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Hermite but not observable

  1. Feb 12, 2010 #1
    Hi.
    I know that operator of observable is Hermitian. Reciprocally, any Hermitian is observable? If not, showing some examples are appreciated.
    Regards.
     
  2. jcsd
  3. Feb 12, 2010 #2
    No, for example consider the identity operator. Certainly Hermitian, but if it were an observable then that would imply that we can directly observe the probability density.
     
  4. Feb 12, 2010 #3

    SpectraCat

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    The identity matrix is one example, so is any (real) constant times the identity matrix.

    Another example is the so called "number operator" [tex]aa^{\dagger}[/tex], which is the product of lowering (annihilation) and raising (creation) operators. It is called the number operator because it returns the quantum number of the state it is applied to.
     
  5. Feb 12, 2010 #4
    Hi. Thanks Monocles and SpectraCat.
    Definition of observable in Dirac II-10 is "We call a real dynamical variable whose eigensates form a complete set an observable."
    Identity operator satisfy eigenvalue eauation Iψ=1ψ and these {ψ} form a complete set.
    Thus identity operator is observable. Am I wrong?
    Regards.
     
  6. Feb 12, 2010 #5
    I disagree. The identity operator is 1 in the case of a particle existing and 0 in the case of it not existing. Therefore, it is an observable value. A value being observable doesn't mean that its entire distribution is observable. Only eigenvalues and average values need to be observable. In the case of the identity operator, every state has the same eigenvalue, so the operator doesn't tell you much.

    The number operator is also observable since it is H/(hw)-1/2, and that is a measurable value for the quantum harmonic oscillator.

    ----

    I'd be hard pressed to say that every Hermitian matrix in Quantum Mechanics corresponds to something measurable... But if a hermitian matrix appears as part of the Hamiltonian, then I'd say it's something observable.

    Basically, if you can build a device that depends on the value of an operator, then that operator can be said to be "observable", either directly or indirectly.
     
  7. Feb 13, 2010 #6
    I agree with LukeD.

    Also if [tex]A[/tex] and [tex]B[/tex] are hermitian, [tex]AB+BA[/tex] and [tex]i(AB-BA)[/tex] are also hermitian. If, say, [tex]A=\hat{x}[/tex] and [tex]B=\hat{p}[/tex], It'd be hard to come up with the measuring devices, let alone the meaning of the observables.

    Since you read Dirac, I have to add that the postulate that, to each observable there corresponds a Hermitian operator, is motivated because of the spectral theorem that the set of all eigenvectors of a Hermitian operator forms an orthonormal basis. (I remember Dirac didn't even mention linear algebra in the book.)

    Edit: What I wanted to say in the last paragraph is that, Dirac was right to say that we need a complete set of eigenstates, but he didn't have this theorem to back it up rigorously.
     
    Last edited: Feb 13, 2010
  8. Feb 13, 2010 #7
    Hi, LukeD and Truecrimson.
    Now I tend to consider any hermitian is observable.
    Thank you.
     
  9. Feb 13, 2010 #8

    f95toli

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    The number operator is certainly an observable, there are actually quite a few ways to experimentally count the number of photons in a number state (and you don't have to do it "indirectly" by measuring the total energy of the system).
     
  10. Feb 13, 2010 #9

    SpectraCat

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    Ok, I guess I see what you are saying ... if we look for a particle, and it is there, by definition we have just applied the identity operator, so it is kind of by default included in any measurement we make. Is that another way to put it?

    Yeah, that was a mistake on my part. I was thinking in terms of building a device to measure just the quantum number, but of course if we measure the energy of an HO eigenstate, we immediately infer the quantum number.

    So, I guess I don't know any examples of Hermitian operators that don't correspond to observables. I guess what Truecrimson said about being able to compose complicated Hermitian operators out of other Hermitian operators might work, but that seems like a practical approach that misses the OP's point.
     
  11. Feb 13, 2010 #10
    Now I think I see what you mean, reminding me of what Ballentine says in his book.

    So yes, sweet springs' conclusion that "any hermitian is observable" is right in a mathematical sense.
     
