eqn:qftLecture7:260}
0 = \int d^d x
\partial_\mu \lr{ \frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta \phi – J_\epsilon^\mu },

or $$\partial_\mu j^\mu = 0$$ provided
\label{eqn:qftLecture7:280}
\boxed{
j^\mu =
\frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta_\epsilon \phi – J_\epsilon^\mu.
}

Integrating the divergence of the current over a space time volume, perhaps that of cylinder (time up, space out) is also zero. That is
\label{eqn:qftLecture7:300}
\begin{aligned}
0
&=
\int d^4 x \, \partial_\mu j^\mu \\
&=
\int d^3 \Bx dt \, \partial_\mu j^\mu \\
&=
\int d^3 \Bx dt \, \partial_t j^0 –
\int d^3 \Bx dt \spacegrad \cdot \Bj \\
&=
\int d^3 \Bx dt \, \partial_t j^0 ,
\end{aligned}

where the spatial divergence is zero assuming there’s no current leaving the volume on the infinite boundary
(no $$\Bj$$ at spatial infinity.)

We write
\label{eqn:qftLecture7:560}
Q = \int d^3x \partial_t j^0,

and call this the on-shell charge associated with the symmetry.

Spacetime translation.

A spacetime translation has the form
\label{eqn:qftLecture7:320}
x^\mu \rightarrow {x’}^\mu = x^\mu + a^\mu,

\label{eqn:qftLecture7:340}
\phi(x) \rightarrow \phi'(x’) = \phi(x)

(contrast this to a Lorentz transformation that had the form $$x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu$$).

If $$\phi'(x + a) = \phi(x)$$, then
\label{eqn:qftLecture7:360}
\phi'(x) + a^\mu \partial_\mu \phi'(x) =
\phi'(x) + a^\mu \partial_\mu \phi(x) =
\phi(x),

so
\label{eqn:qftLecture7:380}
\phi'(x)
= \phi(x) – a^\mu \partial_\mu \phi'(x)
= \phi(x) + \delta_a \phi(x),

or
\label{eqn:qftLecture7:580}
\delta_a \phi(x) = – a^\mu \partial_\mu \phi(x).

Under $$\phi \rightarrow \phi – a^\mu \partial_\mu \phi$$, we have
\label{eqn:qftLecture7:400}
\LL(\phi) \rightarrow \LL(\phi) – a^\mu \partial_\mu \LL.

Let’s calculate this with our scalar theory Lagrangian
\label{eqn:qftLecture7:420}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi)

The Lagrangian variation is
\label{eqn:qftLecture7:440}
\begin{aligned}
\evalbar{\delta \LL}{\phi \rightarrow \phi + \delta \phi, \delta\phi = – a^\mu \partial_\mu \phi}
&=
(\partial_\mu \phi) \delta (\partial^\mu \phi) – m^2 \phi \delta \phi – \PD{\phi}{V} \delta \phi \\
&=
(\partial_\mu \phi)(-a^\nu \partial_\nu \phi \partial^\mu \phi) + m^2 \phi a^\nu \partial_\nu \phi + \PD{\phi}{V} a^\nu \partial_\nu \phi \\
&=
– a^\nu \partial_\nu \lr{ \inv{2} \partial_\mu \partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi) } \\
&=
– a^\nu \partial_\nu \LL.
\end{aligned}

So the current is
\label{eqn:qftLecture7:600}
\begin{aligned}
j^\mu
&=
(\partial^\mu \phi) (-a^\nu \partial_\nu \phi) + a^\nu \LL \\
&=
-a^\nu \lr{ \partial^\mu \phi \partial_\nu \phi – \LL }
\end{aligned}

We really have a current for each $$\nu$$ direction and can make that explicit writing
\label{eqn:qftLecture7:460}
\begin{aligned}
\delta_\nu \LL
&= -\partial_\nu \LL \\
&= – \partial_\mu \lr{ {\delta^\mu}_\nu \LL } \\
&= \partial_\mu {J^\mu}_\nu
\end{aligned}

we write
\label{eqn:qftLecture7:480}
{j^\mu}_\nu = \PD{x_\mu}{\phi} \lr{ – \PD{x^\nu}{\phi} } + {\delta^\mu}_\nu \LL,

where $$\nu$$ are labels which coordinates are translated:
\label{eqn:qftLecture7:500}
\begin{aligned}
\partial_\nu \phi &= – \partial_\nu \phi \\
\partial_\nu \LL &= – \partial_\nu \LL.
\end{aligned}

We call the conserved quantities elements of the energy-momentum tensor, and write it as
\label{eqn:qftLecture7:520}
\boxed{
{T^\mu}_\nu = -\PD{x_\mu}{\phi} \PD{x^\nu}{\phi} + {\delta^\mu}_\nu \LL.
}

Incidentally, we picked a non-standard sign convention for the tensor, as an explicit expansion of $$T^{00}$$, the energy density component, shows
\label{eqn:qftLecture7:540}
\begin{aligned}
{T^0}_0
&=
-\PD{t}{\phi}
\PD{t}{\phi}
+\inv{2}
\PD{t}{\phi}
\PD{t}{\phi}
– \frac{m^2}{2} \phi^2 – V(\phi) \\
&=
-\inv{2} \PD{t}{\phi} \PD{t}{\phi}
– \frac{m^2}{2} \phi^2 – V(\phi).
\end{aligned}

Had we translated by $$-a^\mu$$ we’d have a positive definite tensor instead.

Complex Klein-Gordon field: dead ends take III (or more?)

September 29, 2018 phy2403 No comments

Working on a problem from Peskin and Schroeder (and Zee too actually), I’ve hit dead ends in many different ways. The last dead end looks impressive just sitting there on the table where I abandoned it:

… however, after getting results that weren’t expected a number of different ways, I think I finally understand the fundamental assumption that I had wrong.  Take >= IV.

PHY2403H Quantum Field Theory. Lecture 6: Canonical quantization, Simple Harmonic Oscillators, Symmetries. Taught by Prof. Erich Poppitz

DISCLAIMER: Very rough notes from class, with some additional side notes (QM SHO review, …).

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

Quantization of Field Theory

We are engaging in the “canonical” or Hamiltonian method of quantization. It is also possible to quantize using path integrals, but it is hard to prove that operators are unitary doing so. In fact, the mechanism used to show unitarity from path integrals is often to find the Lagrangian and show that there is a Hilbert space (i.e. using canonical quantization). Canonical quantization essentially demands that the fields obey a commutator relation of the following form
\label{eqn:qftLecture6:20}
\antisymmetric{\pi(\Bx, t)}{\phi(\By, t)} = -i \delta^3(\Bx – \By).

We assumed that the quantized fields obey the Hamiltonian relations
\label{eqn:qftLecture6:160}
\begin{aligned}
\ddt{\phi} &= i \antisymmetric{H}{\phi} \\
\ddt{\pi} &= i \antisymmetric{H}{\pi}.
\end{aligned}

We were working with the Hamiltonian density
\label{eqn:qftLecture6:40}
\mathcal{H} =
\inv{2} (\pi(\Bx, t))^2
+
+
\frac{m^2}{2} \phi^2
+
\frac{\lambda}{4} \phi^4,

which included a mass term $$m$$ and a potential term ($$\lambda$$). We will expand all quantities in Taylor series in $$\lambda$$ assuming they have a structure such as
\label{eqn:qftLecture6:180}
\begin{aligned}
f(\lambda) =
c_0 \lambda^0
+ c_1 \lambda^1
+ c_2 \lambda^2
+ c_3 \lambda^3
+ \cdots
\end{aligned}

We will stop this \underline{perturbation theory} approach at $$O(\lambda^2)$$, and will ignore functions such as $$e^{-1/\lambda}$$.

Within perturbation theory, to leaving order, set $$\lambda = 0$$, so that $$\phi$$ obeys the Klein-Gordon equation (if $$m = 0$$ we have just a d’Lambertian (wave equation)).

We can write our field as a Fourier transform
\label{eqn:qftLecture6:60}
\phi(\Bx, t) = \int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx} \tilde{\phi}(\Bp, t),

and due to a Hermitian assumption (i.e. real field) this implies
\label{eqn:qftLecture6:80}
\tilde{\phi}^\conj(\Bp, t) = \tilde{\phi}(-\Bp, t).

We found that the Klein-Gordon equation implied that the momentum space representation obey Harmonic oscillator equations
\label{eqn:qftLecture6:100}
\begin{aligned}
\ddot{\tilde{\phi}}(\Bp, t) &= – \omega_\Bp \tilde{\phi}(\Bp, t) \\
\omega_\Bp &= \sqrt{\Bp^2 + m^2}.
\end{aligned}

We may represent the solution to this equation as
\label{eqn:qftLecture6:120}
\tilde{\phi}(\Bq, t) = \inv{\sqrt{2 \omega_\Bq}} \lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} b_\Bq^\conj
}.

