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In Quantum Mechanics, path integrals are used to calculate the matrix element:

$$ \langle x_1, t_1|x_2, t_2\rangle_J=\int e^{i(S[x(t)]+\int\!J x(t))/\hbar} d[x(t) ].\tag{1}$$

If we naively try to extend this idea to QFT, we will naturally get:

$$\langle \phi_1, t_1|\phi_2 , t_2\rangle_J=\int e^{i(S[\phi(x^{\mu})]+\int\!J\phi(x^{\mu}))/\hbar} d[\phi (x^{\mu})].\tag{2}$$

$\phi_1$ and $\phi_2$ refer to initial and final field configurations. It seems like this sort of path integral will only be relevant in the wavefunctional formulation, which we don't care about in scattering applications.

Instead, the path integral that is relevant in applications is:

$$Z[J]=\int e^{i(S[\phi]+\int\!J\phi)/\hbar} d[\phi].\tag{3}$$

Differentiating this path integral gives us the correlation functions.

My question is, is this path integral conceptually related to the previous path integral which was designed for wavefunctional evolution? Or is it that this path integral is completely unrelated and we've unfortunately given the name "path integral" to two different ideas?

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    $\begingroup$ I didn't get the question. (1) is a special case of (2) in the sense that QM is QFT in 0+1 dimension. You can as well define generating functional in QM. $\endgroup$
    – paul230_x
    Commented Nov 15, 2022 at 11:39
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    $\begingroup$ @KP99 i mean (3) vs (2) or (1), not (2) vs (1) $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 11:41
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    $\begingroup$ Each answer helped with slightly different things. I just selected what I thought was the most relevant one. $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 14:40

4 Answers 4

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They are in the same framework, but we usually use them in different ways, which explains the differences you are noticing.

In general, the path integral should include factors of the wavefunction to describe the initial and final states \begin{equation} Z = \int D \phi \Psi_{in}[\phi] \Psi^\star_{out}[\phi] e^{i S[\phi]} \end{equation} For the quantum mechanics path integral you wrote down, the initial and final states are delta functions, which have the effect of fixing the end points of the path integral.

Typically in QFT we are interested in vacuum correlation functions. Therefore, in the standard QFT path integral, the initial and final states are always the vacuum. The vacuum is not a field eigenstate, so the "end points" are not fixed. However, you might still wonder why we don't have vacuum wavefunctions in the path integral. In fact, the $i \epsilon$ trick has the effect of reproducing the same result as including the vacuum wavefunctions. So the wave function factors are there, but "hidden" in the formalism of QFT.

You are correct that, in general, we should include the wave function factors in the path integral.

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    $\begingroup$ Combining yours and KP99's answer, what I understand is: The formula $(2)$ first needs to be generalised to initial and final states that are a superposition, because the vacuum is a superposition. But we don't use the generalisation because some $i\epsilon$ trick takes care of that. And the purpose of $J$ in $(3)$ is to initially treat the interacting Quantum field as a classical field. When we differentiate wrt $J$, we sort of take the $J\rightarrow 0$ limit of the vacuum transition amplitude. This limit corresponds to a quantum $J$ field. Is this correct? $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 13:12
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    $\begingroup$ @RyderRude I agree with the first half of your statement, up to "some $i\epsilon$ trick takes care of that." (The $i\epsilon$ trick is the $i\epsilon$ that appears in the definition of the Feynman propagator in case it isn't clear). I don't understand your statements about $J$. What I would say is that we introduce the classical source $J$ as a trick, that allows us to formally define correlation functions in terms of derivatives of the partition function, in a similar way to what is done in statistical mechanics. This turns out to be a useful way to set up perturbation theory. $\endgroup$
    – Andrew
    Commented Nov 15, 2022 at 15:06
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    $\begingroup$ I agree about $J$. I got confused initially but KP99 clarified it. $J$ is just a trick, which allows us to define "generating functional" of the probabilities. And generating functions are known to produce expectation values in their derivatives, and the expectation value is what we need for the the LSZ formula $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 15:31
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Quantum mechanics is just a special case of QFT when the base manifold is one-dimensional. The path integral framework is the same.