  12. Feb 13, 2010 #11
    Very good points - I learned something new!
     
  13. Feb 13, 2010 #12

    Fredrik

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    If we define "observable" to mean "bounded self-adjoint operator", then of course every hermitian operator is an observable. We defined them to be. There is however another approach. We define our "observables" operationally, by describing the devices that are supposed to measure them. The question is then, what sort of mathematical object should we use to represent observables mathematically? This sort of thing is discussed in detail in books on the mathematics of QM, e.g. "An introduction to the mathematical structure of quantum mechanics", by F. Strocchi, and "Mathematical theory of quantum fields" by H. Araki.

    I don't know this stuff myself, but I get that the basic idea is to start with a C*-algebra, define a "state" as a positive linear functional on the C*-algebra of observables, and then invoke the appropriate mathematical theorems to prove that abelian C*-algebras give us classical theories and non-abelian C*-algebras give us quantum theories. (The C*-algebra is then isomorphic to the algebra of bounded self-adjoint operators on a complex separable Hilbert space).

    In this approach, it's not the case that every member of the C*-algebra corresponds to a measuring device, but I don't really have any more information on that. Perhaps someone can read those books and tell the rest of us. :smile:
     
    Last edited: Feb 13, 2010
  14. Feb 14, 2010 #13
    Thank you Fredrik!
     
  15. Feb 20, 2010 #14
    Hi,
    I have a new wondering.

    In 9.2 of my old textbook Mathematical Methods for Physicists, George Arfken states,
    ------------------------------------------------------
    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.
    ------------------------------------------------------
    The * suggests that not all Hermitian are OBSERVABLE. Can anyone suggest me some examples?
    Definition of observable in Dirac II-10 is "We call a real dynamical variable whose eigensates form a complete set an observable."
    Regards.
     
  16. Feb 20, 2010 #15

    Hurkyl

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    Wait a moment -- isn't the identity operator supposed to correspond to a "trivial" measurement that, no matter what is going on in reality, always returns the number "1"?
     
  17. Feb 20, 2010 #16
    Hi, Hurkyl

    Let me confirm my understanding.
    IΨ=1Ψ
    Any state is eigenvector of identity operator I with eigenvalue 1. I is both Hermitian and Observable.

    I am interested in some examples if any, of operators that is Hermite but not observable in the sense that whose eigenvectors do not form a complete set.

    Regards.
     
    Last edited: Feb 20, 2010
  18. Feb 20, 2010 #17

    Fredrik

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    How do you define this? Isn't it just that the identity can be expressed as [itex]\sum_a|a\rangle\langle a|[/itex] where the sum is over the eigenvalues of some operator A and the eigenvector corresponding to the eigenvalue a is written as |a>? In that case, any projection operator for a closed proper subspace will do. (Its eigenvalues are 0 or 1, so any sum over its eigenvalues would only contain two terms). Such an operator is however a mathematical representation of an operationally defined observable. (It corresponds to a device that measures A and only outputs 1 if the result is in the given range, and 0 if it's not). The simplest example is |a><a|.

    I agree with Hurkyl about the identity operator.
     
  19. Feb 20, 2010 #18
    this is an interesting post, i'd like to keep myself updated......[:biggrin:
     
  20. Feb 20, 2010 #19
    Hi, Fredrik

    Let me ask you a question in order to check if I understand your point.

    {|x>} is a complete set of eigenvectors of coordinate. ∫|x><x| dx =I. Then

    a) Operator say Q=∫x>0 |x><x| dx is Hermitian but not Observable, because its eigenvector set does not contain |x> x<0. Eigenstates of Q do not form a complete set.

    b) Q|x> = 0|x> for x<0. The subspace composed of {|x>|x<0 } is eigensubspace of Q with eigenvalue 0. So eigenstates of Q form a complete set.

    Which one (or another one?) is the right answer?

    Regards.
     
  21. Feb 21, 2010 #20

    Fredrik

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    I wouldn't define an observable that way. (I don't think anyone does). Your a) is essentially correct apart from the terminology. (I also wouldn't use the position operator as an example, because it doesn't actually have any eigenvectors. It takes some fairly sophisticated mathematics to make sense of what |x> means).

    Your b) is wrong, if the definition of of "complete set" is that the identity can be expressed as a sum of projection operators for 1-dimensional subspaces with exactly one term for each distinct eigenvalue.
     