This is a general solution, but imposing $$a_\Bq = b_{-\Bq}$$ ensures \ref{eqn:qftLecture6:80} is satisfied.
This leaves us with
\label{eqn:qftLecture6:140}
\tilde{\phi}(\Bq, t) = \inv{\sqrt{2 \omega_\Bq}} \lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\conj
}.

We want to show that iff
\label{eqn:qftLecture6:200}
\antisymmetric{a_\Bq}{a^\dagger_\Bp} = \lr{ 2 \pi }^3 \delta^3(\Bp – \Bq),

then
\label{eqn:qftLecture6:220}
\antisymmetric{\pi(\By, t)}{\phi(\Bx, t)} = -i \delta^3(\Bx – \By)

where everything else commutes (i.e. $$\antisymmetric{a_\Bp}{a_\Bq} = \antisymmetric{a_\Bp^\dagger}{a_\Bq^\dagger} = 0$$).
We will only show one direction, but you can go the other way too.

\label{eqn:qftLecture6:240}
\phi(\Bx, t)
=
\int \frac{d^3 p}{(2\pi)^3 \sqrt{2 \omega_\Bp}} e^{i \Bp \cdot \Bx}
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}

\label{eqn:qftLecture6:260}
\pi(\Bx, t) = \dot{\phi}
=
i \int \frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}} \omega_\Bq e^{i \Bq \cdot \Bx}
\lr{
-e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}

The commutator is
\label{eqn:qftLecture6:280}
\begin{aligned}
\antisymmetric{\pi(\By, t)}{\phi(\Bx, t)}
&=
i \int \frac{d^3 p}{(2\pi)^3
\sqrt{2 \omega_\Bp}}
\frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}}
\omega_\Bq
e^{i \Bp \cdot \By + i \Bq \cdot \Bx}
\antisymmetric
{
-e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}
{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
} \\
&=
i \int \frac{d^3 p}{(2\pi)^3
\sqrt{2 \omega_\Bp}}
\frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}}
\omega_\Bq
e^{i \Bp \cdot \By + i \Bq \cdot \Bx}
\lr{
– e^{i (\omega_\Bp -\omega_\Bq) t}
\antisymmetric { a_\Bq } { a_{-\Bp}^\dagger }
+
e^{i (\omega_\Bq -\omega_\Bp) t}
\antisymmetric{a_{-\Bq}^\dagger}{ a_\Bp }
} \\
&=
i \int \frac{d^3 p}{(2\pi)^3
\sqrt{2 \omega_\Bp}}
\frac{d^3 q}{(2\pi)^3 \sqrt{2 \omega_\Bq}}
\omega_\Bq (2\pi)^3
e^{i \Bp \cdot \By + i \Bq \cdot \Bx}
\lr{
– e^{i (\omega_\Bp -\omega_\Bq) t} \delta^3(\Bq + \Bp)
– e^{i (\omega_\Bq -\omega_\Bp) t} \delta^3(-\Bq -\Bp)
} \\
&=
– 2 i \int \frac{d^3 p}{(2\pi)^3
2 \omega_\Bp}
\omega_\Bp
e^{i \Bp \cdot (\By – \Bx)} \\
&=
-i \delta^3(\By – \Bx),
\end{aligned}

which is what we wanted to prove.

Free Hamiltonian

We call the $$\lambda = 0$$ case the “free” Hamiltonian.

\label{eqn:qftLecture6:300}
\begin{aligned}
H
&= \int d^3 x \lr{ \inv{2} \pi^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m^2}{2} \phi^2 } \\
&=
\inv{2} \int d^3 x
\frac{d^3 p}{(2\pi)^3}
\frac{d^3 q}{(2\pi)^3}
\frac{
e^{i (\Bp + \Bq)\cdot \Bx}
}{\sqrt{2 \omega_\Bp}
\sqrt{2 \omega_\Bq}}
\lr{

(\omega_\Bp)
(\omega_\Bq)
\lr{
-e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
-e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}
}

\Bp \cdot \Bq
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}
+
m^2
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
e^{-i \omega_\Bq t} a_\Bq
+
e^{i \omega_\Bq t} a_{-\Bq}^\dagger
}.
\end{aligned}

An immediate simplification is possible by identifying a delta function factor $$\int d^3 x e^{i(\Bp + \Bq) \cdot \Bx}/(2\pi)^3 = \delta^3(\Bp + \Bq)$$, so
\label{eqn:qftLecture6:1060}
\begin{aligned}
H
&=
\inv{2}
\int
\frac{d^3 p}{(2\pi)^3}
\frac{1
}{2 \omega_\Bp}
\lr{

(\omega_\Bp)^2
\lr{
-e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
-e^{-i \omega_\Bp t} a_{-\Bp}
+
e^{i \omega_\Bp t} a_{\Bp}^\dagger
}
}
+
(\Bp^2 + m^2)
\lr{
e^{-i \omega_\Bp t} a_\Bp
+
e^{i \omega_\Bp t} a_{-\Bp}^\dagger
}
\lr{
e^{-i \omega_\Bp t} a_{-\Bp}
+
e^{i \omega_\Bp t} a_{\Bp}^\dagger
} \\
&=
\inv{2} \int \frac{d^3 p}{(2 \pi)^3} \inv{2 \omega_\Bp}
\lr{
a_\Bp a_{-\Bp}
\lr{
{-\omega_\Bp^2 e^{-2 i \omega_\Bp t}}
+
{\omega_\Bp^2 e^{-2 i \omega_\Bp t}}
}
+
a_{-\Bp}^\dagger a_{\Bp}^\dagger
\lr{
-{\omega_\Bp^2 e^{2 i \omega_\Bp t}}
+
{\omega_\Bp^2 e^{2 i \omega_\Bp t}}
}
+
a_\Bp a^\dagger_{\Bp} \omega^2_\Bp(1 + 1)
+
a^\dagger_{-\Bp} a_{-\Bp} \omega^2_\Bp(1 + 1)
}
\end{aligned}

When all is said and done we are left with
\label{eqn:qftLecture6:320}
H =
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp}{2} \lr{
a^\dagger_{-\Bp}
a_{-\Bp}
+
a_{\Bp}
a^\dagger_{\Bp}
}

and finally with a $$\Bp \rightarrow -\Bp$$ transformation ($$\iiint_{-R}^R d^3 p \rightarrow (-1)^3 \iiint_R^{-R} d^3 p’ = (-1)^6 \iiint_{-R}^R d^3 p’$$) in the first integral our free Hamiltonian ($$\lambda = 0$$) is
\label{eqn:qftLecture6:340}
\boxed{
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp}{2} \lr{
a^\dagger_{\Bp}
a_{\Bp}
+
a_{\Bp}
a^\dagger_{\Bp}
}
}

From the commutator relationship \ref{eqn:qftLecture6:200} we can write
\label{eqn:qftLecture6:360}
a_\Bp a^\dagger_\Bq
=
a^\dagger_\Bq
a_\Bp
+ (2 \pi)^3 \delta^3(\Bp – \Bq),

so

\label{eqn:qftLecture6:380}
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
\lr{
a^\dagger_{\Bp}
a_{\Bp}
+
\inv{2} (2 \pi)^3 \delta^3(0)
}

The delta function term can be interpreted using
\label{eqn:qftLecture6:400}
(2 \pi)^3 \delta^3(\Bq)
= \int d^3 x e^{i \Bq \cdot \Bx},

so when $$\Bq = 0$$
\label{eqn:qftLecture6:420}
(2 \pi)^3 \delta^3(0) = \int d^3 x = V.

We can write the Hamiltonian now in terms of the volume
\label{eqn:qftLecture6:440}
\boxed{
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
a_{\Bp}
+ V_3
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp }{2} \times 1
}

QM SHO review

In units with $$m = 1$$ the non-relativistic QM SHO has the Hamiltonian
\label{eqn:qftLecture6:460}
H
= \inv{2} p^2 + \frac{\omega^2}{2} q^2.

If we define a position operator with a time-domain Fourier representation given by
\label{eqn:qftLecture6:480}
q = \inv{\sqrt{2\omega}} \lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} },

where the Fourier coefficients $$a, a^\dagger$$ are operator valued, then the momentum operator is
\label{eqn:qftLecture6:500}
p = \dot{q} =
\frac{i\omega}{\sqrt{2\omega}} \lr{ -a e^{-i \omega t} + a^\dagger e^{i \omega t} },

or inverting for $$a, a^\dagger$$
\label{eqn:qftLecture6:1080}
\begin{aligned}
a &= \sqrt{\frac{\omega}{2}} \lr{ q – \inv{i \omega} p } e^{-i \omega t} \\
a^\dagger &= \sqrt{\frac{\omega}{2}} \lr{ q + \inv{i \omega} p } e^{i \omega t}.
\end{aligned}

By inspection it is apparent that the product $$a^\dagger a$$ will be related to the Hamiltonian (i.e. a difference of squares). That product is
\label{eqn:qftLecture6:1120}
\begin{aligned}
a^\dagger a
&=
\frac{\omega}{2}
\lr{ q + \inv{i \omega} p }
\lr{ q – \inv{i \omega} p } \\
&=
\frac{\omega}{2} \lr{ q^2 + \inv{\omega^2} p^2 – \inv{i \omega} \antisymmetric{q}{p} } \\
&= \inv{2 \omega} \lr{ p^2 + \omega^2 q^2 – \omega },
\end{aligned}

or
\label{eqn:qftLecture6:1140}
H = \omega \lr{ a^\dagger a + \inv{2} }.

