Is energy of a free particle observable?

[sweet springs]

Is Hamiltonian of a particle in free space H=P^2/2m OBSERVABLE ?
-Yes, we can surely observe energy in some manner.
-No, ∫de|e><e| is not identical operator I, thus |e>s does not form a complete set.
As an example, energy eigenstate |e> degenerates, as |e=p^2/2m> = α|p> + β|-p>, according to the directions of momentum.
∫de|e><e|p> = α*|e=p^2/2m>　does not give |p>.
∫de|e><e|-p> = β*|e=p^2/2m> does not give |-p>.

Your advice on the correct use of the word OBSERVABLE is appreciable.
Regards.

 

[SpectraCat]

Is Hamiltonian of a particle in free space H=P^2/2m OBSERVABLE ?
-Yes, we can surely observe energy in some manner.
-No, ∫de|e><e| is not identical operator I, thus |e>s does not form a complete set.
As an example, energy eigenstate |e> degenerates, as |e=p^2/2m> = α|p> + β|-p>, according to the directions of momentum.
∫de|e><e|p> = α*|e=p^2/2m>　does not give |p>.
∫de|e><e|-p> = β*|e=p^2/2m> does not give |-p>.

Your advice on the correct use of the word OBSERVABLE is appreciable.
Regards.

Yes, there are issues with defining *time-independent* wavefunctions for free particles, because the momentum operator eigenstates are not suitable as wavefunctions, since they are not square-integrable. Since the Hamiltonian commutes with the momentum operator in the case of a free-particle, this is also true for the energy eigenstates.

Probably the best way to think about a "real" free particle is with a wave-packet, a superposition of momentum eigenstates, which can then be propagated in time by solving the time-dependent Schrödinger equation. Such a solution would not however represent an eigenstate of the time-independent Schrödinger equation (which I think is what your post is about).

Regarding observables, they correspond to the expectation values of measurable quantities. The result of any single measurement of the energy of a free-particle wavepacket will yield $$p^{2}/2m$$, where p corresponds to one of the momentum eigenstates in the superposition. However the expectation value of the energy will be the weighted average of $$p^2/2m$$ over all of the states comprising the wavepacket. Since the distribution of momentum states in the wavepacket is fixed (in the absence of a perturbing potential), for a given wave-packet the expectation value of the energy will be a constant of the motion, and so is consistent with classical physics.

 

[sweet springs]

Thanks SpectraCat for your physical analysis.
My concern is rather on mathematical definition. Operator P^2 allowing degeneracy + and - p is not complete but observable?
Regards.

 

[SpectraCat]

Thanks SpectraCat for your physical analysis.
My concern is rather on mathematical definition. Operator P^2 allowing degeneracy + and - p is not complete but observable?
Regards.

I guess I don't understand what you are getting at; "observable" is a physical concept. A postulate of QM is that all single measurements (i.e. observable quantities) correspond to eigenvalues of Hermetian operators, and the mean value of a series of measurements corresponds to the expectation value of that operator, which will in general *not* be an eigenvalue, unless the system happens to be in an eigenstate of the operator.

For a free particle, the operators $$\hat{p}$$ and $$\hat{K}=\hat{p^{2}}/2m$$ commute, therefore they share the same set of eigenstates, which is a complete set of states.

EDIT: fixed incorrect statement pointed out by peteratcam in post #7

 

[sweet springs]

I know the defitition of observable in Dirac II-10 "We call a real dynamical variable whose eigensates form a complete set an observable."
Thanks to SpectraCat P form a complete set, but P^2 doesn't due to degeneracy of + and -.
So P^2, or energy of free particle, is not observable in this definition. Correct?
Regards.

 

[kadas]

hi, coincidentally I am also confused with the concept of free particle. From what i google i found out that there is someone who wrote a contradicting statement from what griffith has written.
here's the link.

folk.uio.no/trondson/freeparticle.pdf

sorry, but i am still a beginner in quantum mechanics, but can you guys point out which one is the more correct. thanks.

