## PHY2403H Quantum Field Theory. Lecture 15: Perturbation ground state, time evolution operator, time ordered product, interaction. 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, taught by Prof. Erich Poppitz, fall 2018.

## Review

We developed the interaction picture representation, which is really the Heisenberg picture with respect to $$H_0$$.

Recall that we found
\label{eqn:qftLecture15:20}
U(t, t’) = e^{i H_0(t – t_0)} e^{-i H(t – t’)} e^{-i H_0(t’ – t_0)},

with solution
\label{eqn:qftLecture15:200}
U(t, t’)
=
T \exp{\lr{ -i \int_{t’}^t H_{\text{I,int}}(t”) dt”}},

\label{eqn:qftLecture15:220}
\begin{aligned}
U(t, t’)^\dagger
&=
T \exp{\lr{ i \int_{t’}^{t} H_{\text{I,int}}(t”) dt”}} \\
&=
T \exp{\lr{ -i \int_{t}^{t’} H_{\text{I,int}}(t”) dt”}} \\
&= U(t’, t),
\end{aligned}

and can use this to calculate the time evolution of a field
\label{eqn:qftLecture15:40}
\phi(\Bx, t)
=
U^\dagger(t, t_0)
\phi_I(\Bx, t)
U(t, t_0)

and found the ground state ket for $$H$$ was
\label{eqn:qftLecture15:60}
\ket{\Omega}
=
\evalbar{
\frac{ U(t_0, -T) \ket{0} }
{
e^{-i E_0(T – t_0)} \braket{\Omega}{0}
}
}{T \rightarrow \infty(1 – i \epsilon)}.

### Question:

What’s the point of this, since it is self referential?

We will see, and also see that it goes away. Alternatively, you can write it as
\begin{equation*}
\ket{\Omega} \braket{\Omega}{0}
=
\evalbar{
\frac{ U(t_0, -T) \ket{0} }
{
e^{-i E_0(T – t_0)}
}
}{T \rightarrow \infty(1 – i \epsilon)}.
\end{equation*}

We can also show that
\label{eqn:qftLecture15:80}
\bra{\Omega}
=
\evalbar{
\frac{ \bra{0} U(T, t_0) }
{
e^{-i E_0(T – t_0)} \braket{0}{\Omega}
}
}{T \rightarrow \infty(1 – i \epsilon)}.

Our goal is still toe calculate
\label{eqn:qftLecture15:100}
\bra{\Omega} T \phi(x) \phi(y) \ket{\Omega}.

Claim: the “LSZ” theorem (a neat way of writing this) relates this to S matrix elements.

Assuming $$x^0 > y^0$$

\label{eqn:qftLecture15:120}
\bra{\Omega} \phi(x) \phi(y) \ket{\Omega}
=
\frac{
\bra{0}
U(T, t_0)
U^\dagger(x^0, t^0)
\phi_I(x)
U(x^0, t^0)
U^\dagger(y^0, t^0)
\phi_I(y)
U(y^0, t^0)
U(t_0, -T)
\ket{0}
}
{
e^{-i 2 E_0 T} \Abs{\braket{0}{\Omega}}^2
}

Normalize $$\braket{\Omega}{\Omega} = 1$$, gives

\label{eqn:qftLecture15:140}
\begin{aligned}
1
&=
\frac{\bra{0} U(T, t_0) U(t_0, -T) \ket{0}}
{
e^{-i 2 E_0 T} \Abs{\braket{0}{\Omega}}^2
} \\
&=
\frac{\bra{0} U(T, -T) \ket{0}}
{
e^{-i 2 E_0 T} \Abs{\braket{0}{\Omega}}^2
},
\end{aligned}

so that
\label{eqn:qftLecture15:240}
\bra{\Omega} \phi(x) \phi(y) \ket{\Omega}
=
\frac{
\bra{0}
U(T, t_0)
U^\dagger(x^0, t^0)
\phi_I(x)
U(x^0, t^0)
U^\dagger(y^0, t^0)
\phi_I(y)
U(y^0, t^0)
U(t_0, -T)
\ket{0}
}
{
\bra{0} U(T, -T) \ket{0}
}

For $$t_1 > t_2 > t_3$$
\label{eqn:qftLecture15:280}
\begin{aligned}
U(t_1, t_2) U(t_2, t_3)
&=
T e^{-i \int_{t_2}^{t_1} H_I}
T e^{-i \int_{t_3}^{t_2} H_I} \\
&=
T \lr{
e^{-i \int_{t_2}^{t_1} H_I}
e^{-i \int_{t_3}^{t_2} H_I}
} \\
&=
T(
e^{-i \int_{t_3}^{t_1} H_I}
),
\end{aligned}

with an end result of
\label{eqn:qftLecture15:320}
U(t_1, t_2) U(t_2, t_3) = U(t_1, t_3).

(DIY: work through the details — this is a problem in [1])

This gives
\label{eqn:qftLecture15:300}
\bra{\Omega} \phi(x) \phi(y) \ket{\Omega}
=
\frac{
\bra{0}
U(T, x^0)
\phi_I(x)
U(x^0, y^0)
\phi_I(y)
U(y^0, -T)
\ket{0}
}
{
\bra{0} U(T, -T) \ket{0}
}.

If $$y^0 > x^0$$ we have the same result, but the $$y$$’s will come first.

### Claim:

\label{eqn:qftLecture15:340}
\bra{\Omega} \phi(x) \phi(y) \ket{\Omega}
=
\frac{
\bra{0}
T\lr{
\phi_I(x)
\phi_I(y)
e^{-i \int_{-T}^T H_{\text{I,int}}(t’) dt’}
}
\ket{0}
}
{
\bra{0}
T ( e^{-i \int_{-T}^T H_{\text{I,int}}(t’) dt’} )
\ket{0}
}.

More generally
\label{eqn:qftLecture15:360}
\boxed{
\bra{\Omega}
\phi_I(x_1) \cdots
\phi_I(x_n)
\ket{\Omega}
=
\frac{
\bra{0}
T\lr{
\phi_I(x_1) \cdots
\phi_I(x_n)
e^{-i \int_{-T}^T H_{\text{I,int}}(t’) dt’}
}
\ket{0}
}
{
\bra{0}
T ( e^{-i \int_{-T}^T H_{\text{I,int}}(t’) dt’} )
\ket{0}
}.
}

This is the holy grail of perturbation theory.

In QFT II you will see this written in a path integral representation
\label{eqn:qftLecture15:380}
\bra{\Omega}
\phi_I(x_1) \cdots
\phi_I(x_n)
\ket{\Omega}
=
\frac
{
\int [\mathcal{D} \phi] \phi(x_1) \phi(x_2) \cdots \phi(x_n) e^{-i S[\phi]}
}
{
\int [\mathcal{D} \phi] e^{-i S[\phi]}
}.

## Unpacking it.

\label{eqn:qftLecture15:400}
\begin{aligned}
\int_{-T}^T H_{\text{I,int}}(t)
&=
\int_{-T}^T
\int d^3 \Bx \frac{\lambda}{4} \lr{ \phi_I(\Bx, t) }^4 \\
&=
\int d^4 x
\frac{\lambda}{4} \lr{ \phi_I }^4
\end{aligned}

so we have
\label{eqn:qftLecture15:420}
\frac{
\bra{0}
T\lr{
\phi_I(x_1) \cdots
\phi_I(x_n)
e^{-i \frac{\lambda}{4} \int d^4 x \phi_I^4(x) }
}
\ket{0}
}
{
\bra{0}
T
e^{-i \frac{\lambda}{4} \int d^4 x \phi_I^4(x) }
\ket{0}
}.

The numerator expands as
\label{eqn:qftLecture15:440}
\bra{0} T\lr{ \phi_I(x_1) \cdots \phi_I(x_n) } \ket{0}
-i \frac{\lambda}{4} \int d^4 x
\bra{0} T\lr{ \phi_I(x_1) \cdots \phi_I(x_n) \phi_I^4(x) }
+
\inv{2}
\lr{-i \frac{\lambda}{4}}^2 \int d^4 x d^4 y
\bra{0} T\lr{ \phi_I(x_1) \cdots \phi_I(x_n)
\phi_I^4(x)
\phi_I^4(y)
} \ket{0}
+ \cdots

so we see that the problem ends up being the calculation of time ordered products.

## Calculating perturbation

Let’s simplify notation, dropping interaction picture suffixes, writing $$\phi(x_i) = \phi_i$$.

Let’s calculate $$\bra{0} T\lr{ \phi_1 \cdots \phi_n } \ket{0}$$. For $$n = 2$$ we have

\label{eqn:qftLecture15:n}
\bra{0} T\lr{ \phi_1 \cdots \phi_n } \ket{0}
= D_F(x_1 – x_2) \equiv D_F(1-2)

### TO BE CONTINUED.

The rest of the lecture was very visual, and hard to type up. I’ll do so later.

# References

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

## PHY2403H Quantum Field Theory. Lecture 14: Time evolution, Hamiltonian perturbation, ground state. 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, taught by Prof. Erich Poppitz, fall 2018.

## Review

Given a field $$\phi(t_0, \Bx)$$, satisfying the commutation relations
\label{eqn:qftLecture14:20}
\antisymmetric{\pi(t_0, \Bx)}{\phi(t_0, \By)} = -i \delta(\Bx – \By)

we introduced an interaction picture field given by
\label{eqn:qftLecture14:40}
\phi_I(t, x) = e^{i H_0(t- t_0)} \phi(t_0, \Bx) e^{-iH_0(t – t_0)}

related to the Heisenberg picture representation by
\label{eqn:qftLecture14:60}
\phi_H(t, x)
= e^{i H(t- t_0)} \phi(t_0, \Bx) e^{-iH(t – t_0)}
= U^\dagger(t, t_0) \phi_I(t, \Bx) U(t, t_0),

where $$U(t, t_0)$$ is the time evolution operator.
\label{eqn:qftLecture14:80}
U(t, t_0) =
e^{i H_0(t – t_0)}
e^{-i H(t – t_0)}

We argued that
\label{eqn:qftLecture14:100}
i \PD{t}{} U(t, t_0) = H_{\text{I,int}}(t) U(t, t_0)

We found the glorious expression
\label{eqn:qftLecture14:120}
\boxed{
\begin{aligned}
U(t, t_0)
&= T \exp{\lr{ -i \int_{t_0}^t H_{\text{I,int}}(t’) dt’}} \\
&=
\sum_{n = 0}^\infty \frac{(-i)^n}{n!} \int_{t_0}^t dt_1 dt_2 \cdots dt_n T\lr{ H_{\text{I,int}}(t_1) H_{\text{I,int}}(t_2) \cdots H_{\text{I,int}}(t_n) }
\end{aligned}
}

However, what we are really after is
\label{eqn:qftLecture14:140}
\bra{\Omega} T(\phi(x_1) \cdots \phi(x_n)) \ket{\Omega}

Such a product has many labels and names, and we’ll describe it as “vacuum expectation values of time-ordered products of arbitrary #’s of local Heisenberg operators”.