We can glean some of the properties of $$a, a^\dagger$$ by computing the commutator of $$p, q$$, since that has a well known value
\label{eqn:qftLecture6:520}
\begin{aligned}
i
&= \antisymmetric{q}{p} \\
&=
\frac{i\omega}{2 \omega} \antisymmetric
{ a e^{-i \omega t} + a^\dagger e^{i \omega t} }
{ -a e^{-i \omega t} + a^\dagger e^{i \omega t} } \\
&=
\frac{i}{2} \lr{
\antisymmetric{a}{a^\dagger} –
\antisymmetric{a^\dagger}{a} } \\
&=
i
\antisymmetric{a}{a^\dagger},
\end{aligned}

so
\label{eqn:qftLecture6:1100}
\antisymmetric{a}{a^\dagger} = 1.

The operator $$a^\dagger a$$ is the workhorse of the Hamiltonian and worth studying independently. In particular, assume that we have a set of states $$\ket{n}$$ that are eigenstates of $$a^\dagger a$$ with eigenvalues $$\lambda_n$$, that is
\label{eqn:qftLecture6:1160}
a^\dagger a \ket{n} = \lambda_n \ket{n}.

The action of $$a^\dagger a$$ on $$a^\dagger \ket{n}$$ is easy to compute
\label{eqn:qftLecture6:1180}
\begin{aligned}
a^\dagger a a^\dagger \ket{n}
&=
a^\dagger \lr{ a^\dagger a + 1 } \ket{n} \\
&=
\lr{ \lambda_n + 1 } a^\dagger \ket{n},
\end{aligned}

so $$\lambda_n + 1$$ is an eigenvalue of $$a^\dagger \ket{n}$$. The state $$a^\dagger \ket{n}$$ has an energy eigenstate that is one unit of energy larger than $$\ket{n}$$. For this reason we called $$a^\dagger$$ the raising (or creation) operator.
Similarly,
\label{eqn:qftLecture6:1200}
\begin{aligned}
a^\dagger a a \ket{n}
&=
\lr{ a a^\dagger – 1 } a \ket{n} \\
&=
(\lambda_n – 1) a \ket{n},
\end{aligned}

so $$\lambda_n – 1$$ is the energy eigenvalue of $$a \ket{n}$$, having one less unit of energy than $$\ket{n}$$.
We call $$a$$ the annihilation (or lowering) operator.
If we argue that there is a lowest energy state, perhaps designated as $$\ket{0}$$ then we must have
\label{eqn:qftLecture6:560}
a\ket{0} = 0,

by the assumption that there are no energy eigenstates with less energy than $$\ket{0}$$.
We can think of higher order states being constructed from the ground state from using the raising operator $$a^\dagger$$
\label{eqn:qftLecture6:580}
\ket{n} = \frac{(a^\dagger)^n}{\sqrt{n!}} \ket{0}

Discussion

We’ve diagonalized in the Fourier representation for the momentum space fields. For every value of momentum $$\Bp$$ we have a quantum SHO.

For our field space we call our space the Fock vacuum and
\label{eqn:qftLecture6:620}
a_\Bp\ket{0} = 0,

and call $$a_\Bp$$ the “annihilation operator”, and
call $$a^\dagger_\Bp$$ the “creation operator”.
We say that $$a^\dagger_\Bp \ket{0}$$ is the creation of a state of a single particle of momentum $$\Bp$$ by $$a^\dagger_\Bp$$.

We are discarding the volume term, a procedure called “normal ordering”. We define
\label{eqn:qftLecture6:640}
\frac{a^\dagger a + a a^\dagger}{2}
:\equiv
a^\dagger a

We are essentially forgetting the vacuum energy as some sort of unobservable quantity, leaving us with the free Hamiltonian of
\label{eqn:qftLecture6:660}
H_0 =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
a_{\Bp}

Consider
\label{eqn:qftLecture6:680}
\begin{aligned}
H_0
a^\dagger_\Bq \ket{0}
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
a_{\Bp}
a^\dagger_\Bq
\ket{0} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp}
\lr{
a^\dagger_\Bq a_\Bp + (2 \pi)^3 \delta^3(\Bp – \Bq)
}
\ket{0} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp
a^\dagger_{\Bp} \lr{
a^\dagger_\Bq {a_\Bp \ket{0}}
+ (2 \pi)^3 \delta^3(\Bp – \Bq) \ket{0}
} \\
&=
\omega_\Bq a^\dagger_\Bq \ket{0}.
\end{aligned}

Question:

Is it possible to modify the Lagrangian or Hamiltonian that we start with so that this vacuum ground state is eliminated? Answer: Only by imposing super-symmetric constraints (that pairs this (Bosonic) Hamiltonian to a Fermonic system in a way that there is exact cancellation).

We will see that the momentum operator has the form
\label{eqn:qftLecture6:700}
\mathcal{P}
=
\int \frac{d^3 p}{(2 \pi)^3} \Bp a^\dagger_\Bp a_\Bp.

We say that $$a^\dagger_\Bp a^\dagger_\Bq \ket{0}$$ a two particle space with energy $$\omega_\Bp + \omega_q$$, and $$(a^\dagger_\Bp)^m (a^\dagger_\Bq)^n \ket{0} \equiv (a^\dagger_\Bp)^m \ket{0} \otimes (a^\dagger_\Bq)^n \ket{0}$$, a $$m + n$$ particle space.

There is a connection to statistical mechanics that is of interest

\label{eqn:qftLecture6:720}
\begin{aligned}
\expectation{E}
&= \inv{Z} \sum_n E_n e^{-E_n/\kB T} \\
&= \inv{Z} \sum_n \bra{n} e^{-\hat{H}/\kB T} \hat{H} \ket{n},
\end{aligned}

so for a SHO Hamiltonian system
\label{eqn:qftLecture6:740}
\begin{aligned}
\expectation{E}
&= \inv{Z} \sum_n e^{-E_n/\kB T} \bra{n} \hat{H} \ket{n} \\
&= \inv{Z} \sum_n e^{-E_n/\kB T} \bra{n} \omega a^\dagger a \ket{n} \\
&= \frac{\omega}{e^{\omega/\kB T} – 1 } \\ \\
&= \expectation{ \omega a^\dagger a }_{\kB T}
\end{aligned}

the $$\kB T$$ ensemble average energy for a SHO system. Note that this sum was evaluated by noting that $$\bra{n} a^\dagger a \ket{n} = n$$ which leaves sums of the form
\label{eqn:qftLecture6:1220}
\begin{aligned}
\frac{\sum_{n = 0}^\infty n a^n }{\sum_{n = 0}^\infty a^n}
&=
a \frac{\sum_{n = 1}^\infty n a^{n-1} }{\sum_{n = 0}^\infty a^n} \\
&=
a (1 – a) \frac{d}{da} \lr{ \inv{1 – a} } \\
&=
\frac{a}{1 – a}.
\end{aligned}

If we consider a real scalar field of mass $$m$$ we have $$\omega_\Bp = \sqrt{ \Bp^2 + m^2 }$$, but for a Maxwell field $$\BE, \BB$$ where $$m = 0$$, our dispersion relation is $$\omega_\Bp = \Abs{\Bp}$$.

We will see that for a free Maxwell field (no charges or currents) the Hamiltonian is
\label{eqn:qftLecture6:760}
H_{\text{Maxwell}} =
\sum_{i = 1}^2
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp {a^i}^\dagger_\Bp {a^i}_\Bp,

where $$i$$ is a polarization index.

We expect that we can evaluate an average such as \ref{eqn:qftLecture6:740} for our field, and operate using the analogy

\label{eqn:qftLecture6:780}
\begin{aligned}
a a^\dagger &= a^\dagger a + 1 \\
a_\Bp a_\Bp^\dagger &= a_\Bp^\dagger a_\Bp + V_3.
\end{aligned}

so if we rescale by $$\sqrt{V_3}$$
\label{eqn:qftLecture6:800}
a_\Bp = \sqrt{V_3} \tilde{a}_\Bp

Then we have commutator relations like standard QM
\label{eqn:qftLecture6:820}
\tilde{a} \tilde{a}^\dagger = \tilde{a}^\dagger \tilde{a} + 1.

