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

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 \psi_k(x)= \begin{cases} Ae^{ik_1x}+Be^{-ik_1x}, & 0 < x < c \\ Ce^{ik_2x}+De^{-ik_2x}, & c < x < a, \end{cases} }

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 k_1=\frac{\sqrt{2mE}}{\hbar}} 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 k_2=\frac{\sqrt{2m(E-V_0)}}{\hbar}.}

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

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 \psi_k(x)=e^{ikx}u_k(x).\!}

Continuity of 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 \psi_k(x)\!} 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 \psi'_k(x)\!} at 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 x=c\!} requires 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 Ae^{ik_1c}+Be^{-ik_1c}=Ce^{ik_2c}+De^{-ik_2c}\!}

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 ik_1(Ae^{ik_1c}-Be^{-ik_1c})=ik_2(Ce^{ik_2c}-De^{-ik_2c}).\!}

The periodicity of 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 u_k(x)\!} and continuity of 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 \psi_k(x)\!} 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 \psi'_k(x)\!} at 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 x=a\!} gives us

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 Ce^{ik_2a}+De^{-ik_2a}=e^{ika}(A+B)\!}

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 ik_2(Ce^{ik_2a}-De^{-ik_2a})=ik_1e^{ika}(A-B).\!}

We have thus obtained four linear equations in 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 A,\!} 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,\!} 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 $D\!$ and then determine when the resulting $2\times 2\!$ system has nontrivial solutions; this yields the condition,

$\cos {ka}=\cos {k_{1}c}\cos {k_{2}b}-{\frac {k_{1}^{2}+k_{2}^{2}}{2k_{1}k_{2}}}\sin {k_{1}c}\sin {k_{2}b},$

where $b=a-c\!$ is the width of the "barrier" parts of the potential. This, along with the equation,

$k_{1}^{2}-k_{2}^{2}={\frac {2mV_{0}}{\hbar ^{2}}},$

yields the energy spectrum of the system.

If we take the limit $V_{0}\rightarrow \infty \!$ and $b\rightarrow 0\!$ in such a way as to keep $V_{0}b\!$ finite, then we can obtain:

$-ik_{2}b=-i{\sqrt {{\frac {2m}{\hbar ^{2}}}(E-V_{0})b^{2}}}\approx {\sqrt {{\frac {2mb}{\hbar ^{2}}}(V_{0}b)}}\ll 1$

In this limit,

$\sin {k_{2}b}\approx k_{2}b$

and

$\cos {k_{2}b}\approx 1.$

Our equations then reduce to, noting that $c=a\!$ in this limit,

$\cos(ka)=\cos(k_{1}a)+{\frac {mV_{0}ab}{\hbar ^{2}}}{\frac {\sin(k_{1}a)}{k_{1}a}}.$

This is just the equation that we obtained for the Dirac comb potential; note that, here, $V_{0}b\!$ stands for the $V_{0}\!$ in the Dirac comb problem described earlier.

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