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In Feynman's book "Quantum Mechanics and Path Integrals" Feynman states that

the probability $P(b,a)$ to go from point $x_a$ at time $t_a$ to the point $x_b$ at the time $t_b$ is $P(b,a) = \|K(b,a)\|^2$ of an amplitude $K(b,a)$ to go from $a$ to $b$. This amplitude is the sum of contributions $\phi[x(t)]$ from each path. $$ K(b,a) = \sum_{\text{paths from $a$ to $b$}} \phi[x(t)]$$ The contributions of a path has a phase proportional to the action $S$: $$ \phi[x(t)] = \text{const}\ e^{(i/\hbar)S[x(t)]}$$

Why must the contribution of a path be $\sim e^{(i/\hbar)S[x(t)]}$? Can this be somehow derived or explained? Why can't the contribution of a path be something else e.g. $\sim \frac{S}{\hbar}$, $\sim \cos(S/\hbar)$, $\log(S/\hbar)$ or $e^{- (S[x(t)]/\hbar)^2}$ ?

Edit: I have to admit that in the first version of this question, I didn't exclude the possibility to derive the contribution of a path directly from Schrödinger's equation. So answers along this line are valid although not so interesting. I think when Feynman developed his formalism his goal was to find a way to quantize systems, which cannot be treated by Schrödinger's equation, because they cannot be described in terms of a Hamiltonian (e.g. the Wheeler-Feynman absorber theory). So I think a good answer would explain Feynman's Ansatz without referring to Schrödinger's equation, because I think Schrödinger's equation can only handle a specific subset of all the systems that can be treated by Feynman's more general principle.

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It's chosen in such a way that for actions that have a classical interpretation, you recover classical mechanics and the variational principle. It's motivated theoretically by the correspondence principle. But it is really just because that's the way nature seems to work. –  Raskolnikov Apr 14 '11 at 20:56
@Raskolnikov It is not chosen that way. Rather, this is a result from computing the quantity $K(b,a)$ rigorously using the Schrödinger equation and the time evolution operator, together with the Trotter-Kato formula. –  Lagerbaer Apr 14 '11 at 23:13
@Lagerbaer: but the Schrödinger equation is yet another way of formulating QM, just like the path-integral method. It's obvious that there is a correspondence between them, but both essentially have to be derived from the correspondence-principle (to classical physics) and the match of experiments (Aharonov-Bohm, dual-slit etc.). –  BjornW Apr 14 '11 at 23:22
@Bjorn Wesen: the classical-quantum correspondence does not always exist (see various comments by Lubos on this site). Last time I checked, the canonical (Hilbert spaces, operators, etc.) formalism was still the correct one (i.e. correct in all limits). For the cases when a path-integral formalism can be found, the latter has to agree with the canonical formulation. –  genneth Apr 15 '11 at 7:35
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5 Answers

up vote 10 down vote accepted

There are already several good answers. Here I will only answer the very last question, i.e., if the Boltzmann factor in the path integral is $f(S(t_f,t_i))$, with action $S(t_f,t_i)=\int_{t_i}^{t_f} dt \ L(t)$, why is the function $f:\mathbb{R}\to\mathbb{C}$ an exponential function, and not something else?

Well, since the Feynman "sum over histories" propagator should have the group property

$$ K(x_3,t_3;x_1,t_1) = \int_{-\infty}^{\infty}\mathrm{d}x_2 \ K(x_3,t_3;x_2,t_2) K(x_2,t_2;x_1,t_1),$$

one must demand that

$$f(S(t_3,t_2)f(S(t_2,t_1)) = f(S(t_3,t_1)) = f(S(t_3,t_2)+S(t_2,t_1)),$$

$$f(S(t_1,t_1)) = 1.$$

So the question boils down to: How many continuous functions $f:\mathbb{R}\to\mathbb{C}$ satisfy $f(s)f(s^{\prime})=f(s+s^{\prime})$ and $f(0)=1$?

Answer: The exponential function!

Proof (ignoring some mathematical technicalities): If $s$ is infinitesimally small, then one may Taylor expand

$$f(s) = f(0) + f^{\prime}(0)s +{\cal O}(s^{2}) = 1+cs+{\cal O}(s^{2}), $$

with some constant $c:=f^{\prime}(0)$. Then one calculates

$$ f(s)=\lim_{n\to\infty}f(\frac{s}{n})^n =\lim_{n\to\infty}\left(1+\frac{cs}{n}+o(\frac{1}{n})\right)^n =e^{cs}, $$

i.e., the exponential function.

