Proving Non-Degeneracy in Position-Momentum Basis of Quantum Mechanics

[yogeshbua]

Homework Statement

We claim that $$\{|\mathbf{x'}\rangle\}$$ and $$\{|\mathbf{p'}\rangle\}$$ form a basis for our space. (Position and momentum basis.)
In J J Sakurai's Modern Quantum Mechanics, it is said (indirectly) that the eigenvectors of the corresponding operators, x and p, which form the sets given above, are all such that each eigen-value is non-degenerate...
What, pray, is the proof of the non-degeneracy?

Homework Equations

We know that each of the operators has a continuous spectrum.
We know that each operator is such that x|x'>=x'|x'> (and similarly for p operator.)
And that $$\int dx'|\mathbf{x'}\rangle\langle\mathbf{x'}| = I$$ , the Identity operator; the latter point being the quantification of 'completeness + orthonormality'.
All of the above three are assumptions...

The Attempt at a Solution

I tried a lot, but decided to ask you all when I noticed that when we said that the eigen set of either operator is complete, we said it is equivalent to <x'|x''> = delta(x' - x'')... which itself entails non-degeneracy! What more fundamental equations/ properties of the eigen set of the operators do we have to work with, to prove the non-degeneracy?

Modified: I'd said '...we said it is equivalent to <x'|x''> = delta(x' - x'')... which itself entails non-degeneracy...'
I'm sorry. This statement is NOT equivalent to completeness... I repeat, this statement is NOT equivalent to completeness.
Orthonormality and non-degeneracy, together, are sufficient to imply this statement... Completeness doesn't get into the picture.

Please correct anything you think is wrong...
Thanks.

 

[Dick]

You already said it in the first line. Position space is DEFINED as the basis for wavefunctions. So the state vector for a wavefunction f(x) is integral f(x')|x'>dx'. If the x operator were degenerate that would mean there were two different states |x1> and |x2> with the same eigenvalue c. I.e. x|x1>=c|x1> and x|x2>=c|x2>. But x|x1>=x1|x1> and x|x2>=x2|x2>. So c=x1=x2. The states are DEFINED to be nondegenerate. You don't have to prove it.

 

[yogeshbua]

You already said it in the first line. Position space is DEFINED as the basis for wavefunctions. So the state vector for a wavefunction f(x) is integral f(x')|x'>dx'. If the x operator were degenerate that would mean there were two different states |x1> and |x2> with the same eigenvalue c. I.e. x|x1>=c|x1> and x|x2>=c|x2>. But x|x1>=x1|x1> and x|x2>=x2|x2>. So c=x1=x2. The states are DEFINED to be nondegenerate. You don't have to prove it.

So may we say that 'We assume the eigenvalues are nondegenerate'.
Or does non-degeneracy follow from the assumption that the eigen-ket set is a basis? (It should not, for one can have an eigen-basis which includes degenerate eigen-values; at least for finite dimensional spaces. Here, it's an infinite dimension space. So what is it that happens differently in case of infinite dimensions that makes the eigenvalues non-degenerate? Or, is it an assumption?)

Oh! And I forgot... Thank you!

Another thing... I accept that c=x1=x2. But this does NOT imply that |x1>=|x2> unless we have assumed non-degeneracy... Is this what you're trying to tell? That the mere fact that we use the notation |x1> and |x2> entails non-degeneracy? Otherwise, we couldn't have labelled them as |x1> and |x2>?

Thanks again...

 

[Dick]

Yes, I guess you could say it's implicit in the notation. The states are labeled by their eigenvalues. I suppose you could construct an extended space by artificially adding another label |x,n>, so now you could say x|x1,n1>=x1|x1,n1> and x|x1,n2>=x1|x1,n2> so |x1,n1> and |x1,n2> are degenerate. But why? We are trying to define what position space is. Not some exotic extension.

BTW, once you've defined position space the nondegeneracy of momentum eigenvectors also follows from a unique solution to an ODE, right? You could also work the other way around, by defining momentum space first and deriving the position eigenvectors.

 

[yogeshbua]

Yes, I guess you could say it's implicit in the notation. The states are labeled by their eigenvalues. I suppose you could construct an extended space by artificially adding another label |x,n>, so now you could say x|x1,n1>=x1|x1,n1> and x|x1,n2>=x1|x1,n2> so |x1,n1> and |x1,n2> are degenerate. But why? We are trying to define what position space is. Not some exotic extension.

BTW, once you've defined position space the nondegeneracy of momentum eigenvectors also follows from a unique solution to an ODE, right? You could also work the other way around, by defining momentum space first and deriving the position eigenvectors.

Agreed... Thank you, again.
Do I need to mark the question as solved/ whatever? How?
Cheers.

 

[Dick]

Agreed... Thank you, again.
Do I need to mark the question as solved/ whatever? How?
Cheers.

You're welcome! It might be under 'Thread Tools', I'm not sure, I haven't done it. If you can't find it, don't worry about it. It's not obligatory.