Noether’s theorem

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

October 14, 2018 phy2403 No comments , , , , , , , , , , , , , ,

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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
\begin{equation}\label{eqn:qftLecture8:20}
\phi(x) \rightarrow \phi(x) + \delta \phi(x),
\end{equation}
such that the transformation of the Lagrangian is only changed by a total derivative
\begin{equation}\label{eqn:qftLecture8:40}
\LL(\phi, \partial_\mu \phi) \rightarrow
\LL(\phi, \partial_\mu \phi)
+ \partial_\mu J_\epsilon^\mu,
\end{equation}
then there is a conserved current
\begin{equation}\label{eqn:qftLecture8:60}
j^\mu = \PD{(\partial_\mu \phi)}{\LL} \delta_\epsilon \phi – J_\epsilon^\mu.
\end{equation}
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
\begin{equation}\label{eqn:qftLecture8:80}
\phi(x) \rightarrow \phi(x) – a^\mu \partial_\mu \phi
\end{equation}
\begin{equation}\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 }
\end{equation}
Here \( J^\mu_\epsilon = \evalbar{J^\mu_\epsilon}{\epsilon = a^\nu} \), and the current is
\begin{equation}\label{eqn:qftLecture8:120}
J^\mu = (\partial^\mu \phi)(-a^\nu \partial_\nu \phi) + {\delta^{\mu}}_\nu a^\nu \LL.
\end{equation}
In particular, we have one such current for each \( \nu \), and we write
\begin{equation}\label{eqn:qftLecture8:140}
{T^\mu}_\nu =
-(\partial^\mu \phi)(\partial_\nu \phi) + {\delta^{\mu}}_\nu \LL.
\end{equation}
By Noether’s theorem, we must have
\begin{equation}\label{eqn:qftLecture8:160}
\partial_\mu
{T^\mu}_\nu = 0, \quad \forall \nu.
\end{equation}

Check:

\begin{equation}\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}
\end{equation}

Example: our potential Lagrangian

\begin{equation}\label{eqn:qftLecture8:180}
\LL = \inv{2} \partial^\mu \phi \partial_\nu \phi – \frac{m^2}{2} \phi^2 – \frac{\lambda}{4} \phi^4
\end{equation}
Written with upper indexes
\begin{equation}\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}
\end{equation}

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
\begin{equation}\label{eqn:qftLecture8:220}
Q^\nu = \int d^3 x T^{0 \nu}.
\end{equation}
We call
\begin{equation}\label{eqn:qftLecture8:240}
Q^0 = \int d^3 x T^{0 0},
\end{equation}
the energy density, and call
\begin{equation}\label{eqn:qftLecture8:260}
P^i = \int d^3 x T^{0 i},
\end{equation}
(i = 1,2,3) the momentum density.

writing this out explicitly the energy density is
\begin{equation}\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}
\end{equation}
and
\begin{equation}\label{eqn:qftLecture8:300}
T^{0i} = \partial^0 \phi \partial^i \phi,
\end{equation}
\begin{equation}\label{eqn:qftLecture8:320}
P^{i} = -\int d^3 x\partial^0 \phi \partial^i \phi
\end{equation}
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
\begin{equation}\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,
\end{equation}
and
\begin{equation}\label{eqn:qftLecture8:360}
P^{i} = \int d^3 x\partial^0 \phi \partial^i \phi
\end{equation}

As an operator we have
\begin{equation}\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
}.
\end{equation}
\begin{equation}\label{eqn:qftLecture8:400}
\hatP^{i} = \int d^3 x \hat{\pi} \partial^i \phi
\end{equation}

We showed that
\begin{equation}\label{eqn:qftLecture8:420}
\ddt{\hatO} = i \antisymmetric{\hatH}{\hatO}
\end{equation}
This implied that \( \phihat, \hat{\pi} \) obey the classical EOMs
\begin{equation}\label{eqn:qftLecture8:440}
\ddt{\phihat} = i \antisymmetric{\hat{H}}{\phihat} = \ddt{\hat{\pi}}
\end{equation}
\begin{equation}\label{eqn:qftLecture8:460}
\ddt{\hat{\pi}} = i \antisymmetric{\hatH}{\hat{\pi}} = …
\end{equation}

In terms of creation and annihilation operators (for the \( \lambda = 0 \) free field), up to a constant
\begin{equation}\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}
\end{equation}
Can show that:

\begin{equation}\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}
\end{equation}
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
\begin{equation}\label{eqn:qftLecture8:520}
\hatU = e^{i \Ba \cdot \hat{\Bp}} = e^{\Ba \cdot \spacegrad}
\end{equation}

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

In one dimension
\begin{equation}\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}
\end{equation}
This uses the Baker-Campbell-Hausdorff formula.

