# I About the premises behind the Schrödinger equations

1. Dec 25, 2017

### snoopies622

Long ago I read on these forums that one cannot derive the Schrödinger equations because they're fundamental scientific laws. But I have noticed that I can generate them by making the following physical assumptions and then doing a trivial amount of substitution and differentiation:
(1) $$\frac {\partial ^2 \psi }{\partial x^2} + k^2 \psi = 0$$
(2) $\lambda = h/p$
(3) Total energy = PE + $p^2/2m$
and (4) $E=h \nu$
where (4) is only needed for the time dependent form.

What bothers me is that (1) assumes a $\psi$ with a definite wavelength — that is — a momentum eigenfunction, and (4) was arrived at for photons, and of course the S.E.'s are used to deal with wavefunctions that are not necessarily momentum eigenfunctions and for particles that are not photons.

Thoughts?

Last edited: Dec 25, 2017
2. Dec 26, 2017

### Staff: Mentor

You derived the Schrödinger equation limited to free particles with fixed momentum. It turns out that the equation is more general than that.

3. Dec 26, 2017

### snoopies622

I did not derive it at all here, I'm saying that putting (1),(2) and (3) together it is easy to arrive at
$$\frac {\partial ^2 \phi} {\partial x ^2} + \frac {8 \pi ^2 m }{h^2}(E - PE) = 0$$
and then adding (4) we can get
$$i \hbar \frac {\partial \phi} {\partial t } = E \phi$$
and these forms are not limited to momentum eigenfunctions, nor are they for photons, yet (1) and (4) are connected to those physical premises.

In case you're wondering, I'll show you explicitly what I mean in an upcoming post.

Last edited: Dec 26, 2017
4. Dec 26, 2017

### snoopies622

$$\phi = \phi_0 e^{i(kx-\omega t)}$$
and take the derivative with respect to x twice and rearrange terms, we get (1) in the original post.
Then if we plug in $k=2 \pi / \lambda$ , $\lambda = h/p$, $KE = p^2 /2m$ and E = KE + PE, we get the time independent form I cited in entry #3 above.

$$\phi = \phi_0 e^{i(kx-\omega t)}$$
and take the time derivative once, then substitute $\omega = 2 \pi f$, f=E/h, $\hbar = h/ 2 \pi$ and rearrange terms, we get the time dependent form I cited in entry #3 above.

Thus we start with a $\psi$ with a definite wavelength and definite speed and yet end up with the general (one dimensional) Schrödinger equations.

Last edited: Dec 26, 2017
5. Mar 1, 2018

### snoopies622

Bumping an old thread from a couple months ago, hope y'all don't mind. I've given this matter more thought but still find it odd that one can produce the Schrödinger equations by making assumptions that are not entirely consistent with their general use. How did Schrödinger himself arrive at them? I have not seen his original paper.

6. Mar 1, 2018

### Staff: Mentor

Can you give a specific link to a post? This claim seems highly dubious to me since AFAIK it is perfectly possible to derive the Schrodinger equation as a non-relativistic limit of an appropriate quantum field theory, and in any case the Schrodinger equation, being non-relativistic, is obviously an approximation only.

7. Mar 1, 2018

### ZapperZ

Staff Emeritus

http://www.pnas.org/content/pnas/110/14/5374.full.pdf

Zz.

8. Mar 1, 2018

9. Mar 1, 2018

### snoopies622

Thanks all. "Inspired guess" is what I figured. The mechanics-optics connection reminds me of something I read many years ago in a book on the history of physics — that had Hamilton lived long enough, he himself would have invented quantum mechanics.

10. Mar 3, 2018

### dumpling

If you wish to derive quantum theory as Schrödinger did, there is really only two fundamental assumptions ( with the addition that you would need to derive S-equations for each class of Hamiltonian).

1) Quantum theory is not classical, that is, the Hamilton-Jacobi equation is not true. A scalar, that measures "nonclassicality" for all the spacetime, can be introduced. The quantum theory is different then all other theories, that is, the scalar has an extrema (typically a minimum). This is philosophically the same as the principle of least action.

2) The theory that guides the "wavefunction" must be linear. The wavefunction is introduced intuitively, through the action.
Linearity is a very strong limitation, and with a correct choice of Hamiltonian, you can get back the S-equation with some variational calculus.

11. Mar 3, 2018

### Staff: Mentor

Like so much of physics its real basis is symmetry - but you must read what I mentioned, then think. BTW Wigner got a Nobel for figuring it all out.

Thanks
Bill

12. Mar 3, 2018

### Staff: Mentor

For the detail see:
https://arxiv.org/pdf/1204.0653.pdf

He even made two errors that cancelled each other out: Schrodinger’s notes [49] show that he was well aware that the solution of (8.12) gives correctly the bound state energies of the hydrogen atom before introducing in the anstaz concerning the hypothetical quantity J. This artifice compensates for the physically incorrect ansatz. The constant K should actually be the pure imaginary quantity −ih in which case (8.5) becomes the correct equation (7.27). Repeating Schrodinger’s stationary algorithm starting with the correct relation S = −ih¯ ln ψ would then give the incorrect equation (8.5)! Indeed, since the H-J equation and the properties of the generating function S already follow from Hamilton’s equations which in turn are a consequence of Hamilton’s Principle the condition that the action S should be stationary for arbitrary variations of space-time path it would seem that Schrodinger is attempting here to close a door that is already shut

Thanks
Bill

Last edited: Mar 3, 2018