So we can immediately evaluate the energy expectation for our quantized fields
\label{eqn:qftLecture6:840}
\begin{aligned}
\expectation{H_0}
&=
\expectation{
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp a_\Bp^\dagger a_\Bp
} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp V_3 \expectation{ \tilde{a}^\dagger_\Bp a_\Bp } \\
&=
V_3
\int \frac{d^3 p}{(2 \pi)^3} \frac{\omega_\Bp}{e^{\omega_\Bp/\kB T} – 1}.
\end{aligned}

Using this with the Maxwell field, we have a factor of two from polarization
\label{eqn:qftLecture6:860}
U^{\text{Maxwell}} = 2 V_3
\int \frac{d^3 p}{(2 \pi)^3} \frac{\Abs{\Bp}}{e^{\omega_\Bp/\kB T} – 1},

which is Planck’s law describing the blackbody energy spectrum.

Switching gears: Symmetries.

The question is how to apply the CCR results to moving frames, which is done using Lorentz transformations. Just like we know that the exponential of the Hamiltonian (times time) represents time translations, we will examine symmetries that relate results in different frames.

Examples.

For scalar field(s) with action
\label{eqn:qftLecture6:880}
S = \int d^d x \LL(\phi^i, \partial_\mu \phi^i).

For example, we’ve been using our massive (Boson) real scalar field with Lagrangian density
\label{eqn:qftLecture6:900}
\LL = \inv{2} \partial_\mu \phi\partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi).

Internal symmetry example

\label{eqn:qftLecture6:920}
H = J \sum_{\expectation{n, n’}} \BS_n \cdot \BS_{n’},

where the sum means the sum over neighbouring indexes $$n, n’$$ as sketched in

fig. 1. Neighbouring spin cells.

Such a Hamiltonian is left invariant by the transformation $$\BS_n \rightarrow -\BS_n$$ since the Hamiltonian is quadratic.

Suppose that $$\phi \rightarrow -\phi$$ is a symmetry (it leaves the Lagrangian unchanged). Example

\label{eqn:qftLecture6:940}
\phi =
\begin{bmatrix}
\phi^1 \\
\phi^2 \\
\vdots
\phi^n \\
\end{bmatrix}

the Lagrangian
\label{eqn:qftLecture6:960}
\LL = \inv{2} \partial_\mu \phi^\T \partial^\mu \phi – \frac{m^2}{2} \phi^\T \phi – V(\phi^\T \phi).

If $$O$$ is any $$n \times n$$ orthogonal matrix, then it is symmetry since
\label{eqn:qftLecture6:980}
\phi^\T \phi \rightarrow \phi^\T O^\T O \phi = \phi^\T \phi.

O(2) model, HW, problem 2. Example for complex $$\phi$$
\label{eqn:qftLecture6:1000}
\phi \rightarrow e^{i \phi} \phi,

\label{eqn:qftLecture6:1020}
\phi = \frac{\psi_1 + i \psi_2}{\sqrt{2}}

\label{eqn:qftLecture6:1040}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
\rightarrow
\begin{bmatrix}
\cos\alpha & \sin\alpha \\
-\sin\alpha & \cos\alpha
\end{bmatrix}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}

PHY2403H Quantum Field Theory. Lecture 5: Klein-Gordon equation, Hamilton’s equations, SHOs, momentum space representation, raising and lowering operators. Taught by Prof. Erich Poppitz

DISCLAIMER: Very rough notes from class. Some additional side notes, but otherwise barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

Canonical quantization

Last time we introduced a Lagrangian density associated with the Klein-Gordon equation (with a quadratic potential coupling)
\label{eqn:qftLecture5:20}
L = \int d^3 x
\lr{
\inv{2} \lr{\partial_0 \phi}^2 – \inv{2} \lr{\spacegrad \phi}^2 – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4
}.

This Lagrangian density was related to the action by
\label{eqn:qftLecture5:40}
S = \int dt L = \int dt d^3 x \LL,

with momentum canonically conjugate to the field $$\phi$$ defined as
\label{eqn:qftLecture5:60}
\Pi(\Bx, t) = \frac{\delta \LL}{\delta \phidot(\Bx, t) } = \PD{\phidot(\Bx, t)}{\LL}

The Hamiltonian defined as
\label{eqn:qftLecture5:80}
H = \int d^3 x \lr{ \Pi(\Bx, t) \phidot(\Bx, t) – \LL },

led to
\label{eqn:qftLecture5:680}
H
= \int d^3 x
\lr{ \inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4 }.

Like the Lagrangian density, we may introduce a Hamiltonian density $$\mathcal{H}$$ as
\label{eqn:qftLecture5:100}
H = \int d^3 x \mathcal{H}(\Bx, t).

For our Klein-Gordon system, this is
\label{eqn:qftLecture5:120}
\mathcal{H}(\Bx, t) =
\inv{2} \Pi^2 + (\spacegrad \phi)^2 + \inv{2} m^2 \phi^2 + \frac{\lambda}{4} \phi^4.

Canonical Commutation Relations (CCR)

:

We quantize the system by promoting our fields to Heisenberg-Picture (HP) operators, and imposing commutation relations
\label{eqn:qftLecture5:140}
\antisymmetric{\hat{\Pi}(\Bx, t)}{\hat{\phi}(\By, t)} = -i \delta^3 (\Bx – \By)

This is in analogy to
\label{eqn:qftLecture5:160}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

To choose a representation, we may map the $$\Psi$$ of QM $$\rightarrow$$ to a wave functional $$\Psi[\phi]$$
\label{eqn:qftLecture5:180}
\hat{\phi}(\By, t) \Psi[\phi] = \phi(\By, t) \Psi[\phi]

This is similar to the QM wave functions
\label{eqn:qftLecture5:200}
\begin{aligned}
\hat{q}_i \Psi(\setlr{q}) &= q_i \Psi(q) \\
\hat{p}_i \Psi(\setlr{q}) &= -i \PD{q_i}{} \Psi(p)
\end{aligned}

Our momentum operator is quantized by expressing it in terms of a variational derivative
\label{eqn:qftLecture5:220}
\hat{\Pi}(\Bx, t) = -i \frac{\delta}{\delta \phi(\Bx, t)}.

(Fixme: I’m not really sure exactly what is meant by using the variation derivative $$\delta$$ notation here), and to
quantize the Hamiltonian we just add hats, assuming that our fields are all now HP operators
\label{eqn:qftLecture5:240}
\hat{\mathcal{H}}(\Bx, t)
=
\inv{2} \hat{\Pi}^2 + (\spacegrad \hat{\phi})^2 + \inv{2} m^2 \hat{\phi}^2 + \frac{\lambda}{4} \hat{\phi}^4.

QM SHO review

Recall the QM SHO had a Hamiltonian
\label{eqn:qftLecture5:260}
\hat{H} = \inv{2} \hat{p}^2 + \inv{2} \omega^2 \hat{q}^2,

where
\label{eqn:qftLecture5:280}
\antisymmetric{\hat{p}}{\hat{q}} = -i,

and that
HP time evolution operators $$O$$ satisfied
\label{eqn:qftLecture5:700}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}.

In particular
\label{eqn:qftLecture5:300}
\begin{aligned}
\ddt{\hat{p}}
&= i \antisymmetric{\hat{H}}{\hatp} \\
&= i \frac{\omega^2}{2} \antisymmetric{\hat{q}^2}{\hatp} \\
&= i \frac{\omega^2}{2} (2 i \hat{q}) \\
&= -i \omega^2 \hat{q},
\end{aligned}

and
\label{eqn:qftLecture5:320}
\begin{aligned}
\ddt{\hat{q}}
&= i \antisymmetric{\hat{H}}{\hat{q}} \\
&= i \inv{2} \antisymmetric{\hatp^2}{\hat{q}} \\
&= \frac{i}{2}(-2 i \hatp ) \\
&= \hatp.
\end{aligned}

Applying the time evolution operator twice, we find
\label{eqn:qftLecture5:340}
\frac{d^2}{dt^2}{\hat{q}}
= \ddt{\hat{p}}
= – \omega^2 \hat{q}.

We see that the Heisenberg operators obey the classical equations of motion.

Now we want to try this with the quantized QFT fields we’ve promoted to operators
\label{eqn:qftLecture5:360}
\begin{aligned}
\ddt{\hat{\Pi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\Pi}(\Bx, t)} \\
&=
i \int d^3 y \inv{2} \antisymmetric{ \lr{\spacegrad \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
+
i \int d^3 y \frac{m^2}{2} \antisymmetric{ \phihat(\By)^2 }{\hat{\Pi}(\Bx) }
+
i \frac{\lambda}{4} \int d^3 \antisymmetric{ \phihat(\By)^4 }{\hat{\Pi}(\Bx) }
\end{aligned}

Starting with the non-gradient commutators, and utilizing the HP field analogues of the relations $$\antisymmetric{\hat{q}^n}{\hatp} = n i \hat{q}^{n-1}$$, we find
\label{eqn:qftLecture5:780}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^2 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 2 i \phihat(\By) \delta^3(\Bx – \By)
= 2 i \phihat(\Bx).

