Schrödinger solutions are usually if not always of the type: $\psi=\operatorname{T}(t)*\operatorname{X}(x)$ (we use the separation of variables method to arrive at the time independent Schroedinger equation).

I was trying to find a non-separable solution. To this purpose I tried the following method: compose the product T(t)*X(x) into another function. For example: $\sinh(\operatorname{T}(t) \cdot \operatorname{X}(x))$ or $\ln(\operatorname{T}(t) \cdot \operatorname{X}(x))$ or $(\operatorname{T}(t) \cdot \operatorname{X}(x))^2$ and so on.

Then I try first time derivative of $\psi = \mathrm{second}$ position derivative of $\psi$ (trying to find a solution for the null potential. In addition I know I need a constant but it's for simplification).

I arrive at a differential equation. I try simple solutions for one of the functions like $\operatorname{T}(t)=t$. I get to a very difficult differential equation that can't be solved even with a look at: http://www.amazon.com/Handbook-Solutions-Ordinary-Differential-Equations/dp/1584882972

If there any obvious non separable solution that I am missing?

  • $\begingroup$ I asked a friend of mine and told me this was a consequence of the cauchy thoerem for partial differential equations. There are many cauchy theorems. Can someone give the url for a page for this theorem please? $\endgroup$
    – Pedro
    Jan 28, 2014 at 10:53
  • $\begingroup$ Also, physics.stackexchange.com/questions/255251/… $\endgroup$
    – Cheng
    Dec 2, 2022 at 7:21

3 Answers 3


In general every linear combinations of a separable solution is still a solution (superposition principle), so you can take the simplest separable solution, put it in a linear combination with arbitrary coefficient (complex numbers, phases) and you obtained a solution of Sc. equation NON-separable.

  • 1
    $\begingroup$ But does this cover all cases of non-separable solutions? Or are there solutions that cannot be represented as a linear combination of simple separable solutions? $\endgroup$
    – Nick
    Jan 5, 2015 at 23:07
  • 2
    $\begingroup$ @Nick Since the Hamiltonian is a self-adjoint operator on the Hilbert space, its eigenfunctions form a complete basis for the space. So any solution to the Schrodinger equation can be expressed as a linear combination of separable stationary-state solutions. $\endgroup$
    – tparker
    Dec 15, 2018 at 21:57

Well, solutions of the Schroedinger equation are usually (but not always) written in the form $\psi = \sum_n T_n(t) \ast X_n(x)$ -- i.e. as a sum of separable solutions. In fact, it is always possible to write the solution in this form. It follows from the fact that the Hamiltonian is Hermitian and all Hermitian operators are diagonalizable. (I'm ignoring operators with continuous spectra, but this is not a serious complication.)

Solutions which are not of the form $T(t)\ast X(x)$ are still sometimes useful though -- coherent states are one example.

I can give more details on any of the above if you ask.

  • $\begingroup$ Yes. Like $e^{-iE_1t } c_1 |1 \rangle + e^{-iE_2t } c_2 |2 \rangle $, where $E_1 \neq E_2$ $\endgroup$
    – user26143
    Jan 28, 2014 at 7:18
  • $\begingroup$ How does it follow from the fact tha the Hamiltonean is an Hermitian operator that the wavefunction can be written as the sum you provided? That is different than claiming that there is a complete set of eigenvectors that generate all functions... $\endgroup$
    – Pedro
    Jan 28, 2014 at 7:19
  • $\begingroup$ " That is different than claiming that there is a complete set of eigenvectors that generate all functions... " Well you're right that this is what is needed, and in fact it is a result from functional analysis that tells us that its exactly what we get with a self adjoint operator. en.m.wikipedia.org/wiki/Self-adjoint_operator $\endgroup$ Jan 28, 2014 at 8:19
  • $\begingroup$ @Pedro if H has a complete set of eigenvectors $\psi_n(x)$ then you can write any initial state as $\sum_n c_n \psi_n$ and its time evolution $\sum_n e^{-i E_n t}c_n \psi_n$ is a generic solution of the Schrödinger equation. $\endgroup$
    – fqq
    Jan 28, 2014 at 11:07
  • $\begingroup$ I would like a proper explanation why this is the case. $\endgroup$ Dec 15, 2018 at 3:07

You get nonseperable solutions to the time-dependent Schrodinger equation, which is when the potential depends on time.

Edit: As the commentor suggested, an example would be if you had a magnetic field originating from a changing current, so a field changing with time.

  • $\begingroup$ Could you elaborate in what situations this is the case? I think this would be helpful. $\endgroup$ Dec 15, 2018 at 3:07
  • $\begingroup$ I think my edit is clearer now? $\endgroup$ Dec 15, 2018 at 20:58

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