# Expectation values in path integral formalism

In quantum field theory, it is often assumed that the expectation value $$\langle A\rangle$$ of an operator $$A$$ can be written in the path integral formalism in the following way:

$$\langle A\rangle = \frac{\int \mathcal{D}\phi\, A\, e^{\frac{i}{\hbar}S[\phi]}}{\int \mathcal{D}\phi\, e^{\frac{i}{\hbar}S[\phi]}}.$$

How can we reach this conclusion? Is this always true for all operators $$A$$, or is it only true in some cases?

• The derivation of the formula has been given in all textbooks I've read on field theory. I'm not sure I would call it an assumption. See Wikipedia where it also appears. Nov 24, 2021 at 21:28
• Do you have any example of a book it is proved in? I have Srednicki's book, and in that it is not really shown. For the harmonic oscillator with position coordinate $q$, I guess it is kind of trivially shown if $A$ is a time ordered combination of $Q$ operators, i.e., $Q(t_n)Q(t_{n-1})\dots Q(t_1)$, if $t''>t_n>\dots >t_1>t'$, where $t'$ and $t''$ are the start and stop times for you path integral, respectively. But if it is anything more complicated that that I'm having problems showing it. But then on the other hand, the action $S$ is not over a field $\phi$, but over a path $q$. Nov 25, 2021 at 15:42
• @MariusLadegårdMeyer In this document, it seems like they define it to be that way (on page 8, the first equation in section 3.2). Nov 25, 2021 at 17:25
• Atland and Simons' Condensed matter field theory for instance. Nov 25, 2021 at 20:27

If you are looking for a proof then the recommendations given in the comments appear to be quite helpful. There is a ''physics'' type proof in $$\underline{Peskin\;\&\;Schroeder's}$$ textbook where they first state the formula, and then show that we obtain the same result for a scalar propagator in the canonical formulation (commutators).

However, if you would like to know the motivation for the expectation value of an operator in field theory, then we can look back at examples from introductory quantum physics. For example, the classical formula to compute the average value for energy values over a distribution is the following, $$\bar{E} = \frac{\int_0^\infty E*P(E)\, dE}{\int_0^\infty P(E)dE},$$ which is just a Maxwell-Boltzmann distribution. The numerator is just the energy of the system, weighted by the probability distribution; while the denominator is the integral of finding the system with any energy.

• So do we need to assume that the system is Maxwell-Boltzmann distributed in order to get this formula, i.e., do we need to use statistical mechanics rather than using only pure quantum mechanics or quantum field theory? Jan 9, 2022 at 23:12
• Yes and no. Yes, the system behaves as a Maxwell-Boltzmann statistics as do practically all quantum mechanical & thermodynamic systems do. But, this formulation is just another way to analyze physics in general. This is why I used the example of how to calculate the average energy as it is practically identical to the way we compute expectation values (which are averages in a sense). I actually wrote some notes on the functional formalism and its use in physics which can be found here: alexcassem.net. Jan 10, 2022 at 1:06
• They are non-formal, and are solely to see how path integral calculations behave. For instance, I did problem 9.2 of Peskin and Schroeder's text which relates the path integral to the partition function of statistical mechanics, and then to distributions. Jan 10, 2022 at 1:09
1. On one hand, the formal connection between the operator formalism in the Heisenberg picture and the Hamiltonian phase space path integral is \begin{align} &{}_J\langle Q_f,t_f |TF[\hat{Q},\hat{P}] |Q_i,t_i\rangle_J \cr ~=~&{}_J\langle Q_f,0| T F[\hat{Q},\hat{P}]\exp\left\{ - \frac{i}{\hbar}e^{-i\epsilon}\int_{t_i}^{t_f}\!dt~ H_J(\hat{Q},\hat{P})\right\} |Q_i,0\rangle_J \cr ~=~&\int_{Q(t_i)=Q_i}^{Q(t_f)=Q_f}\! {\cal D}Q~{\cal D}P~F[Q,P]\exp\left\{ \frac{i}{\hbar}S_J[Q,P]\right\} , \end{align} \tag{1} cf. e.g. this Phys.SE post. We have inserted Feynman's $$i\epsilon$$ prescription to ensure that an otherwise oscillatory Boltzmann-factor is exponentially damped.

2. On the other hand, we can insert a complete set of energy eigenstates \begin{align} {}_J \langle Q_f,t_f|n\rangle_J ~=~&{}_J \langle Q_f,0|T \exp\left\{ - \frac{i}{\hbar}e^{-i\epsilon}\int_{0}^{t_f}\!dt~ H_J(\hat{Q},\hat{P})\right\} | n\rangle_J\cr ~=~& \exp\left\{ - \frac{i}{\hbar}e^{-i\epsilon}t_f E_n \right\} {}_J \langle Q_f,0| n\rangle_J , \end{align} \tag{2} and \begin{align} {}_J \langle m| Q_i,t_i\rangle_J ~=~&{}_J \langle m|T \exp\left\{ \frac{i}{\hbar}e^{-i\epsilon}\int_{t_i}^{0}\!dt~ H_J(\hat{Q},\hat{P})\right\} | Q_i,t_i\rangle_J\cr ~=~& \exp\left\{ \frac{i}{\hbar}e^{-i\epsilon}t_i E_n \right\} {}_J \langle m| Q_i,t_i\rangle_J . \end{align} \tag{3} Therefore the excited energy states wash out of the overlap \begin{align} &{}_J\langle Q_f,t_f |TF[\hat{Q},\hat{P}] |Q_i,t_i\rangle_J \cr ~=~&\sum_{n,m=0}^{\infty} {}_J\langle Q_f,t_f|n\rangle_J {}_J\langle n |TF[\hat{Q},\hat{P}] |m\rangle_J {}_J\langle m |Q_i,t_i\rangle_J\cr ~\stackrel{(2)+(3)}{\sim}&\exp\left\{ -\frac{i}{\hbar}e^{-i\epsilon}(t_f-t_i) E_0 \right\} \langle Q_f,t_f|0\rangle_J {}_J\langle 0 |TF[\hat{Q},\hat{P}] |0\rangle_J {}_J\langle 0 |Q_i,t_i\rangle_J \cr &\quad\text{for}\quad -t_i,t_f~\to~\infty. \end{align} \tag{4}

3. In conclusion, the quotient leads to OP's sought-for formula \begin{align} &\frac{\int_{Q(t_i)=Q_i}^{Q(t_f)=Q_f}\! {\cal D}Q~{\cal D}P~F[Q,P]\exp\left\{ \frac{i}{\hbar}S_J[Q,P]\right\}}{\int_{Q(t_i)=Q_i}^{Q(t_f)=Q_f}\! {\cal D}Q~{\cal D}P~\exp\left\{ \frac{i}{\hbar}S_J[Q,P]\right\}}\cr ~\stackrel{(1)}{=}~&\frac{{}_J\langle Q_f,t_f |TF[\hat{Q},\hat{P}] |Q_i,t_i\rangle_J}{{}_J\langle Q_f,t_f |Q_i,t_i\rangle_J}\cr ~\stackrel{(4)}{\sim}~&\frac{{}_J\langle 0 |TF[\hat{Q},\hat{P}] |0\rangle_J}{{}_J\langle 0 |0\rangle_J} \quad\text{for}\quad -t_i,t_f~\to~\infty. \end{align} \tag{5}

References:

1. M. Srednicki, QFT, 2007; Chapter 6. A prepublication draft PDF file is available here.