\label{eqn:qftLecture5:740}
\int d^3 y \antisymmetric{ \lr{ \phihat(\By) }^4 }{\hat{\Pi}(\Bx) }
=
\int d^3 y 4 i \phihat(\By)^3 \delta^3(\Bx – \By)
= 4 i \phihat(\Bx)^3.

For the gradient commutators, we have more work. Prof Poppitz blitzed through that, just calling it integration by parts. I had trouble seeing what he was doing, so here’s a more explicit dumb expansion required to calculate the commutator
\label{eqn:qftLecture5:720}
\begin{aligned}
\int d^3 y (\spacegrad \phihat(\By))^2 \hat{\Pi}(\Bx)
&=
\int d^3 y
&=
\int d^3 y
\lr{ \spacegrad (\phihat(\By) \hat{\Pi}(\Bx)) } \\
&=
\int d^3 y
\lr{ \spacegrad (\hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By)) } \\
&=
\int d^3 y
\Biglr{
+ i
} \\
&=
\int d^3 y
\Biglr{
\spacegrad \lr{ \hat{\Pi}(\Bx) \phihat(\By) + i \delta^3(\Bx – \By) } \cdot \spacegrad \phihat(\By)
+ i
} \\
&=
\int d^3 y
\hat{\Pi}(\Bx)
\lr{
}
+ 2 i
\int d^3 y
&=
\int d^3 y
+
2 i
\int d^3 y

2 i
\int d^3 y
\delta^3(\Bx – \By) \spacegrad^2 \phihat(\By) \\
&=
\int d^3 y
+
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

\end{aligned}

Here we take advantage of the fact that the derivative operators $$\spacegrad = \spacegrad_\By$$ commute with $$\hat{\Pi}(\Bx)$$, and use the identity
$$\spacegrad \cdot (a \spacegrad b) = (\spacegrad a) \cdot (\spacegrad b) + a \spacegrad^2 b$$, so the commutator is
\label{eqn:qftLecture5:800}
\begin{aligned}
&=
2 i
\int_\partial d^2 y
\delta^3(\Bx – \By)

&=

\end{aligned}

where the boundary integral is presumed to be zero (without enough justification.) All the pieces can now be put back together
\label{eqn:qftLecture5:820}
\ddt{} \hat{\Pi}(\Bx, t)
=

m^2 \phihat(\Bx, t)

\lambda \phihat^3(\Bx, t).

Now, for the $$\phihat$$ time evolution, which is much easier
\label{eqn:qftLecture5:380}
\begin{aligned}
\ddt{\hat{\phi}}(\Bx, t)
&= i \antisymmetric{\hatH}{\hat{\phi}(\Bx, t)} \\
&= i \inv{2} \int d^3 y \antisymmetric{\hat{\Pi}^2(\By)}{\hat{\phi}(\Bx)} \\
&= i \inv{2} \int d^3 y (-2 i) \hat{\Pi}(\By, t) \delta^3(\Bx – \By) \\
&= \hat{\Pi}(\Bx, t)
\end{aligned}

\label{eqn:qftLecture5:400}
\frac{d^2}{dt^2}{\hat{\phi}}(\Bx, t)
=
-m^2 \phi – \lambda \phihat^3.

That is
\label{eqn:qftLecture5:420}
\ddot{\phihat} – \spacegrad^2 \phihat + m^2 \phihat + \lambda \phihat^3 = 0,

which is the classical Euler-Lagrange equation, also obeyed by the Heisenberg operator $$\phi(\Bx, t)$$. When $$\lambda = 0$$ this is the Klein-Gordon equation.

Momentum space representation.

Dropping hats, we now consider the momentum space representation of our operators, as determined by Fourier transform pairs
\label{eqn:qftLecture5:440}
\begin{aligned}
\phi(\Bx, t) &= \int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx} \tilde{\phi}(\Bp, t) \\
\tilde{\phi}(\Bp, t) &= \int d^3 x e^{-i \Bp \cdot \Bx} \phi(\Bx, t)
\end{aligned}

We can discover a representation of the delta function by applying these both in turn
\label{eqn:qftLecture5:480}
\tilde{\phi}(\Bp, t)
= \int d^3 x e^{-i \Bp \cdot \Bx} \int \frac{d^3 q}{(2 \pi)^3} e^{i \Bq \cdot \Bx} \tilde{\phi}(\Bq, t)

so
\label{eqn:qftLecture5:500}
\boxed{
\int d^3 x e^{i \BA \cdot \Bx} = (2 \pi)^3 \delta^3(\BA)
}

Also observe that $$\phi^\conj(\Bx, t) = \phi(\Bx, t)$$ iff $$\tilde{\phi}(\Bp, t) = \tilde{\phi}^\conj(-\Bp, t)$$.

We want the EOM for $$\tilde{\phi}(\Bp, t)$$ where the operator obeys the KG equation
\label{eqn:qftLecture5:520}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi(\Bx, t) = 0

Inserting the transform relation \ref{eqn:qftLecture5:440} we get
\label{eqn:qftLecture5:540}
\int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx}
\lr{
\ddot{\tilde{\phi}}(\Bp, t) + \lr{ \Bp^2 + m^2 }
\tilde{\phi}(\Bp, t)
}
= 0,

or
\label{eqn:qftLecture5:580}
\boxed{
\ddot{\tilde{\phi}}(\Bp, t) = – \omega_\Bp^2 \,\tilde{\phi}(\Bp, t),
}

where
\label{eqn:qftLecture5:560}
\omega_\Bp = \sqrt{ \Bp^2 + m^2 }.

The Fourier components of the HP operators are SHOs!

As we have SHO’s and know how to deal with these in QM, we use the same strategy, introducing raising and lowering operators
\label{eqn:qftLecture5:600}
\tilde{\phi}(\Bp, t) = \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{-i \omega_\Bp t } a_\Bp + e^{i \omega_\Bp t} a^\dagger_{-\Bp}
}

Observe that
\label{eqn:qftLecture5:840}
\begin{aligned}
\tilde{\phi}^\dagger(-\Bp, t)
&= \inv{\sqrt{2 \omega_\Bp}} \lr{ e^{i \omega_\Bp t } a^\dagger_{-\Bp} + e^{-i \omega_\Bp t} a_{\Bp} } \\
&=
\tilde{\phi}(\Bp, t),
\end{aligned}

or
\label{eqn:qftLecture5:620}
\tilde{\phi}^\dagger(\Bp, t) = \tilde{\phi}(-\Bp, t),

so $$\phi(\Bp, t)$$ has a real representation in terms of $$a_\Bp$$.

We will find (Wednesday) that
\label{eqn:qftLecture5:640}
\antisymmetric{a_\Bq}{a^+_\Bp} = \delta^3(\Bp – \Bq) (2 \pi)^3.

These are equivalent to
\label{eqn:qftLecture5:660}
\antisymmetric{\hat{\Pi}(\By, t)}{\tilde{\phi}(\Bx, t)} = -i \delta^3(\Bx – \By)

Notes so far for phy2403, Quantum Field Theory I

September 23, 2018 phy2403 No comments

Here’s an aggregate collection of notes for QFT I that I’ll update as the term progresses.  It contains very rough notes for the first 4 lectures (6 hours of material).

These notes also contain my (ungraded) problem set I solution from a few years ago.  I was originally going to take the class and had started preparing for it with some independent reading of the Professor’s class notes (Prof. Luke), and by doing problems he’d posted for a previous year.  I ended up changing my plans and took something else instead, probably to satisfy the graduation requirements for the M.Eng program.  It’s been so long since I’ve done those problems that I don’t even remember all the material that the problems covered (some of which we haven’t gotten to yet in class).

PHY2403H Quantum Field Theory. Lecture 4: Scalar action, least action principle, Euler-Lagrange equations for a field, canonical quantization. Taught by Prof. Erich Poppitz

DISCLAIMER: Very rough notes from class. May have some additional side notes, but otherwise probably barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory I, taught by Prof. Erich Poppitz fall 2018.

Principles (cont.)