 

[peteratcam]

I know the defitition of observable in Dirac II-10 "We call a real dynamical variable whose eigensates form a complete set an observable."
Thanks to SpectraCat P form a complete set, but P^2 doesn't due to degeneracy of + and -.
So P^2, or energy of free particle, is not observable in this definition. Correct?
Regards.

What are the eigenvalues of the 2x2 identity matrix?
What are the eigenvectors of the 2x2 identity matrix?
Can you form a complete basis out of eigenvectors of the 2x2 identity matrix?
What relevance do these questions have to your question?

 

[peteratcam]

A postulate of QM is that all measurements (i.e. observable quantities) correspond to expectation values of Hermetian operators.

I think I've picked you up on this one before. The result of a measurement must be an eigenvalue of the hermitian operator associated with the observable. The expectation isn't the same as the result of a measurement.

 

[SpectraCat]

I think I've picked you up on this one before. The result of a measurement must be an eigenvalue of the hermitian operator associated with the observable. The expectation isn't the same as the result of a measurement.

I don't know about before, but you definitely caught me here, thanks. I was rushing to write it before leaving work and I changed part of the message at the last minute. That's no excuse of course, and I have fixed it now. (Just for the record, I explained it correctly in my original response in post #2.)

 

[SpectraCat]

I know the defitition of observable in Dirac II-10 "We call a real dynamical variable whose eigensates form a complete set an observable."
Thanks to SpectraCat P form a complete set, but P^2 doesn't due to degeneracy of + and -.
So P^2, or energy of free particle, is not observable in this definition. Correct?
Regards.

No, not correct. The -p and +p eigenstates are distinct eigenstates. Yes, they happen to yield the same eigenvalue for the energy operator, but so what? The 2-s and 2-p orbitals of the hydrogen atom are also degenerate ... does that mean the energy of those states isn't an observable too in your analysis? I really still don't understand what you are getting at here. Are you sure you know what it means for a set of eigenstates to be complete?

 

[sweet springs]

Hi, thanks peteratcam for showing an interesting case.

What are the eigenvalues of the 2x2 identity matrix?

Ans. 1

What are the eigenvectors of the 2x2 identity matrix?

Ans. Any vectors

Can you form a complete basis out of eigenvectors of the 2x2 identity matrix?

Ans. Any vector is eigenvector, including (1,0) and (0,1) that is usual basis. Since I has only one eigenvalue, 1, two-fold degeneracy is happening. it is similar to P^2 case.

What relevance do these questions have to your question?

I cannot catch you by your subtle suggestion.
Question from me to catch you. In your view, 2x2 or any identity operator is observable?

Regards.

 

[sweet springs]

Hi. SpectraCat. I have no objection that |p>s form a complete set.
Let me state my question again. Do |e>s form a complete set where |e> is an eigenvector of Hamiltonian of a free particle?
The answer is No, isn't it? We need more than |e> to describe the state fully.
Then following the definition of Dirac book, energy of a free particle is not observable. Right?
Regards.

PS Please remind that I use the word OBSERVABLE following the definition of Dirac Book. I fully agree with you that we CAN OBSERVE 2s or 2p orbit energy.

 

[SpectraCat]

Hi. SpectraCat. I have no objection that |p>s form a complete set.
Let me state my question again. Do |e>s form a complete set where |e> is an eigenvector of Hamiltonian of a free particle?

Yes they do. The states you call |e> are IDENTICAL to the |p> eigenstates. How can the set be complete for one operator and not the other? A complete set is complete! Your logic is completely nonsensical as far as I can tell. I am using the same definition for observable that you are. You keep asserting the the set of |e>'s is incomplete, without any justification beyond your degeneracy argument that several people have pointed out to you is wrong.

You keep coming back to the same point repeatedly, when multiple posters have tried to explain why you are on the wrong track. Please think about our responses carefully and try to comprehend why our explanations are correct. If you are still confused, post a *detailed* mathematical derivation of why you think the set of energy eigenstates for a free particle is incomplete, so that we can pinpoint where your confusion lies.

