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Heisenberg uncertainty derivation

  1. Sep 29, 2010 #1

    I am trying to teach myself some quantum mechanics and here is something I am stuck on. Various derivations of Heisenberg uncertainty start out with two Hermitian operators, usually called A and B to represent position and momentum. Then they define another two operators ∆A and ∆B as:

    ∆A = A − < A > ∆B = B − < B >

    That appears to me to say that ∆A equals operator A minus the expectation value of A. But how can you subtract a number from a matrix? Is an operator like a variable or is it more like a function? If it is like a function, then how can you plug ∆A and ∆B into the Cauchy-Schwartz inequality? This is probably a dumb question, but I don't think I can move on until I sort it out. Example of said derivation at:


    Thanks for any help.
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Sep 29, 2010 #2
    <A> is a short way of writing <A>I - which means a diagonal matrix with the number <A> on the diagonal. Matrix I has number 1 at each entry on the diagonal - the identity matrix. That is similar to when you you see A-0. Here 0 stands for the matrix consisting of 0-s, not just one number 0. In such cases the meaning of the symbol must be read from the context.
    Last edited: Sep 29, 2010
  4. Sep 29, 2010 #3


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    Staff: Mentor

    Think of a numeric constant as the following operator: "multiply [the following state function or matrix] by this number."
  5. Sep 29, 2010 #4
    So it is just <A> multiplied by the identity matrix. Thanks, that makes sense. I am glad I found this forum!!
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