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Square of a hermitian operator in matrix form

  1. May 23, 2012 #1
    If we have a hermitian operator Q and we know it's matrix representation [Q], does that mean that [Q2] = [Q]2?
    For example, I'm pretty sure that's the case for p2 for a harmonic oscillator. We have p=ic(a+-a-) and so
    p2=c2(a+-a-)(-a++a-)*=c2(a+-a-)(a+-a-)=p p
    Which tells us that [p2]=[p]2. But could we do that with any hermitian operator?
     
  2. jcsd
  3. May 23, 2012 #2

    George Jones

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    Yes. In general, if [itex]C = AB[/itex], then [itex]\left[ C \right] = \left[ AB \right] = \left[ A \right] \left[ B \right][/itex] for any [itex]A[/itex] and [itex]B[/itex] (not just for hermitian operators).
     
  4. May 23, 2012 #3
    But if it's not hermitian, then the operator may not commute with its Hermitian transpose (right?), so how would we know whether [C2]=[C] [C]* or [C]* [C], and generally wouldn't there be some commutator relationship in there?
     
  5. May 23, 2012 #4

    Fredrik

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    C2 means CC (the composition of C with C), not C*C or CC*.

    The result [A°B]=[A] follows immediately from the definition of matrix multiplication. Actually, I think matrix multiplication is defined the way it is precisely to ensure that this result holds.

    $$
    \begin{align}
    [A\circ B]_{ij} &=((A\circ B)e_j)_i=(A(Be_j))_i =(A(Be_j)_k e_k)_i =(A_{kj} e_k)_i\\
    &=(Ae_k)_i _{kj}=[A]_{ij}_{kj} = ([A])_{ij}
    \end{align}
    $$

    If you're having difficulties following this, read the post I linked to in the quote below. Ask if it's still not clear.

     
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