Phy5645/Square Wave Potential Problem

2. consider an infinite series of dirac delta function potential wells in one dimension such that:

$~V(x+a)=V(x)$

solve for $~\psi (x+a)$ in terms $\psi (x)$

${\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x)}{dx^{2}}}+V(x)\psi (x)=E\psi (x)$

which satisfies $~\psi (x+a)=e^{-i\kappa a}\psi (x)$

2.1)

for $~E>0$

let $k={\frac {\sqrt {2mE}}{\hbar }}$

then

${\frac {d^{2}\psi }{dx^{2}}}=-k^{2}\psi$

whose general solution is:

$\psi (x)=Asin(kx)+Bcos(kx),~~(0<x<a)$

by bloch's theorem , the wave function in the cell immediately to the left of the origin:

$\psi (x)=e^{-i\kappa a}\left(Asin(k(x+a))+Bcos(k(x+a))\right),(0<x<a)$

at $x=0$ $\psi$ must be continuous across; so:

$B=e^{-i\kappa a}\left(Asin(ka)+Bcos(ka)\right)$

and the derivative of the wave function suffers a discontinuity proportional the "strength" of the delta function:

$ka-e^{-i\kappa a}\left(Acos(ka)+BSin(ka)\right)={\frac {-2m\alpha }{\hbar ^{2}}}B$

therefore

$Asin(ka)=\left(e^{i\kappa a}-cos(ka)\right)B$

the derivative suffers from a discontinuity proportional to the strength of the delta function:

$\rho A-e^{-i\kappa a}\rho \left(Acosh(\rho a)+Bsinh(\rho a)\right)={\frac {2m\alpha }{\hbar ^{2}}}$

which implies

$\left(e^{i\kappa a}-cos(ka)\right)\left(1-e^{-i\kappa a}cos(ka)\right)+e^{-i\kappa a}sin^{2}(ka)={\frac {-2m\alpha }{\hbar ^{2}k}}sin(ka)$

finally

$cos(\kappa a)=cos(ka)+{\frac {m\alpha }{\hbar ^{2}k}}sin(ka)$

2.2) for $~E<0$ and $~0<x<a$

${\frac {-\hbar ^{2}}{2m}}{\frac {d^{2}\psi }{dx^{2}}}=E\psi$

${\frac {d^{2}\psi }{dx^{2}}}=\rho ^{2}\psi$

where

$\rho ={\frac {\sqrt {-2mE}}{\hbar }}$

the general solution is:

$~\psi (x)=Asinh(\rho x)+Bcosh(\rho x)$ for $0<x<a$

by bloch's theorem the solution on $-a<x<0$ is

$\psi (x)=e^{-i\kappa a}\left(Asinh(\rho (x+a))B\cosh(\rho (x+a))\right)$

for $\psi (x)$ to be continuous at $x=0$

$B=e^{-i\kappa a}\left(Asinh(\rho a)+B\cosh(\rho a)\right)$

which implies

$Asinh(\rho a)=B\left(e^{i\kappa a}-cosh(\rho a)\right)$

which implies

$A\left(1-e^{-i\kappa a}cosh(\rho a)\right)=B\left({\frac {2m\alpha }{\hbar ^{2}\rho }}+e^{-i\kappa a}sinh(\rho a)\right)$

by substitution:

$(e^{i\kappa }-cosh(\rho a))(1-e^{-i\kappa a}cosh(\rho a))={\frac {2m\alpha }{\hbar ^{2}\rho }}sinh(\rho a)+e^{-i\kappa a}sinh^{2}(\rho a)$

$e^{i\kappa a}-2cosh{\rho a}+e^{-i\kappa a}cosh^{2}(\rho a)-e^{-i\kappa a}sinh^{2}(\rho a)={\frac {2m\alpha }{\hbar ^{2}\rho }}sinh(\rho a)$

$e^{i\kappa a}+e^{-i\kappa a}=2cosh(\rho a)+{\frac {2m\alpha }{\hbar ^{2}\rho }}sinh(\rho a)$

$cos(\kappa a)=cosh(\rho a)+{\frac {m\alpha }{\hbar ^{2}\rho }}sinh(\rho a)$