## Perturbation

Following section 4.2, [1].

\label{eqn:qftLecture14:160}
\begin{aligned}
H &= \text{exact Hamiltonian} = H_0 + H_{\text{int}}
\\
H_0 &= \text{free Hamiltonian.
}
\end{aligned}

We know all about $$H_0$$ and assume that it has a lowest (ground state) $$\ket{0}$$, the “vacuum” state of $$H_0$$.

$$H$$ has eigenstates, in particular $$H$$ is assumed to have a unique ground state $$\ket{\Omega}$$ satisfying
\label{eqn:qftLecture14:180}
H \ket{\Omega} = \ket{\Omega} E_0,

and has states $$\ket{n}$$, representing excited (non-vacuum states with energies > $$E_0$$).
These states are assumed to be a complete basis
\label{eqn:qftLecture14:200}
\mathbf{1} = \ket{\Omega}\bra{\Omega} + \sum_n \ket{n}\bra{n} + \int dn \ket{n}\bra{n}.

The latter terms may be written with a superimposed sum-integral notation as
\label{eqn:qftLecture14:440}
\sum_n + \int dn
=
{\int\kern-1em\sum}_n,

so the identity operator takes the more compact form
\label{eqn:qftLecture14:460}
\mathbf{1} = \ket{\Omega}\bra{\Omega} + {\int\kern-1em\sum}_n \ket{n}\bra{n}.

For some time $$T$$ we have
\label{eqn:qftLecture14:220}
e^{-i H T} \ket{0} = e^{-i H T}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n \ket{n}\braket{n}{0}
}.

We now wish to argue that the $${\int\kern-1em\sum}_n$$ term can be ignored.

### Argument 1:

This is something of a fast one, but one can consider a formal transformation $$T \rightarrow T(1 – i \epsilon)$$, where $$\epsilon \rightarrow 0^+$$, and consider very large $$T$$. This gives
\label{eqn:qftLecture14:240}
\begin{aligned}
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i H T(1 – i \epsilon)} \ket{0}
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i H T(1 – i \epsilon)}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n \ket{n}\braket{n}{0}
} \\
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i E_0 T – E_0 \epsilon T}
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n e^{-i E_n T – \epsilon E_n T} \ket{n}\braket{n}{0} \\
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i E_0 T – E_0 \epsilon T}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n e^{-i (E_n -E_0) T – \epsilon T (E_n – E_0)} \ket{n}\braket{n}{0}
}
\end{aligned}

The limits are evaluated by first taking $$T$$ to infinity, then only after that take $$\epsilon \rightarrow 0^+$$. Doing this, the sum is dominated by the ground state contribution, since each excited state also has a $$e^{-\epsilon T(E_n – E_0)}$$ suppression factor (in addition to the leading suppression factor).

### Argument 2:

With the hand waving required for the argument above, it’s worth pointing other (less formal) ways to arrive at the same result. We can write
\label{eqn:qftLecture14:260}
sectionumInt \ket{n}\bra{n} \rightarrow
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \ket{\Bp, k}\bra{\Bp, k}

where $$k$$ is some unknown quantity that we are summing over.
If we have
\label{eqn:qftLecture14:280}
H \ket{\Bp, k} = E_{\Bp, k} \ket{\Bp, k},

then
\label{eqn:qftLecture14:300}
e^{-i H T} sectionumInt \ket{n}\bra{n}
=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \ket{\Bp, k} e^{-i E_{\Bp, k}} \bra{\Bp, k}.

If we take matrix elements
\label{eqn:qftLecture14:320}
\begin{aligned}
\bra{A}
e^{-i H T} sectionumInt \ket{n}\bra{n} \ket{B}
&=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \braket{A}{\Bp, k} e^{-i E_{\Bp, k}} \braket{\Bp, k}{B} \\
&=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} e^{-i E_{\Bp, k}} f(\Bp).
\end{aligned}

If we assume that $$f(\Bp)$$ is a well behaved smooth function, we have “infinite” frequency oscillation within the envelope provided by the amplitude of that function, as depicted in fig. 1.
The Riemann-Lebesgue lemma [2] describes such integrals, the result of which is that such an integral goes to zero. This is a different sort of hand waving argument, but either way, we can argue that only the ground state contributes to the sum \ref{eqn:qftLecture14:220} above.

fig. 1. High frequency oscillations within envelope of well behaved function.

### Ground state of the perturbed Hamiltonian.

With the excited states ignored, we are left with
\label{eqn:qftLecture14:340}
e^{-i H T} \ket{0} = e^{-i E_0 T} \ket{\Omega}\braket{\Omega}{0}

in the $$T \rightarrow \infty(1 – i \epsilon)$$ limit. We can now write the ground state as

\label{eqn:qftLecture14:360}
\begin{aligned}
\ket{\Omega}
&=
\evalbar{
\frac{ e^{i E_0 T – i H T } \ket{0} }{
\braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) } \\
&=
\evalbar{
\frac{ e^{- i H T } \ket{0} }{
e^{-i E_0 T} \braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) }.
\end{aligned}

Shifting the very large $$T \rightarrow T + t_0$$ shouldn’t change things, so
\label{eqn:qftLecture14:480}
\ket{\Omega}
=
\evalbar{
\frac{ e^{- i H (T + t_0) } \ket{0} }{
e^{-i E_0 (T + t_0) } \braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) }.

A bit of manipulation shows that the operator in the numerator has the structure of a time evolution operator.

### Claim: (DIY):

\Cref{eqn:qftLecture14:80}, \ref{eqn:qftLecture14:120} may be generalized to
\label{eqn:qftLecture14:400}
U(t, t’) = e^{i H_0(t – t_0)} e^{-i H(t – t’)} e^{-i H_0(t’ – t_0)} =
T \exp{\lr{ -i \int_{t’}^t H_{\text{I,int}}(t”) dt”}}.

Observe that we recover \ref{eqn:qftLecture14:120} when $$t’ = t_0$$.  Using \ref{eqn:qftLecture14:400} we find
\label{eqn:qftLecture14:520}
\begin{aligned}
U(t_0, -T) \ket{0}
&= e^{i H_0(t_0 – t_0)} e^{-i H(t_0 + T)} e^{-i H_0(-T – t_0)} \ket{0} \\
&= e^{-i H(t_0 + T)} e^{-i H_0(-T – t_0)} \ket{0} \\
&= e^{-i H(t_0 + T)} \ket{0},
\end{aligned}

where we use the fact that $$e^{i H_0 \tau} \ket{0} = \lr{ 1 + i H_0 \tau + \cdots } \ket{0} = 1 \ket{0},$$ since $$H_0 \ket{0} = 0$$.

We are left with
\label{eqn:qftLecture14:420}
\boxed{
\ket{\Omega}
= \frac{U(t_0, -T) \ket{0} }{e^{-i E_0(t_0 – (-T))} \braket{\Omega}{0}}.
}

We are close to where we want to be. Wednesday we finish off, and then start scattering and Feynman diagrams.

# References

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

[2] Wikipedia contributors. Riemann-lebesgue lemma — Wikipedia, the free encyclopedia, 2018. URL https://en.wikipedia.org/w/index.php?title=Riemann%E2%80%93Lebesgue_lemma&oldid=856778941. [Online; accessed 29-October-2018].

## My libertarian primer has arrived.

October 26, 2018 Incoherent ramblings No comments

I’ve been listening to the Tom Woods show and the Scott Horton show for a long time, and these books keep getting mentioned.  I’ve finally purchased them:

I probably won’t get to starting them until the new year (when I’m done with the phy2403, Quantum Field Theory I, course that I’m taking).

Btw., check out the awesome postage on the package that “Man, Economy and State” came from.  It took up the whole of the back of the package:

## Hamiltonian for the non-homogeneous Klein-Gordon equation

In class we derived the field for the non-homogeneous Klein-Gordon equation
\label{eqn:nonhomoKGhamiltonian:20}
\begin{aligned}
\phi(x)
&= \int \frac{d^3 p}{(2\pi)^3} \inv{\sqrt{2 \omega_\Bp}}
\evalbar{
\lr{
e^{-i p \cdot x} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }
+
e^{i p \cdot x} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }
}
}
{
p^0 = \omega_\Bp
} \\
&= \int \frac{d^3 p}{(2\pi)^3} \inv{\sqrt{2 \omega_\Bp}}
\lr{
e^{-i \omega_\Bp t + i \Bp \cdot \Bx} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }
+
e^{i \omega_\Bp t – i \Bp \cdot \Bx} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }
}.
\end{aligned}

This means that we have
\label{eqn:nonhomoKGhamiltonian:40}
\begin{aligned}
\pi = \dot{\phi}
&= \int \frac{d^3 p}{(2\pi)^3} \frac{i \omega_\Bp}{\sqrt{2 \omega_\Bp}}
\lr{
– e^{-i \omega_\Bp t + i \Bp \cdot \Bx} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }
+
e^{i \omega_\Bp t – i \Bp \cdot \Bx} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }
} \\
&= \int \frac{d^3 p}{(2\pi)^3} \frac{i p_k}{\sqrt{2 \omega_\Bp}}
\lr{
e^{-i \omega_\Bp t + i \Bp \cdot \Bx} \lr{ a_\Bp + \frac{ i \tilde{j}(p) }{\sqrt{2 \omega_\Bp}} }

e^{i \omega_\Bp t – i \Bp \cdot \Bx} \lr{ a_\Bp^\dagger – \frac{ i \tilde{j}^\conj(p) }{\sqrt{2 \omega_\Bp}} }
},
\end{aligned}

and could plug these into the Hamiltonian
\label{eqn:nonhomoKGhamiltonian:60}
H = \int d^3 p \lr{ \inv{2} \pi^2 + \inv{2} \lr{ \spacegrad \phi}^2 + \frac{m^2}{2} \phi^2 },

to find $$H$$ in terms of $$\tilde{j}$$ and $$a_\Bp^\dagger, a_\Bp$$. The result was mentioned in class, and it was left as an exercise to verify.

