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I've seen a proof that the path integral formulation of quantum mechanics is equivalent to solving Schrodinger's equation. However, it appears to me that the proof actually depended on the Hamiltonian having a particular form. I'm wondering how general is the equivalence.

Let me sketch a handwavy derivation of path integrals from Schrodinger's equation to illustrate where I think the problem might lie.

Suppose we want to find a green function ##G(x,y,t)## with the property that if ##\psi(x,t)## is a solution to Schrodinger's equation, then ##\psi(x,t) =\int G(x,y,(t-t_0)) \psi(y,t_0) dy##.

Formally, we can write down the answer as follows:

##G(x,y,t) = e^{-iHt/\hbar} \delta(x-y)##

where ##H## is the hamiltonian. To see that this solves the equation:

##\int G(x,y,(t-t_0)) \psi(y,t_0) dy = \int e^{-iH(t-t_0)/\hbar} \delta(x-y) \psi(y,t_0) dy = e^{-iH(t-t_0)/\hbar} \int \delta(x-y) \psi(y,t_0) dy = e^{-iH(t-t_0)/\hbar} \psi(x,t_0)##

Now, the question is to evaluate ##e^{-iHt/\hbar} \delta(x-y)##. We can easily prove the following fact about Green functions:

##G(x,y,t) = \int dx_1 dx_2 ... dx_{N-1} G(x, x_{N-1}, \frac{t}{N}) G(x_{N-1}, x_{N-2}, \frac{t}{N}) ... G(x_1,y, \frac{t}{N})##

So we can pick ##N## large enough that ##\delta t \equiv \frac{t}{N}## is small. So the question really becomes expressing ##G(x,y,\delta t)## in the limit as ##\delta t## is small. If we know what ##G## is like for small times, then we can use the above integral to compute it for larger times. Going back to our original expression for ##G##, we have:

##G(x,y,\delta t) = e^{-iH \delta t/\hbar} \delta(x-y)##

We assume that ##H## is an expression ##H(p,x)## depending on the operators ##p## and ##x##. Here's the first approximation (so far, everything has been exact):

We assume that if you have a function ##f(x,t)## which is strongly localized near ##x=y## at time ##t=0##, and you let it evolve under Schrodinger's equation, then after a small amount of time ##\delta t##, the function will still be strongly localized near ##x=y##. In that case, the spatial dependence of the hamiltonian is likely to be irrelevant. If ##H(p,x) = \frac{p^2}{2m} + V(x)##, then as long as the wave function is strongly localized near ##x=y##, then there should be a negligible error if you replace ##V(x)## by ##V(y)##. This can probably be made more rigorous, but I'm not too concerned about it. So our first approximation is this:

##G(x,y,\delta t) = e^{-iH(p,x) \delta t/\hbar} \delta(x-y) \approx e^{-iH(p,y) \delta t/\hbar} \delta(x-y)##

Since the delta function is so strongly localized, it only "feels" the potential near ##x=y##, so we make a negligible error by replacing ##H(p,x)## by ##H(p,y)##. So we're replacing the potential by a constant potential.

At this point, ##H(p,y)## no longer involves ##x##, which suggests that we should evaluate it when applied to momentum eigenstates. We use the usual representation of ##\delta(x-y)##:

##\delta(x-y) = \frac{1}{2\pi \hbar} \int dp e^{i p (x-y)/\hbar}##

Then we can write:

##e^{-iH(p,y)\delta t/\hbar} \delta(x-y) = \frac{1}{2\pi \hbar} \int dp e^{-iH(p,y)\delta t/\hbar} e^{i p (x-y)/\hbar}##

I was a little sloppy in distinguishing between ##\hat{p}##, the operator, and ##p## the variable of integration. However, since

##\hat{p} e^{i p(x-y)/\hbar} = p e^{i p(x-y)/\hbar}##,

bringing the operator ##e^{-iH(p,y)\delta t/\hbar}## inside the integral allows us to replace ##\hat{p}## by ##p##, its eigenvalue.

