(a) An electron moves in a one-dimensional box of length X. Apply the periodic boundary condition $\phi(x)=\phi(x+X)$ to find the electron eigenfunction and eigenvalues.
(b) Now apply a weak periodic potential $V(x)=V(x+L)$ to the system, where $X = N L$ and $N$ is a large positive integer. Using nondegenerate perturbation theory, find the first-order correction to the wave functions and the second-order correction to the eigenenergies.

Question

(a) An electron moves in a one-dimensional box of length X. Apply the periodic boundary condition [latex]\phi(x)=\phi(x+X)[/latex] to find the electron eigenfunction and eigenvalues.
(b) Now apply a weak periodic potential [latex]V(x)=V(x+L)[/latex] to the system, where [latex]X = N L[/latex] and [latex]N[/latex] is a large positive integer. Using nondegenerate perturbation theory, find the first-order correction to the wave functions and the second-order correction to the eigenenergies.

(c) When wave vector k is close to [latex]n\pi/L,[/latex] where n is an integer, the result in (b) is no longer valid. Use two-state degenerate perturbation theory to find the corrected energy values for [latex]k=n\pi((1+\Delta)/L){\mathrm {~and~}}k^{\prime}=n\pi((1-\Delta)/L),[/latex] where [latex]{{\Delta}}[/latex] is small compared with [latex]π/L.[/latex]

(d) Use the results of (b) and (c) to draw the electron dispersion relation[latex] , E(k).[/latex]
(e) If we choose the lowest-frequency Fourier component of the perturbative periodic potential in part (b), then [latex]V(x)=V_{1}^{ }\cos(\pi x/L).[/latex] Repeat (b), (c), and (d) using this potential.

[latex] {\mathrm{Hint:}}\,\,V(x)=V_{0}+\sum\limits_{n eq0} V_{ n}e^{i2\pi n x/L} \,{{\mathrm{~and~\, choose~}}V_{0}=0.}[/latex]

Accepted Answer

(a) Here, we are interested in an electron of mass m that in the unperturbed state is free to move in a one-dimensional box of length X with periodic boundary conditions. In this situation, the potential is zero and the solution to the time-independent Schrödinger equation gives eigenfunctions that are planewaves of the form [latex]\psi_{k}(x)=A e^{i k x}[/latex]. Applying the periodic boundary conditions to the eigenstates,[latex]\psi(x)=\psi(x+X),[/latex] gives [latex]e^{i\mathbf{k}\cdot X} =1 [/latex] and [latex]k_{n}=2n\pi/X[/latex] for integer n in such a way that [latex]n\,=\,1,2,3,\dots.[/latex] Hence, the energy eigenvalues are [latex] E=\hbar^{2}k_{n}^{2}/2m.[/latex]

(b) To find solutions to the time-independent Schrödinger equation in the presence of a periodic potential [latex]V(x) = V(x + L),[/latex] where [latex] X = NL[/latex] and N is a large integer, we will treat the potential as a perturbation. The potential can be decomposed into its Fourier components, so that

[latex]V(x)=V_{0}+\sum\limits_{n eq0}V_{n}e^{i2\pi n x/L}= V_{ 0}+\sum\limits_{n eq0}V_{n}e^{i2\pi n N x/X}[/latex]

Setting [latex]V_{0}= 0[/latex] and recalling that [latex](1/2\pi\,) extstyle\int e^{-i(k-k^{\prime})x}d x=\delta(k-k^{\prime}),[/latex] the first-order correction to the eigenenergies is

[latex]E_{n}^{(1)}=W_{n n}=\frac{2n^{2}\pi^{2}}{m X^{2}}+ \int \limits_{-\infty}^{\infty}A^{2}\sum\limits_{k e0}V_{k}e^{i2\pi k N x/X}d x=0[/latex]

The new eigenstates are to first order

[latex]\phi_{n}=\phi_{n}^{(0)}+\sum\limits_{m eq n}\frac{W_{m n}}{E_{m}^{(0)}-E_{n}^{(0)}}\phi_{m}^{ (0)}[/latex]

[latex] \phi_{ n}=A e^{i2\pi n x/X}+\sum\limits_{m eq n}\frac{m X^{2} A^{2}\;\int \limits_{-\infty}^{\infty} \sum\limits_{k eq 0}^{} V_{k} e^{i2\pi k N x/X}e^{i2\pi(n-m)x/X}d x}{2\pi^{2}(m^{2}-n^{2})}A e^{i2m\pi x/X}[/latex]

where nonzero contributions occur for [latex]n-m=-k N,[/latex] allowing us to write

