# Electron-Phonon interaction

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1. Nov 9, 2015

### kau

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?

2. Nov 10, 2015

### kau

Is there anyone who can give some suggestion on this???

3. Nov 10, 2015

### DrDu

If w is the magnitude of $\vec{W}$, then what is $\vec{w}$. I don't have the article, so maybe you can explain the meaning of all the symbols.

4. Nov 10, 2015

### kau

As far as I can understand w is the magnitude of $\vec{W}$ ...and I think it should sit inside the sum...then it would look like usual plane wave decomposition ...but the way it's written here is not clear to me.. i have attached the file..thanks.

#### Attached Files:

• ###### Frohlic hamiltonian.pdf
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5. Nov 10, 2015

### DrDu

Yes, that's ok. As far as I can see, only $\mathbf{w}$ and w appear in the paper by Froehlich. This still leaves the question what you mean with $\vec{W}$, $\vec{w}$ and w.

6. Nov 10, 2015

### kau

The expression I have written for the quantization of $\vec{B}$ is the same as written in the paper. I copied it from there... The capital $\vec{W}$ sitting in front of the sum over w s is not clear to me..$$\vec{B}=\frac{\vec{W}}{w}(2h/2\pi wnvs')^{1/2}\sum_{w} b_{w} e^{iw.r}$$.I am trying to understand how someone can write something in that form.

7. Nov 12, 2015

### DrDu

I suppose you are right. The relations 2.3 and 2.4 should be formulated in Fourier space.

8. Nov 12, 2015

### samalkhaiat

What is wrong in writing this?
$$\vec{B} = \frac{\vec{k}}{|\vec{k}|} \ N \sum_{\vec{k}} b(\vec{k}) \ e^{i \vec{k} \cdot \vec{r} }$$

9. Nov 13, 2015

### DrDu

The question is how the k vector outside the sum is related to the k vector inside the sum. Admittedly, the expression makes no sense as it stands. I suppose Froehlich had something like this in mind:
$$\vec{B} = \ \ N \sum_{\vec{k}}\frac{\vec{k}}{\omega}b(\vec{k}) \ e^{i \vec{k} \cdot \vec{r} }$$
You also have to read carefully: the denominator is $\omega$, which is constant, and not the absolute value of $\vec{k}$.

10. Nov 13, 2015

### samalkhaiat

Summation label is a dummy index.
Since you seem to know what Froehlich had in mind, can you use this expression of yours to derive equation (2.5) of Froehlich?
Can you tell me what is your dispersion relation here?
Finally, for your information, I do not post in this part of PF. I just responded to a PM from Kau.

