Hermite but not observable

  • #51
Hurkyl
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Well, the heuristic calculation I went through was as follows:...
Now, I will add there's I don't like something about my derivation, but I haven't managed to place my finger on it.
 
  • #52
Hurkyl
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Now, I will add there's I don't like something about my derivation, but I haven't managed to place my finger on it.
I found it -- my T wasn't unitary. I'll have to work up a better toy example.
 
  • #53
strangerep
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[...]
So let's think about what you said in the text I quoted. The definition of "subspace generated by" and the theorem imply that a vector expressed as

[tex]x=\sum_{n=1}^\infty \langle e_n,x\rangle e_n[/tex]

with infinitely many non-zero terms does not belong to the subspace generated by the orthonormal basis. :confused: That's odd. I didn't expect that.

[...]
I get the feeling we've been talking at crossed purposes. When I read something
like "X is a subspace of V", I've been tacitly assuming X the same kind of space
as whatever V is. I.e., if V is a vector space, then X is also a vector space, or if
V is a Hilbert space, then X is also a Hilbert space, etc. But I probably shouldn't
be assuming that's what you meant.

Such ambiguity is probably the source of any misunderstandings.
 
  • #54
Fredrik
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You're right, I didn't think about the fact that a "subspace" of a Hilbert space should be complete. I meant subspace in the vector space sense. I'm a bit busy right now, so I haven't had time to think about how or if that changes the stuff I said.
 
  • #55
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Hi.
Let A and B be each Hermitian and OBSERVABLE in the sense that whose eigensubspaces contain all the maximal orthogonal sets, i.e. basis.
A+B is Hermite. Is A+B OBSERVABLE? e.g. X + h' P^-1.
Can any Hermitian be diagonalized?
Regards.
 
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  • #56
Fredrik
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I assume that you're trying to say that there's a basis for the Hilbert space that only contains eigenvectors of the operator. (If you're going to talk about eigenspaces, you'll have to say that the direct sum of the eigenspaces is the entire Hilbert space).

If the above is true for A and B separately, is it also true for A+B? Yes, it is, because A+B is hermitian too.


(And for the record, I still think that's a bad definition of "observable").
 
  • #57
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Hi, Fredrik. Thank you so much.
With help of your teachings I could confirm that Hamiltonian of a particle in any artificial potential, say H=P^2/2m + V(X), has eigenstates and is OBSERVABLE whatever V(X) is.
Regards.
 
  • #58
Fredrik
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Yes. H must be hermitian because exp(-iHt) must be unitary.
 
  • #59
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Hi.
I still have a little concern on the description below. But I cannot imagine an operator that is Hermitian but not OBSERVABLE i.e. the direct sum of whose eigensubspaces is the whole space. Now I stop wondering about this subject. Thanks a lot to you all. :smile:



In 9.2, Mathematical Methods for Physicists, George Arfken
------------------------------------------------------
1. The eigenvalues of an Hermite operator are real.
2. The eigen functins of an Hermite operator are orthogonal.
3. The eigen functins of an Hermite operator form a complete set.*
* This third property is not universal. It does hold for our linear, second order differential operators in Strum-Liouville (self adjoint) form.
------------------------------------------------------
 
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