There’s an easy way and a dumb way to do this exercise. I did it the dumb way, and then after suffering through two long pages, where the equations were so long that I had to write on the paper sideways, I realized the way I should have done it.

The easy way is to observe that we’ve already done exactly this for the case $$\tilde{j} = 0$$, which had the answer
\label{eqn:nonhomoKGhamiltonian:80}
H = \inv{2} \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger }.

To handle this more general case, all we have to do is apply a transformation
\label{eqn:nonhomoKGhamiltonian:100}
a_\Bp \rightarrow
a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}},

to \ref{eqn:nonhomoKGhamiltonian:80}, which gives
\label{eqn:nonhomoKGhamiltonian:120}
\begin{aligned}
H
&=
\inv{2} \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }^\dagger\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} } +\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }^\dagger } \\
&=
\inv{2} \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} } \lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} } +\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} }
}.
\end{aligned}

Like the $$\tilde{j} = 0$$ case, we can use normal ordering. This is easily seen by direct expansion:
\label{eqn:nonhomoKGhamiltonian:140}
\begin{aligned}
\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} } \lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }
&=
a_\Bp^\dagger a_\Bp
– \frac{i \tilde{j}^\conj(p) a_\Bp}{\sqrt{2 \omega_\Bp}}
+ \frac{ a_\Bp^\dagger i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}}
+ \frac{\Abs{j}^2}{2 \omega_\Bp} \\
\lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }\lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} }
&=
a_\Bp^\dagger a_\Bp
+ \frac{i \tilde{j}^\conj(p) a_\Bp^\dagger}{\sqrt{2 \omega_\Bp}}
– \frac{ a_\Bp i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}}
+ \frac{\Abs{j}^2}{2 \omega_\Bp}.
\end{aligned}

Because $$\tilde{j}$$ is just a complex valued function, it commutes with $$a_\Bp, a_\Bp^\dagger$$, and these are equal up to the normal ordering, allowing us to write
\label{eqn:nonhomoKGhamiltonian:160}
:H: =
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp \lr{ a_\Bp^\dagger – \frac{i \tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}}} \lr{ a_\Bp + \frac{i \tilde{j}(p)}{\sqrt{2 \omega_\Bp}} },

which is the result mentioned in class.

## PHY2403H Quantum Field Theory. Lecture 13: Forced Klein-Gordon equation, coherent states, number density, time ordered product, pole shifting, perturbation theory, Heisenberg picture, interaction picture, Dyson’s formula. 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, taught by Prof. Erich Poppitz, fall 2018.

## Review: “particle creation problem”.

fig. 1. Finite window impulse response.

We imagined that we have a windowed source function $$j(y^0, \By)$$, as sketched in fig. 1, which is acting as a forcing source for the non-homogeneous Klein-Gordon equation

\label{eqn:qftLecture13:20}
\lr{ \partial_\mu \partial^\mu + m^2 } \phi = j

Our solution was
\label{eqn:qftLecture13:40}
\phi(x) = \phi(x_0) + i \int d^4 y D_R( x – y) j(y),

where $$\phi(x_0)$$ obeys the homogeneous equation, and
\label{eqn:qftLecture13:60}
D_r(x – y) = \Theta(x^0 – y^0) \lr{ D(x – y) – D(y – x) },

and $$D(x) = \int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp } \evalbar{ e^{-i p \cdot x} }{p^0 = \omega_\Bp}$$ is the Weightmann function.

For $$x^0 > t_{\text{after}}$$
\label{eqn:qftLecture13:80}
\phi(x)
=
\int \frac{d^3 p}{(2\pi)^3 \sqrt{ 2 \omega_\Bp }}
\evalbar{
\lr{ e^{-i p \cdot x} a_\Bp + e^{i p \cdot x } a_\Bp^\dagger }
}{
p^0 = \omega_\Bp
}
+ i
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp }
\evalbar{
\lr{ e^{-i p \cdot x} \tilde{j}(p) + e^{i p \cdot x} \tilde{j}(p_0, -\Bp) }
}{
p^0 = \omega_\Bp
}

where we have used $$\tilde{j}^\conj(p_0, \Bp) = \tilde{j}(p_0, -\Bp)$$. This gives
\label{eqn:qftLecture13:100}
\phi(x) =
\int \frac{d^3 p}{(2\pi)^3 \sqrt{ 2 \omega_\Bp } }
\evalbar{
\lr{
e^{-i p \cdot x}
\lr{ a_\Bp + i \frac{\tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }
+ e^{i p \cdot x }
\lr{ a_\Bp^\dagger – i \frac{\tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} }
}
}{
p^0 = \omega_\Bp
}

It was left as an exercise to show that given
\label{eqn:qftLecture13:120}
H = \int d^3 p \lr{ \inv{2} \pi^2 + \inv{2} \lr{ \spacegrad \phi}^2 + \frac{m^2}{2} \phi^2 },

we obtain
\label{eqn:qftLecture13:140}
H_{\text{after}} =
\int d^3 x \omega_\Bp
\lr{ a_\Bp^\dagger – i \frac{\tilde{j}^\conj(p)}{\sqrt{2 \omega_\Bp}} }
\lr{ a_\Bp + i \frac{\tilde{j}(p)}{\sqrt{2 \omega_\Bp}} }

System in ground state
\label{eqn:qftLecture13:160}
\bra{0} \hatH_{\text{before}} \ket{0} = \expectation{E}_{\text{before}} = 0.

\label{eqn:qftLecture13:180}
\begin{aligned}
\bra{0} \hatH_{\text{after}} \ket{0} = \expectation{E}_{\text{after}}
&=
\int d^3 x \omega_\Bp
\frac{ \tilde{j}^\conj(p) \tilde{j}(p)}{2 \omega_\Bp} \\
&=
\inv{2} \int d^3 x
\Abs{j(p)}^2.
\end{aligned}

We can identify
\label{eqn:qftLecture13:200}
N(\Bp) =
\frac{\Abs{j(p)}^2}{2 \omega_\Bp},

as the number density of particles with momentum $$\Bp$$.

## Digression: coherent states.

### Defintion: Coherent state.

A coherent state is an eigenstate of the destruction operator
\begin{equation*}
a \ket{\alpha} = \alpha \ket{\alpha}.
\end{equation*}

For the SHO, if we solve for such a coherent state, we find
\label{eqn:qftLecture13:240}
\ket{\alpha} = \text{constant} \times \sum_{n = 0}^\infty \frac{\alpha^n}{n!} \lr{ a^\dagger }^n \ket{0}.

If we assume the existence of a coherent state
\label{eqn:qftLecture13:260}
a_\Bp \ket{
\frac{j(p)}{\sqrt{2 \omega_\Bp}}
}
=
\frac{j(p)}{\sqrt{2 \omega_\Bp}}
\ket{
\frac{j(p)}{\sqrt{2 \omega_\Bp}}
},

then the expectation value of the number operator with respect to this state is the number density identified in \ref{eqn:qftLecture13:200}
\label{eqn:qftLecture13:1200}
\bra{
\frac{j(p)}{\sqrt{2 \omega_\Bp}}
}
a_\Bp^\dagger a_\Bp
\ket{
\frac{j(p)}{\sqrt{2 \omega_\Bp}}
} = \frac{\Abs{j(p)}^2}{2 \omega_\Bp} = N(\Bp).

## Feynman’s Green’s function

\label{eqn:qftLecture13:280}
\begin{aligned}
D_F(x)
&=
\Theta(x^0) D(x) +
\Theta(-x^0) D(-x) \\
&=
\Theta(x^0) \bra{0} \phi(x) \phi(0) \ket{0}
+\Theta(x^0) \bra{0} \phi(-x) \phi(0) \ket{0}
\end{aligned}

Utilizing a translation operation $$U(a) = e^{i a_\mu P^\mu }$$, where $$U(a) \phi(y) U^\dagger(a) = \phi(y + a)$$, this second operation can be written as
\label{eqn:qftLecture13:300}
\begin{aligned}
\bra{0} \phi(-x) \phi(0) \ket{0}
&=
\bra{0} U^\dagger(a) U(a) \phi(-x) U^\dagger(a) U(a) \phi(0) U^\dagger(a) U(a) \ket{0} \\
&=
\bra{0} U(a) \phi(-x) U^\dagger(a) U(a) \phi(0) U^\dagger(a) \ket{0} \\
&=
\bra{0} \phi(-x + a) \phi(a) \ket{0},
\end{aligned}

In particular, with $$a = x$$
\label{eqn:qftLecture13:320}
\bra{0} \phi(-x) \phi(0) \ket{0}
=
\bra{0} \phi(0) \phi(x) \ket{0},

so the Feynman’s Green function can be written
\label{eqn:qftLecture13:340}
D_F(x) =
\Theta(x^0) \bra{0} \phi(x) \phi(0) \ket{0}
+\Theta(x^0) \bra{0} \phi(x) \phi(x) \ket{0}
=
\bra{0}
\lr{
\Theta(x^0)
\phi(x) \phi(0)
+
\Theta(-x^0)
\phi(0) \phi(x)
}
\ket{0}.

We define

### Definition: Time ordered product.

The time ordered product of two operators is defined as
\begin{equation*}
T(\phi(x) \phi(y)) =
\left\{
\begin{array}{l l}
\phi(x)\phi(y) & \quad \mbox{$$x^0 > y^0$$} \\
\phi(y)\phi(x) & \quad \mbox{$$x^0 < y^0$$} \\
\end{array}
\right.,
\end{equation*}
or
\begin{equation*}
T(\phi(x) \phi(y)) =
\phi(x)\phi(y) \Theta(x^0 – y^0)
+
\phi(y)\phi(x) \Theta(y^0 – x^0).
\end{equation*}

Using this helpful construct, the Feynman’s Green function can now be written in a very simple fashion
\label{eqn:qftLecture13:380}
\boxed{
D_F(x) = \bra{0} T(\phi(x) \phi(0)) \ket{0}.
}

### Remark:

Recall that the four dimensional form of the Green’s function was
\label{eqn:qftLecture13:400}
D_F = i \int \frac{d^4 p}{(2 \pi)^4} e^{-i p \cdot x} \inv{ p^2 – m^2 }.