11. Nov 13, 2015

### samalkhaiat

1) $\vec{W}= \pi \vec{n} / V^{1/3}$, $\vec{n}=(n_{1},n_{2},n_{3})$, $n_{i}= 0, \pm 1 , \pm 2, \cdots$ is the wave vector.
2) $W = |\vec{W}|$ is the wave number. This is clear from (2.5) $$H = \sum_{\vec{W}} \hbar W s \ b^{*}_{\vec{W}} \ b_{\vec{W}} .$$ For this to have unit of energy, $Ws$ must be an angular frequency $\omega$. Since $s$ is the speed of sound, $W$ has to be the wave number $|\vec{W}|$.
3) Since the field $\vec{B}$ point in the same direction of the wave vector $\vec{W}$, we can write $$\vec{B} = \frac{\vec{W}}{W} B(\vec{r}),$$ and hence $$\vec{\nabla} \cdot \vec{B} = \frac{1}{W} \vec{W} \cdot \vec{\nabla} B(\vec{r}) .$$
4) The summation labels are dummy indices $$\sum_{\vec{P}} = \sum_{\vec{Q}} = \cdots .$$
5) Write
$$B(\vec{r}) = \sum_{\vec{Q}} N \ b_{\vec{Q}} \ e^{ i \vec{Q}\cdot \vec{r}} .$$
Then
$$\vec{\nabla}B = \sum_{\vec{Q}} N \ ( i \vec{Q}) \ b_{\vec{Q}} \ e^{ i \vec{Q}\cdot \vec{r}}.$$
So,
$$\vec{\nabla} \cdot \vec{B} = \sum_{\vec{Q}} \frac{N}{W} \left(i \vec{W} \cdot \vec{Q} \right) \ b_{\vec{Q}} \ e^{ i \vec{Q}\cdot \vec{r}} ,$$ and
$$(\vec{\nabla} \cdot \vec{B})^{*} = \sum_{\vec{P}} \frac{N}{W} \left(-i \vec{W} \cdot \vec{P} \right) \ b^{*}_{\vec{P}} \ e^{ -i \vec{P}\cdot \vec{r}} .$$
6)
\begin{align*} \int dV |\vec{\nabla}\cdot \vec{B}|^{2} &= \sum_{\vec{Q},\vec{P}} \frac{N^{2}}{W^{2}} \ (\vec{W}\cdot \vec{Q}) \ (\vec{W}\cdot \vec{P}) \ b_{\vec{Q}} b^{*}_{\vec{P}} \int dV e^{i (\vec{Q} - \vec{P}) \cdot \vec{r}} \\ &= \sum_{\vec{Q},\vec{P}}\frac{N^{2}}{W^{2}} \ (\vec{W}\cdot \vec{Q}) \ (\vec{W}\cdot \vec{P}) \ b_{\vec{Q}}b^{*}_{\vec{P}} \ V \ \delta_{\vec{Q},\vec{P}} \\ &= \sum_{\vec{Q}} \frac{N^{2}V}{W^{2}} \ (\vec{W}\cdot \vec{Q})^{2} \ b^{*}_{\vec{Q}} \ b_{\vec{Q}} \end{align*}
7) Contributions to energy come from oscillations in the $\vec{Q} = \vec{W}$ directions
\begin{align*} H &= \frac{1}{2} n M s^{2} \sum_{\vec{W}} N^{2}VW^{2} \ b^{*}_{\vec{W}} \ b_{\vec{W}} \\ &= \frac{1}{2} n M s^{2} \sum_{\vec{W}} (\frac{2\hbar}{nMVWs}) \ V \ W^{2} \ b^{*}_{\vec{W}} \ b_{\vec{W}} \\ &= \frac{1}{2} \sum_{\vec{W}} \hbar \ W \ s \left( b^{\dagger}_{\vec{W}} b_{\vec{W}} + b_{\vec{W}} \ b^{\dagger}_{\vec{W}} \right) . \end{align*}
8) Why are you reading this old paper?

12. Nov 15, 2015

### DrDu

There is a problem, here: $|\vec{W}|$ is a function of r, while $\omega$ in (2.5) can't be a function of r.

13. Nov 15, 2015

### DrDu

Let's try this:$$H_f=\frac{1}{2} \int (M \dot{\mathbf{P}}^2+Ms'^2(\nabla \cdot \mathbf{P})^2)n d\mathbf{r}$$
$$=\frac{1}{2}nM \sum_\mathbf{W} ( \mathbf{\dot{P}}_{-\mathbf{W}}(t) \mathbf{\dot{P}}_{\mathbf{W}}(t)+s'^2_\mathbf{W}(-i\mathbf{W}\cdot \mathbf{P}_{-\mathbf{W}})(i\mathbf{W}\cdot \mathbf{P}_\mathbf{W}))$$
$$=\frac{1}{2}nM \sum_\mathbf{W}s'^2_\mathbf{W} |(i\mathbf{W}\cdot (\frac{1}{i s'_\mathbf{W}\omega}\mathbf{\dot{P}}_\mathbf{W}+ \mathbf{P}_\mathbf{W}))|^2$$, as $\mathbf{\dot{P}}_\mathbf{W} || \mathbf{W}$ as it is a longitudinal vector field. We set $\omega = |\mathbf{W}|$.
So we find that $B_\mathbf{W}(t)= \frac{1}{i \omega s'_\mathbf{W}}\mathbf{P}_\mathbf{W}+\mathbf{P}_\mathbf{W}$

Last edited: Nov 15, 2015