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kau
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I am reading Frohlic's paper on electron-phonon interaction.
Frohlic.http://rspa.royalsocietypublishing.org/content/royprsa/215/1122/291.full.pdf
Here author has introduced the quantization for complex B field in this paper and claimed to have arrived at the diagonalized form of the hamiltonian. I didn't get the quantization part .I think there is some notational problem. He used w to denote wavevectors and at the sametime in the quantization W sits infront of it.. $$ \vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$ I think there also should be a creation operartor part as well and that's the reason we will get the kind of term author have claimed to get (only then we need to impose the commutation relation,if there is only this much in B then probably I don't need to impose that at all)..most likely I am missing something here..can anyone help me with it..
thanks.Edit: I have hamiltonian $$H= \int |{div\vec{B}}|^{2}nMs'.\, d^{3}r $$
Now We want to quantize it.
So Author have introduced following scheme $$ \vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$ Using this and writing the integration in the following form
$$ H=\int(div\vec{B}^{*}div\vec{B}+div\vec{B}div(\vec{B}^{*}) (nMs'/2).\, d^{3}r $$
and then substituting the B we get the following
$$ H=\sum \frac{hs'(\vec{W}.\vec{w})^{2}}{2w^{3}}[b_{w}^{\dagger}b_{w}+b_{w}b_{w}^{\dagger}]$$
Here w is the magnitude of $\vec{W}$... I need the following
$$ H=\sum_{w} \frac{hs'w}{2}[b_{w}^{\dagger}b_{w}+b_{w}b_{w}^{\dagger}]$$
First thing is how someone can decompose
$$ \vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$?? Why $$\vec{w}$$ should sit outside?
Frohlic.http://rspa.royalsocietypublishing.org/content/royprsa/215/1122/291.full.pdf
Here author has introduced the quantization for complex B field in this paper and claimed to have arrived at the diagonalized form of the hamiltonian. I didn't get the quantization part .I think there is some notational problem. He used w to denote wavevectors and at the sametime in the quantization W sits infront of it.. $$ \vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$ I think there also should be a creation operartor part as well and that's the reason we will get the kind of term author have claimed to get (only then we need to impose the commutation relation,if there is only this much in B then probably I don't need to impose that at all)..most likely I am missing something here..can anyone help me with it..
thanks.Edit: I have hamiltonian $$H= \int |{div\vec{B}}|^{2}nMs'.\, d^{3}r $$
Now We want to quantize it.
So Author have introduced following scheme $$ \vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$ Using this and writing the integration in the following form
$$ H=\int(div\vec{B}^{*}div\vec{B}+div\vec{B}div(\vec{B}^{*}) (nMs'/2).\, d^{3}r $$
and then substituting the B we get the following
$$ H=\sum \frac{hs'(\vec{W}.\vec{w})^{2}}{2w^{3}}[b_{w}^{\dagger}b_{w}+b_{w}b_{w}^{\dagger}]$$
Here w is the magnitude of $\vec{W}$... I need the following
$$ H=\sum_{w} \frac{hs'w}{2}[b_{w}^{\dagger}b_{w}+b_{w}b_{w}^{\dagger}]$$
First thing is how someone can decompose
$$ \vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$?? Why $$\vec{w}$$ should sit outside?