• Lorentz (Poincar\’e : Lorentz and spacetime translations)
• locality
• dimensional analysis
• gauge invariance

These are the requirements for an action. We postulated an action that had the form
\label{eqn:qftLecture4:20}
\int d^d x \partial_\mu \phi \partial^\mu \phi,

called the “Kinetic term”, which mimics $$\int dt \dot{q}^2$$ that we’d see in quantum or classical mechanics. In principle there exists an infinite number of local Poincar\’e invariant terms that we can write. Examples:

• $$\partial_\mu \phi \partial^\mu \phi$$
• $$\partial_\mu \phi \partial_\nu \partial^\nu \partial^\mu \phi$$
• $$\lr{\partial_\mu \phi \partial^\mu \phi}^2$$
• $$f(\phi) \partial_\mu \phi \partial^\mu \phi$$
• $$f(\phi, \partial_\mu \phi \partial^\mu \phi)$$
• $$V(\phi)$$

It turns out that nature (i.e. three spatial dimensions and one time dimension) is described by a finite number of terms. We will now utilize dimensional analysis to determine some of the allowed forms of the action for scalar field theories in $$d = 2, 3, 4, 5$$ dimensions. Even though the real world is only $$d = 4$$, some of the $$d < 4$$ theories are relevant in condensed matter studies, and $$d = 5$$ is just for fun (but also applies to string theories.)

With $$[x] \sim \inv{M}$$ in natural units, we must define $$[\phi]$$ such that the kinetic term is dimensionless in d spacetime dimensions

\label{eqn:qftLecture4:40}
\begin{aligned}
[d^d x] &\sim \inv{M^d} \\
[\partial_\mu] &\sim M
\end{aligned}

so it must be that
\label{eqn:qftLecture4:60}
[\phi] = M^{(d-2)/2}

It will be easier to characterize the dimensionality of any given term by the power of the mass units, that is

\label{eqn:qftLecture4:80}
\begin{aligned}
[\text{mass}] &= 1 \\
[d^d x] &= -d \\
[\partial_\mu] &= 1 \\
[\phi] &= (d-2)/2 \\
[S] &= 0.
\end{aligned}

Since the action is
\label{eqn:qftLecture4:100}
S = \int d^d x \lr{ \LL(\phi, \partial_\mu \phi) },

and because action had dimensions of $$\Hbar$$, so in natural units, it must be dimensionless, the Lagrangian density dimensions must be $$[d]$$. We will abuse language in QFT and call the Lagrangian density the Lagrangian.

$$d = 2$$

Because $$[\partial_\mu \phi \partial^\mu \phi ] = 2$$, the scalar field must be dimension zero, or in symbols
\label{eqn:qftLecture4:120}
[\phi] = 0.

This means that introducing any function $$f(\phi) = 1 + a \phi + b\phi^2 + c \phi^3 + \cdots$$ is also dimensionless, and
\label{eqn:qftLecture4:140}
[f(\phi) \partial_\mu \phi \partial^\mu \phi ] = 2,

for any $$f(\phi)$$. Another implication of this is that the a potential term in the Lagrangian $$[V(\phi)] = 0$$ needs a coupling constant of dimension 2. Letting $$\mu$$ have mass dimensions, our Lagrangian must have the form
\label{eqn:qftLecture4:160}
f(\phi) \partial_\mu \phi \partial^\mu \phi + \mu^2 V(\phi).

An infinite number of coupling constants of positive mass dimensions for $$V(\phi)$$ are also allowed. If we have higher order derivative terms, then we need to compensate for the negative mass dimensions. Example (still for $$d = 2$$).
\label{eqn:qftLecture4:180}
\LL =
f(\phi) \partial_\mu \phi \partial^\mu \phi + \mu^2 V(\phi) + \inv{{\mu’}^2}\partial_\mu \phi \partial_\nu \partial^\nu \partial^\mu \phi + \lr{ \partial_\mu \phi \partial^\mu \phi }^2 \inv{\tilde{\mu}^2}.

The last two terms, called \underline{couplings} (i.e. any non-kinetic term), are examples of terms with negative mass dimension. There is an infinite number of those in any theory in any dimension.

Definitions

• Couplings that are dimensionless are called (classically) marginal.
• Couplings that have positive mass dimension are called (classically) relevant.
• Couplings that have negative mass dimension are called (classically) irrelevant.

In QFT we are generally interested in the couplings that are measurable at long distances for some given energy. Classically irrelevant theories are generally not interesting in $$d > 2$$, so we are very lucky that we don’t live in three dimensional space. This means that we can get away with a finite number of classically marginal and relevant couplings in 3 or 4 dimensions. This was mentioned in the Wilczek’s article referenced in the class forum [1]\footnote{There’s currently more in that article that I don’t understand than I do, so it is hard to find it terribly illuminating.}

Long distance physics in any dimension is described by the marginal and relevant couplings. The irrelevant couplings die off at low energy. In two dimensions, a priori, an infinite number of marginal and relevant couplings are possible. 2D is a bad place to live!

$$d = 3$$

Now we have
\label{eqn:qftLecture4:200}
[\phi] = \inv{2}

so that
\label{eqn:qftLecture4:220}
[\partial_\mu \phi \partial^\mu \phi] = 3.

A 3D Lagrangian could have local terms such as
\label{eqn:qftLecture4:240}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \mu^{3/2} \phi^3 + \mu’ \phi^4
+ \lr{\mu”}{1/2} \phi^5
+ \lambda \phi^6.

where $$m, \mu, \mu”$$ all have mass dimensions, and $$\lambda$$ is dimensionless. i.e. $$m, \mu, \mu”$$ are relevant, and $$\lambda$$ marginal. We stop at the sixth power, since any power after that will be irrelevant.

$$d = 4$$

Now we have
\label{eqn:qftLecture4:260}
[\phi] = 1

so that
\label{eqn:qftLecture4:280}
[\partial_\mu \phi \partial^\mu \phi] = 4.

In this number of dimensions $$\phi^k \partial_\mu \phi \partial^\mu$$ is an irrelevant coupling.

A 4D Lagrangian could have local terms such as
\label{eqn:qftLecture4:300}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \mu \phi^3 + \lambda \phi^4.

where $$m, \mu$$ have mass dimensions, and $$\lambda$$ is dimensionless. i.e. $$m, \mu$$ are relevant, and $$\lambda$$ is marginal.

$$d = 5$$

Now we have
\label{eqn:qftLecture4:320}
[\phi] = \frac{3}{2},

so that
\label{eqn:qftLecture4:340}
[\partial_\mu \phi \partial^\mu \phi] = 5.

A 5D Lagrangian could have local terms such as
\label{eqn:qftLecture4:360}
\LL = \partial_\mu \phi \partial^\mu \phi + m^2 \phi^2 + \sqrt{\mu} \phi^3 + \inv{\mu’} \phi^4.

where $$m, \mu, \mu’$$ all have mass dimensions. In 5D there are no marginal couplings. Dimension 4 is the last dimension where marginal couplings exist. In condensed matter physics 4D is called the “upper critical dimension”.

From the point of view of particle physics, all the terms in the Lagrangian must be the ones that are relevant at long distances.

Least action principle (classical field theory).

Now we want to study 4D scalar theories. We have some action
\label{eqn:qftLecture4:380}
S[\phi] = \int d^4 x \LL(\phi, \partial_\mu \phi).

Let’s keep an example such as the following in mind
\label{eqn:qftLecture4:400}
\LL = \underbrace{\inv{2} \partial_\mu \phi \partial^\mu \phi}_{\text{Kinetic term}} – \underbrace{m^2 \phi – \lambda \phi^4}_{\text{all relevant and marginal couplings}}.

The even powers can be justified by assuming there is some symmetry that kills the odd powered terms.

fig. 1. Cylindrical spacetime boundary.

We will be integrating over a space time region such as that depicted in fig. 1, where a cylindrical spatial cross section is depicted that we allow to tend towards infinity. We demand that the field is fixed on the infinite spatial boundaries. The easiest way to demand that the field dies off on the spatial boundaries, that is
\label{eqn:qftLecture4:420}
\lim_{\Abs{\Bx} \rightarrow \infty} \phi(\Bx) \rightarrow 0.

The functional $$\phi(\Bx, t)$$ that obeys the boundary condition as stated extremizes $$S[\phi]$$.

Extremizing the action means that we seek $$\phi(\Bx, t)$$
\label{eqn:qftLecture4:440}
\delta S[\phi] = 0 = S[\phi + \delta \phi] – S[\phi].

How do we compute the variation?
\label{eqn:qftLecture4:460}
\begin{aligned}
\delta S
&= \int d^d x \lr{ \LL(\phi + \delta \phi, \partial_\mu \phi + \partial_\mu \delta \phi) – \LL(\phi, \partial_\mu \phi) } \\
&= \int d^d x \lr{ \PD{\phi}{\LL} \delta \phi + \PD{(\partial_mu \phi)}{\LL} (\partial_\mu \delta \phi) } \\
&= \int d^d x \lr{ \PD{\phi}{\LL} \delta \phi
+ \partial_\mu \lr{ \PD{(\partial_mu \phi)}{\LL} \delta \phi}
– \lr{ \partial_\mu \PD{(\partial_mu \phi)}{\LL} } \delta \phi
} \\
&=
\int d^d x
\delta \phi
\lr{ \PD{\phi}{\LL}
– \partial_\mu \PD{(\partial_mu \phi)}{\LL} }
+ \int d^3 \sigma_\mu \lr{ \PD{(\partial_\mu \phi)}{\LL} \delta \phi }
\end{aligned}

If we are explicit about the boundary term, we write it as
\label{eqn:qftLecture4:480}
\int dt d^3 \Bx \partial_t \lr{ \PD{(\partial_t \phi)}{\LL} \delta \phi }
=
\int d^3 \Bx \evalrange{ \PD{(\partial_t \phi)}{\LL} \delta \phi }{t = -T}{t = T}
– \int dt d^2 \BS \cdot \lr{ \PD{(\spacegrad \phi)}{\LL} \delta \phi }.

but $$\delta \phi = 0$$ at $$t = \pm T$$ and also at the spatial boundaries of the integration region.