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outstanding answer! i was thinking this myself and couldn't come up with a good argument other than the plane wave limit. This makes the deduction rigorous –  lurscher Apr 15 '11 at 16:16
Wow, that was an eye opener for me. If I could, I would upvote it much more. Let me remark, that I read that the exponential function is also a solution even if $s$ and $s'$ are allowed to be matrices, as long as these matrices commute. I don't know if there can be a solution if $s$ and $s'$ don't commute. –  asmaier Apr 15 '11 at 19:52
small comment; the group property of the propagator is implied by demanding time translation invariance of the propagator description (it doesn't imply that the propagator itself needs to be time translation invariant though) –  lurscher Apr 15 '11 at 19:53
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You start by writing down the probability to find a particle at $y$ at time $t$ when it was at $x$ at time $0$, denoted as $K(y,t;x,0)$. You get this by solving the Schrödinger equation with the initial condition $\psi(y,0) = \delta(y-x)$. Then, $K(y,t;x,0) = \psi(y,t)$. Thus, to solve this, we need to know the time development of the initial condition $\psi(y,0)$.

Let us start with the simple example of a free particle. This is easiest solved in momentum-representation, obtained by Fourier-transforming $\psi(y,t)$:

$$\psi(y,t) = \frac{1}{\sqrt{2\pi\hbar}} \int dp \exp(ipy/\hbar) \tilde \psi(p,t)$$ For $\tilde \psi$, the Schrödinger equation gives $$\tilde \psi(p,t) = \frac{1}{\sqrt{2\pi\hbar}} \exp \left(-\frac{i}{\hbar} \left[\frac{p^2 t}{2m} - px\right]\right)$$ This can be inserted back into the equation for $\psi(y,t)$. The integral over $p$ can be solved exactly. The final result is $$K_\text{free}(y,t;x,0) = \sqrt{\frac{m}{2\pi i\hbar t}} \exp\left(\frac{im(x-y)^2}{2\hbar t}\right)$$

Next step: The solution of the Schrödinger equation can generally be written as $$|\psi, t\rangle = \exp\left(-\frac{iHt}{\hbar}\right) |\psi,0\rangle$$ with $H$ being the Hamiltonian of your system. Writing $H = T+V$, the general formula for $K$ becomes $$K(y,t;x,0) = \langle y \mid \exp(-\frac{i(T+V)t}{\hbar}) \mid x \rangle$$ We use the Trotter-Kato Formula (which holds under certain conditions which I won't go into detail at this point. It allows us to write $$K(y,t;x,0) = \lim_{N\rightarrow \infty} \langle y \mid \left[ \exp(-\frac{iTt}{N\hbar}) \exp(-\frac{iVt}{N\hbar})\right]^N \mid x\rangle$$ We insert the unity operator, decomposed as $1 = \int dx | x \rangle \langle x |$ $N-1$ times, which gives us $$K(y,t;x,0) = \int dx_1 dx_2 \dots dx_{N-1} \prod_{j=0}{N-1} \langle x_{j+1} \mid \exp(-iTt/N\hbar) \exp(-iVt/N\hbar) \mid x_j \rangle$$ Note that $V$ as an operator acting on $|x\rangle$ gives just $V(x) |x\rangle$. And $\langle x_{j+1} | \exp(-iTt/N\hbar) | x_j \rangle$ gives us just the contribution of a free particle, i.e. $$\sqrt{\frac{mN}{2\pi i\hbar t}} \exp\left(\frac{imN}{2\hbar t}(x_{j+1} - x_j)\right)^2$$. If we abbreviate $\tau = t/N$, we can write: $$K(y,t;x,0) = \lim_{N\rightarrow \infty} \int dx_1 dx_2 \dots dx_{N-1} \left( \frac{m}{2\pi i\hbar \tau}\right)^{N/2} \times$$ $$\exp \left(\frac{i\tau}{\hbar} \sum_{j=0}^{N-1} \left[ \frac{m}{2}\left(\frac{x_{j+1}-x_j}{\tau}\right)^2 - V(x_j)\right]\right)$$

The next step is to see the values $x_j$ as points of a certain path $x(t')$ evaluated at points $t' = t_j = j\tau = jt/N$. If $\tau$ is small, we write $$\sum_{j=0}^{N-1} \tau f(t_j) \rightarrow \int f(t') dt'$$ $$\frac{x_{j+1} - x_j}{\tau} \rightarrow \dot x(t')$$ where the dot denotes the time-derivative.

The argument of the exponential then becomes $$\frac{i}{\hbar} \int_0^t dt' \left( \frac{m\dot x(t')^2}{2} - V(x(t'))\right)$$ You will have no trouble identifying the integrand as the Lagrangian $L = T-V$. The integral itself, therefore, is the classical action.

Thus, the formula we have for $K$ can be interpreted as the sum over all possible paths from $(x,0)$ to $(y,t)$ of the function $\exp\left(\frac{i}{\hbar} S(t,0)\right)$ of the classical action.

The interpretation of this was given in other answers: The classical path is that which minimizes the action, i.e. the action is stationary for the classical path. In your path-integral formula, this path will have a large contribution, as all paths that vary only slightly from the classical path will still have pretty much the same phase factor as the classical one, leading to constructive interference of those paths. For paths far from the classical path, the action will vary greater among the paths, so that there all possible phases occur, which will ultimately cancel each outer out.

