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Measurements as operators

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    OK, so assuming we have a physical observable with three values, a(1),a(2) and a(3), and there are given matrices for the measurement operators M(a(1))...M(a(3)). How does one actually go about finding a(1),a(2) and a(3) given the matrices?

    3. The attempt at a solution
    These measurement operators have to be hermitian, and thus have real eigenvalues. Are the values of the three observables a(n) just the eigenvalues of the measurement operators? All three are in the same basis, by the way.
  2. jcsd
  3. Feb 22, 2009 #2
    what? what are you asking? how do you find the eigenstates of the operators? how do find the expectation values?
  4. Feb 22, 2009 #3
    Sorry if I wasn't clear. I don't need to know how to procedurally find eigenstates, I just want to know how to determine the actual values of the a's given the matrices of the projection operators on a. If I'm asking it the wrong way, I could just try writing the problem down verbatim.
  5. Feb 22, 2009 #4


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    write down the problem, I have no clue either what you are asking for.

    are you asking how to perform matrix multiplication with a vector?
  6. Feb 22, 2009 #5
    OK sorry for the confusion. Clearly I don't understand the problem well enough myself. Here it is:

    Suppose a physical observable takes on three values: a(1),a(2),and a(3). Further suppose that the matrices of the measurement operators for the three values are:

    1/14( 1 2 3
    2 4 6
    3 6 9 )

    1/10( 9 0 -3
    0 0 0
    -3 0 1)

    1/35( 1 -5 3
    -5 25 -15
    3 -15 9)

    with respect to some basis. If the operator A has the matrix

    A=(11 -3 3
    -3 29 -9
    3 -9 19)
    in the same basis, find a(1),a(2),and a(3).
  7. Feb 22, 2009 #6
    Just to be sure, am I still not phrasing this right, or is nobody quite sure about the answer? I want to be cooperative, so please tell me if I need to be posting more information or anything, thanks.
  8. Feb 22, 2009 #7
    sorry still no clue what you're asking. could be a few different things.
  9. Feb 23, 2009 #8


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    Sorry, what is A's relation to a(1), .. etc ?

    I think you have to find the eigenstates to A, then let the matricies operate on them.

    What kind of question is this? A teacher gave it to you?
  10. Feb 23, 2009 #9
    Well, glad to know I'm not the only one confused by this problem. It's word for word what my teacher assigned. The last part doesn't make sense to me, what's the point of A? I understand the concept of projection operators acting as measurements, so to find the values of the a's I originally thought it would just be the eigenvalues of the M(a). I think I'll just ask my professor to clarify the question...
  11. Feb 23, 2009 #10


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    Yes, that is the best, to ask him. I have NO idea what is meant by all of this.
  12. Feb 23, 2009 #11
    It doesn't help either that we have no textbook for the course. I have found absolutely no online resource to help with this question. Sometimes I think this guy just created his own physics.
  13. Feb 23, 2009 #12


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    one more guy with no quantum mechanics textbook? :O

  14. Feb 24, 2009 #13
    I am pretty sure I'm in the same class as this person...except the no-textbook thing threw me off. (My class has a textbook, albeit one the professor doesn't seem to use.) If we're not in the same class, I am having trouble with the exact same problem, which I didn't think came from a book...I just assumed he made it up. That would be a weird coincidence.

    You asked how the operator A relates to the measurement operators M(a1)...etc. Well, it does say that A is in the same basis...I wonder if that helps?

    I looked in our class notes and couldn't find any examples resembling this so I'm a bit confused.
  15. Feb 24, 2009 #14


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    What confuses me is that the poster seems to be giving matrices (operators) for each measured value when there should be one matrix for all values. The values should be distinct eigen-values of the same matrix which is the single operator corresponding to the observable in question. The way it has been phrased there must be an error somewhere or a misunderstanding.
  16. Feb 24, 2009 #15


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    OK, I've figured out what's going on here.