For the Feynman case, the contour that we were taking around the poles can also be accomplished by shifting the poles strategically, as sketched in fig. 2.

fig. 2. Feynman deformation or equivalent shift of the poles.

This shift can be expressed explicit algebraically by introducing an offset
\label{eqn:qftLecture13:420}
D_F = i \int \frac{d^4 p}{(2 \pi)^4} e^{-i p \cdot x} \inv{ p^2 – m^2 + i \epsilon }

which puts the poles at

\label{eqn:qftLecture13:440}
\begin{aligned}
p^0
&= \pm \sqrt{ \omega_\Bp – i \epsilon } \\
&= \pm \omega_\Bp \lr{ 1 – \frac{i \epsilon}{\omega_\Bp^2} }^{1/2} \\
&= \pm \omega_\Bp \lr{ 1 – \inv{2} \frac{i \epsilon}{\omega_\Bp^2} } \\
&=
\left\{
\begin{array}{l}
+\omega_\Bp – \inv{2} i \frac{\epsilon}{\omega_\Bp} \\
-\omega_\Bp + \inv{2} i \frac{\epsilon}{\omega_\Bp} \\
\end{array}
\right.
\end{aligned}

## Interacting field theory: perturbation theory in QFT.

We perturb the Hamiltonian
\label{eqn:qftLecture13:500}
H = H_0 + H_{\text{int}}

where $$H_0$$ is the free Hamiltonian and $$H_{\text{int}}$$ is the interaction term (the perturbation).

### Example:

\label{eqn:qftLecture13:460}
\begin{aligned}
H_0 &= SHO = \frac{p^2}{2} + \frac{\omega^2 q^2}{2} \\
H_{\text{int}} &= \lambda q^4,
\end{aligned}

i.e. the anharmonic oscillator.

In QFT
\label{eqn:qftLecture13:480}
\begin{aligned}
H_0 &=
\int d^3 x \lr{ \inv{2} \pi^2 + \inv{2} \lr{ \spacegrad \phi}^2 + \frac{m^2}{2} \phi^2 } \\
H_{\text{int}} &=
\lambda \int d^3 x \phi^4.
\end{aligned}

We will expand the interaction in small $$\lambda$$. Perturbation theory is the expansion in a small dimensionless coupling constant, such as

• $$\lambda$$ in $$\lambda \phi^4$$ theory,
• $$\alpha = e^2/4 \pi \sim \inv{137}$$ in QED, and
• $$\alpha_s$$ in QCD.

## Perturbation theory, interaction representation and Dyson formula

\label{eqn:qftLecture13:520}
H = H_0 + H_{\text{int}}

Example interaction
\label{eqn:qftLecture13:540}
H_{\text{int}} = \lambda \int d^3 x \phi^4

We know all there is to know about $$H_0$$ (decoupled SHOs, …)
\label{eqn:qftLecture13:560}
H_0 \ket{0} = \ket{0} E^0_{\text{vac}}

where $$E^0_{\text{vac}} = 0$$. Assume
\label{eqn:qftLecture13:580}
\lr{ H_0 + H_{\text{int}} } \ket{\Omega} = \ket{\Omega} E_{\text{vac}},

where the ground state energy of the perturbed system is zero when $$\lambda = 0$$. That is $$E_{\text{vac}}(\lambda = 0 ) = 0$$.

So for
\label{eqn:qftLecture13:600}
\evalbar{\phi(x) }{x^0 = t_0, \text{some fixed value}}
=
\int \frac{d^3}{(2 \pi)^3 \sqrt{ 2 \omega_\Bp } }
\evalbar{
\lr{
e^{-i p \cdot x} a_\Bp
+ e^{i p \cdot x} a_\Bp^\dagger }
}
{
p^0 = \omega_\Bp
}.

Let’s call $$\phi(\Bx, t_0)$$ the free Schr\”{o}dinger operator, where
$$\phi(\Bx, t_0)$$ is evaluated at a fixed value of $$t_0$$. At such a point, the Schr\”{o}dinger and Heisenberg pictures coincide.
\label{eqn:qftLecture13:620}
\antisymmetric{\phi(\Bx, t_0)}{\pi(\By, t_0)} = i \delta^3(\Bx – \By).

Normally (QM) one defines the Heisenberg operator as
\label{eqn:qftLecture13:640}
O_H = e^{i H(t – t_0)} O_S e^{-i H(t – t_0)},

where $$O_H$$ depends on time, and $$O_S$$ is defined at a fixed time $$t_0$$, usually 0.
From \ref{eqn:qftLecture13:640} we find
\label{eqn:qftLecture13:660}
\ddt{O_H} = i \antisymmetric{H}{O_H}.

The equivalent of \ref{eqn:qftLecture13:640} in QFT is very complicated. We’d like to develop an intermediate picture.

We will define an intermediate picture, called the “interaction representation”, which is equivalent to the Heisenberg picture with respect to $$H_0$$.

### Definition: Intermediate picture operator.

\begin{equation*}
\phi_I(t, \Bx) =
e^{i H_0(t – t_0) }
\phi(t_0, \Bx)
e^{-i H_0(t – t_0) }.
\end{equation*}

This is familiar, and is the Heisenberg picture operator that we had in free QFT
\label{eqn:qftLecture13:700}
\phi_I(t, \Bx) =
\int \frac{d^3}{(2 \pi)^3 \sqrt{ 2 \omega_\Bp } }
\evalbar{
\lr{
e^{-i p \cdot x} a_\Bp
+ e^{i p \cdot x} a_\Bp^\dagger }
}
{
p^0 = \omega_\Bp
},

where $$x_0 = t$$.

The Heisenberg picture operator is
\label{eqn:qftLecture13:720}
\begin{aligned}
\phi_H(t, \Bx)
&=
\phi(t, \Bx) \\
&=
e^{i H(t – t_0) }
e^{-i H_0(t – t_0) }
\lr{
e^{i H_0(t – t_0) }
\phi_S(t_0, \Bx)
e^{-i H_0(t – t_0) }
}
e^{i H_0(t – t_0) }
e^{-i H(t – t_0) } \\
&=
e^{i H(t – t_0) }
e^{-i H_0(t – t_0) }
\phi_I(t, \Bx)
e^{-i H_0(t – t_0) }
e^{i H(t – t_0) }
\end{aligned}

or
\label{eqn:qftLecture13:760}
\phi_H(t, \Bx)
=
U^\dagger(t, t_0)
\phi_I(t_0, \Bx)
U(t, t_0),

where
\label{eqn:qftLecture13:740}
U(t, t_0) =
e^{i H_0(t – t_0) }
e^{-i H(t – t_0) }.

We want to apply perturbation techniques to find $$U(t, t_0)$$ which is complicated.

\label{eqn:qftLecture13:780}
\begin{aligned}
i \PD{t}{} U(t, t_0)
&=
i e^{i H_0(t – t_0) } i H_0
e^{-i H(t – t_0) }
+
i e^{i H_0(t – t_0) }
e^{-i H(t – t_0) } (-i H) \\
&=
e^{i H_0(t – t_0) }
\lr{ -H_0 + H }
e^{-i H(t – t_0) } \\
&=
e^{i H_0(t – t_0) }
H_{\text{int}}
e^{-i H_0(t – t_0) }
e^{i H_0(t – t_0) }
e^{-i H(t – t_0) }
\end{aligned}

so we have
\label{eqn:qftLecture13:800}
\boxed{
i \PD{t}{} U(t, t_0)
=
H_{\text{int}, I}(t) U(t, t_0).
}

For the (Schr\”{o}dinger) interaction $$H_{\text{int}} = \ \lambda \int d^3 x \phi^4(\Bx, t_0)$$, what we really mean by
$$H_{\text{int}, I}(t)$$ is
\label{eqn:qftLecture13:820}
H_{\text{int}, I}(t) = \lambda \int d^3 x \phi_I^4(\Bx, t).

It will be more convenient to remove the explicit $$\lambda$$ factor from the interaction Hamiltonian, and write instead
\label{eqn:qftLecture13:880}
H_{\text{int}, I}(t) = \int d^3 x \phi_I^4(\Bx, t),

so the equation to solve is
\label{eqn:qftLecture13:1220}
i \PD{t}{} U(t, t_0)
=
\lambda H_{\text{int}, I}(t) U(t, t_0).

We assume that
\label{eqn:qftLecture13:900}
U(t, t_0)
=
U_0(t, t_0)
+ \lambda U_1(t, t_0)
+ \lambda^2 U_2(t, t_0)
+ \cdots
+ \lambda^n U_n(t, t_0)

Plugging into \ref{eqn:qftLecture13:880} we have
\label{eqn:qftLecture13:1160}
\begin{aligned}
i &\lambda^0 \PD{t}{}U_0(t, t_0)
+ i \lambda^1 \PD{t}{}U_1(t, t_0)
+ i \lambda^2 \PD{t}{}U_2(t, t_0)
+ \cdots
+ i \lambda^n \PD{t}{}U_n(t, t_0) \\
&=
\lambda H_{\text{int}, I}(t)
\lr{
1
+ \lambda U_1(t, t_0)
+ \lambda^2 U_2(t, t_0)
+ \cdots
+ \lambda^n U_n(t, t_0)
},
\end{aligned},

so
equating equal powers of $$\lambda$$ on each side gives a recurrence relation for each $$U_k, k > 0$$
\label{eqn:qftLecture13:1180}
\PD{t}{}U_k(t, t_0) = -i H_{\text{int}, I}(t) U_{k-1}(t, t_0).

Let’s consider each power in turn.

### $$O(\lambda^0)$$:

Solving \ref{eqn:qftLecture13:800} to $$O(\lambda^0)$$ gives
\label{eqn:qftLecture13:840}
i \PD{t}{} U_0(t, t_0) = 0,

or
\label{eqn:qftLecture13:860}
U(t, t_0) = 1 + O(\lambda).