Theorem: Baker-Campbell-Hausdorff

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

\begin{equation}\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}
\end{equation}

\begin{equation}\label{eqn:qftLecture8:640}
f(0) = A
\end{equation}
\begin{equation}\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}
\end{equation}
\begin{equation}\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}
\end{equation}
From
\begin{equation}\label{eqn:qftLecture8:700}
f(1)
= f(0) + f'(0) + \inv{2} f”(0) + \cdots \inv{n!} f^{(n)}(0)
\end{equation}
we have
\begin{equation}\label{eqn:qftLecture8:720}
e^{B} A e^{-B} = A +
\antisymmetric{B}{A} + \inv{2} \antisymmetric{B}{\antisymmetric{B}{A}} + \cdots
\end{equation}

Example:
\begin{equation}\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}
\end{equation}

Application:

\begin{equation}\label{eqn:qftLecture8:760}
e^{i \text{Hermitian} } = \text{unitary}
\end{equation}
\begin{equation}\label{eqn:qftLecture8:860}
e^{i \text{Hermitian} } \times
e^{-i \text{Hermitian} }
= 1
\end{equation}
So
\begin{equation}\label{eqn:qftLecture8:780}
\hatU(\Ba) =
e^{i a^j \hat{p}^j }
\end{equation}
is a unitary operator representing finite translations in a Hilbert space.

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

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
\begin{equation}\label{eqn:qftLecture8:880}
\hatU = e^{i \text{parameter} \hatQ}.
\end{equation}
These represent the action of the symmetry in the Hilbert space.

Example: spatial translation

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

Example: time translation

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

Classical scalar theory

For \( d > 2 \) let’s look at
\begin{equation}\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}
}
\end{equation}

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
\begin{equation}\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}
\end{equation}
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.

\begin{equation}\label{eqn:qftLecture8:980}
S =
\int d^4 x \lr{
\inv{2} \partial^\mu \phi \partial_\mu \phi
}
\end{equation}
Here we have a scale or dilatation invariance
\begin{equation}\label{eqn:qftLecture8:1000}
x \rightarrow x’ = e^{\lambda} x,
\end{equation}
\begin{equation}\label{eqn:qftLecture8:1020}
\phi(x) \rightarrow \phi'(x’) = e^{-\lambda} \phi,
\end{equation}
\begin{equation}\label{eqn:qftLecture8:1040}
d^4 x \rightarrow d^4 x’ = e^{4\lambda} d^4 x,
\end{equation}

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

so the partial of the field transforms as
\begin{equation}\label{eqn:qftLecture8:1420}
\partial^\mu \phi(x) \rightarrow \PD{x’_\mu}{\phi'(x’)} = e^{-2\lambda} \partial^\mu \phi(x),
\end{equation}
and finally
\begin{equation}\label{eqn:qftLecture8:1060}
(\partial_\mu \phi)^2 \rightarrow e^{-4\lambda} \lr{ \partial_\mu \phi(x) }^2.
\end{equation}

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
\begin{equation}\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}
\end{equation}
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 \( \phi \)
\begin{equation}\label{eqn:qftLecture8:1440}
\delta \partial_\mu \phi(x)
=
\partial_\mu’ \phi'(x) – \partial_\mu \phi(x),
\end{equation}
By \ref{eqn:qftLecture8:1420}
\begin{equation}\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}
\end{equation}
so
\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}
or
\begin{equation}\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}
\end{equation}
The variation in the Lagrangian density is thus
\begin{equation}\label{eqn:qftLecture8:1520}
\delta \LL = \partial_\mu J^\mu_\lambda = \partial_\mu \lr{ -\lambda x^\mu \LL },
\end{equation}
and the current is
\begin{equation}\label{eqn:qftLecture8:1540}
J^\mu_\lambda = -\lambda x^\mu \LL.
\end{equation}

The Noether current is
\begin{equation}\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}
\end{equation}
or after flipping signs
\begin{equation}\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}
\end{equation}