So now we have an expression for ##G(x,y,\delta t)##, valid for small ##\delta t##, which no longer involves any operators at all:

##G(x,y,\delta t) \approx \frac{1}{2\pi \hbar} \int dp e^{-iH(p,y) \delta t/\hbar + i p (x-y)/\hbar}##

Now, to get further, we have to do another approximation to reduce that integral to something doable. We use the method of stationary phase. Write:

##p\ (x-y) - H(p,y) \delta t \approx p_0 (x-y) - H(p_0, y) \delta t + \delta p ((x-y) - \frac{\partial H}{\partial p}|_{p=p_0} \delta t) - \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} \delta t + Q(\delta p, x, y, \delta t)##

I'm writing the expression as a Taylor series about the point ##p=p_0##, and ##Q## represents third-order terms and higher. I'm defining ##\delta p = p - p_0##.

Here's where an unexpected connection with Lagrangian mechanics comes in. If ##H(p,y)## is treated as a classical Hamiltonian, then the corresponding velocity is ##v_c(p) \equiv \frac{\partial H}{\partial p}##. The velocity is implicitly a function of momentum. If we choose ##p_0## so that ##v_c(p_0) = \frac{x-y}{\delta t}##, then the expression becomes

##p\ (x-y) - H(p,y) \delta t \approx (p_0 v_c(p_0) - H(p_0, y)) \delta t - \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} \delta t + Q(\delta p, x, y, \delta t)##

The linear term vanishes. Then amazingly, the first expression is just the classical lagrangian:

##L(y,v) = p v - H##

So we have our expression for ##G##:

##G(x,y,\delta t) \approx e^{i L(y, v) \delta t/\hbar} \frac{1}{2\pi \hbar} \int dp e^{-i\delta t/\hbar \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} - i/\hbar Q(\delta p, x,y, \delta t)}##

We've been completely general so far. This expression simplifies enormously if we assume that ##H## is quadratic in ##p## and that ##\frac{\partial^2 H}{\partial p^2}## is a constant (for classical mechanics, it is just ##\frac{1}{m}##). In that case, the integral is some function of ##\delta t## alone, and we can write:

##G(x,y,\delta t) \approx e^{i L(y,v) \delta t/\hbar} f(\delta t)##

for some unknown function ##f(\delta t)##. Under the assumption that ##H = \frac{p^2}{2m} + V(y)##, the integral becomes a well-known gaussian integral, except for the peculiarity that the exponent is imaginary. If you assume that it can be analytically continued from the real case, you get:

##f(t) = \sqrt{\frac{m}{2\pi i \hbar \delta t}}##

However, if ##H## is not simply quadratic in ##p##, then the integral becomes some complex function:

##G(x,y,\delta t) \approx e^{i L(y, v) \delta t/\hbar} f(x,y,\delta t)##

So for more complex hamiltonians, the path integral cannot simply be written in terms of the Lagrangian.

Putting the expression for ##G## back to get the value for larger times, we have:

##G(x,y,t) \approx \int dx_1 dx_2 ... dx_{N-1} e^{i/\hbar \sum_j L(x_j, v_j) \delta t} \Pi_j f(x_{j}, x_{j-1}, \delta t)##

Which can be written as:

##G(x,y,t) \approx \int dx_1 dx_2 ... dx_{N-1} e^{i/\hbar \int L(x(t), v(t)) dt} \Pi_j f(x_{j}, x_{j-1}, \delta t)##

So that's almost the path integral representation, except for the pesky functions ##f##. I don't see how to evaluate that product, in general, so I don't see how this actually becomes the path integral expression for the Green function (unless the ##f## does not depend on position).

Of course, we can always fold the unknown functions into the definition of the path integral: Define:

##\int \mathcal{D}x = dx_1 dx_2 ... dx_{N-1} \Pi_j f(x_{j}, x_{j-1}, \delta t)##

then we have the expression

##G(x,y,t) \approx \int \mathcal{D}x e^{i/\hbar \int L(x(t), v(t)) dt} ##

but that buries the problem into the expression ##\mathcal{D}x##

Let me sketch a handwavy derivation of path integrals from Schrodinger's equation to illustrate where I think the problem might lie.