### $$O(\lambda^1)$$:

\label{eqn:qftLecture13:940}
\PD{t}{U_1(t, t_0)} = -i H_{\text{int}, I}(t),

which has solution
\label{eqn:qftLecture13:960}
U_1(t, t_0) = -i \int_{t_0}^t H_{\text{int}, I}(t’) dt’.

### $$O(\lambda^2)$$:

\label{eqn:qftLecture13:1000}
\begin{aligned}
\PD{t}{U_2(t, t_0)}
&= -i H_{\text{int}, I}(t) U_1(t, t_0) \\
&= (-i)^2 H_{\text{int}, I}(t)
\int_{t_0}^t H_{\text{int}, I}(t’) dt’,
\end{aligned}

which has solution
\label{eqn:qftLecture13:1020}
\begin{aligned}
U_2(t, t_0)
&= (-i )^2
\int_{t_0}^t H_{\text{int}, I}(t”) dt”
\int_{t_0}^{t”} H_{\text{int}, I}(t’) dt’ \\
&= (-i )^2
\int_{t_0}^t dt”
\int_{t_0}^{t”}
dt’
H_{\text{int}, I}(t”)
H_{\text{int}, I}(t’).
\end{aligned}

### $$O(\lambda^3)$$:

\label{eqn:qftLecture13:1060}
\PD{t}{U_3(t, t_0)}
=
-i
H_{\text{int}, I}(t) U_2(t, t_0)

so
\label{eqn:qftLecture13:1240}
\begin{aligned}
U_3(t, t_0)
&=
-i
\int_{t_0}^t dt”’
H_{\text{int}, I}(t”’) U_2(t”’, t_0) \\
&=
(-i )^3
\int_{t_0}^t dt”’
H_{\text{int}, I}(t”’)
\int_{t_0}^{t”’} dt”
\int_{t_0}^{t”}
dt’
H_{\text{int}, I}(t”)
H_{\text{int}, I}(t’) \\
&=
(-i)^3
\int_{t_0}^t dt”’
\int_{t_0}^{t”’} dt”
\int_{t_0}^{t”} dt’
H_{\text{int}, I}(t”’)
H_{\text{int}, I}(t”)
H_{\text{int}, I}(t’)
\end{aligned}

### Simplifying the integration region.

For the two fold integral, the integration range is the upper triangular region sketched in fig. 3.

fig. 3. Upper triangular integration region.

### Claim:

We can integrate over the entire square, and divide by two, provided we keep the time ordering
\label{eqn:qftLecture13:1040}
U_2(t, t_0)
= \frac{(-i )^2}{2}
\int_{t_0}^t dt”
\int_{t_0}^{t”}
dt’
T(H_{\text{int}, I}(t”) H_{\text{int}, I}(t’) )

Demonstration:
\label{eqn:qftLecture13:1100}
\begin{aligned}
\frac{(-i)^2}{2}
&\int_{t_0}^t dt”
\int_{t_0}^t dt’
T( H_I(t”) H_I(t’) ) \\
&=
\frac{(-i)^2}{2}
\int_{t_0}^t dt”
\int_{t_0}^t dt’
\Theta(t”- t’)
H_I(t”) H_I(t’)
+
\frac{(-i)^2}{2}
\int_{t_0}^t dt”
\int_{t_0}^t dt’
\Theta(t’- t”)
H_I(t’) H_I(t”),
\end{aligned}

but the $$\Theta(t” – t’)$$ function is non-zero only for $$t” – t’ > 0$$, or $$t’ < t”$$, and the $$\Theta(t’ – t”)$$ function is non-zero only for $$t’ – t” > 0$$, or $$t” < t’$$, so we can adjust the integration ranges for
\label{eqn:qftLecture13:1260}
\begin{aligned}
\frac{(-i)^2}{2}
&\int_{t_0}^t dt”
\int_{t_0}^t dt’
T( H_I(t”) H_I(t’) ) \\
&=
\frac{(-i)^2}{2}
\int_{t_0}^t dt”
\int_{t_0}^{t”} dt’
H_I(t”) H_I(t’)
+
\frac{(-i)^2}{2}
\int_{t_0}^{t’} dt”
\int_{t_0}^t dt’
H_I(t’) H_I(t”) \\
&=
\frac{(-i)^2}{2}
\int_{t_0}^t dt”
\int_{t_0}^{t”} dt’
H_I(t”) H_I(t’)
+
\frac{(-i)^2}{2}
\int_{t_0}^t dt”
\int_{t_0}^{t”} dt’
H_I(t”) H_I(t’) \\
&=
U_2(t, t_0),
\end{aligned}

where we swapped integration variables in second integral. We can clearly do the same thing for the higher order repeated integrals, but instead of a $$1/2 = 1/2!$$ adjustment for the number of orderings, we will require a $$1/n!$$ adjustment for an $$n$$-fold integral.

### Summary:

\label{eqn:qftLecture13:1120}
\begin{aligned}
U_0 &= 1 \\
U_1 &= -i \int_{t_0}^t dt_1 H_I(t_1) \\
U_2 &= \frac{(-i)^2}{2}
\int_{t_0}^t dt_1
\int_{t_0}^t dt_2
T( H_I(t_1)
H_I(t_2) ) \\
U_3 &= \frac{(-i)^3}{3!}
\int_{t_0}^t dt_1
\int_{t_0}^t dt_2
\int_{t_0}^t dt_3
T( H_I(t_1)
H_I(t_2)
H_I(t_3)
) \\
U_n &= \frac{(-i)^n}{n!}
\int_{t_0}^t dt_1
\int_{t_0}^t dt_2
\int_{t_0}^t dt_3
\cdots
\int_{t_0}^t dt_n
T( H_I(t_1)
H_I(t_2)
\cdots
H_I(t_n)
) \\
\end{aligned}

Summing we find
\label{eqn:qftLecture13:1140}
\begin{aligned}
U(t, t_0)
&= T \exp\lr{-i
\int_{t_0}^t dt_1 H_I(t’)
} \\
&=
\sum_{n = 0}^\infty
\frac{(-i)^n}{n!} \int_{t_0}^t dt_1 \cdots dt_n T( H_I(t_1) \cdots H_I(t_n) ).
\end{aligned}

This is called Dyson’s formula.

## Next time.

Our goal is to compute: $$\bra{\Omega} T(\phi(x_1) \cdots \phi(x_n)) \ket{\Omega}$$.

## PHY2403H Quantum Field Theory. Lecture 12: Klein-Gordon Green’s function, Feynman propagator path deformation, Weightmann function, Retarded Green’s function. 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, taught by Prof. Erich Poppitz, fall 2018.

## Green’s functions for the forced Klein-Gordon equation.

The problem were were preparing to do was to study the problem of “particle creation by external classical source”.

We continue with a real scalar field, free, massive, but with an interaction with a source
\label{eqn:qftLecture12:20}
S_{\text{int}} = \int d^4 x j(x) \phi(x).

### Modern application:

think of $$\phi$$ has some SM field and think of $$j$$ as due to inflaton (i.e. cosmological inflation interaction) oscillation. In the inflationary model, the process of “reheating” creates all the matter in the universe. We won’t be talking about inflation, but will be considering a toy model that has some similar characteristics to the inflationary theory.

The equation of motion that we end up with is
\label{eqn:qftLecture12:40}
\lr{ \partial_\mu \partial^\mu + m^2} \phi = j,

and we wish to solve this using Green’s function techniques.

## Definition: Klein-Gordon Green’s function.

The QFT conventions for the Klein-Gordon Green’s function is
\begin{equation*}
\lr{ \partial_\mu \partial^\mu + m^2} G(x – y) = -i \delta^4(x – y).
\end{equation*}

As usual, we assume that it is possible to find a solution $$\phi$$ by convolution
\label{eqn:qftLecture12:80}
\phi(x) = i \int d^4 y G(x – y) j(y).

### Check:

\label{eqn:qftLecture12:100}
\begin{aligned}
\lr{ \partial_\mu \partial^\mu + m^2} \phi(x)
&=
i
\lr{ \partial_\mu \partial^\mu + m^2}
\int d^4 y G(x – y) j(y) \\
&=
i \int d^4 y (-i) \delta^4(x – y) j(y) \\
&= j(x).
\end{aligned}

Also, as usual, we take out our Fourier transforms, the power tool of physics, and determine the structure of the Green’s function by inverting the transform equation
\label{eqn:qftLecture12:120}
G(x – y) = \int \frac{d^4 p}{(2 \pi)^4} e^{-i p \cdot (x-y) } \tilde{G}(p).

Operating with KG gives
\label{eqn:qftLecture12:520}
\lr{ \partial_\mu \partial^\mu + m^2}
G(x)
=
\int \frac{d^4 p}{(2 \pi)^4}
\lr{ (-i p_\mu)(-i p^\mu) + m^2 }
e^{-i p \cdot (x-y) } \tilde{G}(p).

This must equal
\label{eqn:qftLecture12:140}
-i \delta^4(x – y) =
-i \int \frac{d^4 p}{(2 \pi)^4} e^{-i p \cdot (x -y)},

or
\label{eqn:qftLecture12:160}
\lr{ m^2 – p_\mu p^\mu } \tilde{G}(p) = -i.

The Green’s function in the momentum domain is
\label{eqn:qftLecture12:180}
\tilde{G}(p) = \frac{i}{p^2 – m^2}.

The inverse transform provides the spatial domain representation of the Green’s function
\label{eqn:qftLecture12:200}
\begin{aligned}
G(x)
&=
\int \frac{d^4 p}{(2 \pi)^4} e^{-i p \cdot x }
\frac{i}{(p^0)^2 – \Bp^2 – m^2} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\int \frac{d p_0}{2 \pi} e^{-i p_0 x^0 }
\frac{i}{(p_0 – \omega_\Bp)(p_0 + \omega_\Bp)}.
\end{aligned}

In the $$p_0$$ plane, we have two poles at $$p_0 = \pm \omega_\Bp$$. There are 4 ways to go around the poles, the retarded time deformation that we used to derive the Green’s function for the harmonic oscillator, as sketched in fig. 1, the advanced time deformation sketched in fig. 2, and mixed deformations.

fig. 1. Retarded time deformations and contours.

