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I want to ask how to deal with the equation on $L/2 < x < L$ region.

The equation has a general form $C e^{-k_2 L/2} + D e^{k_2 L/2}$ on $L/2 < x < L$.

However, as there exists infinite barrier at $x = L$, we do have non-zero $D$.

Then the boundary condition at $x = L/2$ yields

$B e^{i k_1 L / 2} = C e^{-k_2 L/2} + D e^{k_2 L/2}$ ... (1)

and

$i k_1 B e^{i k_1 L / 2} = k_2( - C e^{-k_2 L/2} + D e^{k_2 L/2})$ ... (2)

However, dividing (1) with (2) would not cancel the RHS terms that we usually see in quantum physics textbooks.

Is there anyway to solve the Schrodinger equation in this setting?

gabriel
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    Your proposed wave functions are incorrect. The variable $x$ is missing. Write down the correct wave functions, then apply the boundary conditions. – M. Wind May 30 '23 at 14:37
  • One strategy I like for such problems is to choose the basis functions as conveniently as possible. For instance: on the left interval we know that the wavefunction vanishes at the origin, so the basis ${\sin k_1 x,\cos k_1x}$ is convenient. For the right-hand interval, we can instead use the shifted basis ${\sin[k_2(x-L)],\cos[k_2(x-L)]}$ to enforce $\psi=0$ without much fuss. – Semiclassical May 30 '23 at 15:15
  • Just realized that my last comment assumed $E>V_0$, so not applicable as written. – Semiclassical May 30 '23 at 15:22
  • This is a stock problem in 10% of QM textbooks. Why are you here? – Cosmas Zachos Jun 08 '23 at 19:57

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