# Continuity Conditions (QM)

Answer

first boundary conditions are that $psi(-L)$ and $psi(L)$ are continuous in $-L$ and $+L$

$psi(-L) = 0$ when $x -> -L$ from left and $psi(L) = 0$ when $x -> +L$ from the right

$psi(-L) = A*exp(-i K L) + B*exp(+i K L)$ when $x -> -L$ from right

$psi(L) = A*exp(i K L) +B*exp(-i K L)$ when $x -> +L$ from the left

$A*exp(-i K L) +B*exp(i K L) = 0$

$A*exp(i K L) +B*exp(-i K L) =0$

summing

$A*(exp(i K L) +exp(-i K L)) +B(exp(i K L)*exp(-i K L)) =0$

or $A+B =0$ because exponential is never zero

$psi(x) = A*(exp(i k x) – exp(-i k x)) = 2A*sin(k x)$

at $+L$ and $-L$ one has again

$0 = 2A*sin(2KL)$

$0= 2A*sin(-2KL)$

and from the above one sees that K is restricted to a certain number of values k(n) for which

$k(n) = n*pi/L$

hence $n*pi/L = sqrt{2mE/hbar}$

and $E(n) = n^2*hbar^2*pi^2/(2mL^2)$

For part B)

$K_1 = pi/L$ , $K_2=2pi/L$ , $K_3 =3pi/L$

$psi1(x) = 2A*sin(pi*x/L)$

$psi2(x) =2A*sin(2*pi/L)$

$psi3(x) = 2A*sin(3*pi/L)$