We will evaluate the integral using the “Feynman propagator” contour sketched in fig. 3. Why we use the Feynman contour, and not the retarded contour can be justified by how well this works for the perturbation methods that will be developed later.

fig. 3. Feynman propagator deformation path.

Consider each contour in turn.

### Case I. $$x^0 > 0$$

For this case, we use the lower half plane contour sketched in fig. 4, which vanishes for $$\Im(p_0) < 0, x_0 > 0$$, where $$-i (i \Im(p_0) x_0) < 0$$.

fig. 4. Feynman propagator contour for t > 0.

Here we pick up just the pole at $$p_0 = \omega_\Bp$$, and take a negatively oriented path
\label{eqn:qftLecture12:220}
\begin{aligned}
G_\txtF
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\int \frac{d p_0}{2 \pi} e^{-i p_0 x^0 }
\frac{i}{(p_0 – \omega_\Bp)(p_0 + \omega_\Bp)} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
(-2 \pi i)
\evalbar{\lr{ \frac{e^{-i p_0 x^0 }}{2 \pi}
\frac{i}{p_0 + \omega_\Bp} }}{p_0 = \omega_p} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\frac{-2 \pi i}{2 \pi} \frac{i e^{-i p_0 x^0 } }{2 \omega_\Bp} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\frac{ e^{-i \omega_\Bp x^0 } }{2 \omega_\Bp}.
\end{aligned}

### Case II. $$x^0 < 0$$

For $$x^0 < 0$$ we use an upper half plane contour with the same deformation around the poles. This time

\label{eqn:qftLecture12:240}
\begin{aligned}
G_\txtF
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\int \frac{d p_0}{2 \pi} e^{-i p_0 x^0 }
\frac{i}{(p_0 – \omega_\Bp)(p_0 + \omega_\Bp)} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
(+ 2 \pi i)
\evalbar{\lr{\frac{e^{-i p_0 x^0 }}{2 \pi}
\frac{i}{p_0 – \omega_\Bp}}}{p_0 = -\omega_\Bp} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\frac{+2 \pi i}{2 \pi} \frac{i e^{-i p_0 x^0 } }{-2 \omega_\Bp} \\
&=
\int \frac{d^3 p}{(2\pi)^3} e^{i \Bp \cdot \Bx}
\frac{ e^{i \omega_\Bp x^0 } }{2 \omega_\Bp}.
\end{aligned}

We’ve obtained a piecewise representation of the Green’s function, where the only difference is the sign of the $$i \omega_\Bp x^0$$ exponential.

We can combine \ref{eqn:qftLecture12:220} \ref{eqn:qftLecture12:240} by using $$\Theta$$ functions
\label{eqn:qftLecture12:260}
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp} e^{i \Bp \cdot \Bx}
\lr{
e^{-i \omega_\Bp x^0 } \Theta(x_0)
+
e^{i \omega_\Bp x^0 } \Theta(-x_0)
}.

The first integral (without the $$\Theta$$ factor) is the Weightmann function
\label{eqn:qftLecture12:280}
D(x)
=
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp} \evalbar{e^{-i p \cdot x}}{p^0 = \omega_\Bp}.

For the second integral, we make a change of variables $$\Bp \rightarrow -\Bp$$ leaving
\label{eqn:qftLecture12:300}
\begin{aligned}
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp} e^{i \Bp \cdot \Bx + i \omega_\Bp x^0}
&\rightarrow
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp} e^{-i \Bp \cdot \Bx + i \omega_\Bp x^0} \\
&=
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp} e^{-i p \cdot x} \\
&= D(-x),
\end{aligned}

so
\label{eqn:qftLecture12:340}
\boxed{
G_\txtF (x) = \Theta(x^0) D(x) + \Theta(-x^0) D(-x)
}

## Matrix element representation of the Weightmann function.

Recall that the Weightmann function also had a matrix element representation
\label{eqn:qftLecture12:360}
D(x) = \bra{0} \phi(x) \phi(0) \ket{0}.

This can be shown by expansion.
\label{eqn:qftLecture12:380}
\bra{0} \phi(x) \phi(0) \ket{0}
=
\bra{0}
\int \frac{d^3 p}{(2 \pi)^3} \inv{\sqrt{2 \omega_\Bp}} \evalbar{\lr{ a_\Bp e^{-i p \cdot x} + a_\Bp^\dagger e^{i p \cdot x} }}{p_0 = \omega_\Bp}
\int \frac{d^3 q}{(2 \pi)^3} \inv{\sqrt{2 \omega_\Bq}} \lr{ a_\Bq^\dagger + a_\Bq }
\ket{0}

Since $$a_\Bq \ket{0} = 0 = \bra{0} a_\Bp^\dagger$$, \ref{eqn:qftLecture12:380}
reduces to
\label{eqn:qftLecture12:540}
\begin{aligned}
\bra{0} \phi(x) \phi(0) \ket{0}
&=
\bra{0}
\int
\frac{d^3 p}{(2 \pi)^3}
\frac{d^3 q}{(2 \pi)^3}
\inv{\sqrt{2 \omega_\Bp}}
\inv{\sqrt{2 \omega_\Bq}}
\evalbar{\lr{ a_\Bp a_\Bq^\dagger e^{-i p \cdot x} }}{p_0 = \omega_\Bp}
\ket{0} \\
&=
\bra{0}
\int
\frac{d^3 p}{(2 \pi)^3}
\frac{d^3 q}{(2 \pi)^3}
\inv{\sqrt{2 \omega_\Bp}}
\inv{\sqrt{2 \omega_\Bq}}
\evalbar{\lr{
\lr{
a_\Bp
a_\Bq^\dagger
+
\antisymmetric{
a_\Bp
}{
a_\Bq^\dagger
}
}
e^{-i p \cdot x} }}{p_0 = \omega_\Bp}
\ket{0} \\
&=
\bra{0}
\int
\frac{d^3 p}{(2 \pi)^3}
\frac{d^3 q}{(2 \pi)^3}
\inv{\sqrt{2 \omega_\Bp}}
\inv{\sqrt{2 \omega_\Bq}}
\evalbar{\lr{
\lr{
(2 \pi)^3 \delta^3(\Bp – \Bq)
}
e^{-i p \cdot x} }}{p_0 = \omega_\Bp} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} \evalbar{ \frac{e^{-i p \cdot x}}{2 \omega_\Bp} }
{p_0 = \omega_\Bp}
\end{aligned}

## Retarded Green’s function.

Claim: Retarded Green’s function (bumps up contour) can be written
\label{eqn:qftLecture12:400}
D_R(x) = \theta(x_0) (D(x) – D(-x)).

Proof: The upper half plane contour ($$x_0 < 0$$) is zero since it encloses no poles. For the lower half plane contour we have \label{eqn:qftLecture12:420} \begin{aligned} \evalbar{D_R(x)}{x_0 > 0}
&=
i \int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx} \int \frac{dp_0}{2 \pi} e^{-i p_0 x^0 }
\frac{i}{(p_0 – \omega_\Bp)(p_0 + \omega_\Bp)} \\
&=
i \int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx} \frac{(-2 \pi i)}{2 \pi}
\lr{
e^{-i \omega_\Bp x^0 }
\frac{i}{2 \omega_\Bp}
+
e^{i \omega_\Bp x^0 }
\frac{i}{-2 \omega_\Bp}
} \\
&=
\int \frac{d^3 p}{(2 \pi)^3} e^{i \Bp \cdot \Bx}
\frac{1}{2 \omega_\Bp}
\lr{
e^{-i \omega_\Bp x^0 }

e^{i \omega_\Bp x^0 }
} \\
&=
D(x) – D(-x).
\end{aligned}

What does the field look like in terms of the propagator? Assuming that $$\phi_0$$ satisfies the homogeneous equation, we have
\label{eqn:qftLecture12:440}
\begin{aligned}
\phi(x)
&= \phi_0(x) + i \int d^4 y D_R(x – y) j(y) \\
&= \phi_0(x) + i \int d^3 y d y_0 \Theta(x_0 – y_0) \lr{ D(x – y) – D(y – x) } j(y)
\end{aligned}

Imagine that we have a windowed source function $$j(y^0, \By)$$, as sketched in fig. 5.

fig. 5. Finite window impulse response.

\label{eqn:qftLecture12:460}
\evalbar{\phi(x)}{x^0 > t_{\text{after}}}
= \phi_0(x)
+ i \int d^4 y
\lr{
\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp } e^{-i p \cdot (x – y)} j(y)

\int \frac{d^3 p}{(2\pi)^3 2 \omega_\Bp } e^{i p \cdot (x – y)} j(y)
}

define
\label{eqn:qftLecture12:480}
\tilde{j}(p) = \int d^4 y e^{i p \cdot y} j(y),

which gives
\label{eqn:qftLecture12:500}
\evalbar{\phi(x)}{x^0 > t_{\text{after}}}
= \phi_0(x)
+ i \int \frac{d^3 p }{(2 \pi)^3}
\inv{
2 \omega_\Bp }
\evalbar{
\lr{
e^{-i p \cdot x} \tilde{j}(p)
– e^{i p \cdot x} \tilde{j}(-p)
}
}{p_0 = \omega_\Bp}.

We will interpret this in the next lecture, and start in on Feynman diagrams.

