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If A is Bloch function, then <A(k,x)|dA(k',x)/dx>=δ(k,k')?

  1. Oct 17, 2015 #1
    Let A(k,x) be a Bloch function, then is the integral <A(k,x)|dA(k',x)/dx>=δ(k,k') ?
  2. jcsd
  3. Oct 17, 2015 #2
    I'm not sure you understand how bra-ket notation is used. You should not have x as a parameter in your bras and kets unless you mean a wave-packet rather than a particular eigenstate. I believe you are asking about.
    $$\int dx e^{-ikx}u^*_{nk}(x) \frac{d}{dx} e^{ik'x}u_{n'k'}(x) $$ $$= \int dx e^{-i(k-k'
    )x}( -iku^*_{nk}(x)u_{n'k'}(x) + u^*_{nk}(x)\frac{d}{dx} u_{n'k'}(x) ) $$

    I think we get something like ##-ik\delta(k-k')\delta_{nn'}## for the first term, but I can't think of any reason why we would get anything in particular for the second term - it depends entirely on the unspecified nature of the ##u_{nk}(x)## functions.
  4. Oct 21, 2015 #3
    MisterX, thank you very much. I agree with your clarification and explanation.
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