11
$\begingroup$

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 $\langle T \{ \phi(x_1)...\phi(x_n) \} \rangle = \frac{\int{D[x] e^{\frac{i}{\hbar}S}\phi(x_1)...\phi(x_n)}}{\int{D[x] e^{\frac{i}{\hbar}S}}}$.

$\endgroup$
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Apr 15 '13 at 5:17
4
$\begingroup$

"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.)

$\endgroup$
  • $\begingroup$ Ah! I forgot about the factor hidden in the measure! Does that mean that the modulus of the weight of each path is equal and the only difference between each path is the phase? $\endgroup$ – ChickenGod Apr 15 '13 at 1:55
  • 1
    $\begingroup$ @ChickenGod In the simplest and most common by far cases, yes - the weight of all paths are equal and the only difference is the phase. Exceptions exist, such as the nonlinear sigma model where the measure contains a determinant which basically gives the invariant measure on the (curved) space of paths, like the $r^2 \sin(\theta)$ Jacobian to go from cartesian to spherical coordinates. You can find more about these models in Zinn-Justin. $\endgroup$ – Michael Brown Apr 15 '13 at 2:12
  • 1
    $\begingroup$ The overall normalization of the path integral is irrelevant for virtually all physical purposes and the integrand's constancy of the absolute value does indeed imply that the integrand cannot be interpreted as any sort of wave function because that wouldn't converge after squaring the absolute value. This answer is just plain wrong. $\endgroup$ – Luboš Motl Jul 3 '17 at 16:14
6
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ When you write 'coarsegrained histories', do you mean "macroscopically distinct" histories in the thermodynamic sense? So for one such history, the path integral really will sum over a huge number of microscopically distinct histories which then accumulates into a probability amplitude which then can have any magnitude (including complete destructive interference)? $\endgroup$ – BjornW Jul 2 '17 at 10:32
  • $\begingroup$ Which regions of the path integral you have to resum depends on the precise quantity you're interested in. Only the total sum over all histories has a truly physical meaning. But what was relevant here is that even though the absolute value of the integrand is constant, this property disappears due to interference if you clump the histories into nearby "families". I was meaning families enough to change $S$ by $O(1)$ or so - these families may still be much smaller than ensembles you could use in thermodynamics for other purposes. But they're larger than the minimum from quantum mechanics. $\endgroup$ – Luboš Motl Jul 3 '17 at 16:18
  • $\begingroup$ Yes that makes sense, thanks. Reason I asked was because I'm trying to show (to myself) how things like a field history with couplings eventually end up in that (1/137)^n lower-probability bin.. It's another thing that's not intuitively clear with the abs mag integrand, but it's not obvious how it appears out of interference either. I guess food for another question :) $\endgroup$ – BjornW Sep 15 '17 at 9:15
  • $\begingroup$ Dear Bjorn, to get the expansion in terms of Feynman diagrams, you need to analyze the path integral perturbatively, expand it to a power law or Taylor series. You surely can't get the individual terms with different powers of 1/137 - you can't even extract the exponents - by making just an order of magnitude estimate of the whole integral. One point is that every new term in the 1/137 expansion measures a correction to a phase of the previous one. It's mostly phases, and not the absolute values, that are being expanded. $i$ is almost always in the expansion parameter. $\endgroup$ – Luboš Motl Sep 16 '17 at 4:01
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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...

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ just now I do not remember what I had in mind to control the general case, but it seems I thought that Gell-Mann RG transformation could surface during the process for QFT, while for QM I hoped to use the "Tangent Groupoid" of Alain Connes. $\endgroup$ – arivero Aug 24 '15 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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