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

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

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

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

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

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

October 11, 2018 phy2403 No comments , , , , , , , , , , , , , , ,

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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”
    \begin{equation}\label{eqn:qftLecture9:20}
    \hatU(\alpha) = e^{i \alpha \hatQ}
    \end{equation}
    We found
    \begin{equation}\label{eqn:qftLecture9:40}
    \hatU(\alpha) \phihat \hatU^\dagger(\alpha) = \phihat + i \alpha \antisymmetric{\hatQ}{\phihat} + \cdots
    \end{equation}

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”.
    \begin{equation}\label{eqn:qftLecture9:60}
    \hatQ \ket{0} = 0
    \end{equation}
    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
    \begin{equation}\label{eqn:qftLecture9:80}
    \hatQ = \text{“charge” under \( U(1) \)}.
    \end{equation}
    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:
\begin{equation}\label{eqn:qftLecture9:100}
\LL = \partial_\mu \phi^\conj \partial^\mu \phi – V(\Abs{\phi}),
\end{equation}
where
\begin{equation}\label{eqn:qftLecture9:120}
V(\Abs{\phi}) = m^2 \phi^\conj \phi + \frac{\lambda}{4} \lr{ \phi^\conj \phi }^2.
\end{equation}
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
\begin{equation}\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}
\end{equation}
or simply
\begin{equation}\label{eqn:qftLecture9:780}
V(\phi)
=
\frac{\lambda}{4} \lr{ \phi^\conj \phi – v^2 }^2 + \text{const}.
\end{equation}
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
\begin{equation}\label{eqn:qftLecture9:160}
\phi(x) = v \lr{ 1 + \frac{\rho(x)}{v} } e^{i \alpha(x)/v },
\end{equation}
so
\begin{equation}\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}
\end{equation}

\begin{equation}\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}
\end{equation}

so
\begin{equation}\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}
\end{equation}
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:
\begin{equation}\label{eqn:qftLecture9:240}
j_\mu = \partial_\mu \alpha \lr{ 1 + \text{higher dimensional \( \rho \) terms} }.
\end{equation}

When we quantize
\begin{equation}\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
\end{equation}
\begin{equation}\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.
\end{equation}

\begin{equation}\label{eqn:qftLecture9:300}
j^\mu(x) \ket{0} \ne 0,
\end{equation}
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
\begin{equation}\label{eqn:qftLecture9:320}
m_u = m_d = 0.
\end{equation}
\( m_u m_d \ll m_p \) (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

\begin{equation}\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}
\end{equation}
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)
    \begin{equation}\label{eqn:qftLecture9:360}
    \hatQ\ket{0} = 0.
    \end{equation}
  2. “spontaneously broken”
    \begin{equation}\label{eqn:qftLecture9:380}
    j^\mu(x) \ket{0} \ne 0
    \end{equation}
    (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
\begin{equation}\label{eqn:qftLecture9:400}
x^\mu \rightarrow {x’}^\mu = {\Lambda^\mu}_\nu x^\nu,
\end{equation}
We are going to consider infinitesimal Lorentz transformations
\begin{equation}\label{eqn:qftLecture9:420}
{\Lambda^\mu}_\nu \approx
{\delta^\mu}_\nu + {\omega^\mu}_\nu
,
\end{equation}
where \( {\omega^\mu}_\nu \) is small.
A Lorentz transformation \( \Lambda \) must satisfy \( \Lambda^\T G \Lambda = G \), or
\begin{equation}\label{eqn:qftLecture9:800}
g_{\mu\nu} = {{\Lambda}^\alpha}_\mu g_{\alpha \beta} {{\Lambda}^\beta}_\nu,
\end{equation}
into which we insert the infinitesimal transformation representation
\begin{equation}\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}
\end{equation}
The quadratic term can be ignored, leaving just
\begin{equation}\label{eqn:qftLecture9:840}
0 =
\omega_{\nu\mu}
+
\omega_{\mu\nu},
\end{equation}
or
\begin{equation}\label{eqn:qftLecture9:860}
\omega_{\nu\mu} = – \omega_{\mu\nu}.
\end{equation}
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
\begin{equation}\label{eqn:qftLecture9:440}
x^\mu \rightarrow {x’}^\mu \approx x^\mu + \omega^{\mu\nu} x_\nu,
\end{equation}
the field transforms as
\begin{equation}\label{eqn:qftLecture9:460}
\phi(x) \rightarrow \phi'(x’) = \phi(x)
\end{equation}
or
\begin{equation}\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),
\end{equation}
so
\begin{equation}\label{eqn:qftLecture9:480}
\delta \phi = \phi'(x) – \phi(x) =
-\omega^{\mu\nu} x_\nu \partial_\mu \phi.
\end{equation}