[latex]\phi_{n}=A e^{i2\pi n x/X}+\sum\limits_{k e0}\frac{m N^{2}A^{3}V_{k}}{\pi k N(2n-k N)}e^{i2\pi(n+k N)x/X}[/latex]

The second-order correction to the eigenenergy is

[latex]E_{n}^{(2)}=\sum\limits_{m eq n}\frac{|W_{m n}|^{2}}{E_{m}^{(0)}-E_{n}^{(0)}}\phi_{m}^{(0)}=[/latex][latex]\sum\limits_{m eq n}\frac{m X^{2}A^{4}\left|\int\limits_{-\infty}^{\infty}\sum\limits_{k eq0}V_{k}e^{i2\pi k N x/X} e^{i2\pi(n-m)x/X}d x ight|^{2}}{2\pi^{2}(m^{2}-n^{2})}[/latex]

which is nonzero if [latex]n-m=-k N,\;\mathrm{so}[/latex]

[latex]E_{n}^{(2)}=\sum\limits_{k eq0}{\frac{m N^{2}A^{4} V_{k}^{2}}{\pi k N(2n-k N)}}[/latex]

(c) When [latex]k ightarrow n\pi/L,[/latex] the denominator in the previous equation approaches zero. Nondegenerate perturbation theory can no longer be used. To gain some insight into the situation, we use two-state degenerate perturbation theory to find the corrected energy values for [latex]k=n\pi((1+\Delta)/L)\,\mathrm{and} \,k^{\prime} =n\pi((1-\Delta)/L),[/latex] where [latex]{{\Delta}}[/latex] is small compared with [latex]{\mathfrak{\pi}}/L.[/latex] In this case, the perturbed Hamiltonian matrix is

[latex]\left[\begin{matrix} \frac{\hbar^{2}k^{2}}{2 m} & A^{2}V_{n} \ A^{2}V_{n} & \frac{\hbar^{2}k^{2}}{2 m} \end{matrix} ight]\left[\begin{matrix} a_{1} \ a_{2} \end{matrix} ight] =E\left [\begin{matrix} a_{1} \ a_{2} \end{matrix} ight][/latex]

The new eigenenergies (the eigenvalues of the matrix) are
[latex]E_{1,2}=\frac{\hbar^{2}(k^{2}+k^{\prime 2})\pm\sqrt{16m^{2}V_{n}^{2}A^{4}+\hbar^{4}(k^{2}+k^{\prime2})^{2}}}{4m}[/latex]
or
[latex]E_{1,2}=\frac{\hbar^{2}n^{2}\pi^{2}(1+\Delta^{2})\pm2L^{2}\sqrt{\frac{\hbar^{4}n^{4}\pi^{4}\Delta^{2}}{L^{4}}+A^{4}m^{2}V_{n}^{2}}}{2L^{2}m}.[/latex]
(d) The following figure sketches the dispersion relation we might anticipate for electrons perturbed by a periodic potential. The parabolic dispersion of a free electron is modified to include band gaps of value [latex]E_{\mathrm{g}}[/latex] at wave vectors [latex]\pm n\pi/L.[/latex]
(e) If we choose the lowest-frequency Fourier component of the perturbative periodic potential in part (b), then [latex]V(x)=V_{1}\cos(\pi x/L)=-2U_{G}\cos(G x),[/latex] where [latex]G=2\pi/L[/latex] is the reciprocal lattice vector. The following figure sketches the potential.
Each oscillation of V(x) is one “cell” of the crystal. If the electron is displaced by [latex]n L[/latex], where n is an integer, and L is the oscillation period, then the electron must find itself in an identical environment. A perfectly periodic crystal looks the same to a particle displaced by [latex]n L.[/latex]
The unperturbed wave function is [latex]\psi_{k}(x)=A_{0}e^{i k x}[/latex]. The wave functions are plane waves in the absence of the periodic potential perturbation. We impose a periodic boundary condition [latex]\psi_{k}(x+X)=\psi_{k}(x),[/latex] where [latex]X=N L[/latex] is the length of the crystal.
Hence, we can now normalize
[latex]\psi_{k}(x)=|k angle={\frac{1}{\sqrt{X}}}e^{i k x}[/latex]
where [latex]k=(n/N)G{\mathrm{~and~}}n=0,\pm1,\pm2,\ldots.[/latex]
[latex]H_{11}=H_{22}=E_{0}={\frac{\hbar^{2}k^{2}}{2m}}[/latex]
is the unperturbed energy.
To evaluate [latex]H_{12}[/latex] and [latex]H_{21}[/latex] we find the matrix element
[latex]H_{12}=\langle k_{1}|V|k_{2} angle[/latex]

