# Maximum # of mutually orthonormal eigenvectors

1. Aug 19, 2004

### RedX

The maximum # of mutually orthonormal eigenvectors of a Hermitian operator must be equal to 2*(number of coordinate basis vectors), where the 2 is for say spin 1/2. Now when solving the eigenvalue problem for the Hamiltonian operator, one of the boundary conditions is that the vector obtained belongs to the Hilbert space (normalizable to unity or Dirac delta). So we discard all the vectors which don't belong to the Hilbert space. However, because of this, the space spanned by our energy eigenvectors is now smaller than 2*(number of coordinate basis vectors), so our world evidently has shrunk - some combinations of position and spin cannot be obtained with linear combinations of the energy eigenvectors that we kept.

So if you take the free particle without spin, which has 2 eigenvectors for each energy value greater than 0, then say the dimension is 1+2*infinity(E>0). Energies less than 0 are tossed out because they don't belong to the Hilbert space.

Now put it in a square well. The eigenvalues become discrete at energies below the wall potential. So you'd think that the space spanned by the eigenvectors of this Hamiltonian would be smaller than the free-particle case. However, now energies below 0 are admissible, but those energies are still discrete.

My question is are the spaces spanned by the eigenvectors of both cases equal in dimension?

Apologies if what I'm saying doesn't make sense (I'm self-studying with Shankar's book).

2. Aug 19, 2004

### zefram_c

Eh? I'm not sure I understand. I also don't see much point for comparing the Hilbert spaces of two different problems - all that's important is to get a complete set to span the solution space to a particular problem. Still, I suppose the answer to your question is that the spaces are not equal in dimension. The free particle allows an uncountable solution set, and the particle in the well allows only a countable solution set. So the HS for the free particle has a higher dimensionality than the HS of a square well.

3. Aug 19, 2004

### RedX

infinities

I think the dimensionality of the Hilbert space has to be the same for both cases - the potential well and the free particle - and has to be equal to the the dimensionality of the set of position eigenvectors (barring spin).

But for the potential well you get discrete spectrum for energies less than the wall potential, but you get your continuum for energies above the wall potential (also if you lower the potential wall to a negative value you can get discrete negative energies), so the dimensionality of the Hilbert space goes to infinity.

For the free particle, you get a continuum for nonnegative energies and the dimensionality of the Hilbert space goes to infinity.

I'm not sure if you can equate two infinities.

4. Aug 19, 2004

### zefram_c

Comparing infinities is a tricky business, but the mathematicians here can assure you that a continuum has a higher cardinality than a countable (discrete) set.

5. Aug 24, 2004

### RedX

Actually, I'm using Shankar's book, and he makes this cryptic remark without proof (which is quite unsual for the book):

Assume, just as in finite dimensions, that eigenfunctions of a Hermitian operator form a complete basis, holds in the Hilbert space.

Now a complete basis would be for an electron with no spin the set of position eigenfunctions - no matter what the Hamiltonian.

So I think the dimensionality of the span of the Hamiltonian eigenfunctions, even when restricted to Hilbert space, equals the dimensionality of the span of the position eigenfunctions.

Shankar's comment was made with such a whisper that I believe I'm not the only one who read it without thinking.

6. Aug 24, 2004

Staff Emeritus
Shankar means that even where the distribution of states is continuous, as in the set of possible positions, there is a basis, and you can write any position state as a sum of basis components. Only now the sum will be an integral, the index on the bases will be a measure, and the basis components will be functionals. But it's still linear.

7. Aug 24, 2004

### RedX

But at the end he tacks on "holds in the Hilbert space". For example, when finding the momentum eigenstates:

P|p>=p|p>
<x|P|p>=p<x|p>
-ih dQ/dx = p Q
Q=(h^2/2Pi)^.5 * exp(-ip x/h) (or whatever the coef to get Dirac delta normalization)

In order to keep Hilbert space, we require p is real.

But even when we throw out imaginary p, the momentum eigenfunctions can span anything that the position eigenfunctions can?

Last edited: Aug 24, 2004
8. Aug 25, 2004

### zefram_c

Yes, the momentum basis is just as good as the position basis. In fact you can convert between the two via Fourier transforms. I'm surprised this is not in your text.

9. Aug 25, 2004

Staff Emeritus
Shankar builds up gradually, he wants you to couple to the material as he presents it, rather than solve problems with material he hasn't got to yet. This can be annoying if you already know some QM.

10. Aug 25, 2004

### RedX

Yes, the momentum basis is just as good as the position basis. In fact you can convert between the two via Fourier transforms. I'm surprised this is not in your text.

Right, if the momentum basis is complete, then the identity relation:

Integral ( |p><p| dp)

holds, and

Q(x)=<x|Q>= Integral (<x|p><p|Q> dp)=(some constant)*Integral (e^(ipx/h)*Q(p)dp)

which is a Fourier transform (anyone recommend a good "math for physicists book"?).

