Is energy of a free particle observable?

[SpectraCat]

Hi. SpectraCat.
Yes, I have been in accord that eigenstate |e>=α|p>+β|-p> where |α|^2+|β|^2=1, not a "vector" but a "plane" 2 dimensional complex "eigenplane". Maybe word eigenvector was misleading. Eigenstate or eigensubspace is more correct use of word.But this is mathematics of Hermitian.

No .. that is wrong, you evidently don't know the definition of Hermetian.

Your favorite two-state example from above is a perfect example of a vector that is linearly dependent on, and therefore not orthogonal to, the other eigenvectors |p> and |-p>. I pointed this out to you in my last post ... you just refuse to read my posts carefully and think this through. I am almost starting to think that you are intentionally baiting me with your apparent obtuseness.

To choose independent specific "vectors" from "plane", you need help of other operator than E, P for example.
I call it eigenvectors of P form a complete set but eigenstates or eigensubspaces of E do not form a complete set.

We are in progress now.
Regards.

Not really .. you are still using the same old flawed logic. You are saying things that make no sense and are inconsistent with other statements that you made yourself. I have pointed out to you that there is a subset of eigenvectors of H (I am tired of you calling the Hamiltonian operator E ... the well established symbol is H), which is complete. You have agreed that is complete. That should end the discussion .. I'll try one last time.

1) The set of eigenvectors of the momentum operator p form a complete basis
2) The eigenvectors of p are also eigenvectors of H
3) Therefore, there exist a set of eigenvectors of H that form a complete basis

QED ... it does not matter that, due to degeneracy, there exist an infinite number of other eigenvectors of H that are not part of that complete basis. Once a basis is complete, it is complete forever and can be used for resolution of the identity.

Do yourself (and us) a favor and look up the definitions of the following terms in the context of Q.M. ... I believe that you do not know their proper definitions and you are definitely not using them correctly.

Hermetian, orthogonal, complete basis, linear-dependence, resolution of the identity

 

[sweet springs]

Hi.

Thanks for teachings. Let me write my understanding now.

*************************
Operator whose eigenvectors form a complete set is observable.

Here, the conditon ★ each basis must be labelled by different eigenvalue ★　does not apply.

example 1 energy of free particle P^2/2m, P^2/2m ψ= p^2/2m ψ
There are two independent ψs that are labelled by the same eigenvalue p^2/2m.

example 2 Identity operator, Iψ=1ψ
There are as many as dimensions of Space independent ψs that are labelled by the same eigenvalue 1.

Considering there are d(e) indepentdent basis labelled by eigenvalue e, projection to bases is
{∫de and/or Σe｝Σi:from 1 to d(e) ｜e_i><e_i| = I
*************************

I was haunted by the idea ★. Thanks a lot.

 

[sweet springs]

Hi. I write down my understanding again.
*******
Operator E is observable when direct sum of eigenspaces labeled by e of Eψ=eψ compose the whole space.
Space = subspace(e') + subspace(e") + ..., where subspace(e') is composed of {ψ| Eψ=e'ψ} so called eigenspace. one eigenspace is orthogonal to the others.
Probability to get obeserved value e' of E is length of the projection of the state on the eigenspace(e') squared.
*******
Regards.

 

[SpectraCat]

Hi. I write down my understanding again.
*******
1) Operator E is observable when direct sum of eigenspaces labeled by e of Eψ=eψ compose the whole space.

2) Space = subspace(e') + subspace(e") + ..., where subspace(e') is composed of {ψ| Eψ=e'ψ} so called eigenspace. one eigenspace is orthogonal to the others.

3) Probability to get obeserved value e' of E is length of the projection of the state on the eigenspace(e') squared.
*******
Regards.

I don't know ... this doesn't look quite right to me. In point 1, I don't know what the words "eigenspace labeled by e" means. I also don't think that your statement has anything to do with it being "an observable". The concept of an observable in quantum mechanics is connected to one of the fundamental postulates, which says that any physically measurable quantity (i.e. observable) has a corresponding operator within the framework of Q.M. There are various theorems based on this postulate (and others) which prove that such operators are Hermetian, and have a complete basis of eigenstates. You should look them up in a standard Q.M. text. In other words, the completeness of the basis set for *any* operator corresponding to a physical observable is complete *by definition* ... it doesn't need to be proved.

I think you also still have significant confusion between the resolution of the identity, which is valid for any complete basis of states, and "the set of all of the eigenstates" of an operator, which allows for degeneracies, and therefore may contain states that do not satisfy an orthogonality condition, which is (usually) required for resolution of the identity.

In point 2, I am not sure what you mean when you say an eigenspace "is orthogonal to" another eigenspace, unless it is the trivial case where all the vectors in one space are orthogonal to all of the vectors in the other space. Finally, I don't see how 3 can possibly be correct ... first of all, I don't know what you mean by "projection of the state on the eigenspace" ... is that the sum of the projections of the state on all the eigenvectors in the space? If so, point 3 definitely seems wrong.

 

[sweet springs]

Hi. thanks for comment.

In point 1, I don't know what the words "eigenspace labeled by e" means. I also don't think that your statement has anything to do with it being "an observable".

"eigenspace labeled by e" is subspace composed of ｛ψ}　that satisfy Eψ=eψ.
I expect that statement #37 and #38 on observable are equivalent although the latter is not popular. It is just an expectation and I have not seen a mathematical proof.

In point 2, I am not sure what you mean when you say an eigenspace "is orthogonal to" another eigenspace, unless it is the trivial case where all the vectors in one space are orthogonal to all of the vectors in the other space. Finally, I don't see how 3 can possibly be correct ... first of all, I don't know what you mean by "projection of the state on the eigenspace" ... is that the sum of the projections of the state on all the eigenvectors in the space? If so, point 3 definitely seems wrong.

"a subspace is orthogonal to other subspace" means any vector in an subspace is orthogonal to any vector in other subspace. For example in XYZ space, a subspace Z axis is orthogonal to the other subspace XY plane. It means that any position vector starting from origin along Z axis, (0,0,z) is orthogonal to any position vector starting from origin on XY plane, (x,y,0). Projection of a position vector starting from origin (x,y,z) to subspace XY plane gives the projected length sqrt(x^2 +y^2)=sqrt(x'^2+y'^2) where x' and y' are coordinates of different chosen X and Y axis. Maybe these are trivial things to you.
Regards.

PS In http:　//www.quantiki.org/wiki/index.php/Observables_and_measurements
Observables with degenerate spectra was helpful to write this memo.

 

[sweet springs]

PS^2 There was confusion of eigenspace with eigenvector in my previous posts.
Thanks again for your teachings.