# References

## New aggregate notes collection for UofT phy2403 Quantum Field Theory I

October 21, 2018 phy2403 No comments , ,

I’ve uploaded a new aggregate notes collection of my UofT phy2403 Quantum Field Theory I class notes (taught by Prof. Erich Poppitz), which now includes up to Wed Oct 17th’s lecture 11 (but doesn’t have my problem set I solution)

• 1 Introduction
• 1.1 What is a field?
• 1.2 Scales.
• 1.2.2 Compton wavelength.
• 1.2.3 Relations.
• 2 Units, scales, and Lorentz transformations.
• 2.1 Natural units.
• 2.2 Gravity.
• 2.3 Cross section.
• 2.4 Lorentz transformations.
• 3 Lorentz transformations and a scalar action.
• 3.1 Determinant of Lorentz transformations.
• 3.2 Field theory.
• 3.3 Actions.
• 3.4 Problems.
• 4 Scalar action, least action principle, Euler-Lagrange equations for a field, canonical quantization.
• 4.1 Principles cont.
• 4.2 d = 2 .
• 4.3 d = 3 .
• 4.4 d = 4 .
• 4.5 d = 5 .
• 4.6 Least action principle (classical field theory)
• 4.7 Canonical quantization.
• 5 Klein-Gordon equation, SHOs, momentum space representation, raising and lowering operators.
• 5.1 Canonical quantization.
• 5.2 Momentum space representation.
• 6 Canonical quantization, Simple Harmonic Oscillators, Symmetries
• 6.1 Quantization of Field Theory.
• 6.2 Free Hamiltonian.
• 6.3 QM SHO review.
• 6.4 Discussion.
• 6.5 Switching gears: Symmetries.
• 7 Symmetries, translation currents, energy momentum tensor.
• 7.1 Symmetries.
• 7.2 Spacetime translation.
• 8 1st Noether theorem, spacetime translation current, energy momentum tensor, dilatation current.
• 8.1 1st Noether theorem.
• 8.2 Unitary operators.
• 8.3 Continuous symmetries.
• 8.4 Classical scalar theory.
• 9 Unbroken and spontaneously broken symmetries, Higgs Lagrangian, scale invariance, Lorentz invariance, angular momentum quantization
• 9.1 Last time.
• 9.2 Examples of symmetries.
• 9.3 Scale invariance.
• 9.4 Lorentz invariance.
• 10 Lorentz boosts, generator of spacetime translation, Lorentz invariant field representation.
• 10.1 Lorentz transform symmetries.
• 10.2 Transformation of momentum states.
• 11 Microcausality, Lorentz invariant measure, retarded time SHO Green’s function.
• 11.1 Relativistic normalization.
• 11.2 Spacelike surfaces.
• 11.3 Condition on microcausality.
• 11.4 Harmonic oscillator.
• 11.5 Field theory (where we are going).
• 12 Independent study problems
• Appendices
• A Useful formulas and review
• Index
• Bibliography

## PHY2403H Quantum Field Theory. Lecture 11: Momentum matrix elements, spacelike surfaces, microcausality, Lorentz invariant measure, wave function Green’s function, retarded time contour, advanced time contour. 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.

## Relativistic normalization.

We will continue looking at the generator of spacetime translation $$\hatU(\Lambda)$$, which has the property
\label{eqn:qftLecture11:40}
\hatU(\Lambda) \ket{0} = \ket{0},

That is
\label{eqn:qftLecture11:760}
\hatU(\Lambda) = \mathbf{1} + \text{operators that anhillate the vacuum state}.

The action on a field was
\label{eqn:qftLecture11:60}
\hatU(\Lambda)
\phihat(x) \hatU^\dagger(\Lambda)
= \phihat(\Lambda x),

and the action on the anhillation operator was
\label{eqn:qftLecture11:300}
\hatU(\Lambda)
\sqrt{ 2 \omega_\Bp } \hat{a}_\Bp
\hatU^\dagger(\Lambda)
=
\sqrt{ 2 \omega_{\Lambda \Bp} } \hat{a}_{\Lambda \Bp}.

If $$\ket{\Bp_1}$$ is the one particle state with momentum $$\Bp_1$$, then that momentum state can be generated from the ground state with the following normalized creation operation
\label{eqn:qftLecture11:780}
\ket{\Bp_1} = \sqrt{ 2 \omega_{\Bp_1} } \hat{a}_{\Bp_1}^\dagger \ket{0}.

We can compute the matrix element between two matrix states using the creation operator representation
\label{eqn:qftLecture11:80}
\begin{aligned}
\braket{\Bp_2}{\Bp_1}
&=
\sqrt{ 2 \omega_{\Bp_1} }
\sqrt{ 2 \omega_{\Bp_2} }
\bra{0}
\hat{a}_{\Bp_2}
\hat{a}_{\Bp_1}^\dagger
\ket{0} \\
&=
\sqrt{ 2 \omega_{\Bp_1} }
\sqrt{ 2 \omega_{\Bp_2} }
\bra{0}
\lr{
\hat{a}_{\Bp_1}^\dagger
\hat{a}_{\Bp_2}
+
i (2 \pi)^3 \delta^3(\Bp – \Bq)
} \\
&=
\sqrt{ 2 \omega_{\Bp_1} }
\sqrt{ 2 \omega_{\Bp_2} }
(2 \pi)^3 \delta^3(\Bp_1 – \Bp_2) \\
&=
2 \omega_{\Bp_1}
(2 \pi)^3 \delta^3(\Bp_1 – \Bp_2).
\end{aligned}

## Spacelike surfaces.

fig. 0. Constant spacelike surface.

If $$x^\mu, p^\mu$$ are four vectors, then $$p^\mu x_\mu = \text{invariant} = {p’}^\mu x’_\mu$$. The light cone is the surface $$p_0^2 = \Bp^2$$, whereas timelike four-momentum form a parabaloid surface $$p_0^2 – \Bp^2 = m^2$$ (i.e. $$E = \sqrt{ m^2 c^4 + \Bp^2 c^2 }$$).
The surface for constant spacelike points (i.e. all related by a Lorentz transformation) is illustrated in fig. 0. A boost moves a point up or down that surface along the energy axis. It is therefore possible to use a sequence of boost and rotation to transform a point $$(E, \Bp) \rightarrow (-E, \Bp) \rightarrow (-E, -\Bp)$$. That is, any spacelike four-vector $$x$$ may be transformed to $$-x$$ using a Lorentz transformation.

## Condition on microcausality.

We defined operators $$\phihat(\Bx)$$, which was a Hermitian operator for the real scalar field. For the complex scalar field we used $$\phihat(\Bx) = (\phihat_1 + \phihat_2)/\sqrt{2}$$, where each of $$\phihat_1, \phihat_2$$ were Hermitian operators. i.e. we can think of these operators as “observables”, that is $$\phihat(\Bx) = \phihat^\dagger(\Bx)$$.

We now want to show that these operators commute at spacelike separations, and see how this relates to the question of causality. In particular, we want to see that an observation of one operator, will not effect the measurement of the other.

The condition of microcausality is
\begin{equation*}
\antisymmetric{\phihat(x)}{\phihat(y)} = 0
\end{equation*}
if $$x \sim y$$, that is $$(x – y)^2 < 0$$. That is, $$x, y$$ are spacelike separated.

We wrote

\label{eqn:qftLecture11:160}
\phihat(x)
=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{
e^{-i p \cdot x} }{p^0 = \omega_\Bp} \hat{a}_\Bp
+
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{
e^{i p \cdot x} }{p^0 = \omega_\Bp} \hat{a}^\dagger_\Bp
,

or $$\phihat(x) = \phihat_{-}(x) + \phihat_{+}(x)$$, where
\label{eqn:qftLecture11:180}
\begin{aligned}
\phihat_{-}(x) &=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{
e^{-i p \cdot x} }{p^0 = \omega_\Bp} \hat{a}_\Bp \\
\phihat_{+}(x) &=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{
e^{i p \cdot x} }{p^0 = \omega_\Bp} \hat{a}^\dagger_\Bp
\end{aligned}

Compute the commutator
\label{eqn:qftLecture11:200}
\begin{aligned}
D(x)
&= \antisymmetric{\phihat_{-}(x)}{\phihat_{+}(0)} \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{ e^{-i p \cdot x} }{p^0 = \omega_\Bp}
\int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_\Bk}}
\evalbar{ e^{i k \cdot 0} }{k^0 = \omega_\Bk}
\antisymmetric{\hat{a}_\Bp }{\hat{a}_\Bk^\dagger } \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{ e^{-i p \cdot x} }{p^0 = \omega_\Bp}
\int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_\Bk}}
(2 \pi)^3 \delta^3(\Bp – \Bk),
\end{aligned}

\label{eqn:qftLecture11:800}
\boxed{
D(x)
=
\int \frac{d^3 p}{(2 \pi)^3 2 \omega_\Bp}
\evalbar{ e^{-i p \cdot x} }{p^0 = \omega_\Bp}.
}

Now about the commutator at two spacetime points
\label{eqn:qftLecture11:220}
\begin{aligned}
\antisymmetric{\phihat(x)}{\phihat(y)}
&=
\antisymmetric{\phihat_{-}(x) + \phihat_{+}(x)}{\phihat_{-}(y) + \phihat_{+}(y)} \\
&=
\antisymmetric{\phihat_{-}(x)}{\phihat_{+}(y)}
+
\antisymmetric{\phihat_{+}(x)}{\phihat_{-}(y)} \\
&=
-D(y – x) + D(x – y)
\end{aligned}

Find
\label{eqn:qftLecture11:240}
\begin{aligned}
\antisymmetric{\phihat(x)}{\phihat(y)} &= D(x – y) – D(y – x) \\
\antisymmetric{\phihat(x)}{\phihat(0)} &= D(x) – D(- x)
\end{aligned}

Let’s look at $$D(x)$$, \ref{eqn:qftLecture11:800}, a bit more closely.

### Claim:

D(x) is Lorentz invariant (has the same value for all $$x^\mu, {x’}^\mu$$

We can see this by writing this out as
\label{eqn:qftLecture11:280}
D(x)
=
\int \frac{d^3 p}{(2 \pi)^3 } dp^0
\delta( p_0^2 – \Bp^2 – m^2) \Theta(p^0)
e^{-i p \cdot x}

The exponential is Lorentz invariant, and the delta function has been put into a Lorentz invariant form.

### Claim 1:

$$D(x) = D(x’)$$ where $$x^2 = {x’}^2$$.

### Claim 2:

$$x^\mu, -x^\mu$$ are related by Lorentz transformations if $$x^2 < 0$$.

From the figure, we see that $$D(x) = D(-x)$$ for a spacelike point, which implies that $$\antisymmetric{\phihat(x)}{\phihat(0)} = 0$$ for a spacelike point $$x$$.

We’ve shown this for free fields, but later we will see that this is the case for interacting fields too.

## Harmonic oscillator.

\label{eqn:qftLecture11:320}
L = \inv{2} \qdot^2 – \frac{\omega^2}{t} q^2 – j(t) q

The term $$j(t)$$ shifts the origin in a time dependent fashion (graphical illustration in class wiggling a hockey stick, as a sample of a harmonic oscillator).