Since \( \LL \) is a scalar
\begin{equation}\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}
\end{equation}
since \( \partial_\nu x_\mu = g_{\nu\mu} \) is symmetric, and \( \omega \) is antisymmetric.
Our current is
\begin{equation}\label{eqn:qftLecture9:520}
J^\mu_\omega
=

\omega^{\mu\nu} x_\mu \LL
.
\end{equation}
Our Noether current is
\begin{equation}\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}
\end{equation}
We identify
\begin{equation}\label{eqn:qftLecture9:560}

{T^\nu}_\mu =
-\partial^\nu \phi \partial_\mu \phi + {\delta^{\nu}}_\mu \LL,
\end{equation}
so the current is
\begin{equation}\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}
\end{equation}
Define
\begin{equation}\label{eqn:qftLecture9:600}
j^{\nu\mu\rho} = \inv{2} \lr{ x^\rho T^{\nu\mu} – x^{\mu} T^{\nu\rho} },
\end{equation}
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 \( \omega_{\rho\mu} \)
\begin{equation}\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}
\end{equation}
which matches \ref{eqn:qftLecture9:580} as desired.

Example. Rotations \( \mu\rho = ij \)

\begin{equation}\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}
\end{equation}
Observe that this has the structure of \( (\Bx \cross \Bp)_k \), where \( \Bp \) is the momentum density of the field.
Let
\begin{equation}\label{eqn:qftLecture9:640}
L_k \equiv Q_k = \int d^3 x J^{0ij} \epsilon_{ijk}.
\end{equation}
We can now quantize and build a generator
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}
\end{equation}
(up to a sign in the exponent which doesn’t matter)
\begin{equation}\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)
\spacegrad \phihat(\Bx) \cross \Bx \\
&=
\phihat(\By) +
\Balpha \cdot \lr{ \spacegrad \phihat(\By ) \cross \By}
\end{aligned}
\end{equation}
Explicitly, in coordinates, this is
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}
\end{equation}
or in matrix form
\begin{equation}\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}.
\end{equation}

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

October 3, 2018 phy2403 No comments , , , , , ,

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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
    \begin{equation}\label{eqn:qftLecture7:20}
    \phi^i =
    \begin{bmatrix}
    \psi_1 \\
    \psi_2 \\
    \vdots \\
    \psi_N \\
    \end{bmatrix}
    \end{equation}
    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
    \begin{equation}\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)
    \end{equation}

  • 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
    \begin{equation}\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},
    \end{equation}
    or if \( \alpha \sim 0 \)
    \begin{equation}\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}
    \end{equation}
    In index notation we write
    \begin{equation}\label{eqn:qftLecture7:100}
    \phi^i \rightarrow \phi^i + \alpha e^{ij} \phi^j,
    \end{equation}
    where \( \epsilon^{12} = +1, \epsilon^{21} = -1 \) is the completely antisymmetric tensor. This can be written in more general form as
    \begin{equation}\label{eqn:qftLecture7:120}
    \phi^i \rightarrow \phi^i + \delta \phi^i,
    \end{equation}
    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
\begin{equation}\label{eqn:qftLecture7:160}
S = \int d^d x \LL.
\end{equation}
EOMs are invariant if under \( \phi(x) \rightarrow \phi'(x) = \phi(x) + \delta_\epsilon \phi(x)\), we have
\begin{equation}\label{eqn:qftLecture7:180}
\LL(\phi) \rightarrow \LL'(\phi’) = \LL(\phi) + \partial_\mu J_\epsilon^\mu(\phi) + O(\epsilon^2).
\end{equation}
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
\begin{equation}\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}
\end{equation}
However from \ref{eqn:qftLecture7:180}
\begin{equation}\label{eqn:qftLecture7:220}
\delta_\epsilon \LL = \partial_\mu J_\epsilon^\mu(\phi, \partial_\mu \phi),
\end{equation}
so after equating these variations we fine that
\begin{equation}\label{eqn:qftLecture7:240}
\delta S = \int d^d x \delta_\epsilon \LL = \int d^d x \partial_\mu J_\epsilon^\mu,
\end{equation}
or
\begin{equation}\label{eqn:qftLecture7:260}
0 = \int d^d x
\partial_\mu \lr{ \frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta \phi – J_\epsilon^\mu },
\end{equation}
or \( \partial_\mu j^\mu = 0 \) provided
\begin{equation}\label{eqn:qftLecture7:280}
\boxed{
j^\mu =
\frac{\delta \LL}{\delta(\partial_\mu \phi)} \delta_\epsilon \phi – J_\epsilon^\mu.
}
\end{equation}

