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

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The discussion centers on the integral involving Bloch functions A(k,x) and the application of bra-ket notation in quantum mechanics. The integral $$\int dx e^{-ikx}u^*_{nk}(x) \frac{d}{dx} e^{ik'x}u_{n'k'}(x)$$ is analyzed, leading to the conclusion that the first term yields ##-ik\delta(k-k')\delta_{nn'}##. The second term's contribution remains unspecified due to the nature of the functions ##u_{nk}(x)##. The clarification provided by user MisterX is acknowledged as valuable.

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Let A(k,x) be a Bloch function, then is the integral <A(k,x)|dA(k',x)/dx>=δ(k,k') ?
 
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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.
 
MisterX, thank you very much. I agree with your clarification and explanation.
 

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