 

[sweet springs]

Hi. SpectraCat.
Please be patient to coming back to my first comment.

|e=p^2/2m> = α|p> + β|-p>, according to the directions of momentum.
∫de|e><e|p> = α*|e=p^2/2m>　does not give |p>.
∫de|e><e|-p> = β*|e=p^2/2m> does not give |-p>.

∫de|e><e| does not equal to identity operator I, does it?
So we cannot say |e>s form complete set, can we?
Of course I admit |p>s or { |e,+> |e,-> } form complete set.
Is the distinction above by degeneracy is meaningless?

Regards.

 

[SpectraCat]

Hi. SpectraCat.
Please be patient to coming back to my first comment.

Ok, but this is my last reply in this thread unless you post something with new content. I'll give it one last shot. It doesn't seem like you are carefully considering what I (and others) have posted before. I have no idea what you are trying to represent with those lines you quoted from your original post ... they don't seem correct to me. I have no idea what you are trying to signify by placing an equal sign inside your ket vector description.

∫de|e><e| does not equal to identity operator I, does it?

Yes, of course it does. The eigenstates of any Hermetian operator *by definition* form a complete set. In this case, the eigenstates of the kinetic energy operator, as I keep pointing out in every post, are IDENTICAL to the eigenstates of the momentum operator.

You say you have no problem with ∫dp|p><p|=1, well then since the two sets are identical, ∫de|e><e|=∫dp|p><p|=1. I can't make it any more explicit than that.

EDIT: Well maybe I can ... consider the expansion:

|p'> = ∫de|e><e|p'> = ∫dp|p><p|p'> = ∫dp|p>delta(p-p') = |p'>, where "delta" is the Dirac delta function. That is correct (but trivial) way to use the resolution of the identity in this case.

So we cannot say |e>s form complete set, can we?

Yes, we can (see above).

Of course I admit |p>s [STRIKE]or { |e,+> |e,-> }[/STRIKE] form complete set.

Good, but I have no idea what you are trying to represent by the vectors in the bracket.

Is the distinction above by degeneracy is meaningless?

Yes! Degeneracy has no bearing on completeness of a basis .. that is what peteratcam was trying to get you to realize with his post.

 

[sweet springs]

Hi. SpectraCat. Let me write my understanding of QM.

The eigenstates of any Hermetian operator *by definition* form a complete set.

Any real physical variable forming complete set is Hermitian, but not all the Hermitian form complete set.

Momentum P form complete set and is Hermitian.

Energy of free particle P^2 is Hermitian but does not form complete set.
Here completeness means projection operator ∫de or Σe |e><e|=I where e is eigenvalue of operator.

PS1
As for Identity operator I, there's only one eigenvalue, 1, thus one eigenstate |1> so if it formed complete set, projection should be |1><1|. Projection |1><1|any vector>∝|1> thus vector does not come back to itself but fall into |1>. It means that identity operator is of course Hermitian but does not form complete set, the case of ultimate degeneracy.

Identity operator I is expressed by help of any observable E as ∫de or Σe |e><e| where e is eigenvalue of E. Similarly by held of |p>s, P^2 is expressed as ∫dp |p>p^2<p|. These examples show that I or P^2 does not form a complete set but a complete set helps representation of I or P^2.

Am I right, peteratcam? I appreciate your teaching.

PS2
I apologize my vague notion of { |e,+>, |e,-> }. Here + or - is the direction of motion, introduced parameter to dissolve degeneracy.

Regards.

 

[peteratcam]

As for Identity operator I, there's only one eigenvalue, 1, thus one eigenstate |1> so if it formed complete set, projection should be |1><1|. Projection |1><1|any vector>∝|1> thus vector does not come back to itself but fall into |1>. It means that identity operator is of course Hermitian but does not form complete set, the case of ultimate degeneracy.

