• Support PF! Buy your school textbooks, materials and every day products Here!

Solving for the wave function

  • Thread starter serverxeon
  • Start date
  • #1
101
0

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

  • #2
DrClaude
Mentor
7,269
3,424
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?
 
  • #3
2
0
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.
 
  • #4
101
0
er, nope.
some pointers pls?
 
  • #5
DrClaude
Mentor
7,269
3,424
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.
 
  • #6
DrClaude
Mentor
7,269
3,424
er, nope.
some pointers pls?
$$\cos(x + \pi/2) = -\sin(x)$$
Your two solutions are the same
 

Related Threads on Solving for the wave function

  • Last Post
Replies
1
Views
450
Replies
9
Views
8K
Top