Once a quantum partition function is in path integral form, does it contain any operators?

Once a quantum partition function is in path integral form, does it contain any operators?

I.e. The quantum partition function is $Z=tr(e^{-\beta H})$ where H is an operator, the Hamiltonian of the system.

But if I put this into the path integral formalism so that we have something like $Z= \int D(\bar{\gamma},\gamma) e^{-\int_0^\beta d\tau H(\bar{\gamma},\gamma)}$, is the $H(\bar{\gamma},\gamma)$ an operator?

Thanks!

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Your expression for the partition function is correct. Calling it a "path integral", however, is unconventional usage. The traditional notion of a path integral is associated with the calculation of scatting amplitudes (or other quantities) for a field theory. Here the "paths" are the trajectories of particles through space-time. In your usage the "paths" are trajectories of the system through phase space. –  user346 Apr 27 '11 at 12:19
... Perhaps the partition function can be though of as measuring the "scattering amplitude" for a system to transition from one point in phase space to another. I'm sure that someone has done work along these lines though I haven't this analogy mentioned explicitly before. –  user346 Apr 27 '11 at 12:19
Actually, your expression for the path integral is missing the $p\dot{q}$ term if $H(\bar{\gamma},\gamma)$ is supposed to be the Hamiltonian function. –  Qmechanic Apr 27 '11 at 14:17

Nope, Feynman's path integral formulation of quantum mechanics is a method to directly calculate the complex probability amplitudes and all objects that appear in its formalism - not counting proofs of equivalence with other approaches to quantum mechanics - are $c$-numbers representing classical observables.

In particular, the exponent in the path integral - which should be $iS$ ($i$ times the action i.e. $i$ times the integrated Lagrangian), not the Hamiltonian - is a $c$-number-valued function of the "classical observables", the same function that is relevant for the classical (non-quantum) theory. So the path integral is an infinite-dimensional integral over otherwise "ordinary classical variables" that produces some probability amplitudes - the same ones that may be (but don't have to be) obtained from the operator formalism.

Uncertainty principle in path integral

By the way, some sorts of the path integral include integration over both positions and momenta, $\int Dx(t)\, Dp(t)$. How it is possible that both of them are treated "classically" as $c$-numbers? Doesn't it violate the uncertainty principle?

The answer is that it doesn't violate the uncertainty principle. One may still deduce that $xp-px=i\hbar$ from the path integral as long as she is careful about putting the right values of the time $t$ (the argument). The quantities $x(t)p(t-\epsilon)$ and $p(t)x(t-\epsilon)$ differ. A necessary condition for this difference to exist is the fact that "most" trajectories contributing to the path integral are discontinuous.

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Thanks! Great answer - exactly what I wanted. –  Jane Apr 27 '11 at 10:52
It was a pleasure. –  Luboš Motl Apr 27 '11 at 12:11
Lubos--- you mean "not differentiable" not "discontinuous". The trajectories contributing to the path integral are continuous in quantum mechanics, becoming discontinuous only if you have Levy quantum mechanics or a field theory path integral. –  Ron Maimon Aug 23 '11 at 4:40
On rereading, I realize you are talking about the p-q path integral, where the p part is certainly discontinuous--- sorry, strike the previous comment. The lack of differentiability is what gives the same commutation relation once you integrate out p. –  Ron Maimon Jun 5 '12 at 20:01
The second part of this answer expanded in LM blog motls.blogspot.com.es/2012/06/… –  arivero Aug 24 at 17:45

I am not going to disagree with Lubos, because his answer is mostly correct, but the quantities in the path integral can be also be interpreted as operators on the Hilbert space of states, if you like. They are classical quantities on each individual trajectory of the path integral (for bosonic fields), but they become operators after you integrate, when they are sitting inside the integral sign.

The state space of a path integral is defined by superpositions on the boundary conditions. If yuo multiply by some insertion A(x,t) in the integral, you are mixing up the superpositions when the integral hits that time by multiplying by a different quantity on each path. The mixing up is a linear operator on the boundary conditions, and it is exactly the linear operator A(x,t) in Heisenberg picture quantum mechanics.

For fermionic fields, they are always "operators" in some sense, because they are always anticommuting. But their anticommutation relation is independent of the dynamics in the path integral expansion, and reduces to classical Grassman variables. Multiplying by the Grassman field inside the path integral has the same effect on Grassman coherent states as the corresponding Heisenberg picture operator.

To give an example, consider the operator X(t). This is an operator in quantum mechanics, and it obeys the canonical commutation relation:

$$[X(t),P(t)] = i$$

Inside the path integral, X(t) is just a number on each trajectory, and P(t) is also a (divergent) number in the Lagrangian path integral. A quantum state $\psi(x)$ at time t_0 is described by the superposition over initial states

$$\psi(z,t) = \int dy \psi(y) \int_{x(t_0)=y}^{x(t)=z} e^{i\int_{t_0}^t {1\over 2} \dot{x}^2 - V(x)} Dx$$

Multiplying by X(t_0) has the effect of rearranging the initial condition wavefunction into

$$\int dy X(y)\psi(y) \int_{x(t_0)=y} r^{i\int_{t_0}^t {1\over 2} \dot{x}^2 - V(x) } Dx$$

And this is exactly the same as multiplying by the operator X. To recover the commutation relation, notice that

$$X(t)V(t)$$

is ambiguous, because it depends on the time order which you use to resolve the product:

$$X(t)V(t+\epsilon) = \hat{V}(t)\hat{X}(t)$$

where the right hand side is the operator product as matrix elements, and this is justified because you multiply the initial conditions by X(t) first, then later you multiply them by P(t),

$$X(t)V(t-\epsilon) = \hat{X}(t) \hat{V}(t)$$

Where again the right hand side is an operator product, and the left side are the matrix elements of this product. The difference between the two is nonzero, because the paths are not differentiable, $\Delta X^2$ is proportional to $\epsilon$, not $\Delta X$. So that:

$$X(t+\epsilon)V(t) - X(t)V(t) = {(X(t+\epsilon) - X(t))^2\over \epsilon} = i$$

Where the "i" is a 1 in Euclidean space, the velocity is a forward (Ito) difference, and so is always slightly ahead in time, and the last equality is a weak equality, valid only in the sense that the average over a small interval of the left and right side are equal (or equal in the sense of distributions), and valid only in the limit of approaching real time from Euclidean time, so that the oscillatory integrals are controlled.

The lack of differentiability is the same as in stochastic processes derived from Brownian motion, the square of the deviation is proportional to $\epsilon$, not like for differentiable functions, where the deviation itself is proportional to $\epsilon$.

This way of looking at things, where the quantities inside the path integral are operators, was used by Schwinger, and he liked it because it incorporated fermions naturally. Today we use Grassman integrals for the same purpose. The non-commutativity of the products is always there, however, and must be taken into account.

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Maybe it's me, but it is quite difficult to understand your notations. How about something like $\psi(x',t)=\int dx_0\psi(x_0)\int_{x(t_0)=x_0}^{x(t)=x'}Dx\,e^{iS[x(t)]}$ ? –  Kostya Sep 13 '11 at 15:14

NO, H is now a function of complex (or Grassmann) numbers rather than operators

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