Author Topic: Halloween Challenge 4  (Read 160 times)

Victor Ivrii

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XinYu Zheng

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Re: Halloween Challenge 4
« Reply #1 on: October 23, 2016, 06:22:30 PM »
Using the hint that all eigenvalues are between $-H$ and $0$, we know we are seeking for bound states of the particle. Let $\lambda=-k^2$. Inside the well, we have $u_{xx}+(\lambda+H)u=0$. Note that since $\lambda+H>0$, the solutions are sinusoidal. This makes sense. From quantum mechanics, we know that whenever the energy of the particle is less than the potential, its wavefunction is sinusoidal. Let $\alpha^2=\lambda+H$. The solution inside the well is then $u(x)=A\sin(\alpha x)+B\cos(\alpha x)$.
Now consider outside the well. We have $V=0$ and thus the equation is $u_{xx}+\lambda u=u_{xx}-k^2 u=0$. Now the solution is exponential: $u(x)=Ce^{kx}+De^{-kx}$.
We are looking for physical solutions. We demand that the wavefunction of the particle must vanish at $\pm \infty$. Thus for $x>L$ we take $C=0$ and for $x<-L$ we take $D=0$.
First we consider even wavefunctions. Inside the well, the even wavefunctions are $u(x)=A\cos(\alpha x)$. The wavefunction outside must be $u(x)=De^{-kx}$ for $x>L$ and $u(x)=Ce^{kx}$ for $x<-L$. However, since we are looking for even wavefunctions, we must have $D=C$. Now we apply boundary conditions.

Continuity of the wavefunction at $x=\pm L$ yields $A\cos(\alpha L)=C e^{-kL}$.
Continuity of the first derivative at $x=\pm L$ yields $-\alpha A\sin(\alpha L)=-kCe^{-kL}$.
Now taking the ratio of these two equations, we obtain the relation
$$\alpha \tan(\alpha L)=k$$
This is a transcendental equation that cannot be solved analytically, but graphically we can solve this for $\lambda$.

Now consider the odd wavefunctions. Inside the well we have $u(x)=B\sin(\alpha x)$ and outside we have $u(x)=De^{-kx}$ for $x>L$ and $u(x)=Ce^{kx}$ for $x<-L$. But since the wavefunction is odd, we have $D=-C$ this time.

Continuity at $x=\pm L$ yields $B\sin(\alpha L)=De^{-kL}$.
Continuity of the first derivative at $x=\pm L$ yields $B\alpha\cos(\alpha L)=-kDe^{-kL}$.
Taking the ratio of these two equations, we obtain
$$\alpha \cot(\alpha L)=-k$$

Once again, this equation can be solved graphically for $\lambda$.
« Last Edit: October 24, 2016, 01:57:23 PM by XinYu Zheng »