In one dimension there are no degenerate bound states?

[epsilonjon]

Hi.

In the book I'm reading I've come to a question regarding degenerate states in one dimension. It says that in one dimension there are no degenerate bound states.

But say I have a stationary state with some energy E, and assume that it is normalizable. You can easily show that the complex conjugate of this state also solves the time-independent Schrödinger eq with the same energy E. The conjugate state must also be normalizable, and as far as I can tell the two are linearly independent?

Does that not disprove the statement? Where am I going wrong in my thinking here?

Thanks,
Jon.

 

[SpectraCat]

Hi.

In the book I'm reading I've come to a question regarding degenerate states in one dimension. It says that in one dimension there are no degenerate bound states.

But say I have a stationary state with some energy E, and assume that it is normalizable. You can easily show that the complex conjugate of this state also solves the time-independent Schrödinger eq with the same energy E. The conjugate state must also be normalizable, and as far as I can tell the two are linearly independent?

Does that not disprove the statement? Where am I going wrong in my thinking here?

Thanks,
Jon.

A wavefunction and it's complex conjugate differ only in their phase. For a stationary state, the phase is rotating with time anyway, so the complex conjugate is just a time-shift of \frac{\pi}{2} relative to the original state. In other words, there is no distinction between the state and its complex conjugate in the context you are describing. They are effectively the same state, not two degenerate states.

 

[epsilonjon]

A wavefunction and it's complex conjugate differ only in their phase. For a stationary state, the phase is rotating with time anyway, so the complex conjugate is just a time-shift of \frac{\pi}{2} relative to the original state. In other words, there is no distinction between the state and its complex conjugate in the context you are describing. They are effectively the same state, not two degenerate states.

Okay, I think...

I don't see how the complex conjugate of a stationary state is just a time shift of \frac{\pi}{2} though? I am thinking of stationary state as just a vector on an Argand diagram, rotating with speed given by its energy? If I take the wavefunction and then think of another one \frac{\pi}{2} ahead or behind it, the two aren't conjugate are they?

Sorry if I'm misunderstanding.

 

[kith]

ψ and ψ* rotate in different directions in the complex plane, so the phase difference is not constant.

The essence of SpectraCat's answer remains unchanged: as long as two wavefunctions differ only in phase, they represent the same state.

 

[SpectraCat]

Okay, I think...

I don't see how the complex conjugate of a stationary state is just a time shift of \frac{\pi}{2} though? I am thinking of stationary state as just a vector on an Argand diagram, rotating with speed given by its energy? If I take the wavefunction and then think of another one \frac{\pi}{2} ahead or behind it, the two aren't conjugate are they?

Sorry if I'm misunderstanding.

The point of my answer is that you can obtain \psi^* from a simple phase rotation of \psi, therefore since the phase of a stationary state is rotating with time anyway, there is no way to tell the difference between \psi and \psi^* in a given problem.

In other cases where you are actually dealing with degenerate states, there is always a way to break the degeneracy (e.g. by application of an external field), so you can prove that there are actually two distinct states involved.

 

[genneth]

I have to jump in and be pendantic and point out that in general, the complex conjugate is not necessarily a valid solution of the Schrödinger equation. In the cases where it is, there is always some extra symmetry which causes it. The degenerate case where the wavefunction is/can be entirely real is special, as SpectraCat has tried to explain, since then they are the same function, modulo some phase.

 

[epsilonjon]

Thanks guys. I get the point you're making now.