Is energy of a free particle observable?

[peteratcam]

Define |e>

E|e>=e|e>

Regards.

At this point, I give up.

 

[SpectraCat]

Thanks peteratcam, SpectraCat.

Maybe it's good to confirm the points we share:
***********************************
E: energy operator
P: momentum operator

eigenset of E: all {|e>} that satisfy the equation E|e>=e|e>.

Ok, fine, but there are an infinite number of equations that satisfy that relation for any single value of e. So you can't use those vectors as an expansion basis

subspaces specified by different eigenvalue (eigensatate) are orthogonal.
<e'|e>=δ(e-e'), <p'|p>=δ(p-p').

Nope this is wrong ... or at least there exist |e> vectors that satisfy your "eigenset" definition that are not orthogonal.

degeneracy
<p|e>=<-p|e>=0 for e≠p^2/2m
<p|e>+<-p|e>=1 for e=p^2/2m

I have no idea what you are trying to say with the above relation. It happens to be true for certain choices of |e> as you defined it originally. So what? There are an infinite number of cases where it does not hold.

A form a complete set {|a>} :
{|a>} satisfy the equation A|a>=a|a> and
projection { ∫da and/or Σa }|a><a|=I identity operator.

This is not correct, you are missing the orthonormality condition. I showed you the proper way to express what you are trying to express in at least one of my previous posts.

Go back and read them until you understand them .. we are not getting anywhere with this.

 

[sweet springs]

Hi, SpectraCat.

Ok, fine, but there are an infinite number of equations that satisfy that relation for any single value of e. So you can't use those vectors as an expansion basis

As you are well aware eigenset of coordinate {|x>} or eigenset of momentum {|p>} is another popular example of expansion basis with continuous so infinite number of eigenvalues.

Nope this is wrong ... or at least there exist |e> vectors that satisfy your "eigenset" definition that are not orthogonal.

There cannot exist. As you are well aware for eigenset of Hermitian H: all {|h>} that satisfy the equation H|h>=h|h>, <h'|h>=δ(h-h') for continuous eigenvalue or δh'h discrete case, otrthogonality.

It happens to be true for certain choices of |e> as you defined it originally. So what? There are an infinite number of cases where it does not hold.

Sorry, I found my mistake in the formula.
<p|e>+<-p|e>=1 should be replaced to |<p|e>|^2+|<-p|e>|^2=1
Then for any choices of |e>, say α|p>+β|-p>.
|<p|e>|^2+|<-p|e>|^2=α*α +β*β=1.

This is not correct, you are missing the orthonormality condition.

This is the definition of eigenvectors forming a complete set. Your improvement, if there is, should be appreciated.

Readers, I would like to know your reply to the question "Do eigenvectors of E form a complete set?"
Mine is "NO".

Regards.

 

[SpectraCat]

Hi, SpectraCat.
As you are well aware eigenset of coordinate {|x>} or eigenset of momentum {|p>} is another popular example of expansion basis with continuous so infinite number of eigenvalues.

But that is not what I said. I said there are an infinite number of eigenvectors satisfying your "eigenset" definition for a *single* eigenvalue of E.

There cannot exist. As you are well aware for eigenset of Hermitian H: all {|h>} that satisfy the equation H|h>=h|h>, <h'|h>=δ(h-h') for continuous eigenvalue or δh'h discrete case, otrthogonality.

Read more carefully and think about what you are saying ... you are assuming that you are right and we are wrong. I assure you this is not the case, we have provided numerous mathematical explanations of why this is so. What you wrote above is not a true statement. I can provide any number of counter-examples .. can you think of one?

Sorry, I found my mistake in the formula.
<p|e>+<-p|e>=1 should be replaced to |<p|e>|^2+|<-p|e>|^2=1
Then for any choices of |e>, say α|p>+β|-p>.
|<p|e>|^2+|<-p|e>|^2=α*α +β*β=1.

I don't understand how you can write the above statement, and still not understand our other points. Is the |e> vector you gave above orthogonal to |p> and |-p>? Do the three vectors |e>, |p> and |-p>, above all satisfy the "eigenset" definition for E? Do you see your mistake yet?

This is the definition of eigenvectors forming a complete set. Your improvement, if there is, should be appreciated.

No it is not. I have already given you the correct definition in another post. You go find it .. I will not repeat myself any more.

Readers, I would like to know your reply to the question "Do eigenvectors of E form a complete set?"

Well, that is the first time you asked *that* particular question, so at least we are making progress, even if it is agonizingly slow. I would say that it is possible to choose a sub-set of the eigenvectors of E that form a complete basis. In fact, there are an infinite number of subsets that can be chosen as complete bases. One such choice, as I pointed out in my original answer, is the eigenvectors of the momentum operator.

Mine is "NO".

And you are WRONG. Since we can find a sub-set of eigenvectors of E that is complete, the set of all eigenvectors of E must also be complete, although there are linear dependencies among some of the members.

 

[sweet springs]

Hi. SpectraCat.

I said there are an infinite number of eigenvectors satisfying your "eigenset" definition for a *single* eigenvalue of E.

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.

What you wrote above is not a true statement. I can provide any number of counter-examples .. can you think of one?

But this is mathematics of Hermitian. And E is Hermitian.
Anyway I would like to hear you.

Is the |e> vector you gave above orthogonal to |p> and |-p>? Do the three vectors |e>, |p> and |-p>, above all satisfy the "eigenset" definition for E? Do you see your mistake yet?

I would say that it is possible to choose a sub-set of the eigenvectors of E that form a complete basis. In fact, there are an infinite number of subsets that can be chosen as complete bases. One such choice, as I pointed out in my original answer, is the eigenvectors of the momentum operator.

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.

 

[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.