1
$\begingroup$

If we consider two independent harmonic oscillators (identical too a two dimensional harmonic oscillator), the hamilton operator is

$$ H = \frac{p_1^2}{2m_1}+\frac{1}{2}m_1\omega_1^2x_1^2 + \frac{p_2^2}{2m_2}+\frac{1}{2}m_2\omega_2^2x_2^2 = H_1(x_1)+H_2(x_2) $$

and the solution of the Schrödinger equation can be written as $\psi(x_1,x_2)_{n,m}=\psi(x_1)_n\psi(x_2)_m$ with $\psi(x_i)$ the usual solution for the quantum harmonic oscillator, so that $H_i\psi_j(x_j)=\delta_{ij}E\psi_j(x_j)$ (for all quantum numbers) as the solution of the stationary Schrödinger equation.

But why is that the case? And why can the solution be written in that form?

I understand, that $p_i\psi_j(x_j)=0$ for $i \neq j$, because its the derivative in $x_i$, which is 0 if $\psi$ is not a function of $x_i$, but what about $x_i\psi_j(x_j)$ (operator $x_i$), isn't it just the multiplication with $x_i$ ?, so why should it be 0 for $i \neq j$ ? (which is necessary to write the solution in that form)

$\endgroup$
  • 1
    $\begingroup$ Hint: write the full thing out as a two-dimensional PDE and reduce it to multiplications of single-variable ODEs times the eigenfunction over the other variable. $\endgroup$ – Emilio Pisanty May 22 '18 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.