But for the identity relation to hold, the basis must be complete, and I know for sure that they're complete if we include imaginary momentum eigenvalues, but evidently they're also complete even if only consider real momentum
eigenvalues. Actually I remember now that Shankar made the comment that when you go to an infinite vector space, Hermicity also depends on the elements of the vector space, not just that all operators are their adjoints.

As for how Shankar teaches, I looked at Sakurai, and it looks tougher. To me Shankar is like Feynman's QM book lectures #III, relaxed, but more rigorous than Feynman. I made the mistake of reading Feynman #3 as my first QM book.

11. Aug 26, 2004

### Eye_in_the_Sky

Posing your question in terms of the "MAGNITUDE of dimensionality" is not well-defined. But posing it in terms of the "SPAN of eigenvectors" is well-defined.

So here are two distinct questions based on your question:

(Q1) Do the eigenkets of the Hamiltonian in each of the two scenarios span the same Hilbert space?

(Q2) Do the eigenkets of the Hamiltonian in either scenario span the "full" Hilbert space L2(R) of square-integrable functions RC?

(A1) Yes.

(A2) Yes. Both do.

And here is another question you might like to ask:

(Q3) What about the eigenkets of the Hamiltonian for a simple harmonic oscillator, whose spectrum of eigenvalues is completely discrete – do those eigenkets span the "full" Hilbert space L2(R)?

(A3) Yes.

So, now comes Shankar's remark:
Now, those last two words "Hilbert space" are obviously a reference to the infinite-dimensional case. And that, I find rather curious. What does Shankar purport – that the finite-dimensional case is not also called by the name "Hilbert space"?

I am not familiar with Shankar's book. Have you provided an exact quotation, or just a (sloppy) paraphrase? (Also, I find the syntax of that sentence to be somewhat disturbing.)

Nevertheless, if Shankar has already defined "Hermitian operator" and has previously asserted that the eigenvectors of such an operator are complete in the finite-dimensional case, then Shankar's remark is not at all "cryptic". ... Shankar is merely telling the reader to make an assumption. However, a reader (like you, yourself) may be 'troubled' by having to make such an assumption without being given any grounds whatsoever for its validity.

Now, what if Shankar had put it this way?

In the infinite-dimensional case, we are forced to reconsider and refine our definition of "Hermitian". This leads to the similar but more restrictive concept of a linear operator which is "self-adjoint". By theorem, it turns out that any such operator will have a complete set of eigenvectors, and as it happens, the Hermitian operators of quantum mechanics are (almost) exclusively self-adjoint.

Would that have been any better? ... If so, does it suffice? ... Perhaps an additional comment with regard to 'exceptions' would make it sufficient. (I am now thinking about the composition of such a comment.)

12. Aug 26, 2004

### RedX

Wow, thanks everyone, especially eye_in_the_sky.

Your questions #1 and #2 is what I suspected, but the proof wasn't in the book.

As for your question #3, I haven't gotten to the harmonic oscillator yet, but since V->infinity as |x|->infinity, you will get all discrete energy levels. Ah but you definitely have to have an infinite amount of eigenfunctions to span all square-integrable functions.

Now, those last two words "Hilbert space" are obviously a reference to the infinite-dimensional case. And that, I find rather curious. What does Shankar purport – that the finite-dimensional case is not also called by the name "Hilbert space"?

Shankar defines the Hilbert space as the space of all vectors normalizable to unity or Dirac delta, so that does not exclude finite dimensions.

I am not familiar with Shankar's book. Have you provided an exact quotation, or just a (sloppy) paraphrase? (Also, I find the syntax of that sentence to be somewhat disturbing.)

That's pretty much what Shankar said. I did a little editing, but here's the real quote:

We will assume that the theorem proved for finite dimensions, namely, that the eigenfunctions of a Hermitian operator form a complete basis, holds in the Hilbert space.

Now, what if Shankar had put it this way?

In the infinite-dimensional case, we are forced to reconsider and refine our definition of "Hermitian". This leads to the similar but more restrictive concept of a linear operator which is "self-adjoint". By theorem, it turns out that any such operator will have a complete set of eigenvectors, and as it happens, the Hermitian operators of quantum mechanics are (almost) exclusively self-adjoint.

Would that have been any better? ... If so, does it suffice? ... Perhaps an additional comment with regard to 'exceptions' would make it sufficient. (I am now thinking about the composition of such a comment.)

That's much better. But still the trouble that we discard possible eigenvectors because they are not normalizable to unity or Dirac delta. Remember the momentum eigenfunctions e^(ipx/h)? Well, if p is imaginary, then e^(ipx/h) is still a solution to the eigenvalue problem, but we are not using it because it's not Hilbert. Can it still span all square-integrable functions? That's what I would add, that the answer is yes it can!