How can one make sense of the idea of extremizing a Hamilton-Jacobi equation?

In Schrödinger's paper "Quantisation as a Problem of Proper Values I" (Annalen der Physik (4), vol. 79, 1926, p. 1, available e.g. in Collected papers on wave mechanics, AMS publishing) he begins with the classical time-independent Hamilton-Jacobi equation

$$H(q,\frac{\partial S}{\partial q}) = E$$

and instead of assuming an additive solution for separation of variables he assumes a multiplicative solution thus defining $S = K\ln(\psi)$ to get

$$H(q,\frac{K}{\psi} \frac{\partial \psi}{\partial q}) = E.$$

Instead of solving it he talks about how this last equation "can always be transformed so as to become a quadratic form (of $\psi$ & its first derivatives) equated to zero", neglecting the relativistic variation of mass.

(I'm assuming that transforming this to a quadratic form means a Taylor expansion in powers of $\psi$, though I'm not sure about that).

Then he seeks "a function $\psi$, such that for any arbitrary variation of it the integral of the said quadratic form, taken over the whole of coordinate space (I am aware this formulation is not entirely unambiguous), is stationary, $\psi$ being everywhere real, single-valued, finite, and continuously differentiable up to the second order."

In other words, he basically extremizes

$$I = \iiint\left(\frac{K^2}{2m}\left[\left(\frac{\partial \psi}{\partial x}\right)^2 \ + \ \left(\frac{\partial \psi}{\partial y}\right)^2 \ + \ \left(\frac{\partial \psi}{\partial z}\right)^2\right] + (V - E)\psi^2\right)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$

and ends up with the time independent Schrödinger equation, but the issue is extremizing the Hamilton-Jacobi equation. Where is the justification for this?

I can only find a reference to it here (in the ps.gz link), so it does seem to be something more than one person has done, but what is the theory behind this? Does it function as another method of solving Hamilton's equations? If so, why?

  • 1
    $\begingroup$ Think you meant $\partial\psi/\partial q$ in the second equation. Is this anything other than saying you can find the solution of $f(x)=0$ by minimizing $f(x)^2$? $\endgroup$
    – Michael
    Sep 13, 2013 at 6:59
  • $\begingroup$ Thanks, fixed. I'm honestly not sure, is it? $\endgroup$
    – bolbteppa
    Sep 13, 2013 at 7:03
  • $\begingroup$ That's what it looks like. Don't know if there is any deeper meaning to his argument or not. $\endgroup$
    – Michael
    Sep 13, 2013 at 7:33
  • $\begingroup$ I think you might be right, but how does one say this explicitly? If your Hamiltonian to begin with is what's inside that triple integral, then how are you minimizing the square of that? Maybe this is the meaning behind his quadratic form comment? $\endgroup$
    – bolbteppa
    Sep 13, 2013 at 7:48
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    $\begingroup$ The thing under the integral isn't the Hamiltonian, it's $\psi^2 (H - E)$ which is (A) (apparently) positive semidefinite for arbitrary $\psi$ and (B) zero for solutions of the Hamilton-Jacobi equation. So by minimizing the integral you get solutions of $H=E$. The subtlety is whether (A) actually holds or not. It's clear for a $q^2$ potential, but $-1/q$? ... $\endgroup$
    – Michael
    Sep 13, 2013 at 8:29

1 Answer 1


While one may use different representations for quantum mechanics (path integrals, Schrodinger equation, Heinsenberg representation, etc...), one cannot "demonstrate" quantum mechanics from classical mechanics principles.

One needs specific new postulates.

From the classical time-independent Hamilton-Jacobi equation, using $S = K \ln \psi$, with $\psi$ real, you will get a time-independent classical quadratic equation, :

$K ~\partial^i \psi ~\partial_i \psi+ (V-E)~\psi^2 = 0 \tag{1}$

And now, in the context of classical mechanics, you are stucked, you can't do anything more.

If you decide to add the necessity of extremizing the integral :

$$I = \iiint (K ~\partial^i \psi ~\partial_i \psi+ (V-E)~\psi^2) \tag{2}$$ this is a new postulate, and you cannot obtain this new postulate from classical mechanics principles.

  • $\begingroup$ Do you have any references or links that go into more detail on the implications of this new postulate, or what it means to add this, or basically any information at all on it? $\endgroup$
    – bolbteppa
    Sep 13, 2013 at 9:32
  • $\begingroup$ @bolbteppa : Sorry, no. I just don't see any postulate of classical mechanics which authorizes you to demand extremizing $I$ $\endgroup$
    – Trimok
    Sep 13, 2013 at 9:58
  • $\begingroup$ @bolbteppa : If we had taken a imaginary $\psi$, and $K = \hbar^2$, with $I(\psi^*, \psi)$, extremizing $I$ will give you Schrodinger equation, so, in some sense, the new postulate, is some "real" form of a quantum mechanics postulate. $\endgroup$
    – Trimok
    Sep 13, 2013 at 10:03
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    $\begingroup$ Yes I now see this is a major assumption. Schrodinger was extremizing the H-J equation & not even assuming the final result contained the solution to the H-J equation, so it looks as though he was doing something completely new. If you follow through on what he does in his paper he ends up with the possibility of complex eigenfunctions even though he begins by assuming his functions are real, thus assuming that integral is stationary must somehow be sneaking in complex numbers, & so I wonder if it's equivalent to allowing complex numbers, which I may have an interesting question on, thanks. $\endgroup$
    – bolbteppa
    Sep 14, 2013 at 17:33

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