Let me deal with a $d$-dimensional base manifold, $Y$, and some field $\phi\in\mathcal{F}(Y)$, where $\mathcal{F}(Y)$ is a suitable space of functions on $Y$. Let me assume that the manifold $Y$ has an incoming boundary $\Sigma_-$ and an outgoing boundary $\Sigma_+$, so $\partial Y = \Sigma_-\sqcup\Sigma_+$. On each of $\Sigma_\pm$ one associates a Hilbert space of states $\mathcal{H}_{\Sigma_\pm}$. Performing the path integral on $Y$, with boundary conditions $$\phi\Big|_{\Sigma_{\pm}} = \phi_\pm$$ yields the integral kernel of the evolution operator $$\langle\phi_+|\widehat{K}_Y|\phi_-\rangle \equiv K_Y[\phi_-,\phi_+] := \int_{\phi_{\pm}} \mathrm{D}\phi\ \exp(-S[\phi]),\tag{a.1}$$ where $\widehat{K}_Y\in\mathcal{H}_{\Sigma_-}^*\otimes\mathcal{H}_{\Sigma_+}$.

To this you can obviously add a $J\cdot\phi$ term and compute the evolution operator in the presence of a source: $$\langle\phi_+|\widehat{K}_Y[J]|\phi_-\rangle \equiv K_Y[J;\phi_-,\phi_+] := \int_{\phi_{\pm}} \mathrm{D}\phi\ \exp(-S[\phi]+\mathrm{i}\;J\cdot \phi).\tag{a.2}$$

When $\Sigma_-=\varnothing$, but $\Sigma_+\neq \varnothing$ (or the opposite) this prepares a state $|\Psi_Y[J]\rangle$ (or $\langle\Psi_Y[J]|$), while when both $\Sigma_+$ and $\Sigma_-$ are empty this computes a $\mathbb{C}$-number $$K_Y[J] \equiv Z[J] = \int \mathrm{D}\phi\ \exp(-S[\phi]+\mathrm{i}\;J\cdot \phi).\tag{a.3}$$

Another way to say the above is that when $\Sigma_\pm = \varnothing$, the Hilbert space that corresponds to it is $\mathcal{H}_{\varnothing} = \mathbb{C}$. This is a one-dimensional Hilbert space, and hence it only has one state; the vacuum. The path integral on a closed manifold, then, prepares the in-state to be the vacuum in and the out-state to be again the vacuum. Inserting operators simply computes the expectation values of these guys in the vacuum. Indeed, however, in the presence of boundaries you don't compute vacuum expectation values. This is clear in my (a.2) equation where it is an expectation value between non-trivial in- and out-states, a bit akin to @Andrew's answer.

All of this is trivially true in quantum mechanics, just take $d=1$ and $Y$ is a $1$-dimensional manifold, whose incoming and outgoing boundaries are just points in time: $\partial Y = \{t_-\}\sqcup\{t_+\}$. Then my equation (a.1) reduces to your equation (1) and my equation (a.3) gives you exactly what you want in quantum mechanics, in the absence of boundaries, e.g. when $Y$ is a circle $Y=\mathbb{S}^1$. My equation (a.2) gives you the most general thing$^{(*)}$, an external source and boundaries.


$^{(*)}$you can get something even more general by including insertions of arbitrary operators $\mathcal{O}[\phi]$. Then the external source is a special case, the inclusion of the operator $\mathcal{O}[\phi]=\mathrm{e}^{\mathrm{i}J\cdot \phi}$.

