I am troubled for understanding P&S's problem 4.1 (a) on p. 126. Consider a Hamiltonian for the creation of Klein-Gordon particles by a classical source: $$H=H_0+\int d^3 x(-j(t, \mathbf{x}) \phi(x)).\tag{p.126} $$ In 4.1 (a), the book said the probability that the source create no particles is given by $$P(0)=\left|\left\langle 0\left|T\left\{\exp \left[i \int d^4 x j(x) \phi_I(x)\right]\right\}\right| 0\right\rangle\right|^2.\tag{p.126} $$ So why they use the free ground state $| 0\rangle$, not $| \Omega\rangle$? Since according to the book's eq.(4.31): $$\langle\Omega|T\{\phi(x) \phi(y)\}| \Omega\rangle=\lim _{T \rightarrow \infty(1-i \epsilon)} \frac{\left\langle 0\left|T\left\{\phi_I(x) \phi_I(y) \exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle}{\left\langle 0\left|T\left\{\exp \left[-i \int_{-T}^T d t H_I(t)\right]\right\}\right| 0\right\rangle} . \tag{4.31}$$ what they show about $P(0)$ is just the denominator of eq.(4.31).

  • $\begingroup$ Your hamiltonian is time dependent, so your groundstate is too. Typically, you assume the sources to be compactly supported in time, so far enough in the future and past the ground state is the free ground state. $\endgroup$
    – LPZ
    Oct 26, 2022 at 12:34

1 Answer 1


I recommend you to look and understand the subsection related to the S-Matrix. There, it is explained that the overlap between an initial state of the interacting theory $|k_1 k_2\rangle_{\mathrm{in}}$ of particles of momentum $k_1,k_2$ with a final interacting state $|p_1 p_2 ...\rangle_{\mathrm{out}}$ with momentum $p_1, p_2,...$ can be computed with the corresponding momentum of the free theory, using the relation $$ \sideset{_{\mathrm{out}}}{} \langle p_1 p_2 ... |k_1 k_2 \sideset{}{_{\mathrm{in}}} \rangle = \langle p_1 p_2 ... | S | k_1 k_2 \rangle = \left \langle p_1 p_2 ... \left | \ \mathrm{exp} \left(T \left \{ - i \int dt H_I (t) \right \} \right ) \ \right | k_1 k_2 \right \rangle $$ where now $|k_1 k_2\rangle$ are momentum states of the free theory, and $S$ is the $S$-matrix. The probability associated with the transition from $k_1 k_2$ to $p_1 p_2 ...$ is then the modulus square of the above expression.

This relation also holds for the vacuum in which we are interested : $$ \sideset{_{\mathrm{out}}}{} \langle \Omega | \Omega \sideset{}{_{\mathrm{in}}} \rangle = \langle 0 | S | 0 \rangle $$

So the probability that the far-past vacuum $|\Omega \rangle_{\mathrm{in}}$ goes into the far-future vacuum $|\Omega \rangle_{\mathrm{out}}$ is $$ P(0) = |\sideset{_{\mathrm{out}}}{} \langle \Omega | \Omega \sideset{}{_{\mathrm{in}}} \rangle |^2 = \left | \left \langle 0 \left | \ \mathrm{exp} \left(T \left \{ - i \int dt H_I (t) \right \} \right ) \ \right | 0 \right \rangle \right |^2 $$

  • $\begingroup$ Thank you very much! Actually, P & S propose this in eq.(4.90), and proved it in Section 7.2. I thought eq.(7.45) is important, the L.H.S of (7.45) is about momentum in full interacting eigenstate, while the R.H.S are our free input momentum. Would my understanding right? Thanks! $\endgroup$
    – Daren
    Oct 26, 2022 at 12:58

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