Energy in infinite square well

[w3390]

Homework Statement

Find the energy of a particle of mass m in an infinite square well with one end at x=-L/2 and the other at x=L/2.

Homework Equations

Schrodinger Equation

The Attempt at a Solution

To save time, I won't type the solving of the differential equation which results in the sin+cos terms. I eventually get to the point where I have found that $$\Psi$$(x)=Asin(2n*PI*x/L). Using boundary conditions, I found k=2n*PI/L.

From my differential equation, I have that k^2 = 2mE/h(bar)^2.

When I plug in k=2n*PI/L to the k^2 equation, it simplifies to:

E = (n^2*h^2)/(2mL^2)

My question is whether or not this is correct because when the infinite well has one end at x=0 and the other at x=L, the result is that:

E = (n^2*h^2)/(8mL^2)

It shouldn't matter where the well is placed as long as the width is the same. Why am I getting two different answers?

Any help would be greatly appreciated.

 

[cepheid]

Hmm, yeah, you should get the same answer and you've probably made an algebraic error somewhere. I would carefully go over the application of the boundary conditions in the -L/2 to +L/2 case.

 

[w3390]

I don't think it's an algebraic error. I have worked the problem through, once using endpoints x=0 and x=L and again using x=-L/2 and x=L/2 and I don't get the same answer.

In the case where I am going from 0 to L, I get that the energy is E=(n^2*h^2)/8mL^2, which is correct.

In the case where I am going from -L/2 to L/2, I have the factor of 1/2 that ends up that ends up getting squared and reducing the eight in the denominator to a 4.

 

[w3390]

Any ideas?

 

[benjy1]

I have a very similar problem with yours but from -a to a. So if you have any idea or something we could help each other