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

1. Oct 17, 2015

### zhanhai

Let A(k,x) be a Bloch function, then is the integral <A(k,x)|dA(k',x)/dx>=δ(k,k') ?

2. Oct 17, 2015

### MisterX

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.

3. Oct 21, 2015

### zhanhai

MisterX, thank you very much. I agree with your clarification and explanation.