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Are the matrices Hermitian?

  1. May 26, 2015 #1
    1. The problem statement, all variables and given/known data

    Are the following matrices hermitian, anti-hermitian or neither

    a) x^2

    b) x p = x (hbar/i) (d/dx)

    2. Relevant equations

    3. The attempt at a solution

    For a) I assume it is hermitian because it is just x^2 and you can just move it to get from <f|x^2 g> to <f x^2|g> but I am not sure how to do it for b) because there is the x term and the d/dx term I can't just differntiate it like normal for these problems.

    Any help would be appreciated.
  2. jcsd
  3. May 27, 2015 #2
    Hello ma18, I agree with a part, if your functions are in configuration space yes x^2 as an operator is real and multiplies your function.
    About b part here's my proposed solution, I hope to help you :)

    <f | i x d/dx | g>= ∫ f* (i x dg/dx) dx Now we use the commutator

    [x,p]g=ihbar g= -ihbar x dg/dx +ihbar d(gx)/dx
    And replace in the integral,, the ihbar x dg/dx =ihbar d(gx)/dx - ihbar g
    So you have
    =i hbar ( ∫ f* ( d(gx)/dx ) dx - ∫ f* g dx)
    the integral of f*g is one (1), if the functions are normalized or a constant if they are not. suppose now that they are normalized.
    so now you apply integration by parts in the first integral setting f* as u and d(gx)/dx as v. And you have

    = i hbar ∫ df*/dx g x dx -i hbar, but this is , = - <i xd f* /dx | g> -i hbar

    Please check my signs , but whatever the sign is it cant be hermitian because you have an extra constant so xp is not a hermitian and therefore cant be mesured in the lab.
  4. May 28, 2015 #3
    Thanks for your help Lisa!
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