Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on function of operators

  1. Aug 5, 2009 #1
    I don't unterstand de function [itex]F(\hat{J})[/itex] where J is the operator
    [tex]\hat{J}=(\hat{J_1},\hat{J_2},\hat{J_3})[/tex]
    and the components of J do not commute. In case when F a function of only one component we have the definition
    [itex]F(\hat{J_1})|m>=F(m)|m>[/itex] where [itex]\hat{J_1}|m>=m|m>[/itex], but
    how do you define the action action of [itex]F(\hat{J})[/itex] on a ket of the state space?
     
  2. jcsd
  3. Aug 5, 2009 #2
    A function can be decomposed into a power series:

    [tex]F(x)=\sum_{i=0}^{\infty}{a_i x^i}[/tex]

    So you can simply plug in an operator [tex]A[/tex] to give:

    [tex]F(A)=\sum_{i=0}^{\infty}{a_i A^i}[/tex]

    Now, if you apply an eigenvector of [tex]A[/tex], such that [tex]A|m>=m|m>[/tex], you get:

    [tex]F(A)|m>=\sum_{i=0}^{\infty}{a_i A^i|m>}=\sum_{i=0}^{\infty}{a_i m^i|m>}=F(m)|m>[/tex]

    So [tex]F(A)|m>=F(m)|m>[/tex] is not really a definition, it's the result of a simplification.

    Hope that answers the question.

    Edit: Oh, I just realized that you want to know what to do when you have a vector of operators. I have no answer for this, sorry...
     
  4. Aug 5, 2009 #3
    First, you have to examine what the other operators do to an eigenstate of an operator they don't commute with. Using your notation, if you have eigenstates defined by

    [tex]\hat{J}_1 |m\rangle = m | m\rangle[/tex]
    [tex]\hat{J}_2 |n\rangle = n | n\rangle[/tex]

    and the operators J_i are Hermitian, then to evaluate

    [tex]\hat{J}_2 |m\rangle[/tex]

    you insert a complete set of states from J_2's eigenvectors in order to evaluate what J_2 does:

    [tex]\hat{J}_2 |m\rangle = \hat{J}_2 \left(\sum_n |n\rangle\langle n| \right) |m\rangle[/tex]
    [tex]\hat{J}_2 |m\rangle = \sum_n |n\rangle n\langle n|m\rangle [/tex]

    which is not too transparent, but we see that there is a contribution from each of J_2's eigenvalues proportionate to the projection of its eigenvector on the state of interest (m). So we project this back onto some other state in the basis of J_1 (to use the jargon, we are calculating matrix elements of J_2 in the basis of J_1):

    [tex]\langle m' | \hat{J}_2 |m\rangle = \sum_n n \langle m'|n\rangle \langle n|m\rangle [/tex]

    Now we see what this does; J_2 acting on some eigenstate of J_1 creates a new state which can have non-zero projections on other eigenstates of J_1, in other words, the action of J_2 does something non-trivial to an eigenvector of J_1.

    You can go through this same analysis for higher powers of J_2, you will find that you just get a higher power of the eigenvalue in the resulting expression. You can use this to the Taylor series that GPPaille mentioned for a more general operator.

    For a vector of operators, you just wind up with a vector of kets:

    [tex]F(J)|m\rangle = (J_1,J_2,J_3) |m\rangle = (J_1 |m\rangle, J_2 |m\rangle, J_3 |m\rangle)[/tex]

    (a vector of operators generally is a vector in a different space than the Hilbert space of your kets, so you just wind up with an "outer product" of these spaces, or rather, each component of your vector operator acting on a ket is its own equation that you are evaluating.)
     
  5. Aug 6, 2009 #4
    Thank you for your answers
    According to Dirac the "definition" [tex]A|m>=m|m>[/tex] is more general because it is valid even when A(x) can not be expanded in power series

    Acoording to kanato's explanation we finally have to expand the function [itex]F(J_1,J_2,J_3)[/itex] in a Taylor series , however it is still not clear to me how to do this because, it seems to me, we can express de power series expansion in many different ways, of course, when the operators J_i do commute all this "different ways" end up with the same result but, how do we choose de correct expansion when the J_i do not commute? Are we supposed to make same kind of symmetrization of the terms in the expansion?
     
  6. Aug 6, 2009 #5

    strangerep

    User Avatar
    Science Advisor

    [tex]A|m>=m|m>[/tex] just expresses a definition of eigenvector(s) and associated
    eigenvalue(s) for an operator A. Extending to [tex]F(A)|m>=F(m)|m>[/tex] (and more
    generally) has deep roots in something called the spectral theorem, actually the
    spectral representation theorem(s). If you really want to know more, try this
    as a starting point:

    http://en.wikipedia.org/wiki/Spectral_theorem

    Yes, you can express it in different ways, but usually one adopts conventions from
    the theory of universal enveloping algebra of such Lie algebras such that the basic
    algebraic elements are ordered a particular way (which is up to you, provided you're
    consistent everywhere).
     
  7. Aug 7, 2009 #6
    Thank you strangerep.
    However I don't see why standard texts make reference to such function as [itex]F(J_x,J_y,J_z)[/itex] without any explanation on this point, they only explain the meaning of functions of one opeator such as F(A). It is very frustrating indeed.
     
  8. Aug 8, 2009 #7

    strangerep

    User Avatar
    Science Advisor

    Consider an ordinary function f(x,y,z) of 3 ordinary (commuting) variables.
    Suppose I give you one specific such function:
    [itex]
    f(x,y,z) ~:=~ x + y + xy + yzx + yx
    [/itex]
    You might then say:
    "oh, that's really the same as [itex]f(x,y,z) ~:=~ x + y + 2xy + xyz[/itex]"

    My point is there's nothing particularly mysterious about a
    function of operators. You just have to work a bit harder (using
    commutation relations, etc) to figure out whether two
    seemingly-different functions are in fact the same.

    HTH.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question on function of operators
  1. Operator is it function? (Replies: 12)

  2. Operator question (Replies: 15)

Loading...