In one dimension there are no degenerate bound states?

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Discussion Overview

The discussion centers around the concept of degenerate bound states in one-dimensional quantum systems, specifically addressing a claim from a book that states there are no degenerate bound states in one dimension. Participants explore the implications of complex conjugate wavefunctions and their relationship to degeneracy.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Jon questions the assertion that there are no degenerate bound states in one dimension, presenting the idea that a stationary state and its complex conjugate are both normalizable and may be linearly independent.
  • Some participants argue that the complex conjugate of a stationary state does not represent a distinct state but rather a phase-shifted version of the original state, suggesting they are effectively the same state.
  • Another participant clarifies that while the complex conjugate can be viewed as a phase rotation, it does not imply that the two states are degenerate, as degeneracy typically requires distinct states that can be separated under certain conditions.
  • One participant emphasizes that the complex conjugate is not always a valid solution of the Schrödinger equation unless specific symmetries are present, indicating that the case where wavefunctions are entirely real is a special scenario.

Areas of Agreement / Disagreement

Participants express differing views on the nature of complex conjugate wavefunctions and their implications for degeneracy. There is no consensus on whether the complex conjugate can be considered a distinct state or if it merely represents a phase shift of the original state.

Contextual Notes

Participants note that the discussion hinges on the assumptions regarding the nature of wavefunctions and the conditions under which complex conjugates are valid solutions to the Schrödinger equation. The implications of symmetry and the specific characteristics of one-dimensional systems are also highlighted.

epsilonjon
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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.
 
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epsilonjon said:
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.
 
SpectraCat said:
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.
 
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.
 
epsilonjon said:
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.
 
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.
 
Thanks guys. I get the point you're making now.
 

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