# Finding the odd parity bound state wave function for a particle in one dimension

Let a particle of mass $m$ and energy $E$ be moving on the $x$-axis in a potential $V(x)$ given by

$$V(x)=\left\{\begin{matrix} -V_{0}, & -a<x<a\\ 0, & otherwise \end{matrix}\right.$$

where $a>0, V_{0}>0$ are constants and energies $E$ satisfying $0<-E<V_{0}$.

I'm then asked to show that the odd parity bound state wave functions have energies $E$ which satisfy

$$\eta^{2}+\xi^{2}=\frac{2mV_{0}a^{2}}{\hbar^{2}}, \eta=-\xi cot(\xi)$$

where $\eta=\sqrt{-2mEa^{2}/\hbar^{2}}$ and $\xi=\sqrt{2m(V_{0}+E)a^{2}/\hbar^{2}}$

So, I think I'm nearly there but I have a feeling I might be misunderstanding exactly how to get to the odd parity bound state. My understanding of "odd parity" is that $\psi(x)=-\psi(-x)$ as stated here Definite Parity of Solutions to a Schrödinger Equation with even Potential? and my understanding of "bound state" is just that the wave function is normalizable (i.e. $\int_{-\infty}^{\infty}|\psi(x)|^{2}dx<\infty$). My working so far gives:

For $-a<x<a$: $$\frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}-V_{0}\psi=E\psi \\ \Rightarrow \frac{d^{2}\psi}{dx^{2}}=\frac{-2m(E+V_{0})}{\hbar^{2}}\psi \\ \Rightarrow \psi=Acos(\lambda x)+Bsin(\lambda x), \lambda=\sqrt{2m(E+V_{0})}/\hbar$$

Otherwise: $$\frac{-\hbar^{2}}{2m}\frac{d^{2}\psi}{dx^{2}}=E\psi \\ \Rightarrow \frac{d^{2}\psi}{dx^{2}}=\frac{-2mE}{\hbar^{2}}\psi \\ \Rightarrow \psi=Cexp(\mu x)+Dexp(-\mu x), \mu=\sqrt{-2mE}/\hbar$$

I then used the fact that both the wave function and its derivative must be continuous to get: $$Acos(\lambda a)+Bsin(\lambda a)=Cexp(\mu a)+Dexp(-\mu a)\\ Acos(\lambda a)-Bsin(\lambda a)=Cexp(-\mu a)+Dexp(\mu a) \\ -A \lambda sin(\lambda a)+B \lambda cos(\lambda a)=C \mu exp(\mu a) - D \mu exp(-\mu a) \\ -A \lambda sin(\lambda a)+B \lambda cos(\lambda a)=C\mu exp(-\mu a) - D \mu exp(\mu a)$$

Right, now this is where I run into difficulty. Normally, I'd try and use this system of equations to get rid of a some of the constants. But that doesn't seem to work. I feel like I'm supposed to impose the fact that I want $\psi$ to be normalizable and have odd parity. I thought maybe I could just assume that it was and then work from there? But that seemed quite an arbitrary assumption to make. But then I thought it might not be so arbitrary as the solutions to the Schrodinger equation are elements of the Hilbert space and could we just be looking for the specific one that has odd parity?

• "bound state" are states with a negative energy (and of course normalizable too) – Antonio Ragagnin Apr 17 '14 at 12:41
• Ah, okay. But since we're given that the energy is negative, does that change the problem much? – user37154 Apr 17 '14 at 12:45
• nope (replying..) – Antonio Ragagnin Apr 17 '14 at 12:48
• Haha, cheers. Also, any idea then why a previous question had been "What does it mean to say that a stationary state wave function describes a bound state (or, equivalently, is normalizable)?" suggesting that the two mean the same? – user37154 Apr 17 '14 at 12:52

As you said, you want that your wave function is normalized.

So, it is not correct that the solution outside the wall is $$\psi=Cexp(\mu x)+Dexp(-\mu x).$$ In fact this function diverges for $x\rightarrow\pm\infty.$

In order to keep your wave function normalized, you must use one solution for $x<-a$ and another one for $x>a.$

For $x<-a$, you have

$$\psi_-=Cexp(\mu x).$$

While for $x>a$ you get

$$\psi_+=Dexp(-\mu x).$$

$$\begin{cases} x\in[-\infty,-a] & \psi(x)=Cexp(\mu x) \\ x\in[-a,a] & \psi(x)=Acos(\lambda x)+Bsin(\lambda x) \\ x\in[a,\infty] & \psi(x)=Dexp(-\mu x) \end{cases}$$ In this way you imposed the normalizability request and you can just ask continuity and derivability in $x=\pm a.$
• And then, since our function must have odd parity, can we say that $\psi=\begin{cases} x\in[-\infty,-a] & \psi(x)=Cexp(\mu x) \\ x\in[-a,a] & \psi(x)=Bsin(\lambda x) \\ x\in[a,\infty] & \psi(x)=-Cexp(-\mu x) \end{cases}$? – user37154 Apr 17 '14 at 13:33