Showing two wavefunctions are proportional to one another

Asked Nov 17 '15 at 01:03

Active Nov 17 '15 at 01:38

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I am struggling to answer the following question:

Let ψ₁(x) and ψ₂(x) be normalisable energy eigenfunctions for a particle of mass m in one dimension moving in a potential V(x). Suppose that ψ₁ and ψ₂ have the same energy eigenvalue E. By considering ψ₁ψ₂' - ψ₂ψ₁', show that these wavefunctions must be proportional to one another.

I tried differentiating the Schrödinger equations and multiplying them together appropriately, but this just yielded a lot of messy algebra with no useful outcome.

I'm also wondering if there is a physical interpretation of this? Does it mean that the wavefunction for a normalisable energy eigenstate in one dimension always be chosen to be real?

edited Nov 17 '15 at 01:38

Gert

35,289

asked Nov 17 '15 at 01:03

David Smith

$\psi_{1}'$ or $\psi_{2}'$ does not imply 'differentiating the Schrödinger equations', they are simply the first derivatives of these wave functions. Just try $\psi_{1}= a\psi_{2}$ with $a$ a constant. The wave functions are degenerate: they have the same energy eigenvalue. – Gert Nov 17 '15 at 01:42
Degenerate solutions don’t exist in one dimension – Alfred Centauri Nov 17 '15 at 02:04
@AlfredCentauri: thanks for that. The question seems incomplete, I think it should have read: $\psi_1 \psi_2'-\psi_1' \psi_2=C$. – Gert Nov 17 '15 at 02:47

Showing two wavefunctions are proportional to one another

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