This point is at the heart of your confusion. The eigenstates of the identity operator must form a complete set!
Physicists love to abuse notation. One such abuse is to label eigenstates of an operator by the associated eigenvalue. This is fine without degeneracy, but is ambiguous with degenerate eigenstates. So for example, $$|\mathbf p\rangle$$ isn't ambiguous, whereas $$|E\rangle$$ is ambiguous. You might benefit from the discussion of quantum numbers in https://www.physicsforums.com/showthread.php?t=375171. A more transparent definition would be to say: a hermitian operator A has eigenvalues $$\lambda_1,\lambda_2,\lambda_3,\lambda_4,\ldots$$ not necessarily distinct and we choose deliberately an orthonormal set of associated eigenvectors $$|a_1\rangle,|a_2\rangle,|a_3\rangle,\ldots$$.


So the point is that $$\int dE | E \rangle \langle E |$$ is meaningless because the $$|E\rangle$$ are not unambiguously defined. We could include an extra quantum number, for example the direction of momentum ( in 1D).
$$\sum_{p = +,-}\int dE | E,p \rangle \langle E,p |$$
And actually, this is still wrong. States are equally spaced in momentum space, so there should be a density of states factor when writing as an integral over energy. Continuum limits are always a bit fiddly.

Peter

 

[sweet springs]

Hi. petratcam. I almost agree with you except the point that

The eigenstates of the identity operator must form a complete set!

#1
I|any vector> = 1|any vector>
How can you generate a complete set from this relation?

#2
| any vector > is expressed by using any observable A of eigenvale a as
|any vector >= ∫da or Σa ψ(a) |a> where ψ(a)=<any vector|a>

I am afraid you are confusing #1 Generation (or Formation) of complete set from I with #2 Representation of |any vector> by a complete set generated by A not I.

Identity Operator Does Not Form or Generate a Complete Set.
I will appreciate it if you correct my mistakes by showing the way of generation or formation of a complete set from I.

If it does not generate a complete set then it is NOT OBSERVABLE according to the definition of Dirac text.

Regards.

 

[SpectraCat]

Hi. SpectraCat. Let me write my understanding of QM.



Any real physical variable forming complete set is Hermitian, but not all the Hermitian form complete set.

Energy is a real physical variable ... so your example falls into case 1 (I guess), but your language is *very* imprecise here .. I have corrected it below (my text in red).

The eigenstates of the momentum operator p form a complete set and p is Hermitian.

This means that the momentum eigenvalues are real, and the momentum eigenstates satisfy the relations:

$$\hat{\textbf{p}}\left|\vec{p}\:\right\rangle=p\left|\vec{p}\:\right\rangle$$ (eigenvalue/eigenvector relation)

$$\left\langle\vec{p}\:'|\vec{p}\:\right\rangle = \delta\left(\vec{p}\:'-\vec{p}\:\right)$$ (orthonormality)

$$\int|d\vec{p}\:\left|\vec{p}\:\right\rangle\left\langle\vec{p}\:'\right| = 1$$ (completeness, or closure)

$$\left\langle\vec{p}\:'\right|\hat{\textbf{p}}\left|\vec{p}\:\right\rangle = \left\langle\vec{p}\:\right|\hat{\textbf{p}}\left|\vec{p}\:'\right\rangle^{*}$$ (Hermicity of operator)

where $$\delta$$ is the Dirac delta function, and I have used vector notation to help distinguish eigenstates and eigenvalues.

If we apply the Hamiltonian operator for the free particle, $$\hat{\textbf{p}}^{2}/2m$$, to one of these states, it is easy to see they are also eigenstates of that operator (i.e. they are energy eigenstates). Of course the orthonormality and completeness relations still hold, since we are talking about the same states.

Energy of free particle P^2 is Hermitian but does not form complete set.
Here completeness means projection operator ∫de or Σe |e><e|=I where e is eigenvalue of operator.

No, as I told you many times, even if you corrected the language to state what you are trying to say correctly, the content is wrong. I gave you explicit justification above why it is wrong. Yet you keep claiming that it is true without any justification or proof. Why do you think the eigenstates of the free-particle energy operator do not form a complete set? Show a detailed proof.

PS1
As for Identity operator I, there's only one eigenvalue, 1, thus one eigenstate |1> so if it formed complete set, projection should be |1><1|. Projection |1><1|any vector>∝|1> thus vector does not come back to itself but fall into |1>. It means that identity operator is of course Hermitian but does not form complete set, the case of ultimate degeneracy.

