# Solving for the wave function

## 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?

## Answers and Replies

DrClaude
Mentor
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?

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.

er, nope.
some pointers pls?

DrClaude
Mentor
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
Mentor
er, nope.
some pointers pls?
$$\cos(x + \pi/2) = -\sin(x)$$
Your two solutions are the same