Suppose we want to find a green function ##G(x,y,t)## with the property that if ##\psi(x,t)## is a solution to Schrodinger's equation, then ##\psi(x,t) =\int G(x,y,(t-t_0)) \psi(y,t_0) dy##.

Formally, we can write down the answer as follows:

##G(x,y,t) = e^{-iHt/\hbar} \delta(x-y)##

where ##H## is the hamiltonian. To see that this solves the equation:

##\int G(x,y,(t-t_0)) \psi(y,t_0) dy = \int e^{-iH(t-t_0)/\hbar} \delta(x-y) \psi(y,t_0) dy = e^{-iH(t-t_0)/\hbar} \int \delta(x-y) \psi(y,t_0) dy = e^{-iH(t-t_0)/\hbar} \psi(x,t_0)##

Now, the question is to evaluate ##e^{-iHt/\hbar} \delta(x-y)##. We can easily prove the following fact about Green functions:

##G(x,y,t) = \int dx_1 dx_2 ... dx_{N-1} G(x, x_{N-1}, \frac{t}{N}) G(x_{N-1}, x_{N-2}, \frac{t}{N}) ... G(x_1,y, \frac{t}{N})##

So we can pick ##N## large enough that ##\delta t \equiv \frac{t}{N}## is small. So the question really becomes expressing ##G(x,y,\delta t)## in the limit as ##\delta t## is small. If we know what ##G## is like for small times, then we can use the above integral to compute it for larger times. Going back to our original expression for ##G##, we have:

##G(x,y,\delta t) = e^{-iH \delta t/\hbar} \delta(x-y)##

We assume that ##H## is an expression ##H(p,x)## depending on the operators ##p## and ##x##. Here's the first approximation (so far, everything has been exact):

We assume that if you have a function ##f(x,t)## which is strongly localized near ##x=y## at time ##t=0##, and you let it evolve under Schrodinger's equation, then after a small amount of time ##\delta t##, the function will still be strongly localized near ##x=y##. In that case, the spatial dependence of the hamiltonian is likely to be irrelevant. If ##H(p,x) = \frac{p^2}{2m} + V(x)##, then as long as the wave function is strongly localized near ##x=y##, then there should be a negligible error if you replace ##V(x)## by ##V(y)##. This can probably be made more rigorous, but I'm not too concerned about it. So our first approximation is this:

##G(x,y,\delta t) = e^{-iH(p,x) \delta t/\hbar} \delta(x-y) \approx e^{-iH(p,y) \delta t/\hbar} \delta(x-y)##

Since the delta function is so strongly localized, it only "feels" the potential near ##x=y##, so we make a negligible error by replacing ##H(p,x)## by ##H(p,y)##. So we're replacing the potential by a constant potential.

At this point, ##H(p,y)## no longer involves ##x##, which suggests that we should evaluate it when applied to momentum eigenstates. We use the usual representation of ##\delta(x-y)##:

##\delta(x-y) = \frac{1}{2\pi \hbar} \int dp e^{i p (x-y)/\hbar}##

Then we can write:

##e^{-iH(p,y)\delta t/\hbar} \delta(x-y) = \frac{1}{2\pi \hbar} \int dp e^{-iH(p,y)\delta t/\hbar} e^{i p (x-y)/\hbar}##

I was a little sloppy in distinguishing between ##\hat{p}##, the operator, and ##p## the variable of integration. However, since

##\hat{p} e^{i p(x-y)/\hbar} = p e^{i p(x-y)/\hbar}##,

bringing the operator ##e^{-iH(p,y)\delta t/\hbar}## inside the integral allows us to replace ##\hat{p}## by ##p##, its eigenvalue.

