Quantum 1D box obtain an expression for the normalization constant

mike232
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Homework Statement


An electron in a one-dimensional box with walls at x =(o,a) is in the quantum state
psi = A o<x<a/2
psi = -A a/2<x<a
A) obtain an expression for the normalization constant, A.
B) What is the lowest energy of the electron that will be measured in this state?

Homework Equations


Not given anything. But its a chapter on Hermitian operators, and Hamiltonian.

The Attempt at a Solution


So for part a I think I am just supposed to normalize, so 1=integral of A*A ... and I get A=(a)^(-1/2) which i think is my normalization constant.

But is it asking not for the constant but to work back and find an equation?

For part b,,, I just operated on psi with the Hamiltonian and because all I had were constants, I got zero... which is boring if true, but I think isn't the answer I am looking for.

If anyone could help out, many thanks.
 
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mike232 said:
So for part a I think I am just supposed to normalize, so 1=integral of A*A ... and I get A=(a)^(-1/2) which i think is my normalization constant.

But is it asking not for the constant but to work back and find an equation?
It is only asking for A, in order to normalize the wave function, such that you can do part B.

mike232 said:
For part b,,, I just operated on psi with the Hamiltonian and because all I had were constants, I got zero... which is boring if true, but I think isn't the answer I am looking for.
That is indeed not the correct approach. What you have to do is decode what
mike232 said:
B) What is the lowest energy of the electron that will be measured in this state?
actually means. What possible results can you get if you measure the energy of the electron?
 
DrClaude said:
It is only asking for A, in order to normalize the wave function, such that you can do part B.That is indeed not the correct approach. What you have to do is decode what

actually means. What possible results can you get if you measure the energy of the electron?
Don't you find the energy by using the Hamiltonian operator? If this isn't I have no idea what to do with this.
 
What do the postulates of QM say about measurements?
 
The only measurable results are the eigenvalues associated with the operator... but I don't know how that helps because zero is a valid result.
 
mike232 said:
The only measurable results are the eigenvalues associated with the operator...
Yes, and given the wave function, what is the probability of finding a given eigenvalue?
 
Do I find the expansion coefficent?
 
Yes. You basically need to calculate the coefficients, starting from the ground state and going up in energy, to find the first one that is not 0. Then you can get the lowest possible measured energy.
 
DrClaude said:
Yes, and given the wave function, what is the probability of finding a given eigenvalue?

So
DrClaude said:
Yes. You basically need to calculate the coefficients, starting from the ground state and going up in energy, to find the first one that is not 0. Then you can get the lowest possible measured energy.
OK so I haven't done the work yet. But just intuition it's something like the second state?
 
  • #10
mike232 said:
OK so I haven't done the work yet. But just intuition it's something like the second state?
Good intuition. The ground state is even with respect to the center of the box, while ψ is odd, so the integral will be zero. You should get a non-zero coefficient with the first excited state (n = 2).
 
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