Given a potential well of V=0 on the interval (0,L) and V=∞ outside the well, I am working to solve the Time Independent Schrodinger Equation d2dx2ψ=−2mEℏ2ψ=−k2ψ.
With the ansatz ψ∝exp(αx) we can write α2=−k2 so α=±ik and the general solution is ψ(x)=Aeikx+Be−ikx.
From the boundary condition ψ(0)=0, we see that ψ(0)=Ae0+Be0=A+B=0⇒B=−A
I am unsure of how to evaluate this last part. A=0 is not valid, so what is the solution to this last equation?
Answer
Solving eikL−e−ikL=0
From Euler's Formula, eiθ=cos(θ)+isin(θ) the solution is that k is quantized: k=nπL for positive integer n.
Thus ψ(x)=A(eikx−e−ikx)=√2a(eikx−e−ikx)
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