In the book "Quantum Mechanics and Path Integrals" Feynman & Hibbs state that

The probability $P(b,a)$ to go from point $x_a$ at the time $t_a$ to the point $x_b$ at the time $t_b$ is the absolute square $P(b,a) = \|K(b,a)\|^2$ 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. $$ K(b,a) = \sum_{\text{paths from $a$ to $b$}} \phi[x(t)].\tag{2-14}$$ The contributions of a path has a phase proportional to the action $S$: $$ \phi[x(t)] = \text{const}\ e^{(i/\hbar)S[x(t)]}.\tag{2-15}$$

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.


7 Answers 7


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 $$S(t_f,t_i)~=~\int_{t_i}^{t_f} dt \ L(t)\tag{1},$$ 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

$$\begin{align} K(x_3,t_3;x_1,t_1) ~=~&\cr \int_{-\infty}^{\infty}\mathrm{d}x_2 \ K(x_3,t_3;x_2,t_2)& K(x_2,t_2;x_1,t_1),\end{align}\tag{2}$$

one must demand that

$$\begin{align}f(S(t_3,t_2)&f(S(t_2,t_1)) \cr~=~& f(S(t_3,t_1)) \cr~=~& f(S(t_3,t_2)+S(t_2,t_1)).\end{align}\tag{3}$$ In the last equality of eq. (3) we used the additivity of the action (1). Eq. (3) implies that

$$f(0)~=~f(S(t_1,t_1)) ~=~ 1.\tag{4}$$ (The other possibility $f\equiv 0$ is physically un-acceptable.)

So the question boils down to: How many continuous functions $f:\mathbb{R}\to\mathbb{C}$ satisfy $$f(s)f(s^{\prime}) ~=~f(s+s^{\prime})\quad\text{and}\quad f(0) ~=~1~?\tag{5}$$

Answer: The exponential function!

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

$$\begin{align}f(s) ~=~& f(0) + f^{\prime}(0)s +{\cal O}(s^{2}) \cr ~=~& 1+cs+{\cal O}(s^{2}) \end{align}\tag{6}$$

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

$$\begin{align} f(s) ~=~&\lim_{n\to\infty}f(\frac{s}{n})^n\cr ~=~&\lim_{n\to\infty}\left(1+\frac{cs}{n}+o(\frac{1}{n})\right)^n\cr ~=~&e^{cs},\end{align} \tag{7}$$

i.e., the exponential function! $\Box$

  • $\begingroup$ 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 $\endgroup$
    – lurscher
    Apr 15, 2011 at 16:16
  • $\begingroup$ 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. $\endgroup$
    – asmaier
    Apr 15, 2011 at 19:52
  • $\begingroup$ 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) $\endgroup$
    – lurscher
    Apr 15, 2011 at 19:53
  • $\begingroup$ @Qmechanic Shouldn't we also sum over $t_2$ in the group property equation? $\endgroup$ Oct 30, 2014 at 12:43
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    $\begingroup$ @LoveLearning: No. See e.g. Feynman & Hibbs, eq. (2-31). $\endgroup$
    – Qmechanic
    Oct 30, 2014 at 12:51

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 :)

  • 1
    $\begingroup$ +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. $\endgroup$
    – qftme
    Apr 15, 2011 at 13:38
  • $\begingroup$ 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. $\endgroup$
    – Lagerbaer
    Apr 15, 2011 at 14:37

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: $$M \sim \sum e^{i*\text{phase}}$$

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 $\text{phase} \sim S[x(t)]$. 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.


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.


As the OP asks to try to avoid Schrödinger equation, I think that it could be worthwhile to exhibit some variation of my answer from the similar post https://physics.stackexchange.com/a/202298/1335:

The great thing of the exponential measure in Feynman path integral is that when we evaluate the probability it transforms to a sort of derivative of the action, which is what we really want for the classical limit.

