schroedinger equation - Classical limit of a quantum system

If we have a one dimensional system where the potential

$$V~=~\begin{cases}\infty & |x|\geq d, \\ a\delta(x) &|x|

where $a,d >0$ are positive constants, what then is the corresponding classical case -- the approximate classical case when the quantum number is large/energy is high?

Answer

Here we derive the bound state spectrum from scratch. Not surprisingly, the conclusion is that the Dirac delta potential doesn't matter in the semi-classical continuum limit, in accordance with Spot's answer.

The time-independent Schrödinger equation reads for positive $E>0$,

$$ -\frac{\hbar^2}{2m}\psi^{\prime\prime}(x) ~=~ (E-V(x))\psi(x), \qquad V(x)~:=~V_0\delta(x)+\infty \theta(|x|-d), \qquad V_0~>~0, $$

with the convention that $0\cdot \infty=0$. Define

$$v(x) ~:=~ \frac{2mV(x)}{\hbar^2}, \qquad e~:=~\frac{2mE}{\hbar^2}~>~0 \qquad k~:=~\sqrt{e}~>~0\qquad v_0 ~:=~ \frac{2mV_0}{\hbar^2}. $$

Then

$$ \psi^{\prime\prime}(x) ~=~ (v(x)-e)\psi(x). $$

We know that the wave function $\psi$ is continuous with boundary conditions

$$\psi(x)~=0 \qquad {\rm for}\qquad |x|\geq d.$$

Also the derivative $\psi^{\prime}$ is continuous for $0<|x|

$${\lim}_{\epsilon\to 0^+}[\psi^{\prime}(x)]^{x=\epsilon}_{x=-\epsilon} ~=~v_0\psi(x=0). $$

We get $$\psi_{\pm}(x)~=~A_{\pm}\sin(k(x\mp d))\qquad {\rm for } \qquad 0 \leq \pm x \leq d.$$

1. 
   

   $\underline{\text{Case} ~\psi(x=0)=0}$. Then $$n~:=~\frac{kd}{\pi}~\in~ \mathbb{N}.$$ We get an odd wave function $$\psi_n(x)~\propto~\sin(kx).$$ In particularly, the odd wave functions do not feel the presence of the Dirac delta potential.

   
   


2. 
   

   $\underline{\text{Case} ~\psi(x=0)\neq 0}$. Then continuity at $x=0$ implies that the wave function is even $A_{+}+A_{-}=0$. Phrased equivalently, $$\psi(x)~=~A\sin(k(|x|-d)).$$ The kink condition at $x=0$ becomes $$ v_0A\sin(-kd)~=~2kA \cos(kd), $$ or equivalently, $$ v_0\tan(kd)~=~-2k.$$ In the semiclassical continuum limit $$k \gg \frac{1}{d}, \qquad k \gg v_0,$$ this becomes $$\frac{kd}{\pi}+\frac{1}{2}~\in ~\mathbb{Z}, $$ i.e., in the semiclassical continuum limit the even wave functions do not feel the presence of the Dirac delta potential as well.