Identity operator I is expressed by help of any observable E as ∫de or Σe |e><e| where e is eigenvalue of E. Similarly by held of |p>s, P^2 is expressed as ∫dp |p>p^2<p|. These examples show that I or P^2 does not form a complete set but a complete set helps representation of I or P^2.

As far as I can tell, everything you wrote above is nonsense, and indicates that you don't have a clear idea what the terms eigenstate, projection operator, observable, Hermetian operator, closure relation, or identity operator mean. I suggest that you go back to the textbook, read carefully, and work some problems so that you understand better what you are talking about. Then you should be able to phrase your question better, if you haven't already realized the answer for yourself.

 

[peteratcam]

Hi. petratcam. I almost agree with you except the point that
#1
I|any vector> = 1|any vector>
How can you generate a complete set from this relation?

That relation shows that any complete set is an eigenset of the identity operator. Any basis you can think of for a vector space V, is composed of eigenvectors of the identity.

#2
| any vector > is expressed by using any observable A of eigenvale a as
|any vector >= ∫da or Σa ψ(a) |a> where ψ(a)=<any vector|a>

No this is wrong. You are using the frequent abuse of notation I referred to previously.
Observable A has a set of eigenvalues $$a_1,a_2,a_3,\ldots$$ (not necessarily distinct) and an associated set of eigenvectors $$|1\rangle,|2\rangle,|3\rangle\ldots$$ (by our choice, necessarily orthonormal).
Any vector:
$$|\psi\rangle = \sum_{n}\psi_n|n\rangle$$
where
$$\psi_n = \langle n | \psi \rangle$$.
The eigenvalues have nothing at all to do with this expansion. However, $$\psi_n$$, does have the interpretation as an amplitude to measure $$a_n$$ (assuming non-degenerate) in an experiment.
Only in special cases is it appropriate to change the sum over labels into an integral over eigenvalues.

I am afraid you are confusing #1 Generation (or Formation) of complete set from I with #2 Representation of |any vector> by a complete set generated by A not I.

Maybe I am because I don't understand the distinction you are trying to make.

 

[sweet springs]

Hi, SpectraCat. Thanks a lot for your corrections and teachings.
I almost agree with you except :

(i.e. they are energy eigenstates).

I agree with you that an eigenstate of P with eigenvalue p belongs to an eigenstate of P^2 with eigenvalue p^2. But the reciprocal, an eigenstate of P^2 with eigenvalue p^2 does not belong to an eigenstate of P with eigenvalue p. It is a linear combination of two eigenstates of P with eigenvalues p and -p.
This shows that the eigenstates of P and the eigenstates of P^2 are not same.

I tell it you in another way. The mathematical principle says One eigenstate for One eigenvalue of Hermitian, doesn't it?
So it can't be TWO ENERGY eigenstates |p> and |-p> for ONE ENERGY eigenvalue p^2. ( I made 1/2m =1 simply.).
We should call it TWO MOMENTUM eigenstates corresponds to one ENERGY eigenvalue.

Why do you think the eigenstates of the free-particle energy operator do not form a complete set? Show a detailed proof.

I hope the above satisfy you.
Energy operator cannot distinguish degenerated two momentum states. We can't call it complete.
For completeness, we must introduce another operator MOMENTUM.

Regards.

 

[sweet springs]

Thank you peteratcam for correction. I would like to show my view different from yours just on one point :

That relation shows that any complete set is an eigenset of the identity operator.

Definition of EIGENSET of the Hermitian E is set of bra {|e>} for all e that satisfy E|e>=e|e>.
Applying this definition, eigenset of the identity operator consists of only ONE bra corresponding to e=1 that is only ONE eigenvalue of identity operator.

Maybe I am because I don't understand the distinction you are trying to make.

"E form a complete set" means the above set {|e>} is a complete set of space. P or X does, P^2 or identity operator doesn't.

Regards.

