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| 2. Consider an infinite series of dirac delta function potential wells in one dimension such that:
| | 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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| <math> ~V(x+a) = V(x) </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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| solve for <math> ~\psi(x+a)</math> in terms <math>\psi(x)</math>
| | 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{\hbar^2}{2m}\frac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)</math>
| | By Bloch's theorem, the full wave function must have the form, |
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| which satisfies <math>~\psi(x+a) = e^{-i\kappa a}\psi(x) </math>
| | :<math>\psi_k(x)=e^{ikx}u_k(x).\!</math> |
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| 2.1)
| | Continuity of <math>\psi_k(x)\!</math> and <math>\psi'_k(x)\!</math> at <math>x=c\!</math> requires that |
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| for <math>~E>0</math>
| | :<math> Ae^{ik_1c}+Be^{-ik_1c}=Ce^{ik_2c}+De^{-ik_2c}\!</math> |
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| let <math> k = \frac{\sqrt{2mE}}{\hbar}</math>
| | and |
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| then
| | :<math> ik_1(Ae^{ik_1c}-Be^{-ik_1c})=ik_2(Ce^{ik_2c}-De^{-ik_2c}).\!</math> |
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| <math>\frac{d^2 \psi}{dx^2}= -k^2 \psi </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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| whose general solution is:
| | :<math>Ce^{ik_2a}+De^{-ik_2a}=e^{ika}(A+B)\!</math> |
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| <math> ~\psi(x) = A sin(kx) +B cos(kx) , (0<x<a) </math>
| | and |
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| by Bloch's theorem , the wave function in the cell immediately to the left of the origin:
| | :<math>ik_2(Ce^{ik_2a}-De^{-ik_2a})=ik_1e^{ika}(A-B).\!</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> | | 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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| at <math>~x=0</math> <math>~\psi</math> must be continuous across; so:
| | :<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> B = e^{-i\kappa a} \left(A sin(k a) + B cos( k a) \right)</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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| and the derivative of the wave function suffers a discontinuity proportional the "strength" of the delta function:
| | <math>k_1^2-k_2^2=\frac{2mV_0}{\hbar^2},</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>
| | yields the energy spectrum of the system. |
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| therefore
| | 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>A sin(ka) = \left(e^{i\kappa a} - cos (ka) \right) B</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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| the derivative suffers from a discontinuity proportional to the strength of the delta function:
| | In this limit, |
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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> | | :<math>\sin{k_2b}\approx k_2b </math> |
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| which implies
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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> | | :<math>\cos{k_2b}\approx 1.</math> |
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| finally
| | Our equations then reduce to, noting that <math>c=a\!</math> in this limit, |
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| <math> cos(\kappa a) = cos(ka) + \frac{m \alpha}{\hbar^2 k}sin(ka)</math> | | :<math>\cos(ka)=\cos(k_1a)+\frac{mV_0ab}{\hbar^2}\frac{\sin(k_1a)}{k_1a}.</math> |
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| 2.2) for <math>~E<0</math> and <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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| <math> \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E \psi </math>
| | Back to [[Motion in a Periodic Potential#Problem|Motion in a Periodic Potential]] |
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| <math>\frac{d^2 \psi}{dx^2} = \rho^2 \psi </math>
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| where
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| <math> \rho = \frac{\sqrt{-2mE}}{\hbar} </math>
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| the general solution is:
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| <math>~\psi(x) = A sinh(\rho x) + B cosh(\rho x) </math> for <math>0<x<a\!</math>
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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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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