- #1
Dizzle
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If you solve the 1D time independent schrodinger equation for a system with V(x)=0, you get a solution like:
Ψ(x) = Aexp(ikx) + Bexp(-ikx)
Applying certain boundary conditions would give you the solution to the Simple harmonic oscillator or infinite well etc. But I often see the wavefunction for a free particle written as
Ψ(x) = Aexp(ikx)
So is there some boundary condition which causes B=0 ?
I have tried the periodic BC Ψ(0) = Ψ(L) and dΨ/dx (x=0) = dΨ/dx (x=L) but this only seems to give me certain values of k, it does nothing with A or B.
I see that each term represents a wave in each direction, but I don't see how the B is lost if we start with the Schrodinger equation. I am under the impression that periodic boundary conditions result in the solution Ψ(x) = Aexp(ikx) but can't get rid of my B..
Thanks
Ψ(x) = Aexp(ikx) + Bexp(-ikx)
Applying certain boundary conditions would give you the solution to the Simple harmonic oscillator or infinite well etc. But I often see the wavefunction for a free particle written as
Ψ(x) = Aexp(ikx)
So is there some boundary condition which causes B=0 ?
I have tried the periodic BC Ψ(0) = Ψ(L) and dΨ/dx (x=0) = dΨ/dx (x=L) but this only seems to give me certain values of k, it does nothing with A or B.
I see that each term represents a wave in each direction, but I don't see how the B is lost if we start with the Schrodinger equation. I am under the impression that periodic boundary conditions result in the solution Ψ(x) = Aexp(ikx) but can't get rid of my B..
Thanks