1. Mar 12, 2017

### dyn

Hi. I'm confused about degeneracy in I-D As far as I understand it ( and please tell me if i'm wrong ) ; a free particle is doubly degenerate with a continuous energy spectrum with eigenfunctions eikx and e-ikx. A particle on a ring in I-D is doubly degenerate but this time the energy is quantized.
My main question concerns a particle in a 1-D infinite well with infinite walls at x=0 and x=a. The eigenfunction is given by ψ = sin (nπx/a) where n = 1,2,3,.... These eigenfunctions are non-degenerate as n only takes positive values but sin kx can be written as sin kx = ( eikx - e-ikx) / 2i which is a superposition of 2 waves travelling in opposite directions. Would this not make the particle in a box doubly degenerate ?

2. Mar 12, 2017

### hilbert2

There's not much room for choice of $A$ and $B$ if you want the function $\psi (x) = Ae^{ikx}+Be^{-ikx}$ to be zero at x=0 and x=a. For most values of k it's not possible at all (except if A and B are both zero).

3. Mar 12, 2017

### Staff: Mentor

Neither of those functions are solutions to the eigenvalue equation for the particle-in-a-box Hamiltonian.

4. Mar 12, 2017

### dyn

The solution to the particle in a box Hamiltonian is sin kx where k = nπ /a and n takes positive integer values. This shows that the energy eigenvalues are non-degenerate. But the part that confuses me is that sin kx can be written as a linear combination of eikx and e-ikx which implies doubly degenerate ?

5. Mar 12, 2017

### hilbert2

The function $e^{ikx}$ is not zero for any value of x. Its absolute value is 1 everywhere.

6. Mar 12, 2017

### Staff: Mentor

No, because, as Nugatory said, $e^{ikx}$ and $e^{-ikx}$ are not solutions of the eigenvalue equation, whereas $\sin kx$ is. Writing a function that is a solution as a linear combination of functions that are not tells you nothing at all about degeneracy; degeneracy implies that you are talking only about functions which are solutions. For each $k$ there is only one such function, hence no degeneracy.

7. Mar 12, 2017

### hilbert2

Yeah, even the harmonic oscillator ground state $\psi (x) = Ae^{-kx^2}$ can be written as $\psi (x) = A\left(\frac{1}{2}e^{-kx^2}+\sin kx\right) + A\left(\frac{1}{2}e^{-kx^2}-\sin kx\right)$ if you want to, but neither of the functions

$\psi (x) = \frac{1}{2}e^{-kx^2}+\sin kx$, or
$\psi (x) =\frac{1}{2}e^{-kx^2}-\sin kx$,

is a solution to the SHO Schrödinger equation.

8. Mar 13, 2017

### dyn

Thanks everyone. Much appreciated !