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It is known that in a box potential, when we set $V = 0$ inside and $V = \infty$ on the boundaries, the solution to the equation $$ - \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \bigg) \psi(x,y,z) = E \psi(x,y,z) $$ is given by $$ \psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}). $$ Here $L$ is the dimension of the box. The energies are $E_n = \frac{\hbar^2 \pi^2 }{2mL^2} n^2 $ with $n^2 = n_x^2 + n_y^2 + n_z^2$. Now, I was wondering, what if the potential inside the box is not zero, but we let it be $V = V_0$, a constant? What is the solution of the Schrödinger equation then? And what are the energy eigenvalues?

Does this amount to adding a phase factor to the solution (the product of sines)?

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  • $\begingroup$ Still the same. Only that $E_n = \frac{\hbar^2n^2\pi^2} {2mL^2} + V_0$ $\endgroup$ – philip_0008 Jun 10 '16 at 20:56
  • $\begingroup$ I see. But how did you deduce that? $\endgroup$ – Kamil Jun 10 '16 at 21:07
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The equation will become $$ - \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \bigg) \psi(x,y,z) = (E-V_0) \psi(x,y,z) $$ And the solutions are the same: $$ \psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}). $$ And Energy: $$(E_n - V_0) = \frac{\hbar^2\pi^2} {2mL^2}n^2$$ $$E_n = \frac{\hbar^2\pi^2} {2mL^2}n^2 + V_0$$
Path to the solution:

By separation of variables: $$\frac{1}{X}\frac{d^2X}{dx^2} + \frac{1}{Y}\frac{d^2Y}{dy^2} + \frac{1}{Z}\frac{d^2Z}{dz^2} = -\frac{2m}{\hbar^2}(E-V_0)$$ $$\frac{d^2X}{dx^2} = -k_x^2X; \frac{d^2Y}{dy^2} = -k_y^2Y; \frac{d^2Z}{dz^2} = -k_z^2Z$$ with $$(E_n - V_0) = \frac{\hbar^2} {2m}(k_x^2 + k_y^2 + k_z^2)$$ Solution:
$X(x) = A_x\sin k_xx + B_x\cos k_xx$ and so on.
as usual, $B = 0$ because $X(0) = 0$ because of infinite potential at boundaries.
also, $X(L) = 0$ (infinite potential) means $\sin k_xL = 0$ or
$k_x = n_x\pi/L$ and so with the others.
So still the same: $$ \psi_{n_x, n_y, n_z} (x,y,z) = A_xA_yA_z \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}). $$ $$ \psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}). $$ $$(E_n - V_0) = \frac{\hbar^2\pi^2} {2mL^2}(n_x^2+n_y^2 + n_z^2)$$

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  • $\begingroup$ This is because $V_0$ is a constant $\endgroup$ – philip_0008 Jun 10 '16 at 21:11
  • $\begingroup$ Very nice little derivation. +10. $\endgroup$ – Gert Jun 10 '16 at 23:05
  • $\begingroup$ I got this from solution manual (Griffiths) :) but without the $V_0$. $\endgroup$ – philip_0008 Jun 11 '16 at 8:35

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