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What are the continuity constraints on the wavefunction of a system when the potential of the system has a discontinuous derivative?

As a concrete example, consider the following potential, which is symmetric about the origin at $x=0$:

$$V(x) = \begin{cases} x^{2}+(x+\frac{d}{2}) &\text{for}\ x < -d/2\\ x^{2} &\text{for}\ -d/2 < x < d/2\\ x^{2}-(x-\frac{d}{2}) &\text{for}\ x > d/2 \end{cases}$$

so that

$$V(x) = \begin{cases} (x+\frac{1}{2})^{2}+\frac{2d-1}{4} &\text{for}\ x < -d/2\\ x^{2} &\text{for}\ -d/2 < x < d/2\\ (x-\frac{1}{2})^{2}+\frac{2d-1}{4} &\text{for}\ x > d/2 \end{cases}$$

and therefore

$$V(x) = \begin{cases} (x+\frac{1}{2})^{2} &\text{for}\ x < -d/2\\ x^{2} &\text{for}\ -d/2 < x < d/2\\ (x-\frac{1}{2})^{2} &\text{for}\ x > d/2 \end{cases}$$

So, the wavefunctions in the left and right sectors are the shifted-in-position harmonic oscillator wavefunctions:

$$\psi_{n}(x) \sim \begin{cases} \text{exp}\Big(-\frac{m\omega}{2\hbar}(x-\frac{1}{2})^{2}\Big)\ H_{n}\Big(\sqrt{\frac{m\omega}{\hbar}}(x-\frac{1}{2})\Big) &\text{for}\ x < -d/2\\ \text{exp}\Big(-\frac{m\omega}{2\hbar}x^{2}\Big)\ H_{n}\Big(\sqrt{\frac{m\omega}{\hbar}}x\Big) &\text{for}\ -d/2 < x < d/2\\ \text{exp}\Big(-\frac{m\omega}{2\hbar}(x+\frac{1}{2})^{2}\Big)\ H_{n}\Big(\sqrt{\frac{m\omega}{\hbar}}(x+\frac{1}{2})\Big) &\text{for}\ x > d/2 \end{cases}$$

But, the wavefunctions at the potential kinks at $x=-d/2$ and $x=d/2$ do not match.

Is there some sorcery of the kink in the potential - perhaps the discontinuity at $V'(-d/2)$ and at $V'(d/2)$ - that is causing this behaviour of the wavefunction?

Qmechanic
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nightmarish
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  • Related: http://physics.stackexchange.com/q/19667/2451 and links therein. – Qmechanic Sep 24 '16 at 22:19
  • Your wave functions in the left and right sectors are NOT the position-shifted harmonic oscillator wave functions, neither is the wave function in the central sector an eigenfunction of a harmonic oscillator. You have disregarded that in no sector a parabolic potential of unlimited extent can be assumed. You can solve the problem only by finding the eigenfunction of the whole system by assuming the appropriate boundary conditions at x=−d/2 and x=d/2. See my answer below. – freecharly Sep 28 '16 at 01:50

1 Answers1

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It seems that you have not incorporated the boundary conditions for the wave functions at ±d/2 in finding the wave function for the whole system. In each of the three regions, you just used the shifted-in-position wave functions for the parabolic potentials assuming they have infinite range.The boundary conditions follow from the continuity conditions of the probability currents at ±d/2. This means that the boundary conditions are that the wave function and its derivative are continuous at ±d/2. So you have to find the general solutions in each of the 3 regions with appropriate indefinite constants then from the boundary conditions you get equations for their determination. From this you will get the wave functions for the whole system and the corresponding eigenvalues for the energies, which will, in general, not be the same as the eigenvalues of the separate harmonic oscillators.

freecharly
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