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Even though double well is a $\mathrm{1D}$ potential and it must not have degenerate spectrum, then why does double well have degenerate energy eigenvalue?

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    $\begingroup$ Why do you think that it does? $\endgroup$
    – J. Murray
    Jul 23 '20 at 1:01
  • $\begingroup$ @J.Murray I dont know sir. Will you tell me? I am a bit confused $\endgroup$ Jul 23 '20 at 1:04
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    $\begingroup$ What I mean is that the bound state spectrum of the double well potential does not have any degeneracy, so I don't understand the meaning of your question. $\endgroup$
    – J. Murray
    Jul 23 '20 at 1:06
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    $\begingroup$ Nowhere on that page does it say that the double well has degeneracy. Can you quote the passage you are referring to, and add it to your question body? $\endgroup$
    – J. Murray
    Jul 23 '20 at 1:11
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    $\begingroup$ almost degenerate is non-degenerate. The 2 levels may be are close but do not have exactly the same energy. $\endgroup$ Jul 23 '20 at 1:19
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As per the quoted page,

The important point is that there are two almost degenerate energy eigenstates for every one energy eigenstate in the corresponding infinite square well [...]

If two states are almost degenerate, then they are not degenerate. The corresponding energy eigenvalues are very close to being the same, but are not the same.


To address your follow-up questions,

Why is it the case that they have identical (even though not matching with the exact ground state value but by looking them one can say they are degenerate) degenerate values.

They don't have identical values, and they aren't degenerate. The reason that they are nearly degenerate is as follows: Note that for an energy eigenstate $\psi$,

$$E = \int_{-L/2}^{L/2} -\frac{\hbar^2}{2m}\psi^*(x) \cdot \psi''(x) + V(x) \psi^*(x) \psi(x)\ dx$$

enter image description here Source

The two wavefunctions differ (up to a minus sign) only in a small neighborhood of $x=0$. In that neighborhood, the second derivative of the symmetric state (the red curve in the figure) is positive, but the second derivative of the antisymmetric state vanishes. As a result, the kinetic part of the Hamiltonian $$\int - \frac{\hbar^2}{2m}\psi^* \cdot \psi'' dx$$

makes a negative contribution to the symmetric state and no contribution to the antisymmetric state. We would therefore expect the symmetric state to have a slightly lower energy than the antisymmetric state, which is indeed what we see.

[...] he quote a question by saying that for 3 well one expect the eigenvalues will have 4 almost same value but fermion principle violates it so you must convince it must have 3 independent level and 1 must go up. So what is the point even though they are not degenerate but we still are applying paulis principle and why it has 3 same value.

This argument is a bit subtle. The Pauli exclusion principle says that only one electron can occupy each energy state (we are ignoring spin for the moment). If there is only one atom in the universe, then you can calculate its energy levels and imagine placing an electron in each one.

If you have two atoms separated by a large distance then we typically treat them as though they were totally isolated. Naively, we imagine that there are two ground states, one localized around each electron. But this is wrong - as the exercise shows, there is a single ground state and one excited state which is extremely close in energy, and it is these two states which are occupied by the electrons in reality.

If you have three atoms which are separated by large distances, experience suggests that we can treat them as effectively isolated, which would mean that we could put a single electron in the "ground state" of each atom. Again, this is not quite right - there is a single ground state and two nearby excited states which are occupied by those three electrons, and the Pauli principle is satisfied.

The author notes that you might expect there to be four nearly degenerate states, but this would mean that you could fit another electron somewhere in the three-atom system without going up to the next excited state, which is contrary to what we observe.

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    $\begingroup$ My first question is why is it the case that they have identical (even though not matching with the exact ground state value but by looking them one can say they are degenerate) degenerate values. The 2nd question is -also he quote a question by saying that for 3 well one expect the eigenvalues will have 4 almost same value but fermion principle violates it so you must convince it must have 3 independent level and 1 must go up. So what is the point even though they are not degenerate but we still are applying paulis principle and why it has 3 same value. What logic is the reader using. $\endgroup$ Jul 23 '20 at 1:25
  • $\begingroup$ @AnshulSharma I have updated my answer. As an FYI, you cannot ping another user in a comment thread unless they have commented in the same thread. $\endgroup$
    – J. Murray
    Jul 23 '20 at 2:23

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