How Do You Determine the Correct Wave Function in a Quantum Square Well?

[serverxeon]

Homework Statement

Assume a free particle, V=0, in a infinite potential square well between -L/2 and L/2
solve for the wave function for this particle.

Homework Equations

Time independent schrodinger equation

The Attempt at a Solution

After arriving at the second order differential equation, I get a general solution of
ψ=Acos(kx) + Bsin(kx)

How do I go on to solve for A, B and k?

I've plugged in the boundary conditions, but the solution is still quite indeterminate.

I have to assume A=0, then ψ=Bsin(2n∏x/L)
however, if I assume B=0, I get ψ=Acos([2n+1]∏x/L)

which is correct?
I do need to pick one before going onto normalising right?

 

[DrClaude]

I've plugged in the boundary conditions, but the solution is still quite indeterminate.

I have to assume A=0, then ψ=Bsin(2n∏x/L)
however, if I assume B=0, I get ψ=Acos([2n+1]∏x/L)

which is correct?

Can you find a trigonometric relation between these two solutions?

 

[YOUGI]

You can rewrite the solution to ψ=Aexp(ik)+Bexp(-ik) (k=2pi/λ)

when x→+∞,ψ=0.so A=0.
when x→-∞,ψ=0, so B=0

that means ψ is a segmented function.

your solution is right,and hope my answer can help you to comprenhen it in other way.

 

[serverxeon]

er, nope.
some pointers pls?

 

[DrClaude]

You can rewrite the solution to ψ=Aexp(ik)+Bexp(-ik) (k=2pi/λ)

when x→+∞,ψ=0.so A=0.
when x→-∞,ψ=0, so B=0

The wave function is limited to the range -L/2, L/2 because of the infinite wall.

 

[DrClaude]

er, nope.
some pointers pls?

$$\cos(x + \pi/2) = -\sin(x)$$
Your two solutions are the same