Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  Mephisto May 11 '13 at 1:21
1  
$Z$ is related to e.g. $<A|B>$ by the LSZ reduction formula. Maybe someone can explain this formula... –  innisfree May 11 '13 at 1:30
add comment

2 Answers

up vote 5 down vote accepted

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 $$ \langle \psi_b|U(t_a-t_a)|\psi_a\rangle =\int dx_a\int dx_b\langle \psi_b|x_b\rangle\langle x_a|\psi_a\rangle \int \mathcal Dx\mathcal Dp\, e^{\frac{i}{\hbar}S} $$

share|improve this answer
    
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? –  Mephisto May 11 '13 at 2:16
    
@Mephisto Basically yes. –  joshphysics May 11 '13 at 2:23
    
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! –  Mephisto May 11 '13 at 2:36
add comment

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.

share|improve this answer
1  
Why has this been downvoted, is there a mistake in it I dont see? It would be nice if somebody could explain it ... :-/? –  Dilaton May 11 '13 at 11:14
1  
I don't know which bird dropped that shit here with no comments. It is specially annoying in this case, since any comment would help me understanding this answer. $lnZ$ is like the generator of a Lie group or something, right? I still lack the appropiate knowledge. –  Mephisto May 11 '13 at 15:40
    
Exactly! This answer is perfectly fine! +1! –  Dimensio1n0 Aug 4 '13 at 14:12
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.