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Normalization and expectation

  1. Mar 14, 2015 #1

    Maylis

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    Hello,

    I am very confused how this is true? Where does this come from??
    $$<f| \hat{Q}f> = (\sum_{n}a_{n}^{*} |\psi_{n}>)(\sum_{m}a_{m} \hat{Q} | \psi_{m}>)$$

    thanks
     
  2. jcsd
  3. Mar 14, 2015 #2

    stevendaryl

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    I think your formula is slightly wrong. If you have a complete set of orthonormal basis states [itex]|\psi_m\rangle[/itex], then you can write an arbitrary state [itex]|f\rangle[/itex] as a linear combination of basis states:

    [itex]|f\rangle = \sum_m a_m |\psi_m\rangle[/itex]

    That being the case, you take the adjoint of both sides:

    [itex]|f\rangle^\dagger = \langle f | = \sum_m a_m^* \langle \psi_m|[/itex]

    So as to not mix up the indices, let's relabel [itex] m[/itex] by [itex]n[/itex] in this expression:

    [itex]|f\rangle^\dagger = \langle f | = \sum_n a_n^* \langle \psi_n|[/itex]

    Then it follows automatically that

    [itex]\langle f|\hat{O}|f\rangle = (\sum_n a_n^* \langle \psi_n|)\ \hat{O}\ ( \sum_m a_m |\psi_m\rangle)[/itex]

    which can also be written as:
    [itex]\langle f|\hat{O}|f\rangle = (\sum_n a_n |\psi\rangle)^\dagger\ \hat{O}\ ( \sum_m a_m |\psi_m\rangle)[/itex]

    Maybe you don't know what [itex]\langle f|[/itex] means? Well, you know that for ordinary wave functions, you define [itex]\langle f|g\rangle[/itex] to be the integral:

    [itex]\langle f|g\rangle = \int dx f^*(x) g(x)[/itex]

    Then you can define [itex]\langle f|[/itex] as that operator that acts on [itex]|g\rangle[/itex] to produce [itex]\langle f|g\rangle[/itex]
     
  4. Mar 14, 2015 #3

    Maylis

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    Ok, I was definitely unaware that ##|f \rangle^{ \dagger} = \langle f|##. No wonder I was so confused over the definition of a hermition conjugate, how somehow moving the operator to the bra somehow made it a complex conjugate.You are right, I am very shaky about ##\langle f|##. I understand the ket is a vector, but what is the bra? I know you say its an operator that acts on the vector g to produce the inner product, but that leaves me feeling icky inside. As in no intuition at all.
     
  5. Mar 14, 2015 #4

    stevendaryl

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    Well, the simplest case is when there are only a finite number of states. For example, an electron's state, if you ignore the spatial part, is a two-state system: It can be spin-up or spin-down. So if we use matrices to represent the states, then we can have:

    [itex]|\psi_1\rangle = \left( \begin{array}\\ 1 \\ 0 \end{array} \right)[/itex] which represents spin-up

    [itex]|\psi_2\rangle = \left( \begin{array}\\ 0 \\ 1 \end{array} \right)[/itex] which represents spin-down

    An arbitrary state [itex]|f\rangle[/itex] is a combination:

    [itex]|f\rangle = a_1 |\psi_1\rangle + a_2 |\psi_2\rangle = \left( \begin{array}\\ a_1 \\ a_2 \end{array} \right)[/itex]

    Then [itex]|f\rangle^\dagger[/itex] is just [itex] \left( \begin{array}\\ a_1^* & a_2^* \end{array} \right)[/itex]. You compute [itex]|g\rangle^\dagger |f\rangle[/itex] by matrix multiplication: If [itex]|g\rangle[/itex] is [itex] \left( \begin{array}\\ b_1 \\ b_2 \end{array} \right)[/itex], then

    [itex]|g\rangle^\dagger |f\rangle = \left( \begin{array}\\ b_1^* & b_2^* \end{array} \right)\ \left( \begin{array}\\ a_1 \\ a_2 \end{array} \right) = b_1^* a_1 + b_2^* a_2[/itex]

    To generalize to an infinite, but countable, number of states,
    [itex]|g\rangle^\dagger |f\rangle = \sum_i b_i^* a_i[/itex]

    To generalize to a continuous number of states,

    [itex]|g\rangle^\dagger |f\rangle = \int dx\ g(x)^* f(x)[/itex]
     
  6. Mar 15, 2015 #5

    Maylis

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    Thanks, that clears things up a lot.
     
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