### 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].