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Find the Hermitian conjugates: ##x##, ##i##,##\frac{d}{dx}##

  1. Dec 7, 2016 #1
    Just doing some studying before my final exam later today. I think I've got this question right but wanted to make sure since the problem is from the international edition of my textbook, so I can't find the solutions for that edition online.
    1. The problem statement, all variables and given/known data
    The Hermitian conjugate (or adjoint) of an operator ##\hat{Q}## is the operator ##\hat{Q}## such that $$\left<f | \hat{Q} g \right> = \left<\hat{Q} | g\right>, \ \text{for all }f \text{ and }g$$
    (A Hermitian operator, then, is equal to its Hermitian conjugate: ##\hat{Q} = \hat{Q}##.)
    a. Find the Hermitian conjugates of ##x##, ##i##, and ##\frac{d}{dx}##.
    b. Construct the Hermitian conjugate of the harmonic oscillator raising operator, ##a_+##.
    ##a_+ = \frac{1}{\sqrt{2 \hbar m \omega}} (-ip + m\omega x)##

    2. Relevant equations
    ##\left< f | \hat{Q} g \right> = \left< \hat{Q} f | g \right>##

    3. The attempt at a solution
    For ##x##: ##\left< f | x g \right> = \int_{-\infty}^{\infty} f^{*} x g dx##
    Also, ##\left< x f | g \right> = \int_{-\infty}^{\infty} (x f)^{*} g dx##
    If ##x## is real, then ##x^{*} = x##
    So ##\int_{-\infty}^{\infty} (x f)^{*} g dx = \int_{-\infty}^{\infty} x f^{*} g dx = \int_{-\infty}^{\infty} f^{*} x g dx##
    So ##x## is a Hermitian operator; its Hermitian conjugate is itself.

    For ##i##: ##\left< f | i g \right> = \int_{-\infty}^{\infty} f^{*} i g dx##
    Also, ##\left< i f | g \right> = \int_{-\infty}^{\infty} (i f)^{*} g dx = \int_{-\infty}^{\infty} f^{*} (-i) g dx##
    So ##\left< f | i g \right> = \left< (-i) f | g \right>## which means that the Hermitian conjugate of ##i## is ##-i##.

    For ##\frac{d}{dx}##: (this is the one I'm unsure about)
    ##\left< f | \frac{d}{dx} g \right> = \int_{-\infty}^{\infty} f^{*} \frac{dg}{dx} dx##
    Integrate by parts: let ##u = f^{*}## and ##dv = \frac{dg}{dx} dx##.
    Then ##du = \frac{df}{dx}^{*} dx## and ##v = g##
    Integrating by parts, ##\int_{-\infty}^{\infty} f^{*} \frac{dg}{dx} dx = f^{*} g |_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{df}{dx}^{*} g dx##
    The term on the left goes to zero since f(x) and g(x) are square integrable, so they must be zero at either extreme.
    So the Hermitian conjugate of ##\frac{d}{dx}## is ##-\frac{d}{dx}##.

    For ##a_+##: (writing this quickly because I have to get back to studying, and formatting equations takes a while)
    ##\left< f | a_+ g \right> = \left< f | \frac{1}{\sqrt{2 \hbar m \omega}}(-ip + m \omega x) g \right>##
    ##p## and ##x## are Hermitian, and the Hermitian conjugate of any constant is just the negative of that constant. So this should be equivalent to
    ##\left< -\frac{1}{\sqrt{2 \hbar m \omega}} ( -i p + m \omega x) f | g \right>##
    Therefore,
    ##(a_+)^\dagger = -\frac{1}{\sqrt{2 \hbar m \omega}}(-ip + m \omega x) = -a_+##.
     
  2. jcsd
  3. Dec 7, 2016 #2

    DrDu

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    The first three proofs are perfect, but you made a mistake with a+.
     
  4. Dec 7, 2016 #3

    BvU

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  5. Dec 7, 2016 #4
  6. Dec 7, 2016 #5

    BvU

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    Yes, ##(a_+)^\dagger = a_-## and vice versa
     
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