Physical Interpretation of the Integrand of the Feynman Path Integral

Question

In quantum mechanics, we think of the Feynman Path Integral $\int{D[x] e^{\frac{i}{\hbar}S}}$ (where $S$ is the classical action) as a probability amplitude (propagator) for getting from $x_1$ to $x_2$ in some time $T$. We interpret the expression $\int{D[x] e^{\frac{i}{\hbar}S}}$ as a sum over histories, weighted by $e^{\frac{i}{\hbar}S}$.

Is there a physical interpretation for the weight $e^{\frac{i}{\hbar}S}$? It's certainly not a probability amplitude of any sort because it's modulus squared is one. My motivation for asking this question is that I'm trying to physically interpret the expression $\langle T \{ \phi(x_1)...\phi(x_n) \} \rangle = \frac{\int{D[x] e^{\frac{i}{\hbar}S}\phi(x_1)...\phi(x_n)}}{\int{D[x] e^{\frac{i}{\hbar}S}}}$.

Answer 1

Up to a universal normalization factor, $\exp(iS_{\rm history}/\hbar)$ is the probability amplitude for the physical system to evolve along the particular history. All the complex numbers in quantum mechanics are "probability amplitudes of a sort".

This is particularly clear if we consider the sum over histories in a short time interval $(t,t+dt)$. In that case, all the intermediate histories may be interpreted as "linear fields" – for example, a uniform motion $x(t)$ – which is why only the straight line contributes and its meaning is nothing else than the matrix elements of the evolution operator.

It may be puzzling why all the histories have probability amplitudes with the same absolute value. But it's true – in the sense of Feynman's path integral – and it's how Nature operates. At the end, some histories (e.g. coarse-grained histories in the sense of the consistent history interpretation of quantum mechanics) are more likely or much more likely than others. But within quantum mechanics, all these differences between the likelihood of different coarse-grained histories are explained by constructive or destructive interference of the amplitudes (and/or from different "sizes of the ensemble of histories")! That's just the quantum mechanics' universal explanation for why some histories are more likely than others. Histories that are not coarse-grained are equally likely!

Answer 2

"It's certainly not a probability amplitude of any sort because it's modulus squared is one." This does not follow... Anyway, an (infinite) normalisation factor is hidden away in the measure. The exponential has the interpretation of an unnormalised probability amplitude. Typically you don't have to worry about the normalisation explicitly because you compute ratios of path integrals, as your example shows. The book about the physical interpretation of path integrals is the original, and very readable, one by Feynman and Hibbs, which now has a very inexpensive Dover edition. I heartily recommend it. :) (Though make sure you get the emended edition as the original had numerous typos.)

Answer 3

One interesting interpretation comes from the Wick rotation, where you interpret $it/\hbar=\beta=1/(k_BT)$ - an imaginary time turns many quantum equations into similar equations of thermodynamics / statistical mechanics.

Since $S = \int L\,dt = \int (T-V)\,dt$ has the dimension of energy times time, this means the exponent can be interpreted as energy times $\beta$, i.e.

$$e^{\frac i\hbar S} \approx e^{-\beta E}$$

which is the summand of the partition function, from which you can then derive other thermodynamic quantities like entropy, heat capacity or the Helmholtz free energy.

Note I've been very hand-wavy about the transition from $L$ to -$E$, you'd have to insert the actual Legendre transformation $L=pq-H$ and really evaluate the path integration. I think we've done that more exactly in lectures at one point, but I can't remember for sure...

Answer 4

At a time when my hammer was the Dirac's delta distribution, I conjectured that the answer was Feynman Integral is a generalization of a Dirac's delta, the use of this delta being to find the extreme of the action.

Given a function $f(x)$, find a Dirac measure $\delta_f$ concentrated in the critical points of $f$. The answer is obviously $ \delta(f'(x))$, and using that $\delta(w)=\frac 1{2\pi}\int e^{iwt} dt$ we can write

$$ < \delta(f'(x)) | g(x)> = \int \int e^{i z f'(x)} g(x) dz dx = \int \int \lim_{y\to x} e^{i z {f(y)-f(x)\over y-x} } g(x) dz dx $$

and substuting $\epsilon= {y - x \over z}$

$$ <\delta_f | g> = \int \int \lim_{\epsilon\to 0} e^{i {1 \over \epsilon} (f(y) - f(x))} g(x) dx {dy \over \epsilon} $$ And if we know that the extreme is unique, we can work with the "halved" expresion

$$<\delta_f^{1/2} | O> = \lim_{\epsilon\to 0} {1 \over \epsilon^{1/2}} \int e^{i {1\over\epsilon} f(x)} O(x) dx$$ from which, by taking modulus square, $ <\delta_f | g> = <\delta_f^{1/2} | O> <\delta_f^{1/2} | O>^* $

But this is only a zero dimensional static argument. It is not even a D=0+1 theory, it is D=0+0. The same argument for Quantum or Classical Mechanics (D=0+1) or for Field Theory (D=3+1, say) should involve to control $$<\delta_L^{h,\epsilon'} | O[\phi] > = \int ... \int {1\over (h \epsilon')^{n/2}} e^{i {1\over h} L^{\epsilon'}_n[\phi_0,x_1,...x_n,\phi_1]} O[\phi] (\Pi dx_i) $$ with some technique similar to a brownian bridge for a Wiener measure, or at least my note of 1998 says that.

The funny thing of this idea was to start from the Lagrangian without any postulate of wuantum mechanics, not even the propagator, which is the cornerstone of the answers in Why is the contribution of a path in Feynmans path integral formalism $\sim e^{(i/\hbar)S[x(t)]}$. I think that pursuing this way I had tripped against the non differentiable paths mentioned in Once a quantum partition function is in path integral form, does it contain any operators? and What type of non-differentiable continuous paths contribute for the path integral in quantum mechanics.

The connection of path integral to classical mechanics is discussed also here What type of non-differentiable continuous paths contribute for the path integral in quantum mechanics and in the paper from Dirac quoted in this answer https://physics.stackexchange.com/a/134215/1335

Answer 5

One of the the numerical values of the weight $\exp{\frac{i S}{\hbar}}$ is going to have a maximum contribution to the Feynman path integral. You've probably seen a probability density plot in 2D or 1D. The classical path is going to be the one that minimizes the action. Think of it as a maximum probability density moving from one most probable position to another. The classical action Lagrangian density is going to contribute the most to the path integral.