For example for the propagator

$$K(x,y;T) = \langle y;T|x;0 \rangle = \int_{f(0)=x}^{f(T)=y} e^{i {S[f]\over \hbar}} Df$$

$$P(x,y;T)^2=K^*(x,y;T)K(x,y;T)=\int_{g(0)=x}^{g(T)=y} e^{-i {S[g]\over \hbar}} Dg\int_{f(0)=x}^{f(T)=y} e^{i {S[f]\over \hbar}} Df$$ $$P(x,y;T)^2=\int_{g(0)=x}^{g(T)=y} \int_{f(0)=x}^{f(T)=y} e^{i {S[f]- S[g]\over \hbar}} Dg Df$$

Or, putting $\epsilon(t)=f(t)-g(t)$:

$$P(x,y;T)^2= \int_{g(0)=x}^{g(T)=y} ( \int_{\epsilon(0)=0}^{\epsilon(T)=0} e^{i {S[g \epsilon]- S[g]\over \hbar}} D\epsilon ) Dg$$

and the idea, that I do not know how to prove, is that the term between parenthesis should be the functional version of distribution $\delta(F'(u))$ with F' a discrete derivative $F(u+\Delta)-F(u) \over \Delta$. At least it works in this way for the usual delta distribution. Consider un potencial $V(x)$. The value of any function $f(x)$ at the minimum of the potential is

$$f(x)|_{V'(x)=0}=<\delta(V') | f>={1 \over 2\pi}\int \int e^{i p V'(x)} f(x) dp dx$$

or, discretized

$$f(x)|_{V'(x)=0}=\lim_{\beta \to 0}{1 \over 2\pi}\int \int e^{i p {V(x+\beta)-V(x) \over \beta}} f(x) dp dx$$

Note that for the classical limit we need both $\epsilon$ and the $\hbar$ go to zero, perhaps we could use other function such that $\hbar \gamma=\epsilon$ but the physical meaning of such function $\gamma(t)$ is obscure.

In conclusion, you need the weight to be complex in order to produce a difference when conjugating, and you need it to be an exponential so that this difference is really measured with the Dirac delta, forcing in this way the classical variational condition ${\delta L \over \delta g }= 0$

  • 1
    $\begingroup$ this answer is severely underappreciated, +1 $\endgroup$
    – lurscher
    Apr 26, 2018 at 17:07

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


This was ment to be posted as a question, but I think that it can help answering yours, and that mine will be answered by looking at the paper calmly.

The result is logical, since it corresponds to the classical path in the limit $\hbar \to 0$ due to the stationary property of the action, you can find this explanation for example on the chapter of "Functional methods" of "An Introduction to Quantum Field Theory" by Peskin and Schroeder. But this explanation is clearly one a posteriori.

I saw that Dirac had proven the existence of a quantum analogue to action playing a role in the transformation of coordinates and momenta from one set to another at a future time on his paper The Lagrangian in Quantum Mechanics, but it seems that the presence of the action as the pase of the wave function is previous, since he says

We have found here the natural extension of the well-known result that the phase of the wave function corresponds to Hamilton's principal function (the time integral of the lagrangian) in classical theory.

Now, in Dirac's book The Principles of Quantum Mechanics, in the section 31, "The motion of wave packets", he considers a system with classical analogue and says

We suppose that the time-dependent wave function in Schrödinger's representation is of the form $$\psi(q, t) = A\mathrm{e}^{iS/\hbar}$$

from this point, using an unitary transformation on the canonical momenta with $S$ as generator, he finds in the classical limit that

$$-\frac{\partial S}{\partial t} = H\left(q_r, \frac{\partial S}{\partial q_r} \right),$$ which allows from Hamilton-Jacobi equation to identify $S$ with the action, but this result is clearly also a posteriori.

I've traced the exponential of the action back to the paper of Schrödinger Quantisation as a Problem of Proper Values (Part II), where he obtains it from the Hamilton-Jacobi equation, but I found his reasoning a bit obscure from the beginning.

I think that using these sources the answer gets clearer. Finally, in his thesis, The Principle of Least Action in Quantum Mechanics, in section III: Least Action in Quantum Mechanics , Feynman shows that the "quantum analogue" of the action present in $\phi$, is in a lot of cases not just an analogue of the action as Dirac said, but exactly the action as the time integral of the Lagrangian.


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