  22. Feb 21, 2010 #21
    Hi, Fredrik
    I repeat your teachings.

    An operator is OBSERVABLE whether it has degeneracy or not. For example squqre of momentum P^2 is observable while eigenspace for eigenvalue p^2 has two-fold degeneracy of |p> and |-p>.
    We cannot (or must not?) find eigenspace for eiganvalue 0. [STRIKE]The idea that ALL THE NOT PROJECTED states degenerate and belong to eigenspace for eigenvalue 0 was wrong.[/STRIKE]

    Thanks.
     
    Last edited: Feb 22, 2010
  23. Feb 21, 2010 #22

    Fredrik

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    I don't understand that last sentece, but think of it this way: Suppose that A has a degenerate spectrum. Then we can't write [itex]\sum_a|a\rangle\langle a|=I[/itex]. But we should be able to find a second operator B that commutes with A, and find state vectors [itex]|ab\rangle[/itex] that are eigenvectors of both A and B. If we now have [itex]\sum_{a,b}|ab\rangle\langle ab|=I[/itex], then {A,B} is a "complete set of commuting observables". If not, then we should be able to find a third operator C that commutes with both A and B, and simultaneous eigenstates [itex]|abc\rangle[/itex]. We can keep finding more and more observables until all the degeneracy is removed. Let's say that the last operator we need is Q. Then our set {A,B,C,...,Q} is a complete set of commuting observables.

    I think that when this happens, there's always an observable X with non-degenerate spectrum such that all of the members in the complete set can be expressed as functions of X, e.g. [itex]A=f_A(X), B=f_B(X),\dots,Q=f_Q(X)[/itex].

    Avodyne's post here explains some of the details.
     
  24. Feb 22, 2010 #23
    You can't write that because [tex]|a\rangle\langle a |[/tex] is not defined for a degenerate spectrum. It's not just that the lhs doesn't equal the right hand side, the left hand side doesn't mean anything.

    That would be a bad definition of a complete set.
    A set (of vectors) is complete if it spans the vector space. The set of eigenvectors of an observable is complete, whether or not there is degeneracy.

    The notion of completeness with regards a complete set of mutually commuting observables is a separate one.
     
  25. Feb 22, 2010 #24
    Hi. I have further questions on eigenvalue 0.

    Here Observable means operator whose eigenstates form a complete set.

    #1
    Null Operator O
    OΨ=0Ψ
    Any state is eigenvector of identity operator O with eigenvalue 0. So O is Hermitian and Observable.

    The both discussion seem to be similar but I feel the former right and the latter wrong. Both are wrong? Can anyone show me the right way?

    #2
    Measurement of some physical variable, for example coordinate X, leads value 0. X|0>=0|0>. |0> is a specific vector, δ(x). This eigenvalue equation looks like above OΨ=0Ψ, Ψ is any vector. Don't we have to distinguish these two types of eigenvalue equations for eigenvalue 0 ?

    Regards.
     
    Last edited: Feb 22, 2010
  26. Feb 22, 2010 #25

    Fredrik

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    Yeah, I should have written something like [itex]|a\sigma\rangle[/itex] instead of just [itex]|a\rangle[/itex], and written the sum in a different way too. I was a bit sloppy there.

    I don't think I've heard anyone define a "complete set of vectors" that way. I've heard terms like "maximal orthonormal set", but not "complete set". That's why I've been asking sweet springs to clarify what he meant. Hm, I guess it would make sense to define "complete set" as one that has a maximal orthonormal set (i.e. a basis) as a subset.

    That case doesn't need to be treated separately.

    And by "complete set" you mean a set that has a basis as a subset? I think it's much better to define observables operationally, and then say that they are represented mathematically by bounded self-adjoint operators. (The set of eigenvectors of a bounded self-adjoint operator is always "complete" in the sense defined above).

    Some of the symbols you're typing don't display properly for me, on either of my two computers. It's very confusing. And as I said before, the position operator isn't the greatest example, since it isn't bounded, and doesn't have eigenvectors. You have to define a "rigged Hilbert space" just to be able to define the "eigenstates" [itex]|x\rangle[/itex].
     
    Last edited: Feb 22, 2010
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