The definition of the path integral

I still have big conceptual questions about the path integral.

According to (24.6) of the book "QFT for the gifted amateur" from Lancaster & Blundell the path integral is equal to

$$Z =\langle \text{no particle}, t=\infty|\text{no particle},t=-\infty\rangle.\tag{24.6}$$

If I consider the path integral as result of a time evolution of the vacuum state like

$$Z =\langle 0, t=\infty | 0, t=-\infty\rangle = \langle 0, t=\infty|U(-\infty,\infty)| 0, t=\infty \rangle$$

so that I would get (I don't know if the operator U can be taken out of the bra-ket, so I do not use the equal sign between $$Z$$ and $$U$$)

$$Z \sim U(-\infty,\infty)$$

I wonder if it makes any sense thinking of a time evolution of a QFT vacuum state. I would rather think of a state with non-zero particle content.

But the biggest conceptual problem I see is that QFT is essentially formulated in the Heisenberg picture where states do not depend on time. The whole dynamics is transfered to the operators. So if the vaccum state is time-independent, then

$$\langle \text{no particle}, t=\infty| t=-\infty, \text{no particle}\rangle =1$$

in particular $$Z=1$$ and any derivative of the path integral would be zero. How can this contradiction be solved ?

• "The path integral" is not $\langle 0\vert 0\rangle$, path integration is a technique, there isn't a single unique object called "the path integral". $Z[J]$ is the partition function usually dependent on a source $J$, and indeed $Z[0]$ is more or less $\langle 0\vert 0\rangle$. What's the precise question about this setup? Specifically, where is the contradiction alluded to at the end of the question? Commented Apr 13 at 13:08
• related/possible dup: physics.stackexchange.com/q/249307/84967 Commented Apr 13 at 13:11
• in absence of sources $J$, the partition function indeed equals $1$. (Up to counterterms, and in flat spacetime, etc.) Commented Apr 13 at 13:11
• Z is a "transition amplitude". Commented Apr 13 at 13:23

The generating of functional of a (scalar) quantum field theory with field operator $$\Phi(x)$$ is defined by $$Z[f]=\langle 0 |e^{i \int d^dx \, \Phi(x) f(x)} |0\rangle, \tag{1} \label{1}$$ where $$f(x)$$ is a classical external source. $$Z[f]$$ has a simple physical interpration: suppose the external source was initially (i.e. for $$t\to -\infty$$) turned off and the system was in its ground state (vacuum state) $$|0\rangle$$t some point, the external source $$f$$ is switched on, "shakes" the system and is finally switched off again. $$Z[f]$$ is nothing else than the transition amplitude that the system is found again in the ground state for $$t\to +\infty$$. For this reason, the generating functional is sometimes written as $$Z[f]=\langle 0 |0\rangle_{f} \tag{2}$$ meaning "vaccum to vacuum transition amplitude in the presence of the external source $$f$$", where $$\langle0|0\rangle_{f=0} =1$$ by definition.
The path integral representation of the generating functional, $$Z[f]=\int [d\phi] \, e^{iS[\phi]+ \phi \cdot f} \quad \text{with} \quad \phi\cdot f =\!\int \!d^dx \, \phi(x) f(x) \quad \text{and} \quad Z[0]=1, \tag{3}$$ is just an alternative method for the computation of the generating functional.
Performing the corresponding calculations for the simple quantum mechanical harmonic oscillator in the presence of an external source $$f(t)$$ (in fact an external time-dependent force) using the operator formalism as well as the path integral technique is highly recommended!
Edit: To be more specific, consider the Hamilton operator $$H= H_0-\underbrace{\int d^3x \, \Phi(t, \vec{x}) f(t, \vec{x})}_{H_1(t)}, \tag{4}$$ where $$H_0$$ refers to some (in general) interacting scalar field theory. The time evolution of the operators in the interaction picture (IP) is given by $$A_{\rm IP}(t)= e^{iH_0t} A_{\rm IP}(0) e^{-iH_0t}, \tag{5}$$ whereas the full time evolution of the expectation value of the corresponding observable in some state $$|\psi_H \rangle$$ is given by $$\langle \psi_H | U^\dagger(t) A_{\rm IP}(t) U(t) |\psi_H\rangle, \qquad U(t)= {\rm T} e^{-i\int\limits_0^T d\tau \, H_{1, \rm IP}(\tau)}. \tag{7}$$ The state vectors in the Heisenberg picture and the interaction picture are related by $$|\psi_{\rm IP}(t)\rangle = U(t) |\psi_H\rangle. \tag{8}$$ We assume that the time dependent term $$H_1(t)$$ vanishes for $$t \to \pm \infty$$ (i.e. the external source tends to zero both in the remote past and the far future). Suppose the motion starts in the ground state $$|0\rangle$$ of the unperturbed system (i.e. $$H_0 |0\rangle=0$$) we have $$\lim\limits_{t \to -\infty} |\psi_{\rm IP}(t) \rangle = |0\rangle, \tag{9}$$ or, equivalently, $$U(-\infty)\! \! \! \! \underbrace{|0, \rm in\rangle}_{\text{Heisenberg state}}\! \! \! \! = |0\rangle, \tag{10}$$ such that $$|0, {\rm in} \rangle = {\rm T} e^{-i \int\limits_{-\infty}^0 \! d\tau \, H_{1, \rm IP}(\tau)} |0\rangle. \tag{11}$$ In the course of time, the state vector moves away from the ground state, $$|\psi_{\rm IP}(t)\rangle = {\rm T} e^{-i \int\limits_{-\infty}^t \! d \tau \, H_{1, \rm IP}(\tau)}|0\rangle \tag{12}$$ until it does not change anymore once the external source is switched off. For $$H_1 \equiv 0$$, the state vector $$|\psi_{\rm IP}(+\infty)\rangle$$ would coincide with the ground state of $$H_0$$ , but in the presence of $$H_1(t)$$ it does in general not represent an eigenstate of $$H_0$$. The perturbation violates time-translation invariance and energy is not conserved. The vector $$|\psi_{\rm IP}(t)\rangle$$ becomes a superposition of eigenstates of $$H_0$$. The probability amplitude that the system is found in the ground state at $$t\to \infty$$ is thus given by $$\langle 0 | {\rm T} e^{-i \int\limits_{-\infty}^\infty \! d\tau \, H_{1,\rm IP}(\tau)} |0\rangle =\langle 0, {\rm out}|0, {\rm in} \rangle_{H_1(t)}, \tag{13}$$ making contact with the discussion above.
• Thank you for the post. But does the source $f\neq 0$ change on the observation that a state in Heisenberg picture has no time evolution ? It should be same at $-\infty$ as well as at $+\infty$. Therefore $Z[f] = <0|0>_f =1$. I guess it is not supposed to be like this, i.e. $<0|0> \neq <0|0>_f$, but why? It also seems that there is no alternative: One starts with $|0 -\infty>$ and ends up inevitably in $|0 +\infty>$. Therefore their projection on each other is always 1. Can the path integral tell me that with a certain probability I end up in a state with certain non-zero number of particles? Commented Apr 13 at 16:23
• The state $|0\rangle$ refers to the ground state of the unperturbed system (i.e. for $f=0$). And, of course, you can end up in a state being a superposition with all possible particle numbers. I'll add an edit to make this clear. Commented Apr 13 at 17:40