Background material:

These are the parts that I can follow.

Previously Peskin & Schroeder have derived already the expression of the interaction ground state $|\Omega\rangle$ in terms of the free vacuum state $|0\rangle$: $$|\Omega\rangle=\lim_{T\to \infty(1-i\epsilon)}\left(e^{-iE_0T}\langle\Omega|0\rangle\right)^{-1}e^{-iHT}|0\rangle\tag{4.27}$$ where $\epsilon $ is a positive real number.

We are postulating a time $t_0$ where the interaction field operator coincides with the free field operator, i.e. when the interaction was just about to be turned on: $$\phi(t,\vec{x})=U^\dagger(t,t_0)\phi_I(t,\vec{x})U(t,t_0)\tag{4.16}$$ where $\phi_I$ is the free field operator, $U(t,t')$ is given by $$U(t,t')=e^{iH_0(t-t_0)}e^{-iH(t-t')}e^{-iH_0(t'-t_0)}.\tag{4.25}$$

We are setting further that the free ground state energy is $0$. The interaction ground state energy is denoted $E_0$. With (4.25) then, we could move $T$ slightly to $T+t_0$ in (4.27) and obtain $$|\Omega\rangle=\lim_{T\to\infty(1-i\epsilon)}\left(e^{-iE_0(t_0-(-T))}\langle\Omega|0\rangle\right)^{-1}U(t_0,-T)|0\rangle.\tag{4.28}$$

Problem to solve:

This is the part I cannot follow. $$\langle\Omega|=\lim_{T\to\infty(1-i\epsilon)}\langle0|U(T,t_0)\left(e^{-iE_0(T-t_0)}\langle 0|\Omega\rangle\right)^{-1}.\tag{4.29}$$

Question I had when trying to solve the problem:

To obtain (4.29), it is certainly wrong to start from (4.28) because 1. we are sending $T$ to an imaginary number, so we need to take $T^*$ but because $\epsilon$ is so small this detour to imaginary part should not matter but even if this is the case 2. $e^{-iE_0(t_0+T)}$ should become $e^{iE_0(t_0+T)}$ which is NOT the expression in (4.29) and 3. $U^\dagger(t_0,-T)=U(-T,t_0)$.

So we better start from one step before (4.27), namely $$e^{-iHT}|0\rangle=e^{-iE_0T}|\Omega\rangle\langle\Omega|0\rangle+\sum_{n\neq 0}e^{-iE_nT}|n\rangle \langle n|0\rangle.\tag{p.86}$$ Here when you move the ket state to its 'dual' in bra space you would get factor $e^{iE_nT}$. Now we need to take the same limit $T\to \infty(1-i\epsilon)$ for later use, but then the real part of power becomes $\mathcal{R}\to\epsilon E_n\infty$. This is not desirable because $\epsilon$ is real and positive and we would like to obtain something like $\mathcal{R}\to-\epsilon E_n\infty$ to ignore all terms containing $|n\rangle$.

So how did Peskin&Schroeder get to (4.29)? I am aware there are books that do differently from what they did, like Srednicki, and I know other approaches which give me Wick theorem but in this case I am just interested in how Peskin&Schroeder did it.


2 Answers 2


The key to derive the bra eq. (4.29) in P&S is to not just take the Hermitian conjugate of the ket eqs. (4.27)-(4.28) on p. 86 but also replace early time with late time $$-T~\to~ T.$$ In particular, the ket eq. before eq. (4.27) then tranform into the following bra eq.
$$\langle 0|e^{-iHT}~=~\langle 0|\Omega\rangle\langle\Omega|e^{-iE_0T} +\sum_{n\neq 0}\langle 0|n\rangle \langle n|e^{-iE_nT}.$$ Moreover, the $i\epsilon$ prescription then again leads to exponential suppression of excited states, as it should.

  • $\begingroup$ Wondering if my answer is wrong or just an alternative, any insights? $\endgroup$
    – JohnA.
    Dec 30, 2023 at 15:19


by taking $T$ to $\infty$ in a slightly imaginary direction: $T → \infty(1 - i\epsilon)$

where $\epsilon >0$.

We can take $T → \infty(1 + i\epsilon)$, $\epsilon >0$ and this is still a slightly imaginary direction. And thus you can continue down the desired path. The pre-factor in front of all the $E_n >0$ on the dual space summation is $$e^{iE_nT(1+i\epsilon)} = e^{-E_n\epsilon T}e^{iE_nT}$$ which has the desired effect of cancelling the higher order terms as $T\rightarrow\infty$ faster than the $E_0$ term.

Further explanation:

When the Epsilon trick is introduced in Srednicki (Ch 6), he concludes

What all this means is that if we use $(1−i\epsilon)H$ instead of $H$, we can be cavalier about the boundary conditions on the endpoints of the path. Any reasonable boundary conditions will result in the ground state as both the initial and final state. Thus we have $$\langle 0|0\rangle_{f,h} = \int \mathscr{D}p\mathscr{D}q\exp\Big[i\int_\infty^\infty dt\big(p\dot{q}-(1-i\epsilon)H+fq+hp\big)\Big]$$

Srednicki does this to $H$ rather that to $T$ but its equivalent, and for the same reason (i.e. getting rid of the additional states).

Down the line, we learn that taking $ H \rightarrow H(1-i\epsilon)$ is the same as taking $ m^2 \rightarrow m^2-i\epsilon$ which results in this Feynman propagator $$\Delta (x-x') = \int \frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-x')}}{k^2+m^2-i\epsilon}$$ and so the loop correction integral in Chapter 14 looks like $$\int \frac{d^d q}{(2\pi)}\int_0^1 dx \big[q^2 +D-i\epsilon\big]^{-2}$$

and so the poles look like this:

enter image description here

However, if we take $ H \rightarrow H(1+i\epsilon)$ and so $ m^2 \rightarrow m^2+i\epsilon$ the poles will shift to

enter image description here

but we have to do the Wick rotation the opposite of shown in the picture (i.e clockwise). Related and Related.


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