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[H,Q] = 0 means d<Q>/dt = 0 proof

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Show any operator Q that commutes with the Hamiltonian, [Q,H] = 0, is conserved in the above sense (d/dt〈Ψ|QΨ〉 = 0).
    The solution to this problem is as follows:

    iħd/dt〈Ψ|QΨ〉 = (iħd/dt〈Ψ|)|QΨ〉 + 〈Ψ|(|Qiħd/dtΨ〉 = –〈HΨ|QΨ〉 + 〈Ψ|QHΨ〉 = 〈Ψ|[QH]Ψ〉 = 0.

    What I am confused about is the second step:

    [tex]
    \[i\hbar \frac{d}{{dt}}\left\langle \psi \right|\left. {Q\psi } \right\rangle = \left\langle {i\hbar \frac{d}{{dt}}\psi } \right|\left. {Q\psi } \right\rangle + \left\langle \psi \right|\left. {Qi\hbar \frac{d}{{dt}}\psi } \right\rangle\]
    [/tex]

    Why is this true? Where did the sum come from? Why is the Hamiltonian before Q in the second term of the sum? What allows you to put it in that order? Does it matter (I assume it does since we're trying to prove a commutation relation here..)
     
  2. jcsd
  3. Apr 13, 2008 #2
    well, poisson brackets satsify the next relation where @ stands for partial derivative:
    df/dt=[f,H]+@f/@t
    so what you wrote in the title is correct only if f isn't explicitly depended on t.
    for your last question obviously we have here the derivative of a multiplication.
    in a unitary space, <psi|Q(psi)>=<Q(psi)|psi>* where * stands for complex conjugate.
     
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