 

[peteratcam]

Definition of EIGENSET of the Hermitian E is set of bra {|e>} for all e that satisfy E|e>=e|e>.
Applying this definition, eigenset of the identity operator consists of only ONE bra corresponding to e=1 that is only ONE eigenvalue of identity operator.

No, no, no, no and no. You are wilfully ignoring the points that Spectracat and I have been making. You can't use the eigenvalue as a label for eigenstates when the eigenvalue is degenerate.

I am absolutely astounded that you have convinced yourself that the eigenvectors of the 2x2 identity matrix don't span the vector space.

 

[sweet springs]

Hi peteratcam. Thanks for your clear statement of No^5.

Again just on one point:

You can't use the eigenvalue as a label for eigenstates when the eigenvalue is degenerate.

Yes, you are right in case you request a label be complete.
Label by eigenset of P^2 or I is like saying I am in US or on Earth,very large area.
If you wish more detailed and ultimate labeling e.g. I am in xx street NY, another eigenset of OBSERVABLE operator e.g. P should be introduced, but this is another and side story.

Regards.

 

[SpectraCat]

Hi, SpectraCat. Thanks a lot for your corrections and teachings.
I almost agree with you except :I agree with you that an eigenstate of P with eigenvalue p belongs to an eigenstate of P^2 with eigenvalue p^2. But the reciprocal, an eigenstate of P^2 with eigenvalue p^2 does not belong to an eigenstate of P with eigenvalue p. It is a linear combination of two eigenstates of P with eigenvalues p and -p.
This shows that the eigenstates of P and the eigenstates of P^2 are not same.

Ok, I think I may see where your confusion arises now. Eigenstates can be degenerate, and furthermore, when you have two degenerate eigenstates, *any* linear combination of them is *also* an eigenstate. However, this does not affect the basis in question (the complete set of momentum eigenstates in this example), because the new eigenstates that you construct from the linear combinations are not *orthogonal* to the old ones. They are linearly dependent on your original set of momentum eigenstates, so you wouldn't include them in the closure relation that seems to be confusing you.

I tell it you in another way. The mathematical principle says One eigenstate for One eigenvalue of Hermitian, doesn't it?

No, it most certainly doesn't ... where did you get that idea? In fact, if you were reading our posts carefully, you would see that peteratcam told you exactly the opposite in post #17. Eigenstates don't have to have unique eigenvalues ... you already know that there can be degeneracies, so it is strange that you would make such a statement.

Energy operator cannot distinguish degenerated two momentum states. We can't call it complete.

Well, who said anything about completeness of an operator? Completeness is a characteristic describing of a set of vectors, as I indicated in post #19. And who cares what the energy operator can "distinguish"? Your problem is that you are focusing on your own linguistically-based interpretations of precisely defined mathematical terms, rather than learning what they are intended to mean in context, and using them appropriately. As I indicated earlier, you need to go to a textbook and study this stuff some more ... learn the standard language so that you can communicate effectively. This is becoming timesome.

 

[sweet springs]

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>.
eigenset of P: all {|p>} that satisfy the equation P|p>=p|p>.

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

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

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.
***********************************
If you find anything wrong or incorrect, I should appreciate your teachings.

Then I would like to restate my question.

E form a complete set or not?

Regards.

 

[peteratcam]

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>.
eigenset of P: all {|p>} that satisfy the equation P|p>=p|p>.

How many times! You continue to wilfully ignore my point that E|e>=e|e> is needlessly misleading notation when there is degeneracy. Until you engage with this point it seems pointless trying to correct any of your other claims.

 

[sweet springs]

Hi, peteratcam.

You continue to wilfully ignore my point that E|e>=e|e> is needlessly misleading notation when there is degeneracy.

Eigenvalue relation E|e>=e|e> holds both with and without degeneracy.

I would like to know your idea to write down energy eigenvalue relation in your not misleading notation.
Regards.

 

[peteratcam]

Eigenvalue relation E|e>=e|e> holds both with and without degeneracy.

Define |e>.

 

[sweet springs]

Define |e>.

E|e>=e|e>

Regards.