\label{eqn:qftLecture11:340}
H = \frac{p^2}{2} + \frac{\omega^2}{t} q^2 + j(t) q

\label{eqn:qftLecture11:360}
\begin{aligned}
i \qdot_H(t) &= \antisymmetric{q_H}{H} = i p_H \\
i \pdot_H(t) &= \antisymmetric{p_H}{H} = -i \omega^2 q_H – i j(t)
\end{aligned}

\label{eqn:qftLecture11:380}
\ddot{q}_H(t) = – \omega^2 q_H(t) – j(t)

or
\label{eqn:qftLecture11:400}
(\partial_{tt} + \omega^2 ) q_H(t) = – j(t)

\label{eqn:qftLecture11:420}
q_H(t) = q_H^0( t ) +
\int G_R(t – t’) j(t’) dt’

This solves the equation provided $$G_R(t – t’)$$ has the property that
\label{eqn:qftLecture11:440}
\boxed{
(\partial_{tt} + \omega^2)
G_R(t – t’)
= – \delta(t – t’)
}

That is
\label{eqn:qftLecture11:460}
(\partial_{tt} + \omega^2)
q_H(t) =
(\partial_{tt} + \omega^2)
q_H^0( t )
+
(\partial_{tt} + \omega^2)
\int G_R(t – t’) j(t’) dt’

This function $$G_R$$ is called the retarded Green’s function. We want to find this function, and as usual, we do this by taking the Fourier transform of \ref{eqn:qftLecture11:440}

\label{eqn:qftLecture11:480}
\begin{aligned}
\int dt e^{i p t}
(\partial_{tt} + \omega^2) G_R(t – t’)
&=
-\int_{-\infty}^\infty dt e^{i p t}
\delta(t – t’) \\
&= – e^{i p t’}
\end{aligned}

Let
\label{eqn:qftLecture11:500}
G(t – t’) = \int \frac{dp }{2 \pi} e^{- i p'(t – t’)} \tilde{G}(p’),

so

\label{eqn:qftLecture11:520}
\begin{aligned}
– e^{i pt’}
&=
\int dt e^{i p t}
(\partial_{tt} + \omega^2)
\int \frac{dp’}{2 \pi} e^{- i p'(t – t’)} \tilde{G}(p’) \\
&=
\int dt e^{i p t} \int
\frac{dp’}{2 \pi} \lr{ -{p’}^2 + \omega^2 } e^{- i p'(t – t’)} \tilde{G}(p’) \\
&=
\int dp’ \lr{ -{p’}^2 + \omega^2 } e^{i p’ t’} \delta(p – p’) \tilde{G}(p’) \\
&=
\lr{ -{p}^2 + \omega^2 } \tilde{G}(p) e^{i p t’}
\end{aligned}

so
\label{eqn:qftLecture11:540}
\tilde{G}(p)
= \inv{p^2 – \omega^2}

Now

\label{eqn:qftLecture11:560}
G(t)
= \int \frac{dp}{2 \pi} e^{-i p t}
\tilde{G}(p)

Let’s write the momentum space Green’s function as
\label{eqn:qftLecture11:580}
\tilde{G}(p)
= \inv{(p – \omega)(p + \omega)}

The solution contained
\label{eqn:qftLecture11:600}
\int G(t – t’) j(t’) dt’.

Suppose $$j(t) = 0$$ for all $$t < t_0$$. We want the effect of $$j(t)$$ to be felt in the future, for example, $$j(t)$$ is an impulse starting at some time. We want $$G(t)$$ to vanish at negative times.

We want the integral
\label{eqn:qftLecture11:620}
G(t)
= \int \frac{dp}{2 \pi} e^{-i p t}
\inv{(p – \omega)(p + \omega)}

to vanish when $$t < 0$$. Start with $$t > 0$$ (that is $$t’ < t$$), so that $$e^{-i p t} = e^{-i p \Abs{t}}$$ which means that we have to integrate over a lower plane contour like fig. 1, because the imaginary part of $$p$$ is negative, but for $$t < 0$$ (that is $$t’ > t$$), we want an upper plane contour like fig. 2.

fig. 1. Lower plane contour.

fig. 2. Upper plane contour.

Question: since we are integrating over the real line, how can we get away with deforming the contour?
Answer: it works. If we do this we get a Green’s function that makes sense (better answer later?)

We add an infinite circle, so that we can integrate over a closed contour, and pick the contour so that it is zero for $$t < 0$$ and non-zero (enclosed poles) for $$t > 0$$.

\label{eqn:qftLecture11:640}
\begin{aligned}
G_R(t > 0)
&= \int_C \frac{dp}{2 \pi} e^{-i p t}
\inv{(p – \omega)(p + \omega)} \\
&=
\inv{2 \pi} (-2 \pi i) \lr{
\frac{e^{-i \omega t}}{2 \omega}

\frac{e^{i \omega t}}{2 \omega}
} \\
&=
-\frac{\sin(\omega t)}{\omega}.
\end{aligned}

Now we write the Green’s function for all time as
\label{eqn:qftLecture11:660}
\boxed{
G_R(t) =
-\frac{\sin(\omega t)}{\omega} \Theta(t).
}

The question of what contour to pick can now be justified by the result, since this satisfies fig. 3). In particular, the bumps up and down contour will be used to derive the “Feynman propagator” that we’ll use later.

fig. 3. All possible deformations around the poles.

## Field theory (where we are going).

We will consider a massive real scalar field theory with an external source with action

\label{eqn:qftLecture11:680}
S = \int d^4 x \lr{
\inv{2} \partial_\mu \phi \partial^\mu \phi – \frac{m^2}{2} \phi^2 + j(x) \phi(x)
}

We don’t have examples of currents that create scalar fields, but to study such as system, recall that
in electromagnetism we added sources to the field by adding a term like
\label{eqn:qftLecture11:700}
\int d^4 x A^\mu(x) j_\mu(x),

to our action.

The equation of motion can be found to be
\label{eqn:qftLecture11:720}
\lr{ \partial_\mu \partial^\mu + m^2 } \phi(x) = j(x).

We want to study the Green’s function of this Klien-Gordon equation, defined to obey
\label{eqn:qftLecture11:740}
\lr{ \partial_\mu \partial^\mu + m^2 }_x G(x – y) = -i \delta^4(x – y),

where the $$-i$$ factor is for convienience.
This is analogous to the Green’s function that we just studied for the QM harmonic oscillator.

## Question: Compute $$D(x-y)$$ from the commutator.

Generalize the derivation \ref{eqn:qftLecture11:800} by computing the commutator at two different space time points $$x, y$$.

Let
\label{eqn:qftLecture11:860}
\begin{aligned}
D(x – y)
&= \antisymmetric{\phihat_{-}(x)}{\phihat_{+}(y)} \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{ e^{-i p \cdot x} }{p^0 = \omega_\Bp}
\int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_\Bk}}
\evalbar{ e^{i k \cdot y} }{k^0 = \omega_\Bk}
\antisymmetric{\hat{a}_\Bp }{\hat{a}_\Bk^\dagger } \\
&=
\int \frac{d^3 p}{(2 \pi)^3 \sqrt{2 \omega_\Bp}}
\evalbar{ e^{-i p \cdot x} }{p^0 = \omega_\Bp}
\int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_\Bk}}
\evalbar{ e^{i k \cdot y} }{k^0 = \omega_\Bk}
(2 \pi)^3 \delta^3(\Bp – \Bk) \\
&=
\int \frac{d^3 p}{(2 \pi)^3 2 \omega_\Bp}
\evalbar{ e^{-i p \cdot (x – y)} }{p^0 = \omega_\Bp}.
\end{aligned}

## Question: Verification of harmonic oscillator Green’s function.

Take the derivatives of a convolution of the Green’s function \ref{eqn:qftLecture11:660} to show that it satisifies \ref{eqn:qftLecture11:440}.

Let
\label{eqn:qftLecture11:880}
\begin{aligned}
q(t)
&= \int_{-\infty}^\infty G(t – t’) j(t’) dt’ \\
&= -\inv{\omega} \int_{-\infty}^\infty \sin(\omega(t – t’)) \Theta(t – t’) j(t’) dt’.
\end{aligned}

We are free to add any $$q_0(t)$$ that satisfies the homogeneous wave equation $$q_0”(t) + \omega^2 q_0(t) = 0$$ to our assumed convolution solution \ref{eqn:qftLecture11:880}, but that isn’t interesting for this exersize.
Since $$\Theta(t – t’) = 0$$ for $$t – t’ < 0$$, or $$t’ > t$$, the convolution can be written as
\label{eqn:qftLecture11:900}
q(t)
= -\inv{\omega} \int_{-\infty}^t \sin(\omega(t – t’)) j(t’) dt’,

which is now in a convient form to take derivatives. We have contributions from the boundary’s time dependence and from the integrand. In particular
\label{eqn:qftLecture11:920}
\ddt{} \int_{a(t)}^{b(t)} g(x, t) dx
=
g(b(t)) b'(t) – g(a(t)) a'(t) + \int_a^b \frac{\partial}{\partial t} g(x, t) dx.

Assuming that $$j(-\infty) = 0$$, this gives
\label{eqn:qftLecture11:940}
\begin{aligned}
\ddt{q(t)}
&=
-\inv{\omega} \evalbar{\sin(\omega(t – t’)) j(t’) }{t’ = t}
-\int_{-\infty}^t \cos(\omega(t – t’)) j(t’) dt’ \\
&=
-\int_{-\infty}^t \cos(\omega(t – t’)) j(t’) dt’.
\end{aligned}

For the second derivative we have
\label{eqn:qftLecture11:960}
\begin{aligned}
q”(t)
&=
– \evalbar{ \cos(\omega(t – t’)) j(t’) }{t’ = t}
+\omega \int_{-\infty}^t \sin(\omega(t – t’)) j(t’) dt’ \\
&=
-j(t) -\omega^2
\int_{-\infty}^t \frac{-\sin(\omega(t – t’))}{\omega} j(t’) dt’,
\end{aligned}

or
\label{eqn:qftLecture11:980}
q”(t) = -j(t) – \omega^2 q(t),

which is our forced Harmonic oscillator equation.

## 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.