Question

(a) Using Equations 5.59 and 5.63, show that the wave function for a particle in the periodic delta-function potential can be written in the form

$\mathbf{\psi }\left(X\right)\mathbf{=}\mathbf{C}\left[\sin \left(\mathrm{kx}\right)+e^{-\mathrm{ika}}\sin \left(a-x\right)\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{\le }\mathbf{x}\mathbf{\le }\mathbf{a}$

(b) There is an exception; At the top of a band where z is an integer multiple of$\mathbf{\pi }\mathbf{}\mathbf{yiels}\mathbf{}\mathbf{\psi }\mathbf{}\left(x\right)\mathbf{=}\mathbf{0}$ yields .

Find the correct wave function for the case. Note what happens to$\mathbf{\psi }$each delta function.

Short answer

(a) Using the equation given in textbook, we derived the wave function for periodic delta potential

(b)The correct wave function for the wave is $\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{Asin}\left(\mathrm{kx}\right)$

Step-by-step solution

Define Schrödinger equation

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.
• The time-dependent Schrödinger equation is represented as

$\mathrm{iħ}\frac{\mathrm{d}}{\mathrm{dt}}\left|\mathrm{\psi }\left(\mathrm{t}\right)\right|>=\hat{\mathrm{H}}\left|\mathrm{\psi }\left(\mathrm{t}\right)\right|$

Showing the wave function

(a)

To show that wave function for a particle in the periodic delta potential is:

$\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{C}\left[\sin \left(\mathrm{kx}\right)+{\mathrm{e}}^{-\mathrm{ika}}\sin \mathrm{k}\left(\mathrm{a}-\mathrm{x}\right)\right]\mathrm{for}0\le \mathrm{x}\le \mathrm{a}$

We start from equations 5.59 and 5.63:

_{$$\begin{gathered} \mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{A}\sin \left(\mathrm{kx}\right)+\mathrm{B}\cos \left(\mathrm{kx}\right) \\ \mathrm{A}\sin \left(\mathrm{ka}\right)=\left[{\mathrm{e}}^{\mathrm{ika}}-\cos \left(\mathrm{ka}\right)\right]\mathrm{B} \\ \mathrm{B}=\frac{\mathrm{A}\sin \left(\mathrm{ka}\right)}{{\mathrm{e}}^{\mathrm{ika}}-\cos \left(\mathrm{ka}\right)} \end{gathered}$$}

We insert previous expression for B in wave function $\mathrm{\psi }\left(\mathrm{x}\right)$:

$$\begin{gathered} \psi \left(x\right)=A\sin \left(kx\right)+\frac{A\sin \left(ka\right)}{e^{ika}-\cos \left(ka\right)}\cos \left(kx\right) \\ =A\frac{e^{ika}\sin \left(kx\right)-\sin \left(kx\right)\cos \left(ka\right)+\sin \left(ka\right)\cos \left(ka\right)}{e^{ika}-\cos \left(ka\right)} \\ =\frac{A{e}^{ika}}{e^{ika}-\cos \left(ka\right)}\left[\sin \left(kx\right)-e^{ika}-\sin \left(kx\right)\cos \left(ka\right)+e^{ika}-\sin \left(ka\right)\cos \left(ka\right)\right] \\ \psi \left(x\right)=C\left[\sin \left(kx+\right)e^{ika}-\sin \left[k\left(a-x\right)\right]\right] \end{gathered}$$

Therefore using the equation given in textbook, we derived the wave function for periodic delta potential.

Observing from graph

(b)

$f\left(z\right)=\cos \left(z\right)+\mathrm{\beta }\frac{\sin \left(\mathrm{z}\right)}{\mathrm{z}}$

For$\mathrm{\beta }=10$ .From equation 5.64 we have:

$\cos \left(ka\right)=\cos \left(kx\right)+\frac{ma}{h^{2}k}\sin \left(ka\right)$

In this case is an integer multiple of$\pi$, then we have$z=ka=n\pi$, where is $n$an integer. This implies:

$$\begin{gathered} \sin \left(\mathrm{ka}\right)=\sin \left(\mathrm{Ka}\right)=0 \\ \cos \left(\mathrm{ka}\right)=\cos \left(\mathrm{Ka}\right) \\ ={\left(-1\right)}^{\mathrm{n}} \\ =\cos \left(\mathrm{Ka}\right)+\mathrm{isin}\left(\mathrm{Ka}\right) \\ ={\mathrm{e}}^{\mathrm{iKa}} \\ ={\left(-1\right)}^{\mathrm{n}} \end{gathered}$$

Constant C from previous task is then $C=\frac{A}{0}$which implies that A=0 or B=0 .So the equation 5.62 is:

$$\begin{gathered} \frac{2\mathrm{m\alpha }}{{\mathrm{h}}^2}\mathrm{B}=\mathrm{kA}-{\mathrm{e}}^{-\mathrm{iKa}}\mathrm{k}\left[\mathrm{A}\cos \left(\mathrm{ka}\right)-\mathrm{B}\sin \left(\mathrm{ka}\right)\right] \\ =\mathrm{kA}-{\left(-1\right)}^{\mathrm{n}}\mathrm{k}\left(\mathrm{A}{\left(1\right)}^2-\mathrm{B}.0\right) \\ =\mathrm{kA}-\mathrm{kA} \\ \Rightarrow \mathrm{B}=0 \end{gathered}$$

So, from equation 5.59, we are left with only $A\sin \left(kx\right)$term:

$\psi \left(x\right)=A\sin \left(kx\right)$

At each delta function$\psi \left(x\right)=0$, so the wave function doesn't "see" any potential.

Hence the correct wave function for the wave is $\mathrm{\psi }\left(\mathrm{x}\right)=\mathrm{Asin}\left(\mathrm{kx}\right)$