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    $\begingroup$ What I don't understand is that we use the path integral in QFT in a very different way. We differentiate the path integral wrt $J$ at $J=0$, and then claim that the result correponds to the vacuum transition amplitude. Traditionally, we'd have needed to explicitly use the vacuum wavefunctional in the path integral, like Andrew wrote. But somehow, this "differentiate wrt $J$" trick works. $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 13:22
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    $\begingroup$ @RyderRude updated answer. Hope it's clear now! $\endgroup$ Commented Nov 15, 2022 at 13:32
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    $\begingroup$ Much thanks. But in traditional QFT, we don't assume the manifold to have a point-like boundary. Also, what we mean by the vacuum state in QFT is a non-zero wavefunctional correponding to the product of vacuums of Harmonic oscillators. $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 13:47
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    $\begingroup$ @RyderRude In usual (perturbative) QFT we take $Y=\mathbb{R}^d$, so $\partial Y = \partial \mathbb{R}^d = \varnothing$ $\endgroup$ Commented Nov 15, 2022 at 14:01
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    $\begingroup$ @RyderRude also, however you describe your vacuum state, since it's the only state it's the vacuum state $\endgroup$ Commented Nov 15, 2022 at 14:03
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  1. Eq. (1) [and implicitly eq. (2)] with source terms included are discussed in Ref. 1, which via an $i\epsilon$ prescription and limit $t_f-t_i\to \infty$ washes out the boundary conditions to a vacuum-to-vacuum amplitude. This is essentially Andrew's answer.

  2. Similarly, the trace in the partition function (3) reduces to the vacuum-to-vacuum amplitude in the aforementioned limit.

References:

  1. M. Srednicki, QFT, 2007; chapter 6. A prepublication draft PDF file is available here.
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    $\begingroup$ I am accepting Andrew's answer but yours was also necessary clarification $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 14:22
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Generating functional (3) can be interpreted as semi-classical quantity containing the quantum field $\phi$ and classical background field $J$ (which is non-dynamical). As it is apparent, in the limit where the strength of background field $J$ goes to zero, we recover transition amplitude (2).

We can make $J$ dynamical by adding the dynamical action ($S_B[J]$) of this field to the original quantum system ($S[\phi,J]=S[\phi]+J\phi$). In the absence of the quantum system, we can recover the equation of motion of background $J$ from the relation : $$\frac{\delta S_B[J]}{\delta J}|_{\phi=0}=0$$ We can define effective action by the relation: $$\exp\{iS_{eff}[J]\}=\int d[\phi]\exp\{iS[\phi,J]+iS_B[J]\}=\exp\{i\Gamma[J]+iS_B[J]\}$$ In presence of the quantum field, the new equation of motion for background $J$ is given by extremizing the effective action $S_{eff}$. The quantity $\frac{\delta \Gamma[J]}{\delta J}$ can be interpreted as vacuum expectation value of the backreaction produced by quantum field $\phi$ which acts as a source for the classical field $J$

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    $\begingroup$ Let me know if this is what you were asking $\endgroup$
    – paul230_x
    Commented Nov 15, 2022 at 12:19
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    $\begingroup$ $J$ is not dynamical, $J$ is still a background field. It would be dynamical if you integrated over it. $\endgroup$ Commented Nov 15, 2022 at 13:13
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    $\begingroup$ Combining yours and Andrew's answer, what I understand is: The formula $(2)$ first needs to be generalised to initial and final states that are a superposition, because the vacuum is a superposition. But we don't use the generalisation because some $i\epsilon$ trick takes care of that. And the purpose of $J$ in $(3)$ is to initially treat the interacting Quantum field as a classical field. When we differentiate wrt $J$, we sort of take the $J\rightarrow 0$ limit of the vacuum transition amplitude. This limit corresponds to a quantum $J$ field. Is this correct? $\endgroup$
    – Ryder Rude
    Commented Nov 15, 2022 at 13:14
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    $\begingroup$ @ɪdɪətstrəʊlə J is background field and is dynamical once $S_B[J]$ is included. Dynamical here just means that there is an equation for motion for $J$. Have a look at section 12.3 of Mukhanov and Winitzki, Introduction toQunatum fields in Classical Backgrounds $\endgroup$
    – paul230_x
    Commented Nov 15, 2022 at 13:19
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    $\begingroup$ @RyderRude J is always the classical field. We introduce the field J just for calculation convenience. But we motivate this inclusion by saying that a realistic system will contain the quantum field + environment (which we approximate by J). J is not required in the original problem, so after doing all necessary manipulation, we set J=0 $\endgroup$
    – paul230_x
    Commented Nov 15, 2022 at 13:27

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