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I am considering a particle within a potential given by $$V(x)=mg|x|$$ and am attempting to find the energy eigenvalues of the system.

Taking $V(x)$ to be defined piecewise, I've solved the Schrödinger equation in a piecewise manner, finding the solutions, when $x>0$: $$ \psi(y_1)=C_1\mathrm{Ai}(y_1)+C_2\mathrm{Bi}(y_1), $$ where $\mathrm{Ai}$ and $\mathrm{Bi}$ are Airy functions and $y_1=\alpha^{1/3}(x-\frac{E}{mg})$, where I've defined $\alpha$ as being such that $\alpha=\frac{2m^2g}{\hbar^2}$, and when $x<0$: $$ \psi(y_2)=C_3\mathrm{Ai}(y_2)+C_4\mathrm{Bi}(y_2), $$ where in this instance $y_2=\alpha^{1/3}(-x-\frac{E}{mg})$, and the other parameters are defined in the same way as previously.

Based on the form of the Airy function of the second kind, $\mathrm{Bi}$, both $C_2$ and $C_4$ should be zero, I think, and the two functions should equal one another when evaluated at $x=0$, or, in terms of $y$, when $y=-\alpha^{1/3}\frac{E}{mg}$.

The other condition which must be met is that their derivatives should equal one another at this point as well. It is beyond this point that I do not know how to proceed.

In the case of an asymmetric triangular potential, where when $x>0$, $V(x)=mgx$, and is $\infty$ otherwise, because the wavefunction must go to zero at $x=0$, one can simply solve for the roots of the Airy function and find the energy eigenvalues.

This isn't the case here, because the condition is that the two wavefunctions must equal one another at $x=0$. So, how can the energy eigenvalues of the symmetric triangular potential herein be found?

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You're over-complicating things.

The potential is even (i.e. $V(-x) = V(x)$), which means that the eigenfunctions need to be either even or odd functions, and that means in turn that you can replace your problem with just the $x>0$ half, considering both possibilities of

  • $\psi(0)=0$ (for odd functions), and
  • $\psi'(0)=0$ (for even functions).

This reduces the problem to just searching for the solutions of $$\mathrm{Ai}(x=0)=0,$$ together with those of $$\mathrm{Ai}'(x=0)=0.$$

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Apart from wave function continuity, there is a condition of its derivative continuity at $x=0$. It follows from the Schrodinger equation integration from $x-\epsilon$ to $x+\epsilon$. This will give you the necessary equation for the eigenvalues: $\psi'(y(x=0))=0$ unless $\psi'(y(x=0))=0$ automatically - due to symmetry.

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  • $\begingroup$ Thank-you for your response. With your condition being the case, does it simply imply that $y$, as I've defined it, must equal the roots of $Ai'$, such that: $E_n=-a_n' \frac{mg}{\alpha^{1/3}}$? $\endgroup$ – T. Zaborniak Nov 18 '18 at 17:51

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