Say you have an electron departing from point A and reaching poing B after a time t.

According to some helping friend, the Partition Function for that electron going from point A to B can be written as

$$Z = \int_{A \to B} [\mathcal{D}x]~ e^{iS[x]}$$

where $\mathcal{D}x$ is the measure that sums up over all paths going from $A$ to $B$, and $e^{iS[x]}$ is the weight of each path, $S[x]$ is the action.

That friend states then "From this partition function all desirable quantities can be obtained."

Not having much idea about Feynman's Path Integral Formulation of Quantum Mechanics, I have looked around a bit, and I would like someone to confirm the following statement I make:

The amplitude for the electron to go from A to B in the time t can be found in terms of that Partition Function $Z$ given above, as

$$\langle B|e^{-iHt}|A\rangle = Z$$

Did I catch it right?

  • $\begingroup$ Of course, if there are several answers, I will choose whichever has more additional information, but in principle, a simple confirmation (better with some reference or link to further reading) is enough. $\endgroup$
    – Mephisto
    May 11, 2013 at 1:21
  • 1
    $\begingroup$ $Z$ is related to e.g. $<A|B>$ by the LSZ reduction formula. Maybe someone can explain this formula... $\endgroup$
    – innisfree
    May 11, 2013 at 1:30

2 Answers 2


For simplicity, let's restrict the discussion to that of a single particle moving in one dimension. Path integrals can be performed in much broader contexts like quantum field theory, but I think that would conceptually obscure the issue at this point.

Let $H$ denote the (time-independent) quantum hamiltonian. Then the time evolution on the system is governed by the operator $U(t) = e^{-itH/\hbar}$ called the time evolution operator. If at a time $t_a$ the particle is in the state $|x_a\rangle$, the eigenstate of the position operator corresponding to position $x_a$, then the probability amplitude for finding the particle in the state $|x_b\rangle$ at some later time $t_b$ is given by $$ K(t_b,x_b;t_a,x_a) = \langle x_b|U(t_b-t_a)|x_a\rangle $$ This object is often referred to as the propagator. It turns out the propagator has the following path integral expression $$ K(t_b,x_b;t_a,x_a) = \int \mathcal Dx\mathcal Dp\, e^{\frac{i}{\hbar}S} $$ where the integral is over all paths $(x(t), p(t))$ in classical phase space satisfying the boundary conditions $x(t_a) = x_a$ and $x(t_b) = x_b$.

Now you could ask, what about if at time $t_a$, the system started in a state $|\psi_a\rangle$ that is not necessarily a position eigenstate, and I want to know what the probability is of finding that particle in the state $|\psi_b\rangle$ at time $t_b$. Can I write this in terms of the path integral? Yes!

What we're looking to compute is the following quantity: $$ \langle \psi_b|U(t_b-t_a)|\psi_a\rangle $$ Now we recall that the we can write the identity operator as $$ I = \int dx\,|x\rangle\langle x| $$ Inserting this expression twice into the amplitude we want to compute, and using integration variables $x_a$ and $x_b$ gives \begin{align} \langle \psi_b|U(t_a-t_a)|\psi_a\rangle &= \int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_b|U(t_a-t_a)|x_a\rangle\langle x_a|\psi_a\rangle \\ &= \int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_a|\psi_a\rangle K(t_b,x_b;t_a,x_a) \end{align} Now we just write the propagator in terms of the path integral as above to obtain $$ \langle \psi_b|U(t_a-t_a)|\psi_a\rangle =\int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_a|\psi_a\rangle \int \mathcal Dx\mathcal Dp\, e^{\frac{i}{\hbar}S} $$

  • $\begingroup$ Thus, that so-called partition function equals the path integral expression of the propagator (at least in the restricted 1-dimensional single-particle case), and the statement in my question is right when $|A>$ and $|B>$ are eigenstates of position, right? $\endgroup$
    – Mephisto
    May 11, 2013 at 2:16
  • $\begingroup$ @Mephisto Basically yes. $\endgroup$ May 11, 2013 at 2:23
  • $\begingroup$ Thanks! I have to see now what is that $\int \mathcal Dx \mathcal Dp$ but I can cope with that, I see in Zee/Srednicki that it is a short notation for an integral with multiple products due to dividing the path in small time steps... I'll have a closer look. Thanks! $\endgroup$
    – Mephisto
    May 11, 2013 at 2:36

What your friend actually meant is that you can obtain all desirable correlation functions. Assuming you're talking about a non-relativistic electron, consider a source term added to your action $$ S'[x] = S[x] + \int dt \, J(t) x(t)\,. $$

Now you can write any correlation function as a derivative of $\ln Z$ calculated at $J = 0$, i.e. $$ \langle B| \hat{x}(t') | A \rangle = \left.\frac{\partial}{\partial J(t')} \ln Z\right|_{J=0}\,, $$ $$ \langle B|\hat{T}\left\{ \hat{x}(t') \hat{x}(t'') \right\}| A \rangle = \left.\frac{\partial^2}{\partial J(t')\partial J(t'')} \ln Z\right|_{J=0}\,, $$ and so on, where $\hat{T}$ is the time ordering operator.

Hope this helps you.

  • 1
    $\begingroup$ Why has this been downvoted, is there a mistake in it I dont see? It would be nice if somebody could explain it ... :-/? $\endgroup$
    – Dilaton
    May 11, 2013 at 11:14
  • $\begingroup$ Exactly! This answer is perfectly fine! +1! $\endgroup$ Aug 4, 2013 at 14:12

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