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Is energy of a free particle observable?

  1. Feb 5, 2010 #1
    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.
  2. jcsd
  3. Feb 5, 2010 #2


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    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 Schrodinger equation. Such a solution would not however represent an eigenstate of the time-independent Schrodinger 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 [tex]p^{2}/2m[/tex], 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 [tex]p^2/2m[/tex] 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.
  4. Feb 5, 2010 #3
    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?
  5. Feb 5, 2010 #4


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    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 [tex]\hat{p}[/tex] and [tex]\hat{K}=\hat{p^{2}}/2m[/tex] 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
    Last edited: Feb 5, 2010
  6. Feb 5, 2010 #5
    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?
  7. Feb 5, 2010 #6
    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.


    sorry, but i am still a beginner in quantum mechanics, but can you guys point out which one is the more correct. thanks.
  8. Feb 5, 2010 #7
    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?
  9. Feb 5, 2010 #8
    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.
  10. Feb 5, 2010 #9


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    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.)
    Last edited: Feb 5, 2010
  11. Feb 5, 2010 #10


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    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?
  12. Feb 5, 2010 #11
    Hi, thanks peteratcam for showing an interesting case.

    Ans. 1

    Ans. Any vectors

    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.

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

    Last edited: Feb 5, 2010
  13. Feb 5, 2010 #12
    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?

    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.
    Last edited: Feb 5, 2010
  14. Feb 5, 2010 #13


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    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.
  15. Feb 6, 2010 #14
    Hi. SpectraCat.
    Please be patient to coming back to my first comment.

    ∫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?

  16. Feb 6, 2010 #15


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    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.

    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.

    Yes, we can (see above).

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

    Yes! Degeneracy has no bearing on completeness of a basis .. that is what peteratcam was trying to get you to realize with his post.
    Last edited: Feb 6, 2010
  17. Feb 6, 2010 #16
    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.

    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.

    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.

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

    Last edited: Feb 6, 2010
  18. Feb 6, 2010 #17
    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, [tex]|\mathbf p\rangle[/tex] isn't ambiguous, whereas [tex]|E\rangle[/tex] 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 [tex]\lambda_1,\lambda_2,\lambda_3,\lambda_4,\ldots[/tex] not necessarily distinct and we choose deliberately an orthonormal set of associated eigenvectors [tex]|a_1\rangle,|a_2\rangle,|a_3\rangle,\ldots[/tex].

    So the point is that [tex]\int dE | E \rangle \langle E |[/tex] is meaningless because the [tex]|E\rangle[/tex] are not unambiguously defined. We could include an extra quantum number, for example the direction of momentum ( in 1D).
    [tex]\sum_{p = +,-}\int dE | E,p \rangle \langle E,p |[/tex]
    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.

  19. Feb 6, 2010 #18
    Hi. petratcam. I almost agree with you except the point that

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

    | 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.

    Last edited: Feb 6, 2010
  20. Feb 6, 2010 #19


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    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).

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

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

    [tex]\left\langle\vec{p}\:'|\vec{p}\:\right\rangle = \delta\left(\vec{p}\:'-\vec{p}\:\right)[/tex] (orthonormality)

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

    [tex]\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^{*}[/tex] (Hermicity of operator)

    where [tex]\delta[/tex] 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, [tex]\hat{\textbf{p}}^{2}/2m [/tex], 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.

    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.

    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.
    Last edited: Feb 6, 2010
  21. Feb 7, 2010 #20
    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.
    No this is wrong. You are using the frequent abuse of notation I referred to previously.
    Observable A has a set of eigenvalues [tex]a_1,a_2,a_3,\ldots[/tex] (not necessarily distinct) and an associated set of eigenvectors [tex]|1\rangle,|2\rangle,|3\rangle\ldots[/tex] (by our choice, necessarily orthonormal).
    Any vector:
    [tex]|\psi\rangle = \sum_{n}\psi_n|n\rangle[/tex]
    [tex]\psi_n = \langle n | \psi \rangle[/tex].
    The eigenvalues have nothing at all to do with this expansion. However, [tex]\psi_n[/tex], does have the interpretation as an amplitude to measure [tex]a_n[/tex] (assuming non-degenerate) in an experiment.
    Only in special cases is it appropriate to change the sum over labels into an integral over eigenvalues.
    Maybe I am because I don't understand the distinction you are trying to make.
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