My attempt :

Let $$V(x) = \begin{cases} V_0 \left(\frac{x^2}{a^2}-1 \right), & \text{for } |x|≤a \\ 0, & \text{for } |x|>a \end{cases}$$

Suppose $\psi$ is an odd eigenfunction with eigenvalue $E \in (-V_0, \ 0)$.

Then $$\psi(x)= \begin{cases} \left(\sum_{k=0}^{\infty}c_k d^k x^k\right) e^{-\frac{d^2x^2}{2}}, & \text{for } |x|≤a \\ \psi(x) = C e^{-\sqrt{\frac{-2mE}{\hbar^2}}x}, & \text{for }x>a \end{cases}$$

where, $$\frac{a_{k+2}}{a_k}=\frac{2k+1-\epsilon}{(k+2)(k+1)} $$

$$d= \left(\frac{2mV_0}{\hbar^2a^2}\right)^{\frac{1}{4}}, \ \epsilon = \sqrt{\frac{2m}{V_0}} \frac{a(V_0+E)}{\hbar}.$$

Now I want to show that $\exists E \in (-V_0, \ 0)$ such that $\psi(a-) =\psi(a+)$ and $\psi'(a-) =\psi'(a+)$.

Above is equal to showing that $\exists E \in (-V_0, \ 0)$ such that $\frac{\sum c_k d^k a^k}{\sum kc_k d^k a^{k-1}-c_k d^{k+2}a^{k+1}}=-\sqrt{\frac{\hbar^2}{-2mE}}$.

How should I proceed? Any hints or reference are appreciated. Thank you.


1 Answer 1


Maybe I'm somewhere wrong, I write it at late night. Anyway, here my thoughts.

Schrödinger equation is a second-order differential equation, so you need to define 2 arbitrary constants, which are $c_0$ and $c_1$.
We know that eigenstates must be odd or even, because $[\hat P,\hat H]=0$ (where $\hat P$ is the parity operator). So we can consider only two possibilities: $$\begin{pmatrix} c_0\\ c_1 \end{pmatrix} = \begin{pmatrix} 0\\ C_m\\ \end{pmatrix} \quad\text{or}\quad \begin{pmatrix} c_0\\ c_1 \end{pmatrix} = \begin{pmatrix} C_m\\ 0 \end{pmatrix} \tag{1}$$ where $C_m$ is normalizing constant, which is different for different eigenstates $\psi_m$.

The first choice gives an odd solution. Indeed, the recursive relation $c_{k+2}(c_k)$ sets $c_{2n}=0\ \forall n \in \mathbb{N}_0$, and the series $\sum_{k=0}^{\infty}c_k(xd)^k=\sum_{n=0}^{\infty}c_{2n+1}(xd)^{2n+1}$, which is an odd function. The function $\exp[-\frac{(xd)^2}{2}]$ is even, so $\psi=\small\text{odd}\normalsize\cdot\small\text{even}$ is odd. Similarly, the second choice in (1) gives an even solution.

For further analysis I used Node Theorem:

If $\psi_0, \psi_1, \psi_2, ..., \psi_n, ...$ are bound states in 1D-potential (and also there are no degeneracy and continuity of the energy spectrum), then $\psi_n$ has $n$ nodes

(See explanation, for example, here)

Let's find out whether the first odd eigenfunction exists (if there is no the first odd function, then there are no any odd functions). So, the first odd eigenfunction $\psi_1$ has to have a one node. It's possible only if all $c_k=0$ except $c_1$, because the number of zeros is determined by the series $\sum_{k=0}^{\infty}c_k(xd)^k$ (the number of zeros equals to the polynomial degree). Then $\psi_1=C_1 \cdot xd\cdot\exp[-\frac{(xd)^2}{2}]$ for $|x|\leqslant a$.

The equation $$\frac{\sum_{k=0}^{\infty}c_k(ad)^k}{\sum_{k=0}^{\infty}c_{k+1}(k+1)a^kd^{k+1}-c_ka^{k+1}d^{k+2}}=-\sqrt{-\frac{\hbar^2}{2mE}}$$ gives $$\frac{a}{1-(ad)^2}=-\sqrt{-\frac{\hbar^2}{2mE_1}}$$ A solution exists only if $(ad)^2>1$, otherwise $\sqrt{-\frac{\hbar^2}{2mE_1}}<0$, which isn't possible. In other words: $$V_0a^2>\frac{\hbar^2}{2m}\tag{2}$$ But we also need to make sure that $E_1<0$, which is the condition on the first odd eigenfunction existing: $$E_1=-\frac{\hbar^2}{2ma^2}((ad)^2-1)^2=-V_0\left(1-\frac{\hbar}{a\sqrt{2mV_0}}\right)^2$$ so, the condition is met. Therefore, an odd bound state exists, if (2) is right. Note that (2) guarantees $E_1>-V_0\quad\Leftrightarrow\quad V_0a^2>\frac{\hbar^2}{8m}$


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