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Given the following Hamiltonian:

$$\hat H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_1^2} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_2^2} + \frac{1}{2}m\omega_1^2x_1^2 + \frac{1}{2}m\omega_2^2x_2^2 \, .$$

Where there is no coupled term $x_1x_2$, is the correct solution to assume a separation ansatz $\Psi_n = \psi_1(x_1)\psi_2(x_2)$, substitute in, divide through by $\psi_1(x_1)\psi_2(x_2)$ and get two equations of the form

$$\frac{\hbar^2}{2m}\frac{\partial^2\psi_i}{\partial x_i^2} + \frac{1}{2}m\omega_1^2x_i^2\psi_i = E_i\psi_i $$

where the solutions are the standard solutions to the quantum harmonic oscillator?

Following this the full solution $\Psi_n$ is the product of two independent solutions $\psi_1\psi_2$ with $E_{tot}=E_1+E_2$, so the total energy is a sum of energies but the discrete wavefunction is a product of discrete wavefunctions?

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    $\begingroup$ Your approach should be correct, are you asking for a proof or you just needed confirmation? $\endgroup$
    – MannyC
    Commented Apr 24, 2019 at 23:24
  • $\begingroup$ Just confirmation, thanks. I'm struggling a bit with the idea of the energies adding but the solution being a product of the independent solutions. For the coupled case (with an $x_1x_2$ term) I'm assuming you'd need to introduce new variables to make it look like the non-coupled case $\endgroup$
    – Tapedeck
    Commented Apr 24, 2019 at 23:33
  • $\begingroup$ yep $\endgroup$
    – MannyC
    Commented Apr 25, 2019 at 0:19
  • $\begingroup$ If you have two independent random variables $X$ and $Y$, the probability distribution of the combined system is $P_{X,Y}(x, y) = P_X(x) P_Y(y)$, and the average total value is $\langle X + Y \rangle = \langle X \rangle + \langle Y \rangle$. So there's really nothing interesting or new going on in the quantum case. $\endgroup$
    – DanielSank
    Commented Apr 25, 2019 at 2:30

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