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If we have a one dimensional system where the potential

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

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?

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What is $V$ when $x \in (-d,0) \cup (0,d)$? –  C.R. Apr 27 '12 at 9:09
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Did you mean "$\infty$ when $|x| > d$"? Also did you mean "$a$ when $x = 0$" i.e. $a\delta(x)$. Finally is $a$ of the order of classical energies or much less? If the latter, the system just looks like a square well with no barrier at classical energies. –  John Rennie Apr 27 '12 at 9:41
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Dear @Sys, it's a virtue and necessity, not a bug, that the delta-function is infinite at $x=0$. If it were finite at a single point (i.e. interval of length zero), like in your example, it would have no impact on the particle because zero times finite is zero. So your potential as you wrote it is physically identical to $V=\infty$ for $|x|<d$ and $0$ otherwise which is just a well with the standing wave energy eigenstates. The finite modification of $V$ at one point, by $a$, plays no role at all. A potential with $a\delta(x)$ in it would be another problem. –  Luboš Motl Apr 27 '12 at 10:44
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@LubošMotl: Thanks, actually the delta function version instead of V=a at x=0 is the right one. What is the classical limit of that? –  Sys Apr 27 '12 at 11:15
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@JohnRennie: I think your comment suggestion was right, that there is a delta function at x=0. –  Sys Apr 27 '12 at 11:17
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2 Answers 2

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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 $0<|x|<d$, and possibly has a kink at $x=0$,

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

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.

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Firstly, it's easy to start off with just the Dirac delta potential and see what that does. Wiki has a nice solution for the Delta fuction potential, and I am lifting off parts of it here.

Consider a potential $V(x) = a\delta (x)$ and consider a scattering like configuration, where a plane wave $e^{ikx}$ is incident from the left.

$$ \psi(x)=\begin{cases}e^{ikx}+re^{-ikx} & x<0 \\ te^{ikx} & x> 0\end{cases} $$

By matching the boundary conditions, like on the wiki page, you get $$ t = 1+r\\ (1-\alpha)t = 1-r $$

where $$ \alpha = \frac{ 2ma}{ik\hbar^2} $$ characterizes the effect of the delta potential. Solving for $r$ and $t$, $$ t = \frac{1}{1-\alpha/2}\\ r=-\frac{\alpha/2}{1-\alpha/2} $$

Now, it is easy to see that for high incident $k$, the only effect of the dirac delta potential is to write a phase discontinuity on the wavefuction. This is because, as $k$ increases, the transmission $|t|^2=1/(1+|\alpha|^2/4)$ approaches 1, but the transmitted wavefunction gets an extra phase given by $$ \text{Arg}(t) = -\tan^{-1}(|\alpha|/2) $$

Getting back to the problem at hand, for a particle in a box (without the delta function), the allowed $k$ vectors are given by forcing the wavefunction to be zero at the walls at $x=-d$ and $x=d$, which gives us the condition

$$ k_n=\frac{\pi n}{2d} $$

If now, we add a delta potential, then for high values of $n$ (or $k$), all the delta function will do is introduce a phase discontinuity at the origin, and consequently what you should expect is that the boundary condition is matched not for $k_n$, but something slightly off $k_n+\delta k_n$, where $\delta k_n$ is a small correction due to the delta function potential. For high values of $n$, this correction would drop, as the phase discontinuity decreases, and for classical like states (very large $n$) you expect to recover 1D box states, as mentioned by John Rennie.

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Thank you, Spot! –  Sys Apr 27 '12 at 17:59
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