## Momentum operator for the Dirac field?

In the borrowed notes I have for last Monday’s lecture (which I missed) I see the momentum operator defined by
\label{eqn:momentumDirac:20}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.

There’s a “use Noether’s theorem” comment associated with this. For the scalar field, using Noether’s theorem, we identified the conserved charge of a spacetime translation as the momentum operator
\label{eqn:momentumDirac:40}
P^i = \int d^3 x T^{0i} = – \int d^3 x \pi(x) \spacegrad \phi(x),

and if we plugged in the creation and anhillation operator representation of $$\pi, \phi$$, out comes
\label{eqn:momentumDirac:60}
\BP =
\inv{2} \int \frac{d^3 q}{(2\pi)^3} \Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger},

(plus $$e^{\pm 2 i \omega_\Bp t}$$ terms that we can argue away.)

It wasn’t clear to me how this worked with the Dirac field, but it turns out that this does follow systematically as expected. For a spacetime translation
\label{eqn:momentumDirac:80}
x^\mu \rightarrow x^\mu + a^\mu,

we find
\label{eqn:momentumDirac:100}
\delta \Psi = -a^\mu \partial_\mu \Psi,

so for the Dirac Lagrangian, we have
\label{eqn:momentumDirac:120}
\begin{aligned}
\delta \LL
&= \delta \lr{ \overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \Psi } \\
&=
(\delta \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \delta \Psi \\
&=
(-a^\sigma \partial_\sigma \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } (-a^\sigma \partial_\sigma \Psi ) \\
&=
-a^\sigma \partial_\sigma \LL \\
&=
\partial_\sigma (-a^\sigma \LL),
\end{aligned}

i.e. $$J^\mu = -a^\mu \LL$$.
To plugging this into the Noether current calculating machine, we have
\label{eqn:momentumDirac:160}
\begin{aligned}
\PD{(\partial_\mu \Psi)}{\LL}
&=
\PD{(\partial_\mu \Psi)}{} \lr{ \overline{\Psi} i \gamma^\sigma \partial_\sigma \Psi – m \overline{\Psi} \Psi } \\
&=
\overline{\Psi} i \gamma^\mu,
\end{aligned}

and
\label{eqn:momentumDirac:180}
\PD{(\partial_\mu \overline{\Psi})}{\LL} = 0,

so
\label{eqn:momentumDirac:140}
\begin{aligned}
j^\mu
&=
(\delta \overline{\Psi}) \PD{(\partial_\mu \overline{\Psi})}{\LL}
+
\PD{(\partial_\mu \Psi)}{\LL} (\delta \Psi)
– a^\mu \LL \\
&=
\overline{\Psi} i \gamma^\mu (-a^\sigma \partial_\sigma \Psi)
– a^\sigma {\delta^{\mu}}_{\sigma} \LL \\
&=
– a^\sigma
\lr{
\overline{\Psi} i \gamma^\mu \partial_\sigma \Psi
+ {\delta^{\mu}}_{\sigma} \LL
} \\
&=
-a_\nu
\lr{
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL
}.
\end{aligned}

We can now define an energy-momentum tensor
\label{eqn:momentumDirac:200}
T^{\mu\nu}
=
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL.

A couple things are of notable in this tensor. One is that it is not symmetric, and there’s doesn’t appear to be any hope
of making it so. For example, the space+time components are way different
\label{eqn:momentumDirac:220}
\begin{aligned}
T^{0k} &= \overline{\Psi} i \gamma^0 \partial^k \Psi \\
T^{k0} &= \overline{\Psi} i \gamma^k \partial^0 \Psi,
\end{aligned}

so if we want a momentum like creature, we have to use $$T^{0k}$$, not $$T^{k0}$$. The charge associated with that current is
\label{eqn:momentumDirac:240}
\begin{aligned}
Q^k
&=
\int d^3 x
\overline{\Psi} i \gamma^0 \partial^k \Psi \\
&=
\int d^3 x
\Psi^\dagger (-i \partial_k) \Psi,
\end{aligned}

or translating from component to vector form
\label{eqn:momentumDirac:260}
\BP =
\int d^3 x

which is the how the momentum operator is first stated in [2]. Here the vector notation doesn’t have any specific representation, but it is interesting to observe how this is directly related to the massless Dirac Lagrangian

\label{eqn:momentumDirac:280}
\begin{aligned}
\LL(m = 0)
&=
\overline{\Psi} i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i (\partial_0 + \gamma_0 \gamma^k \partial_k) \Psi \\
&=
\Psi^\dagger i (\partial_0 – \gamma_0 \gamma_k \partial_k ) \Psi,
\end{aligned}

but since $$\gamma_0 \gamma_k$$ is a $$4 \times 4$$ representation of the Pauli matrix $$\sigma_k$$ Lagrangian itself breaks down into
\label{eqn:momentumDirac:300}
\LL(m = 0)
=
\Psi^\dagger i \partial_0 \Psi
+
\Bsigma \cdot \lr{ \Psi^\dagger (-i\spacegrad) \Psi },

components, and lo and behold, out pops the momentum operator density! There is ambiguity as to what order of products $$\gamma_0 \gamma_k$$, or $$\gamma_k \gamma_0$$ to pick to represent the Pauli basis ([1] uses $$\gamma_k \gamma_0$$), but we also have sign ambiguity in assembling a Noether charge from the conserved current, so I don’t think that matters. Some part of this should be expected this since the Dirac equation in momentum space is just $$\gamma \cdot p – m = 0$$, so there is an intimate connection with the operator portion and momentum.

The last detail to fill in is going from \ref{eqn:momentumDirac:260} to \ref{eqn:momentumDirac:20} using the $$a, b$$ representation of the field. That’s an algebraically messy looking job that I don’t feel like trying at the moment.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

## 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.