    An NxN hermitian matrix A has N real eigenvalues and corresponding eigenvectors,

    [tex]\sum_{j=1}^N A_{ij}x^{(\alpha)}_j = a^{(\alpha)}x^{(\alpha)}_i [/tex]

    The eigenvectors are orthogonal, and can be taken to be normalized, so that we have

    [tex]\sum_{i=1}^N (x^{(\alpha)}_i)^* x^{(\beta)}_i= \delta^{\alpha\beta} [/tex]

    Given an eigenvector, we can form a projection matrix [itex]M^{(\alpha)}_{ij}[/itex] onto that vector:

    [tex]M^{(\alpha)}_{ij}=x^{(\alpha)}_i (x^{(\alpha)}_j)^*[/tex]

    Because of the orthonormality of the the eigenvectors, each of these matrices squares to itself, and one of them times a different one is zero:

    [tex]\sum_{j=1}^NM^{(\alpha)}_{ij}M^{(\beta)}_{jk}=\delta^{\alpha\beta}M^{(\alpha)}_{ik} [/tex]

    This property is true of the M matrices you were given, so this must be what your teacher meant.

    Given the matrix A, you can find its eigenvalues and eigenvectors by the usual procedure. This gives you the three a's. You can then construct, from the three normalized eigenvectors, the three M matrices, and see if they match the ones you were given. (They do.) Then you know which eigenvalues go with which matrices.

    All much easier if you have Mathematica or something similar available ...
  17. Feb 24, 2009 #16


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    Oh, I forgot to say, you can now write the original operator as a sum over the projection matrices times the eigenvalues:

    [tex]A_{ij} = \sum_{\alpha=1}^N a^{(\alpha)}M^{(\alpha)}_{ij}[/tex]

    This is sometimes useful.
  18. Feb 24, 2009 #17


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    Ahhhh.... I think there are density operators floating around here. Let me see ...

    [tex]M(a_1) = \frac{1}{14} \left( \begin{array}{ccc}{1& 2& 3& 2& 4& 6& 3& 6& 9 }\end{array}\right) \equiv \rho_1[/tex]
    [tex] M(a_2)=\frac{1}{10}\left(\begin{array}{ccc}{9& 0& -3& 0& 0& 0& -3& 0& 1}\end{array}\right) \equiv \rho_2[/tex]
    [tex] M(a_3) = \frac{1}{35}\left( \begin{array}{ccc}{1&-5&3&-5&25&-15&3&-15&9}\end{array}\right)\equiv \rho_3[/tex]

    A = \left( \begin{array}{ccc}{11& -3& 3& -3& 29 &-9& 3 &-9& 19}\end{array}\right)[/tex]

    Note that the three "measurement matrices" are hermitian and have trace 1. I'm guessing these are the density operators for the eigen-modes so they are eigen-vectors under the action that the measurement operator on density operators.

    Since A is hermitian (but not unit trace) I will assume it is the observable in question.

    The action of A on an eigen-density operator is:
    [tex] A\rho A^\dag = \lambda \rho[/tex]
    [tex]\lambda = aa^*[/tex]
    and where [tex]a[/tex] is the vector eigen-value.

    So let's try that... Work out:
    [tex] A\rho_1 A^\dag =
    \left( \begin{array}{ccc}{11& -3& 3& -3& 29 &-9& 3 &-9& 19}\end{array}\right)
    \frac{1}{14} \left( \begin{array}{ccc}{1& 2& 3& 2& 4& 6& 3& 6& 9 }\end{array}\right)
    \left( \begin{array}{ccc}{11& -3& 3& -3& 29 &-9& 3 &-9& 19}\end{array}\right)

    You should find that you get a multiple of [tex]\rho_1[/tex] and its square root is +/- the eigen-value of A. You could find this more quickly however by just left multiplication or just by finding the eigen-vectors and eigen-values of A.

    Note you have a 3x3 observable and 3 distinct eigen-values. This means each density operator should be "sharp" i.e. formed from a single eigen-vector. So just taking the left product [tex] A\rho_k = a\rho_k[/tex] should work. (Check to see if each density operator has rows which are multiples of each other).

    This should get you there.
  19. Feb 24, 2009 #18


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    Ah someone beat me to the solution. Yes the "sharp" density operators are also projection operators and unit projectors can also be viewed as density operators. The instructor may not have been thinking "density operator" in this case.

    Well it appears the mystery is solved.
  20. Feb 25, 2009 #19


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    hats off for mr. Avodyne :biggrin:
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