Phy5645/Square Wave Potential Problem

Let us once again confine our attention to the region, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < x < a.\!} The wave function for this region is given by

$\psi _{k}(x)={\begin{cases}Ae^{ik_{1}x}+Be^{-ik_{1}x},&0<x<c\\Ce^{ik_{2}x}+De^{-ik_{2}x},&c<x<a,\end{cases}}$

where $k_{1}={\frac {\sqrt {2mE}}{\hbar }}$ and $k_{2}={\frac {\sqrt {2m(E-V_{0})}}{\hbar }}.$

By Bloch's theorem, the full wave function must have the form,

$\psi _{k}(x)=e^{ikx}u_{k}(x).\!$

Continuity of $\psi _{k}(x)\!$ and $\psi '_{k}(x)\!$ at $x=c\!$ requires that

$Ae^{ik_{1}c}+Be^{-ik_{1}c}=Ce^{ik_{2}c}+De^{-ik_{2}c}\!$

and

$ik_{1}(Ae^{ik_{1}c}-Be^{-ik_{1}c})=ik_{2}(Ce^{ik_{2}c}-De^{-ik_{2}c}).\!$

The periodicity of $u_{k}(x)\!$ and continuity of $\psi _{k}(x)\!$ and $\psi '_{k}(x)\!$ at $x=a\!$ gives us

$Ce^{ik_{2}a}+De^{-ik_{2}a}=e^{ika}(A+B)\!$

and

$ik_{2}(Ce^{ik_{2}a}-De^{-ik_{2}a})=ik_{1}e^{ika}(A-B).\!$

We have thus obtained four linear equations in $A,\!$ $B,\!$ Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C,\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D.\!} To derive the condition under which these equations have a nontrivial solution, we first eliminate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D\!} and then determine when the resulting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\times 2\!} system has nontrivial solutions; this yields the condition,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{ka}=\cos{k_1 c}\cos{k_2 b}-\frac{k_1^2+k_2^2}{2k_1k_2}\sin{k_1 c}\sin{k_2 b},}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b=a-c\!} is the width of the "barrier" parts of the potential. This, along with the equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_1^2-k_2^2=\frac{2mV_0}{\hbar^2},}

yields the energy spectrum of the system.

If we take the limit Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0 \rightarrow \infty \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b \rightarrow 0 \!} in such a way as to keep Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0b\!} finite, then we can obtain:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -ik_2 b=-i\sqrt{\frac{2m}{\hbar^2}(E-V_0)b^2}\approx \sqrt{\frac{2mb}{\hbar^2}(V_0b)}\ll 1}

In this limit,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin{k_2b}\approx k_2b }

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{k_2b}\approx 1.}

Our equations then reduce to, noting that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=a\!} in this limit,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos(ka)=\cos(k_1a)+\frac{mV_0ab}{\hbar^2}\frac{\sin(k_1a)}{k_1a}.}

This is just the equation that we obtained for the Dirac comb potential; note that, here, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0b\!} stands for the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_0\!} in the Dirac comb problem described earlier.

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