This leaves
\label{eqn:qftLecture4:500}
\delta S[\phi] = \int d^d x \delta \phi
\lr{ \PD{\phi}{\LL} – \partial_\mu \PD{(\partial_mu \phi)}{\LL} } = 0 \forall \delta \phi.

That is

\label{eqn:qftLecture4:540}
\boxed{
\PD{\phi}{\LL} – \partial_\mu \PD{(\partial_mu \phi)}{\LL} = 0.
}

This are the Euler-Lagrange equations for a single scalar field.

Returning to our sample scalar Lagrangian
\label{eqn:qftLecture4:560}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \inv{2} m^2 \phi^2 – \frac{\lambda}{4} \phi^4.

This example is related to the Ising model which has a $$\phi \rightarrow -\phi$$ symmetry. Applying the Euler-Lagrange equations, we have
\label{eqn:qftLecture4:580}
\PD{\phi}{\LL} = -m^2 \phi – \lambda \phi^3,

and
\label{eqn:qftLecture4:600}
\begin{aligned}
\PD{(\partial_\mu \phi)}{\LL}
&=
\PD{(\partial_\mu \phi)}{} \lr{
\inv{2} \partial_\nu \phi \partial^\nu \phi } \\
&=
\inv{2} \partial^\nu \phi
\PD{(\partial_\mu \phi)}{}
\partial_\nu \phi
+
\inv{2} \partial_\nu \phi
\PD{(\partial_\mu \phi)}{}
\partial_\alpha \phi g^{\nu\alpha} \\
&=
\inv{2} \partial^\mu \phi
+
\inv{2} \partial_\nu \phi g^{\nu\mu} \\
&=
\partial^\mu \phi
\end{aligned}

so we have
\label{eqn:qftLecture4:620}
\begin{aligned}
0
&=
\PD{\phi}{\LL} -\partial_\mu
\PD{(\partial_\mu \phi)}{\LL} \\
&=
-m^2 \phi – \lambda \phi^3 – \partial_\mu \partial^\mu \phi.
\end{aligned}

For $$\lambda = 0$$, the free field theory limit, this is just
\label{eqn:qftLecture4:640}
\partial_\mu \partial^\mu \phi + m^2 \phi = 0.

Written out from the observer frame, this is
\label{eqn:qftLecture4:660}
(\partial_t)^2 \phi – \spacegrad^2 \phi + m^2 \phi = 0.

With a non-zero mass term
\label{eqn:qftLecture4:680}
\lr{ \partial_t^2 – \spacegrad^2 + m^2 } \phi = 0,

is called the Klein-Gordan equation.

If we also had $$m = 0$$ we’d have
\label{eqn:qftLecture4:700}
\lr{ \partial_t^2 – \spacegrad^2 } \phi = 0,

which is the wave equation (for a massless free field). This is also called the D’Alembert equation, which is familiar from electromagnetism where we have
\label{eqn:qftLecture4:720}
\begin{aligned}
\lr{ \partial_t^2 – \spacegrad^2 } \BE &= 0 \\
\lr{ \partial_t^2 – \spacegrad^2 } \BB &= 0,
\end{aligned}

in a source free region.

Canonical quantization.

\label{eqn:qftLecture4:740}
\LL = \inv{2} \dot{q} – \frac{\omega^2}{2} q^2

This has solution $$\ddot{q} = – \omega^2 q$$.

Let
\label{eqn:qftLecture4:760}
p = \PD{\dot{q}}{\LL} = \dot{q}

\label{eqn:qftLecture4:780}
H(p,q) = \evalbar{p \dot{q} – \LL}{\dot{q}(p, q)}
= p p – \inv{2} p^2 + \frac{\omega^2}{2} q^2 = \frac{p^2}{2} + \frac{\omega^2}{2} q^2

In QM we quantize by mapping Poisson brackets to commutators.
\label{eqn:qftLecture4:800}
\antisymmetric{\hatp}{\hat{q}} = -i

One way to represent is to say that states are $$\Psi(\hat{q})$$, a wave function, $$\hat{q}$$ acts by $$q$$
\label{eqn:qftLecture4:820}
\hat{q} \Psi = q \Psi(q)

With
\label{eqn:qftLecture4:840}
\hatp = -i \PD{q}{},

so
\label{eqn:qftLecture4:860}
\antisymmetric{ -i \PD{q}{} } { q} = -i

Let’s introduce an explicit space time split. We’ll write
\label{eqn:qftLecture4:880}
L = \int d^3 x \lr{
\inv{2} (\partial_0 \phi(\Bx, t))^2 – \inv{2} \lr{ \spacegrad \phi(\Bx, t) }^2 – \frac{m^2}{2} \phi
},

so that the action is
\label{eqn:qftLecture4:900}
S = \int dt L.

The dynamical variables are $$\phi(\Bx)$$. We define
\label{eqn:qftLecture4:920}
\begin{aligned}
\pi(\Bx, t) = \frac{\delta L}{\delta (\partial_0 \phi(\Bx, t))}
&=
\partial_0 \phi(\Bx, t) \\
&=
\dot{\phi}(\Bx, t),
\end{aligned}

called the canonical momentum, or the momentum conjugate to $$\phi(\Bx, t)$$. Why $$\delta$$? Has to do with an implicit Dirac function to eliminate the integral?

\label{eqn:qftLecture4:940}
\begin{aligned}
H
&= \int d^3 x \evalbar{\lr{ \pi(\bar{\Bx}, t) \dot{\phi}(\bar{\Bx}, t) – L }}{\dot{\phi}(\bar{\Bx}, t) = \pi(x, t) } \\
&= \int d^3 x \lr{ (\pi(\Bx, t))^2 – \inv{2} (\pi(\Bx, t))^2 + \inv{2} (\spacegrad \phi)^2 + \frac{m}{2} \phi^2 },
\end{aligned}

or
\label{eqn:qftLecture4:960}
H
= \int d^3 x \lr{ \inv{2} (\pi(\Bx, t))^2 + \inv{2} (\spacegrad \phi(\Bx, t))^2 + \frac{m}{2} (\phi(\Bx, t))^2 }

In analogy to the momentum, position commutator in QM
\label{eqn:qftLecture4:1000}
\antisymmetric{\hat{p}_i}{\hat{q}_j} = -i \delta_{ij},

we “quantize” the scalar field theory by promoting $$\pi, \phi$$ to operators and insisting that they also obey a commutator relationship
\label{eqn:qftLecture4:980}
\antisymmetric{\pi(\Bx, t)}{\phi(\By, t)} = -i \delta^3(\Bx – \By).

References

[1] Frank Wilczek. Fundamental constants. arXiv preprint arXiv:0708.4361, 2007. URL https://arxiv.org/abs/0708.4361.

PHY2403H Quantum Field Theory. Lecture 3: Lorentz transformations and a scalar action. Taught by Prof. Erich Poppitz

September 18, 2018 phy2403 No comments , ,

DISCLAIMER: Very rough notes from class. Some additional side notes, but otherwise barely edited.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz.

Determinant of Lorentz transformations

We require that Lorentz transformations leave the dot product invariant, that is $$x \cdot y = x’ \cdot y’$$, or
\label{eqn:qftLecture3:20}
x^\mu g_{\mu\nu} y^\nu = {x’}^\mu g_{\mu\nu} {y’}^\nu.

