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The path integral provides a method for computing a time evolution by a weighted summing up all possible deviations.

Is there such a method for a system, where one not only knows the initial condition, but also how the system end up? I.e. given is the beginning and the end configuration and what one is interested is the average field configuration in between this interval. Of course the path integral $$\langle \psi_{t_2},\psi_{t_1} \rangle=\int D\psi... $$ gives a value for both states at $\psi_{t_1}$ (start) and $\psi_{t_2}$ (end) fixed, but that's just interpreted as "what is the probability for this state later, if I start out like that". I want to consider a situation, where that is known to be equal to 1 and I'm interested in the evolution in between.

One might think of a transition from one field equilibrium to another here. The point being that in such a situation, it is a priori clear that the summing will 100% end up in one specific field situation. I think knowing the end configuration must sure give new information/restriction to get more out of it.

For example, a naive idea I come up with right now are the possibility that one can consider the time evolution from beginning and end, i.e. one would approach a field configuration in between from both sites, and they would have to coincide.

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3 Answers 3

Suppose you have some action described by a degree of freedom $x$, i.e. $S[x(t)]$. The path integral with fixed boundary condition looks like

$$ \int_{(x_a,t_a)}^{(x_b,t_b)} \mathcal{D}x e^{\frac{i}{\hbar}S[x(t)]}$$

We can write a general path in this path integral as,

$$ x(t) = x_{cl}(t) + y(t)$$

where $x_{cl}$ is the classical solution of the action subject to the boundary conditions, $x_{cl}(t_a) = x_a$ and $x_{cl}(t_b) = x_b$. Furthermore, $y(t)$ obeys $y(t_a) = y(t_b) = 0$, and this function will be the degree of freedom we integrate over. So the effect of the boundary conditions is taken care of by this "classical" piece of the path integration variable $x(t)$. This is probably the answer you are looking for.

Let's suppose the action is quadratic. If we fill this solution for $x$ into the action we can write the action as a sum of three "terms": (1) $S_{cl}$ which is the action evaluated for the classical path $x_{cl}$. (2) The action evaluated with $y$. (3) the terms linear in $y$. This term vanishes, because $x_{cl}$ is the classical solution, hence any linear deviation of the actions vanishes due to the Euler-Lagrange equations. For the path integral we are left with:

$$ e^{\frac{i}{\hbar} S_{cl}}\int_{(0,t_a)}^{(0,t_b)} \mathcal{D}x e^{\frac{i}{\hbar}S[y(t)]} $$

So the effect of the boundary conditions are taken into account through this phase factor in front. The path integral is now an integral over only those paths which vanish at the boundary.

If the action is not quadratic then we first need to resort to perturbation theory. This amounts to writing the non-quadratic piece of the action $e^{i V(x)/\hbar}$ as a series expansion, and treating each term in this expansion as a quadratic action again. So each term in the perturbation theory is in fact a Gaussian integration again, meaning we can apply the trick described above.

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1) I'm not quite sure if this is what OP is asking(v1), but Dirichlet boundary conditions, i.e., where the initial and final positions are fixed, are the standard boundary conditions imposed on a Feynman path integral (FPI) in quantum mechanics. However, there do exist FPIs with more general boundary conditions.

2) As an aside, there exists a Keldysh formalism, where one path-integrates along a Keldysh time integration contour, which starts and ends at $t=-\infty$.

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Okay, maybe I wasn't 100% clear: Of course the path integral $\langle \psi_{t_2},\psi_{t_1} \rangle=\int D\psi... $ gives a value for both states at $\psi_{t_1}$ (start) and $\psi_{t_2}$ (end) fixed, but that's just interpreted as "what is the probability for this state later, if I start out like that". I want to consider a situation, where that is known to be $=1$ and I'm interested in the evolution in between. –  NikolajK Feb 9 '12 at 14:39
    
In elementary PI treatments of quantum mechanics, the integral seems to be performed between initial and final states, but for quantum field theory PIs it seems usually to be done for vacuum->vacuum (with some sources along the way to make it interesting). Is this because it's not possible in general to prepare field eigenstates ? –  twistor59 Feb 9 '12 at 14:42

As Qmechanic said, when you specify the initial and final configurations, you assign a unity probability to them.

Note, specifying a certain initial/final state means many-many measurements that guarantee the state are such and such, as usually in QM.

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