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How to prove that under the potential like$V(x) = -\frac{\nu}{2}x^2 + \frac{\beta}{4}x^4$, in which $\nu, \beta >0$, the system has no degeneracy? That is to say the Schrödinger equation $-\partial^2\Psi/\partial x^2 +V(x) \Psi = E\Psi$ has no degenerate level.

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It is a general fact that in a one dimensional system, bound states cannot be degenerate. To see this, suppose the opposite statement is true. So corresponding to a certain energy eigenvalue $E$, there exist at least two linearly independent states satisfying \begin{align} -\dfrac{\hbar^2}{2m}\dfrac{d^2\psi_1}{dx^2}=(E-V)\psi_1\tag{1}\\ -\dfrac{\hbar^2}{2m}\dfrac{d^2\psi_2}{dx^2}=(E-V)\psi_2\tag{2} \end{align} And $(1)\times \psi_2-(2)\times \psi_1$ yields \begin{equation} \dfrac{d}{dx}(\psi_2\psi_1^\prime-\psi_1\psi_2^\prime)=0\tag{3} \end{equation} because they're bound states, $\psi_1,\psi_2$ and their derivatives vanish at infinity, so $(3)$ says that $$ \psi_2\psi_1^\prime-\psi_1\psi_2^\prime=0 $$ or $$ \mathrm{ln}\psi_1=\mathrm{ln}\psi_2+\mathrm{const}\Longrightarrow \psi_1=\mathrm{const}\cdot \psi_2 $$ which contradicts with our assumption.

Since your potential only allows bound states, the energy levels must be non-degenerate.

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