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