[latex]H_{12}=\int\limits_{0}^{X}\psi_{k_{1}}^{*}(x)V\psi_{k_{2}}(x)d x=\int\limits_{0}^{X}{\frac{1}{\sqrt{X}}}e^{-i k_{1}x}\cdot-2U_{G}\cos(G x)\cdot{\frac{1}{\sqrt{X}}}e^{i k_{2}x}d x[/latex]

[latex]H_{12}=-\frac{U_{G}}{X}\int\limits_{0}^{X}e^{-i k_{1}x} \bigl(e^{G x}+e^{-G x}\bigr)e^{i k_{2}x}d x=-\frac{U_{G}}{X}\int \limits_{0}^{X}\bigl(e^{i(k_{2}-k_{1}+G)x}+e^{i(k_{2}-k_{1}-G)x} \bigr)d x[/latex]

We note that the definition of the delta function is [latex]\delta(x-x^{\prime})=(1/2\pi)\int_{-\infty}^{\infty}e^{i y(x-x^{\prime})}d y.[/latex]

Our periodic boundary condition requires that [latex]H_{12}=0,\; \mathrm{unless},\;\mathrm{e.g.},\;k_{2}-k_{1}=[/latex][latex]\pm G=\pm2\pi/L,[/latex] in which case the integral is a Dirac delta function. Hence, the matrix element connecting two arbitrary plane-wave states [latex]|k_{1} angle[/latex] and [latex]|k_{2} angle [/latex] is zero unless

[latex]H_{12}=H_{21}=-U_{G}\cdot\delta(k_{2}-k_{1}=\pm G)[/latex]

For the case in which k is near [latex]\pm G/2,[/latex] it is useful if we rewrite

[latex]k_{-}=k_{1}=-\frac{G}{2}+k^{\prime}[/latex]

[latex]k_{+}=k_{2}={\frac{G}{2}}+k^{\prime}[/latex]

so that

[latex]\psi(x)=e^{i k^{\prime}x}\cdot\left(a_{+}e^{i{\frac{G}{2}}x} + a_{-}e^{-i{\frac{G}{2}}x} ight)[/latex]

is the perturbed wave function and [latex]|k_{+} angle{\mathrm {~and~}}|k_{-} angle[/latex] are the unperturbed wave functions.

If we consider the limit [latex]k=\pm G/{2},[/latex] then we set [latex]k^{\prime} = 0 [/latex] and [latex]k_{1}=-k_{2}=-G/2.[/latex] There is a degeneracy at the unperturbed energy

[latex]H_{11}=H_{22}=E_{0}=\frac{\hbar^{2}k^{2}}{2m}=\frac {\hbar^{2}G^{2}}{8m}[/latex]

[latex]H_{21}=H_{12}=\left\langle-{\frac{G}{2}}{\Bigg|}V{\Bigg|}{\frac{G}{2}} ight angle=-U_{G}[/latex]

The secular equation for this problem is

[latex]\left|\begin{matrix} H_{11}-E & \,\,\,\, \, H_{12} \ H_{21} & H_{22}-E \end{matrix} ight| = 0[/latex]

which has solution

[latex]E_{\pm}=\frac{(H_{11}+H_{22})}{2}\pm\left(\frac{1}{4}(H_{11}-H_{22})^{2}+H_{12}H_{21} ight)^{1/2}[/latex]

Since [latex]H_{11}=H_{22}\quad\mathrm{and}\quad H_{12} H_{21} =U_{G}^{2}[/latex]

[latex]E_{\pm}=E_{0}\pm{{U}}_{G}[/latex]

The following figure illustrates the effect the lowest-frequency Fourier component of the periodic potential has on electron dispersion.

The following few sketches illustrate the effect the lowest-frequency Fourier component of the periodic potential has on electron dispersion at G/2.

Question 10.7: (a) An electron moves in a one-dimensional box of length X. ......

This Question has been Answered.

Already have an account?

Login

Related Answered Questions

Verified Answer:

In this chapter we solved for the first excited state of a two-dimensional harmonic oscillator subject to perturbation W = κ′ x y. How do the three-fold degenerate energy E = 3ћω and the four-fold degenerate energy E = 4ћω separate due to the same perturbation? ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer: