- #1

omg!

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could someone give me a quick answer on the exact conditions for the hamiltonian to be non degenerate, i.e. to have different eigenvalues?

thanks in advance.

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- Thread starter omg!
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- #1

omg!

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could someone give me a quick answer on the exact conditions for the hamiltonian to be non degenerate, i.e. to have different eigenvalues?

thanks in advance.

- #2

CompuChip

Science Advisor

Homework Helper

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What conditions do you need on a matrix to get different eigenvalues? (Does it even exist?)

- #3

omg!

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What conditions do you need on a matrix to get different eigenvalues? (Does it even exist?)

I have found a proof of:

In a system where there are only bound states, i.e. states that vanish at infinity, there can be no degeneracy

does anybody know more general conditions?

- #4

comote

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the identity matrix has repeated eigenvalues.

I don't think your statement about systems with no scattering states is true. Consider a Harmonic oscillator in dimension n > 1. There is angular momentum eigenstates, this is a form of degeneracy and the Harmonic oscillator does not have any scattering states.

I don't think your statement about systems with no scattering states is true. Consider a Harmonic oscillator in dimension n > 1. There is angular momentum eigenstates, this is a form of degeneracy and the Harmonic oscillator does not have any scattering states.

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- #5

lbrits

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I agree with comote. There's also Kramers theorem.

- #6

omg!

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i think the statement is only true for 1-dim. systems. the proof is written for the 1-dim. case., sorry about the imprecision.

so there can't be any magnetic fields.

- #7

comote

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As for spin-less, time independent Hamiltonians in one dimension, I hesitantly believe the statement although I wouldn't be too surprised if someone provided a counter-example. Would you be able to provide the proof of that statement? At least a link or a reference.

- #8

omg!

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if we have

$-\frac{d^2}{dx^2}\psi_1+V(x)\psi_1=E\psi_1$

$-\frac{d^2}{dx^2}\psi_2+V(x)\psi_2=E\psi_2$

then we obtain after multiplying $\psi_i$ and substracting one equation from the other:

$-\psi_2\frac{d^2}{dx^2}\psi_1+\psi_1\frac{d^2}{dx^2}\psi_2=0$

so that

$\frac{d}{dx}\left(-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2\right)=0$

thus

$-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2=C$.

Since our states fall off to 0 at $\pm\infty$, we deduce that $C=0$

Now comes the step that i don't understand:

Shankar goes on with

$\frac{1}{\psi_1}\frac{d}{dx}\psi_1=\frac{1}{\psi_2}\frac{d}{dx}\psi_2$

and integrates this equation to

$\log(\psi_1)-\log(\psi_2)=c$

and therefore

$\psi_1=e^c\psi_2$,

and that supposedly concludes the proof.

- #9

lbrits

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- #10

omg!

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if we have

[tex]-\frac{d^2}{dx^2}\psi_1+V(x)\psi_1=E\psi_1[/tex]

[tex]-\frac{d^2}{dx^2}\psi_2+V(x)\psi_2=E\psi_2[/tex]

then we obtain after multiplying $\psi_i$ and substracting one equation from the other:

[tex]-\psi_2\frac{d^2}{dx^2}\psi_1+\psi_1\frac{d^2}{dx^2}\psi_2=0[/tex]

so that

[tex]\frac{d}{dx}\left(-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2\right)=0[/tex]

thus

[tex]-\psi_2\frac{d}{dx}\psi_1+\psi_1\frac{d}{dx}\psi_2=C[/tex].

Since our states fall off to 0 at [tex]\pm\infty[/tex], we deduce that [tex]C=0[/tex]

Now comes the step that i don't understand:

Shankar goes on with

[tex]\frac{1}{\psi_1}\frac{d}{dx}\psi_1=\frac{1}{\psi_2}\frac{d}{dx}\psi_2[/tex]

and integrates this equation to

[tex]\log(\psi_1)-\log(\psi_2)=c[/tex]

and therefore

[tex]\psi_1=e^c\psi_2[/tex],

and that supposedly concludes the proof.

- #11

lbrits

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What part do you have trouble with?

- #12

omg!

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- #13

omg!

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more evidently, the equation with the logarithms doesn't hold when the wave function is negative! so i guess shankar is wrong? (this proof can be found on page 176).

edit: i just remembered that the logarithm can have different branches where negative values of the wavefunction (that can be chosen real) would be well-defined (complex analysis was a loong time ago). but anyway, I'm sure there's something wrong with shankar's proof. for one reason, since i couldn't find it in many other popular QM books!

edit: i just remembered that the logarithm can have different branches where negative values of the wavefunction (that can be chosen real) would be well-defined (complex analysis was a loong time ago). but anyway, I'm sure there's something wrong with shankar's proof. for one reason, since i couldn't find it in many other popular QM books!

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- #14

lbrits

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- #15

omg!

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i'm unfamiliar with this method, but wikipedia says that only the reverse of your statement is true in general.

- #16

lbrits

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Normally, when you patch together solutions in different regions (where potentials are defined piecewise), you have matching conditions which ensure that the wavefunction is continuous, and that the discontinuity in [tex]\psi'(x)[/tex] is given by integrating the Schrodinger equation across that point. Thus, in 1D, you specify the WF and it's derivative at 1 point (remember, the SE is a second order equation), and the matching conditions uniquely give you the wavefunction everywhere else.

Therefore, you won't find yourself in a situation where there is an ambiguity in terms of how to patch up the solutions. It is unique.

- #17

omg!

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Normally, when you patch together solutions in different regions (where potentials are defined piecewise), you have matching conditions which ensure that the wavefunction is continuous, and that the discontinuity in [tex]\psi'(x)[/tex] is given by integrating the Schrodinger equation across that point. Thus, in 1D, you specify the WF and it's derivative at 1 point (remember, the SE is a second order equation), and the matching conditions uniquely give you the wavefunction everywhere else.

Therefore, you won't find yourself in a situation where there is an ambiguity in terms of how to patch up the solutions. It is unique.

ok i think i understand your argument using basic ode theory together with the statement on wikipedia that the functions must be linearly dependent in some region, if their wronskian vanishes everywhere. thank you very much for clarifying this point!

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