Explicitly, with coordinate transformations
\label{eqn:qftLecture3:40}
\begin{aligned}
{x’}^\mu &= {\Lambda^\mu}_\rho x^\rho \\
{y’}^\mu &= {\Lambda^\mu}_\rho y^\rho
\end{aligned}

such a requirement is equivalent to demanding that
\label{eqn:qftLecture3:500}
\begin{aligned}
x^\mu g_{\mu\nu} y^\nu
&=
{\Lambda^\mu}_\rho x^\rho
g_{\mu\nu}
{\Lambda^\nu}_\kappa y^\kappa \\
&=
x^\mu
{\Lambda^\alpha}_\mu
g_{\alpha\beta}
{\Lambda^\beta}_\nu
y^\nu,
\end{aligned}

or
\label{eqn:qftLecture3:60}
g_{\mu\nu}
=
{\Lambda^\alpha}_\mu
g_{\alpha\beta}
{\Lambda^\beta}_\nu

multiplying by the inverse we find
\label{eqn:qftLecture3:200}
\begin{aligned}
g_{\mu\nu}
{\lr{\Lambda^{-1}}^\nu}_\lambda
&=
{\Lambda^\alpha}_\mu
g_{\alpha\beta}
{\Lambda^\beta}_\nu
{\lr{\Lambda^{-1}}^\nu}_\lambda \\
&=
{\Lambda^\alpha}_\mu
g_{\alpha\lambda} \\
&=
g_{\lambda\alpha}
{\Lambda^\alpha}_\mu.
\end{aligned}

This is now amenable to expressing in matrix form
\label{eqn:qftLecture3:220}
\begin{aligned}
(G \Lambda^{-1})_{\mu\lambda}
&=
(G \Lambda)_{\lambda\mu} \\
&=
((G \Lambda)^\T)_{\mu\lambda} \\
&=
(\Lambda^\T G)_{\mu\lambda},
\end{aligned}

or
\label{eqn:qftLecture3:80}
G \Lambda^{-1}
=
(G \Lambda)^\T.

Taking determinants (using the normal identities for products of determinants, determinants of transposes and inverses), we find
\label{eqn:qftLecture3:100}
det(G)
det(\Lambda^{-1})
=
det(G) det(\Lambda),

or
\label{eqn:qftLecture3:120}
det(\Lambda)^2 = 1,

or
$$det(\Lambda)^2 = \pm 1$$. We will generally ignore the case of reflections in spacetime that have a negative determinant.

Smart-alec Peeter pointed out after class last time that we can do the same thing easier in matrix notation
\label{eqn:qftLecture3:140}
\begin{aligned}
x’ &= \Lambda x \\
y’ &= \Lambda y
\end{aligned}

where
\label{eqn:qftLecture3:160}
\begin{aligned}
x’ \cdot y’
&=
(x’)^\T G y’ \\
&=
x^\T \Lambda^\T G \Lambda y,
\end{aligned}

which we require to be $$x \cdot y = x^\T G y$$ for all four vectors $$x, y$$, that is
\label{eqn:qftLecture3:180}
\Lambda^\T G \Lambda = G.

We can find the result \ref{eqn:qftLecture3:120} immediately without having to first translate from index notation to matrices.

Field theory

The electrostatic potential is an example of a scalar field $$\phi(\Bx)$$ unchanged by SO(3) rotations
\label{eqn:qftLecture3:240}
\Bx \rightarrow \Bx’ = O \Bx,

that is
\label{eqn:qftLecture3:260}
\phi'(\Bx’) = \phi(\Bx).

Here $$\phi'(\Bx’)$$ is the value of the (electrostatic) scalar potential in a primed frame.

However, the electrostatic field is not invariant under Lorentz transformation.
We postulate that there is some scalar field
\label{eqn:qftLecture3:280}
\phi'(x’) = \phi(x),

where $$x’ = \Lambda x$$ is an SO(1,3) transformation. There are actually no stable particles (fields that persist at long distances) described by Lorentz scalar fields, although there are some unstable scalar fields such as the Higgs, Pions, and Kaons. However, much of our homework and discussion will be focused on scalar fields, since they are the easiest to start with.

We need to first understand how derivatives $$\partial_\mu \phi(x)$$ transform. Using the chain rule
\label{eqn:qftLecture3:300}
\begin{aligned}
\PD{x^\mu}{\phi(x)}
&=
\PD{x^\mu}{\phi'(x’)} \\
&=
\PD{{x’}^\nu}{\phi'(x’)}
\PD{{x}^\mu}{{x’}^\nu} \\
&=
\PD{{x’}^\nu}{\phi'(x’)}
\partial_\mu \lr{
{\Lambda^\nu}_\rho x^\rho
} \\
&=
\PD{{x’}^\nu}{\phi'(x’)}
{\Lambda^\nu}_\mu \\
&=
\PD{{x’}^\nu}{\phi(x)}
{\Lambda^\nu}_\mu.
\end{aligned}

Multiplying by the inverse $${\lr{\Lambda^{-1}}^\mu}_\kappa$$ we get
\label{eqn:qftLecture3:320}
\PD{{x’}^\kappa}{}
=
{\lr{\Lambda^{-1}}^\mu}_\kappa \PD{x^\mu}{}

This should be familiar to you, and is an analogue of the transformation of the
\label{eqn:qftLecture3:340}
=

Actions

We will start with a classical action, and quantize to determine a QFT. In mechanics we have the particle position $$q(t)$$, which is a classical field in 1+0 time and space dimensions. Our action is
\label{eqn:qftLecture3:360}
S
= \int dt \LL(t)
= \int dt \lr{
\inv{2} \dot{q}^2 – V(q)
}.

This action depends on the position of the particle that is local in time. You could imagine that we have a more complex action where the action depends on future or past times
\label{eqn:qftLecture3:380}
S
= \int dt’ q(t’) K( t’ – t ),

but we don’t seem to find such actions in classical mechanics.

Principles determining the form of the action.

• relativity (action is invariant under Lorentz transformation)
• locality (action depends on fields and the derivatives at given $$(t, \Bx)$$.
• Gauge principle (the action should be invariant under gauge transformation). We won’t discuss this in detail right now since we will start with studying scalar fields.
Recall that for Maxwell’s equations a gauge transformation has the form
\label{eqn:qftLecture3:520}
\phi \rightarrow \phi + \dot{\chi}, \BA \rightarrow \BA – \spacegrad \chi.

Suppose we have a real scalar field $$\phi(x)$$ where $$x \in \mathbb{R}^{1,d-1}$$. We will be integrating over space and time $$\int dt d^{d-1} x$$ which we will write as $$\int d^d x$$. Our action is
\label{eqn:qftLecture3:400}
S = \int d^d x \lr{ \text{Some action density to be determined } }

The analogue of $$\dot{q}^2$$ is
\label{eqn:qftLecture3:420}
\begin{aligned}
\lr{ \PD{x^\mu}{\phi} }
\lr{ \PD{x^\nu}{\phi} }
g^{\mu\nu}
&=
(\partial_\mu \phi) (\partial_\nu \phi) g^{\mu\nu} \\
&= \partial^\mu \phi \partial_\mu \phi.
\end{aligned}

This has both time and spatial components, that is
\label{eqn:qftLecture3:440}
\partial^\mu \phi \partial_\mu \phi =

so the desired simplest scalar action is
\label{eqn:qftLecture3:460}
S = \int d^d x \lr{ \dotphi^2 – (\spacegrad \phi)^2 }.

The measure transforms using a Jacobian, which we have seen is the Lorentz transform matrix, and has unit determinant
\label{eqn:qftLecture3:480}
d^d x’ = d^d x \Abs{ det( \Lambda^{-1} ) } = d^d x.

Question: Four vector form of the Maxwell gauge transformation.

Show that the transformation
\label{eqn:qftLecture3:580}
A^\mu \rightarrow A^\mu + \partial^\mu \chi

is the desired four-vector form of the gauge transformation \ref{eqn:qftLecture3:520}, that is
\label{eqn:qftLecture3:540}
\begin{aligned}
j^\nu
&= \partial_\mu {F’}^{\mu\nu} \\
&= \partial_\mu F^{\mu\nu}.
\end{aligned}

Also relate this four-vector gauge transformation to the spacetime split.

\label{eqn:qftLecture3:560}
\begin{aligned}
\partial_\mu {F’}^{\mu\nu}
&=
\partial_\mu \lr{ \partial^\mu {A’}^\nu – \partial_\nu {A’}^\mu } \\
&=
\partial_\mu \lr{
\partial^\mu \lr{ A^\nu + \partial^\nu \chi }
– \partial_\nu \lr{ A^\mu + \partial^\mu \chi }
} \\
&=
\partial_\mu {F}^{\mu\nu}
+
\partial_\mu \partial^\mu \partial^\nu \chi

\partial_\mu \partial^\nu \partial^\mu \chi \\
&=
\partial_\mu {F}^{\mu\nu},
\end{aligned}

by equality of mixed partials. Expanding \ref{eqn:qftLecture3:580} explicitly we find
\label{eqn:qftLecture3:600}
{A’}^\mu = A^\mu + \partial^\mu \chi,

which is
\label{eqn:qftLecture3:620}
\begin{aligned}
\phi’ = {A’}^0 &= A^0 + \partial^0 \chi = \phi + \dot{\chi} \\
\BA’ \cdot \Be_k = {A’}^k &= A^k + \partial^k \chi = \lr{ \BA – \spacegrad \chi } \cdot \Be_k.
\end{aligned}

The last of which can be written in vector notation as $$\BA’ = \BA – \spacegrad \chi$$.