Integrating the divergence of the current over a space time volume, perhaps that of cylinder (time up, space out) is also zero. That is
\begin{equation}\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}
\end{equation}
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
\begin{equation}\label{eqn:qftLecture7:560}
Q = \int d^3x \partial_t j^0,
\end{equation}
and call this the on-shell charge associated with the symmetry.

Spacetime translation.

A spacetime translation has the form
\begin{equation}\label{eqn:qftLecture7:320}
x^\mu \rightarrow {x’}^\mu = x^\mu + a^\mu,
\end{equation}
\begin{equation}\label{eqn:qftLecture7:340}
\phi(x) \rightarrow \phi'(x’) = \phi(x)
\end{equation}
(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
\begin{equation}\label{eqn:qftLecture7:360}
\phi'(x) + a^\mu \partial_\mu \phi'(x) =
\phi'(x) + a^\mu \partial_\mu \phi(x) =
\phi(x),
\end{equation}
so
\begin{equation}\label{eqn:qftLecture7:380}
\phi'(x)
= \phi(x) – a^\mu \partial_\mu \phi'(x)
= \phi(x) + \delta_a \phi(x),
\end{equation}
or
\begin{equation}\label{eqn:qftLecture7:580}
\delta_a \phi(x) = – a^\mu \partial_\mu \phi(x).
\end{equation}
Under \( \phi \rightarrow \phi – a^\mu \partial_\mu \phi \), we have
\begin{equation}\label{eqn:qftLecture7:400}
\LL(\phi) \rightarrow \LL(\phi) – a^\mu \partial_\mu \LL.
\end{equation}
Let’s calculate this with our scalar theory Lagrangian
\begin{equation}\label{eqn:qftLecture7:420}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \frac{m^2}{2} \phi^2 – V(\phi)
\end{equation}
The Lagrangian variation is
\begin{equation}\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}
\end{equation}

So the current is
\begin{equation}\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}
\end{equation}

We really have a current for each \( \nu \) direction and can make that explicit writing
\begin{equation}\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}
\end{equation}
we write
\begin{equation}\label{eqn:qftLecture7:480}
{j^\mu}_\nu = \PD{x_\mu}{\phi} \lr{ – \PD{x^\nu}{\phi} } + {\delta^\mu}_\nu \LL,
\end{equation}
where \( \nu \) are labels which coordinates are translated:
\begin{equation}\label{eqn:qftLecture7:500}
\begin{aligned}
\partial_\nu \phi &= – \partial_\nu \phi \\
\partial_\nu \LL &= – \partial_\nu \LL.
\end{aligned}
\end{equation}
We call the conserved quantities elements of the energy-momentum tensor, and write it as
\begin{equation}\label{eqn:qftLecture7:520}
\boxed{
{T^\mu}_\nu = -\PD{x_\mu}{\phi} \PD{x^\nu}{\phi} + {\delta^\mu}_\nu \LL.
}
\end{equation}

Incidentally, we picked a non-standard sign convention for the tensor, as an explicit expansion of \( T^{00} \), the energy density component, shows
\begin{equation}\label{eqn:qftLecture7:540}
\begin{aligned}
{T^0}_0
&=
-\PD{t}{\phi}
\PD{t}{\phi}
+\inv{2}
\PD{t}{\phi}
\PD{t}{\phi}
– \inv{2} (\spacegrad \phi) \cdot (\spacegrad \phi)
– \frac{m^2}{2} \phi^2 – V(\phi) \\
&=
-\inv{2} \PD{t}{\phi} \PD{t}{\phi}
– \inv{2} (\spacegrad \phi) \cdot (\spacegrad \phi)
– \frac{m^2}{2} \phi^2 – V(\phi).
\end{aligned}
\end{equation}
Had we translated by \( -a^\mu \) we’d have a positive definite tensor instead.