So now we have an expression for ##G(x,y,\delta t)##, valid for small ##\delta t##, which no longer involves any operators at all:

##G(x,y,\delta t) \approx \frac{1}{2\pi \hbar} \int dp e^{-iH(p,y) \delta t/\hbar + i p (x-y)/\hbar}##

Now, to get further, we have to do another approximation to reduce that integral to something doable. We use the method of stationary phase. Write:

##p\ (x-y) - H(p,y) \delta t \approx p_0 (x-y) - H(p_0, y) \delta t + \delta p ((x-y) - \frac{\partial H}{\partial p}|_{p=p_0} \delta t) - \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} \delta t + Q(\delta p, x, y, \delta t)##

I'm writing the expression as a Taylor series about the point ##p=p_0##, and ##Q## represents third-order terms and higher. I'm defining ##\delta p = p - p_0##.

Here's where an unexpected connection with Lagrangian mechanics comes in. If ##H(p,y)## is treated as a classical Hamiltonian, then the corresponding velocity is ##v_c(p) \equiv \frac{\partial H}{\partial p}##. The velocity is implicitly a function of momentum. If we choose ##p_0## so that ##v_c(p_0) = \frac{x-y}{\delta t}##, then the expression becomes

##p\ (x-y) - H(p,y) \delta t \approx (p_0 v_c(p_0) - H(p_0, y)) \delta t - \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} \delta t + Q(\delta p, x, y, \delta t)##

The linear term vanishes. Then amazingly, the first expression is just the classical lagrangian:

##L(y,v) = p v - H##

So we have our expression for ##G##:

##G(x,y,\delta t) \approx e^{i L(y, v) \delta t/\hbar} \frac{1}{2\pi \hbar} \int dp e^{-i\delta t/\hbar \frac{\delta p^2}{2} \frac{\partial^2 H}{\partial p^2}|_{p=p_0} - i/\hbar Q(\delta p, x,y, \delta t)}##

We've been completely general so far. This expression simplifies enormously if we assume that ##H## is quadratic in ##p## and that ##\frac{\partial^2 H}{\partial p^2}## is a constant (for classical mechanics, it is just ##\frac{1}{m}##). In that case, the integral is some function of ##\delta t## alone, and we can write:

##G(x,y,\delta t) \approx e^{i L(y,v) \delta t/\hbar} f(\delta t)##

for some unknown function ##f(\delta t)##. Under the assumption that ##H = \frac{p^2}{2m} + V(y)##, the integral becomes a well-known gaussian integral, except for the peculiarity that the exponent is imaginary. If you assume that it can be analytically continued from the real case, you get:

##f(t) = \sqrt{\frac{m}{2\pi i \hbar \delta t}}##

However, if ##H## is not simply quadratic in ##p##, then the integral becomes some complex function:

##G(x,y,\delta t) \approx e^{i L(y, v) \delta t/\hbar} f(x,y,\delta t)##

So for more complex hamiltonians, the path integral cannot simply be written in terms of the Lagrangian.

Putting the expression for ##G## back to get the value for larger times, we have:

##G(x,y,t) \approx \int dx_1 dx_2 ... dx_{N-1} e^{i/\hbar \sum_j L(x_j, v_j) \delta t} \Pi_j f(x_{j}, x_{j-1}, \delta t)##

Which can be written as:

##G(x,y,t) \approx \int dx_1 dx_2 ... dx_{N-1} e^{i/\hbar \int L(x(t), v(t)) dt} \Pi_j f(x_{j}, x_{j-1}, \delta t)##

So that's almost the path integral representation, except for the pesky functions ##f##. I don't see how to evaluate that product, in general, so I don't see how this actually becomes the path integral expression for the Green function (unless the ##f## does not depend on position).

Of course, we can always fold the unknown functions into the definition of the path integral: Define:

##\int \mathcal{D}x = dx_1 dx_2 ... dx_{N-1} \Pi_j f(x_{j}, x_{j-1}, \delta t)##

then we have the expression

##G(x,y,t) \approx \int \mathcal{D}x e^{i/\hbar \int L(x(t), v(t)) dt} ##

but that buries the problem into the expression ##\mathcal{D}x##

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