Phy5645/Square Wave Potential Problem: Difference between revisions

Revision as of 16:07, 6 August 2013

Let us once again confine our attention to the region, $0<x<a.\!$ 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 $A,\!$ $B,\!$ $C,\!$ and $D.\!$ To derive the condition under which these equations have a nontrivial solution, we first eliminate $C\!$ 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 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.

Back to Motion in a Periodic Potential

@@ Line 1: / Line 1: @@
-'''Case 1:''' <math>~E>0</math>
+Let us once again confine our attention to the region, <math>0 < x < a.\!</math>  The wave function for this region is given by
-let  <math> k = \frac{\sqrt{2mE}}{\hbar}</math>
+<math>\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}
+</math>
-then
+where <math>k_1=\frac{\sqrt{2mE}}{\hbar}</math> and <math>k_2=\frac{\sqrt{2m(E-V_0)}}{\hbar}.</math>
-<math>\frac{d^2 \psi}{dx^2}= -k^2 \psi </math>
+By Bloch's theorem, the full wave function must have the form,
-whose general solution is:
+:<math>\psi_k(x)=e^{ikx}u_k(x).\!</math>
-<math> ~\psi(x) = A sin(kx) +B cos(kx) , (0<x<a) </math>
+Continuity of <math>\psi_k(x)\!</math> and <math>\psi'_k(x)\!</math> at <math>x=c\!</math> requires that
-by Bloch's theorem , the wave function in the cell immediately to the left of the origin:
+:<math> Ae^{ik_1c}+Be^{-ik_1c}=Ce^{ik_2c}+De^{-ik_2c}\!</math>
-<math> \psi(x) = e^{-i\kappa a} \left(A sin(k(x+a)) + B cos(k(x+a)) \right) , ~(0<x<a) </math>
+and
-at <math>~x=0</math> <math>~\psi</math> must be continuous across; so:
+:<math> ik_1(Ae^{ik_1c}-Be^{-ik_1c})=ik_2(Ce^{ik_2c}-De^{-ik_2c}).\!</math>
-<math> B = e^{-i\kappa a} \left(A sin(k a) + B cos( k a) \right)</math>
+The periodicity of <math>u_k(x)\!</math> and continuity of <math>\psi_k(x)\!</math> and <math>\psi'_k(x)\!</math> at <math>x=a\!</math> gives us
-and the derivative of the wave function suffers a discontinuity proportional the "strength" of the delta function:
+:<math>Ce^{ik_2a}+De^{-ik_2a}=e^{ika}(A+B)\!</math>
-<math>ka - e^{-i\kappa a}\left( A cos(k a) + B Sin(ka) \right) = \frac{-2m \alpha}{\hbar^2} B </math>
+and
-therefore
+:<math>ik_2(Ce^{ik_2a}-De^{-ik_2a})=ik_1e^{ika}(A-B).\!</math>
-<math>A sin(ka) = \left(e^{i\kappa a} - cos (ka) \right) B</math>
+We have thus obtained four linear equations in <math>A,\!</math> <math>B,\!</math> <math>C,\!</math> and <math>D.\!</math> To derive the condition under which these equations have a nontrivial solution, we first eliminate <math>C\!</math> and <math>D\!</math> and then determine when the resulting <math>2\times 2\!</math> system has nontrivial solutions; this yields the condition,
-the derivative suffers from a discontinuity  proportional to the strength of the delta function:
+:<math>\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},</math>
-<math> \rho A - e^{-i\kappa a}\rho\left(A cosh(\rho a) + B sinh(\rho a) \right) = \frac{2m \alpha}{\hbar^2} </math>
+where <math>b=a-c\!</math> is the width of the "barrier" parts of the potential.  This, along with the equation,
-which implies
+<math>k_1^2-k_2^2=\frac{2mV_0}{\hbar^2},</math>
-<math> \left(e^{i\kappa a}-cos(ka) \right)\left(1- e^{-i\kappa a}cos(ka)\right) + e^{-i\kappa a}sin^2(ka) = \frac{-2m\alpha}{\hbar^2 k} sin(ka) </math>
+yields the energy spectrum of the system.
-finally
+If we take the limit <math> V_0 \rightarrow \infty \!</math> and <math> b \rightarrow 0 \!</math> in such a way as to keep <math>V_0b\!</math> finite, then we can obtain:
-<math> cos(\kappa a) = cos(ka) + \frac{m \alpha}{\hbar^2 k}sin(ka)</math>
+:<math>-ik_2 b=-i\sqrt{\frac{2m}{\hbar^2}(E-V_0)b^2}\approx \sqrt{\frac{2mb}{\hbar^2}(V_0b)}\ll 1</math>
-'''Case 2:''' <math>~E<0</math> and <math>~0<x<a</math>
+In this limit,
-<math> \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E \psi </math>
+:<math>\sin{k_2b}\approx k_2b </math>
-<math>\frac{d^2 \psi}{dx^2} = \rho^2 \psi </math>
+and
-where
+:<math>\cos{k_2b}\approx 1.</math>
-<math> \rho = \frac{\sqrt{-2mE}}{\hbar} </math>
+Our equations then reduce to, noting that <math>c=a\!</math> in this limit,
-the general solution is:
+:<math>\cos(ka)=\cos(k_1a)+\frac{mV_0ab}{\hbar^2}\frac{\sin(k_1a)}{k_1a}.</math>
-<math>~\psi(x) = A sinh(\rho x) + B cosh(\rho x) </math> for <math>0<x<a\!</math>
+This is just the equation that we obtained for the Dirac comb potential; note that, here, <math>V_0b\!</math> stands for the <math> V_0\!</math> in the Dirac comb problem described earlier.
-by Bloch's theorem the solution on <math>-a<x<0\!</math> is
-<math> \psi(x) = e^{-i\kappa a}\left( A sinh(\rho(x+a)) B \cosh(\rho(x+a)) \right) </math>
-for <math>\psi(x)\!</math> to be continuous  at <math>x=0</math>
-<math> B = e^{-i\kappa a} \left( A sinh(\rho a) + B \cosh(\rho a) \right) </math>
-which implies
-<math> A sinh(\rho a )  = B \left( e^{i\kappa a} - cosh(\rho a) \right) </math>
-which implies
-<math> A \left( 1- e^{-i\kappa a} cosh(\rho a) \right) = B\left( \frac{2 m \alpha}{\hbar^2 \rho} + e^{-i \kappa a}sinh(\rho a) \right)</math>
-by substitution:
-<math> (e^{i \kappa } - cosh(\rho a) )(1- e^{-i \kappa a} cosh(\rho a)) = \frac{ 2m \alpha}{\hbar^2 \rho} sinh(\rho a)+ e^{-i \kappa a}sinh^2(\rho a) </math>
-<math> e^{i\kappa a} - 2 cosh{\rho a} + e^{-i\kappa a} cosh^2(\rho a) - e^{-i \kappa a} sinh^2(\rho a) = \frac{ 2m \alpha}{\hbar^2 \rho} sinh(\rho a) </math>
-<math>e ^{i\kappa a}+e^{-i\kappa a} = 2 cosh(\rho a) + \frac{2m \alpha}{\hbar^2\rho}sinh(\rho a) </math>
-<math>cos(\kappa a) = cosh(\rho a) + \frac{m \alpha}{\hbar^2 \rho} sinh(\rho a) </math>
 Back to [[Motion in a Periodic Potential]]

Phy5645/Square Wave Potential Problem: Difference between revisions

Revision as of 16:07, 6 August 2013

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