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

How to determine the product of two Hermitian operators is Hermitian

  1. Oct 31, 2013 #1
    Let's say we have operator X that is Hermitian and we have operator P that is Hermitian. Is the following true:

    [X,P]=ihbar

    This is the commutator of X and P.
    This particular result is known as the canonical commutation relation.
    Expanding:
    [X,P]=XP-PX=ihbar
    This result indicates that XP[itex]\neq[/itex]PX because XP-PX[itex]\neq[/itex]0
    Because XP[itex]\neq[/itex]PX, XP is not a Hermitian operator.
    Likewise, because PX[itex]\neq[/itex]XP, PX is not a Hermitian operator.

    So to summarize:

    The commutator implies multiplication of operators
    Multiplication of Hermitian operators does not always produce another Hermitian operator.
    If two Hermitian operators do not commute, then their product is not Hermitian.

    Any mistakes here?
     
  2. jcsd
  3. Oct 31, 2013 #2

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    The product of Hermitian operators A,B is Hermitian only if the two operators commute: AB=BA.
    This follows directly from the definition of Hermitian: H*=H. Then using the properties of the conjugate transpose:

    (AB)*= B*A* = BA which is not equal to AB unless they commute.

    The relationship [X,P]=ihbar holds when X,P form a conjugate pair from Lagrangian mechanics - if X is the position operator, then P is the momentum operator conjugate to X.

    For example, see http://en.wikipedia.org/wiki/Canonical_coordinates
     
  4. Oct 31, 2013 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Actually. XP=PX+const*1 tells you 3 things directly:

    1. If the representation is made on a complex separable Hilbert space, this space is infinite dimensional.
    2. At least one of the 2 operators is unbounded.
    3. The Stone-von Neumann uniqueness theorem.

    From these 3 things one can infer that when XP and PX are defined on a common dense everywhere invariant domain of L^2 (R) such as the Schwartz space S(R), X,P are essentially self-adjoint and the operators XP and PX are not even symmetric.
     
  5. Oct 31, 2013 #4

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    I didn't understand any of that ... not how it is connected to the OP, not too clear on 1,2, 3, and not at all on the conclusions.


    And I had QM as an undergrad and in graduate school ...
     
  6. Oct 31, 2013 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    1 is a trivial result.
    2 Is a result of Wintner (1947) and Wielandt (1948).
    3 is a statement of Stone rigorously proved by von Neumann for the first time.
     
  7. Oct 31, 2013 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think this, at least partially, shows the difference between North American and European programs. I also saw none of this in undergrad or grad quantum mechanics (in Canada).

    I think your post #2 is the appropriate answer to the OP.

    For my own interest, I took loads of pure math courses not required for my physics degrees.

    Since all linear operators on finite-dimensional vector spaces are bounded, 1. follows from 2., and 2. is proved here:

     
  8. Oct 31, 2013 #7

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    None of this seems relevant to the OP ... no matter how erudite it is.
     
  9. Oct 31, 2013 #8
    UltrafastPED, thanks for your post. You are right, that is more along the lines of what I was asking. Thanks also George Jones and dexter, it's interesting to hear about concepts beyond what I have studied so far
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to determine the product of two Hermitian operators is Hermitian
  1. Hermitian operators (Replies: 1)

  2. Hermitian Operator (Replies: 3)

  3. Hermitian Operators? (Replies: 34)

Loading...