Question

Suppose we use delta function wells, instead of spikes (i.e., switch the sign ofin Equation 5.57). Analyze this case, constructing the analog to Figure 5.6. this requires no new calculation, for the positive energy solutions (except that $\mathit{\beta }$ is now negative; use $\mathit{\beta }\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{5}$ for the graph), but you do need to work out the negative energy solutions (let$\mathbf{k}\mathbf{\equiv }\sqrt{\mathbf{-}\mathbf{2}\mathbf{mE}}\mathbf{/}\mathbf{}\mathbf{handZ}\mathbf{\equiv }\mathbf{-}\mathbf{ka}\mathbf{,}\mathbf{forE}\mathbf{<}\mathbf{0}\mathbf{)}$ and , for). How many states are there in the first allowed band?

Short answer

The states in first allowed band are

If $$\begin{gathered} E>0\mathrm{then}\cos \left(\mathrm{Ka}\right)=\cos \left(\mathrm{ka}\right)-\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2\mathrm{k}}\sin \left(\mathrm{ka}\right) \\ \mathrm{If}E<0\mathrm{then}\cos \left(\mathrm{Ka}\right)=\cosh \left(\mathrm{ka}\right)-\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2\mathrm{k}}\sinh \left(\mathrm{ka}\right) \end{gathered}$$

Step-by-step solution

Define the Schrodinger 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 Schrodinger equation is represented as

Finding the states in the band

Consider two cases, when particle has positive energy (E > 0) and when it has negative energy$(E>0)$

i) $E>0$: Here $\alpha$is negative:

$$\begin{gathered} \cos \left(Ka\right)=\cos \left(ka\right)-\frac{m\alpha }{h^{2}k}sin\left(ka\right) \\ k=\sqrt{\frac{2mE}{h^2}},K=\frac{2\pi n}{Na} \end{gathered}$$

(ii) $E>0$First write Schrodinger equation for region $0<x<a-\frac{{ħ}^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}=-\left|E\right|$

$\frac{{\partial }^2\psi }{\partial x^2}=\frac{2m\left|E\right|}{{ħ}^2}\psi =k\psi$

Since k is positive, solution is

_{${\psi }_l\left(\mathrm{x}\right)=\mathrm{A}\cosh \left(\mathrm{kx}\right)+\mathrm{B}\sin \mathrm{h}\left(\mathrm{kx}\right)$}

Using Bloch’s theorem, wave function in the “cell” to the left of the origin (in the region ₎,$-a<x<0),\mathrm{is}{\mathrm{\psi }}_{ll}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{ika}}\left[\mathrm{A}\cosh \left(\mathrm{k}\left(\mathrm{x}+\mathrm{a}\right)\right)+\mathrm{B}\sin \mathrm{h}\left(\mathrm{k}\left(\mathrm{x}+\mathrm{a}\right)\right)\right]$

Calculating the first derivation

Now we impose boundary conditions, at x=0 wave function must be continuous but its deviation isn’t because of delta potential.$A=e^{-ika}\left(A\cos h\left(ka\right)+B\sin h\left(ka\right)\right)$

To find the derivation of wave function in point x=0 , we write Schrodinger equation:

$-\frac{h^2}{2m}\frac{{\partial }^2\psi }{\partial x^2}-\alpha \delta \left(x\right)\psi =-E\psi$

To find first derivation we need to integrate previous equation from$-\epsilon$ to $+\epsilon$. After taking the limit $\epsilon \to 0$, RHS of equation vanishes, and we are left with:

$$\begin{gathered} \frac{{\mathrm{h}}^2}{2m}{\left[\frac{\partial \psi }{\partial x}\right]}_{x\to 0^{+}}-{\left[\frac{\partial \psi }{\partial x}\right]}_{x\to 0^{-}}=-\alpha \psi \left(0\right) \\ {\left[\frac{\partial {\psi }_l}{\partial x}\right]}_{x=0}-{\left[\frac{\partial {\psi }_{ll}}{\partial x}\right]}_{x=0}=-\frac{2m\alpha }{{\mathrm{h}}^2}\psi \left(0\right) \\ Bk-e^{-iKa}(A\sin h\left(ka\right)+B\cos h\left(ka\right))=-\frac{2m\alpha }{h^2}A \end{gathered}$$

Calculating the value of B

$B=\frac{A\left(e^{iKa}-\cos h\left(ħa\right)\right)}{\sin h\left(ka\right)}$

Result in expression about it

$$\begin{gathered} Ak\left(e^{ika}-\cos h\left(ka\right)\right)-k{e}^{-ika}\left(A\sin h^2\left(ka\right)+A{e}^{ika}\cos h\left(ka\right)-A\cos h^2\left(ka\right)\right)=-\frac{2m\alpha A}{h^2}\sin h\left(ka\right) \\ k{e}^{ika}-k\cos h\left(ka\right)-k{e}^{-ika}\left(e^{ika}\cos h\left(ka\right)-1\right)=-\frac{2m\alpha A}{h^2}\sin h\left(ka\right) \\ k{e}^{ika}-2k\cos h\left(ka\right)-\frac{2m\alpha A}{h^2}\sin h\left(ka\right)\frac{2m\alpha }{{ħ}^2}\sin h\left(ka\right) \\ \cos \left(Ka\right)=\cos h\left(ka\right)-\frac{m\alpha }{h^{2}k}\sin h\left(ka\right) \end{gathered}$$

In order to graph this function, we introduce substitute:

$$\begin{gathered} z=-ka,\beta =-\frac{m\alpha a}{ħ} \\ \cos \left(ka\right)=f\left(z\right)=\cos h\left(z\right)+\beta \frac{\sin h\left(z\right)}{\left(z\right)} \end{gathered}$$

Because must be in region between and 1 that means:

$Ka=\frac{2\pi n}{Na},n=0,1,2,...,N-1$

Every band has N states.

Therefore the states in first allowed band are

role="math" localid="1658232648734" $$\begin{gathered} \mathrm{If}E>0\mathrm{then}\cos \left(\mathrm{Ka}\right)=\cos \left(\mathrm{ka}\right)-\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2\mathrm{k}}\sin \left(\mathrm{ka}\right) \\ \mathrm{If}\mathrm{E}<0\mathrm{then}\cos \left(\mathrm{Ka}\right)=\cosh \left(\mathrm{ka}\right)-\frac{\mathrm{m\alpha }}{{\mathrm{h}}^2\mathrm{k}}\sin \left(\mathrm{ka}\right) \end{gathered}$$