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

These are notes for the UofT course PHY2403H, Quantum Field Theory, 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

\begin{equation}\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)},

\end{equation}

with solution

\begin{equation}\label{eqn:qftLecture15:200}

U(t, t’)

=

T \exp{\lr{ -i \int_{t’}^t H_{\text{I,int}}(t”) dt”}},

\end{equation}

\begin{equation}\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}

\end{equation}

and can use this to calculate the time evolution of a field

\begin{equation}\label{eqn:qftLecture15:40}

\phi(\Bx, t)

=

U^\dagger(t, t_0)

\phi_I(\Bx, t)

U(t, t_0)

\end{equation}

and found the ground state ket for \( H \) was

\begin{equation}\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)}.

\end{equation}

### Question:

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

### Answer:

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

\begin{equation}\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)}.

\end{equation}

Our goal is still toe calculate

\begin{equation}\label{eqn:qftLecture15:100}

\bra{\Omega} T \phi(x) \phi(y) \ket{\Omega}.

\end{equation}

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

Assuming \( x^0 > y^0 \)

\begin{equation}\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

}

\end{equation}

Normalize \( \braket{\Omega}{\Omega} = 1 \), gives

\begin{equation}\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}

\end{equation}

so that

\begin{equation}\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}

}

\end{equation}

For \( t_1 > t_2 > t_3 \)

\begin{equation}\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}

\end{equation}

with an end result of

\begin{equation}\label{eqn:qftLecture15:320}

U(t_1, t_2) U(t_2, t_3) = U(t_1, t_3).

\end{equation}

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

This gives

\begin{equation}\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}

}.

\end{equation}

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

### Claim:

\begin{equation}\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}

}.

\end{equation}

More generally

\begin{equation}\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}

}.

}

\end{equation}

This is the holy grail of perturbation theory.

In QFT II you will see this written in a path integral representation

\begin{equation}\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]}

}.

\end{equation}

## Unpacking it.

\begin{equation}\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}

\end{equation}

so we have

\begin{equation}\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}

}.

\end{equation}

The numerator expands as

\begin{equation}\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

\end{equation}

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

\begin{equation}\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)

\end{equation}

### 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.