Degeneracy and Symmetrization Requirement

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In summary: I am not familiar with the Pauli Exclusion Principle.Now, the symmetrization requirement says: \psi(x1,x2) = -\psi(x2,x1)for fermions, which is the case above. I feel like this is relevant. But I still see there being two separate states with the same energy. Isn't this the definition of degeneracy?It does get very confusing with the sums and differences and symmetrizations etc, but I wonder if you haven't made the very simple mistake of writing out two states which are the same as each other except for a negative sign? So they're actually the same state.Ah, its as simple
  • #1
hbweb500
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I am encountering two-particle systems for the first time, and I am a little confused about something.

Consider a infinite square well with two noninteracting particles, both fermions (though, disregard spin degrees of freedom, this is a strictly 1-d scenario).

So I go to write down the states of the system corresponding to different energies. Consider the case:

[tex] \psi_{23} = \frac{1}{\sqrt{2}} \left[ \psi_2(x_1) \psi_3(x_2) - \psi_3(x_1) \psi_2(x_2) \right] [/tex]

Where

[tex] \psi_n (x) = \sqrt{\frac{2}{a}} \sin n \pi x / a [/tex]

This corresponds to energy

[tex] E = \frac{13 \hbar^2 \pi^2}{2 m a^2} [/tex]

This is all good. What I don't get is this: why isn't this energy degenerate? In other words, isn't the state:

[tex] \psi_{32} = \frac{1}{\sqrt{2}} \left[ \psi_3(x_1) \psi_2(x_2) - \psi_2(x_1) \psi_3(x_2) \right] [/tex]

Also an energy eigenfunction of the Hamiltonian, with the same eigenvalue. Aren't the two states distinct?

Now, the symmetrization requirement says:

[tex] \psi(x1,x2) = -\psi(x2,x1)[/tex]

for fermions, which is the case above. I feel like this is relevant. But I still see there being two separate states with the same energy. Isn't this the definition of degeneracy?
 
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  • #2
It does get very confusing with the sums and differences and symmetrizations etc, but I wonder if you haven't made the very simple mistake of writing out two states which are the same as each other except for a negative sign? So they're actually the same state.
 
  • #3
Ah, its as simple as that. If two eigenvectors are the same up to a scale factor, they are the same eigenvector. I was forgetting that.

Thanks.
 
  • #4
The two states you wrote down are linearly dependent. They only differ up to an overall minus sign.

[tex]\psi_{23} = -\psi_{32}[/tex]

Two states are considered degenerate when they are not linearly dependent.

You can in fact construct an infinite number of states which are all eigenstates of the Hamiltonian. Namely,

[tex] \psi = e^{i\theta}\psi_{32}[/tex]

where [itex]0\leq \theta< 2\pi[/itex]. All these states are eigenstates, with the values [itex]\theta=0[/itex] and [itex]\theta=\pi[/itex] corresponding to your two functions. But you do not consider these states as "new states", since the overall phase doesn't matter.

EDIT: too slow..
 
  • #5
That was helpful too, I hadn't thought about the phase factor. Thanks!
 
  • #6
hbweb500 said:
I am encountering two-particle systems for the first time, and I am a little confused about something.

Consider a infinite square well with two noninteracting particles, both fermions (though, disregard spin degrees of freedom, this is a strictly 1-d scenario).

I don't understand what you have written. How can you have fermions without considering spin? Are you just saying that you are only going to accept spatial wavefunctions that satisfy the anti-symmetry property of fermionic wavefunctions? That is fine I guess, but the issue with that is there is another class of solutions where the spatial part of the wavefunction is symmetric under exchange, but the antisymmetry is provided by the spin portion of the wavefunction.

All of this is usually worked out in some detail in a standard textbook ... it can often be found under the heading of "Pauli Exclusion Principle" in the context of excited states of the helium atom.
 
  • #7
I understand, this is just an example I am pulling from Griffiths.
 

1. What is degeneracy in the context of symmetrization requirement?

Degeneracy refers to the phenomenon where multiple quantum states have the same energy, resulting in a degenerate energy level. In the context of symmetrization requirement, it means that for a system with identical particles, the wavefunction must be symmetrical or anti-symmetrical in order to properly account for the indistinguishability of the particles.

2. Why is symmetrization requirement important in quantum mechanics?

The symmetrization requirement is important because it takes into account the fundamental principle of indistinguishability in quantum mechanics. Without this requirement, the wavefunction would not accurately describe the behavior of identical particles and could lead to incorrect predictions of physical phenomena.

3. How does the symmetrization requirement affect the properties of a system?

The symmetrization requirement affects the properties of a system by limiting the ways in which identical particles can be arranged. For example, in a system of two identical particles, the wavefunction must be either symmetric or anti-symmetric, which leads to different physical properties such as the spin of the particles.

4. What is the difference between symmetrical and anti-symmetrical wavefunctions?

A symmetrical wavefunction has the property that swapping the positions of two identical particles does not change the overall wavefunction. An anti-symmetrical wavefunction, on the other hand, changes sign when the particles are swapped. This means that the wavefunction is positive for one arrangement of particles and negative for the other, reflecting the indistinguishability of the particles.

5. How does degeneracy affect the behavior of a system?

Degeneracy can affect the behavior of a system in various ways. In some cases, it can lead to increased stability or robustness of the system. In other cases, degeneracy can result in complex energy landscapes and lead to unexpected phenomena. In the context of symmetrization requirement, degeneracy can also impact the allowed wavefunctions and result in different physical properties of the system.

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