Reference A lecture on advanced quantum mechanics given by Prof. Crispin Gardiner. Lecture notes are, unfortunately, not freely available. It was a good lecture :)

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@Lagerbaer I don't think the OP was asking for the derivation which can be found in any text on QFT. –  user346 Apr 15 '11 at 4:45
+1: Nobody who spent their time to genuinely try to answer deserves a negative vote. –  timur Apr 15 '11 at 8:13
@Timur you're right that @Lagerbaer did put in a great deal of effort. However, it was my feeling that the OP asked more for a physical explanation than a mathematical one. The complete derivation is of course nice to see but it does not provide insight into why the appropriate quantity, to associate the phase of the particle's world-line with, is the classical action $S$? In this regard I think first @Bjorn's and secondly @Lurscher's answers do a better job. –  user346 Apr 15 '11 at 11:47
+1 for taking the time to write this up. Whether a direct answer to the OP's question or not, it is certainly related and makes for some interesting reading. –  qftme Apr 15 '11 at 13:38
The OP asked if the form could be derived. If you start from the Schrödinger equation, it can indeed be derived. It wasn't specified where one should start with the derivation. –  Lagerbaer Apr 15 '11 at 14:37
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If you accept that Quantum Mechanics is built upon the fact that you sum complex amplitudes of processes (see this previous Question/Answers about this fact) you would expect that a sum over multiple paths behaves like a sum of different complex phases: $$M \sim \sum e^{i*phase}$$

Applying the variational principle to the phase, you see that the paths which vary their phase the least will contribute the most to the sum (because the others will average each other). Add the fact that you want the classical path to be the main contribution (because we want to match classical physics, this is the correspondence princple), and that the classical path is the path where the action $S$ varies the least, you can identify the phase with the action and get $phase \sim S[x(t)]$. Then you get $1 / \hbar$ as an experimental constant.

I'm not sure if this is a satisfactory answer, but most of the "strangeness" here comes from the QM superposition principle in the first place anyway. Note that the variational principle in classical mechanics was known and used before QM was invented and had the teleological property of "sniffing out" paths of least action. In the QM path-integral method this is at least explained from a more local point of view.

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An approach similar to Lagerbaer’s may be formulated without reference to the probability function. The overlap between states at different times $\langle\psi_{t^\prime}|\psi_t\rangle$ may be written according to a product of $|q_{t+n\delta t}\rangle\langle q_{t+n\delta t}$ at different time slices. The wave overlap is then $$ \langle\psi_{t^\prime}|\psi_t\rangle~=~\lim_{\delta t\rightarrow 0}\lim_{N\rightarrow\infty} \prod_{n=0}^N \int dq_{t+n\delta t}\langle\psi_{t+n\delta t}|q_{t+n\delta t}\rangle\langle q_{t+(n-1)\delta t}|\psi_{t+(n-1)\delta t}\rangle. $$ This description of the overlap is then according to snapshots determined by projectors, where in the limit the time increment vanish they recover the density matrix.

We now focus on a product defined on one particular time slice. Each infinitesimal overlap is written as $$ \psi^*(q,t)\psi(q,t~-~\delta t)~=~ \psi^*(t)\Big(\psi(q,t)~+~\delta t{\frac{d\psi}{dt}}(q,t)~+~O(\delta t^2)\Big). $$ The term to $O(\delta t)$ is easily seen to be $$ \delta t {\frac{d\psi}{dt}}(q,t) ~=~\delta t \Big(\frac{\partial\psi}{\partial t}(q,t)~+~ \frac{dq}{dt}\nabla \psi(q,t)\Big)~=~\frac{i\delta t}{\hbar}\Big(\frac{dq}{dt} p~-~ H\psi(t)\Big). $$ The integrand of the infinitesimal overlap is $$ \psi^*(q,t)\psi(q,t~-~\delta t)~=~ e^{\frac{i\delta t}{\hbar} \big({\dot q} p~-~H\big)}\psi^*(t)\psi(t) $$ This is a way of deriving the probability function above.

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I think the justification goes like this:

first a couple observations from the classical limit:

1) paths that are far from the classical solution are not near an extremum value of the action, which means that the action will have a non-zero variance in all paths that are neighbour to this path.

2) the classical solution itself is an extremum value of the action (either a minimum or a maximum) in the space of classical paths. in the neighbourhood of this path, the variance of the paths will approach zero.

so, an approach to construct a quantum limit would be to think on the double-slit experiment and see that the interference pattern is constructed from taking two plane waves paths that go from the source to each slit, and then from the slit to the a point in the interference screen.

In this case, none of the paths match exactly the classical path. if you write a plane wave you'll see that the argument $p x - E t$ and you'll notice that this is actually the action of a free particle. So you can think of the De Broglie wave as a plane wave with the action of a free particle $e^{i(kx - \omega t)}$

From this, is just a small step to just infer that in general, when paths are not restricted by a double slit, you need to allow for all classical paths possible, and when the action is more complex than a free particle, you need to replace the wavefunction argument by the action of the path

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