# A A question about linear response and conductivity

Tags:
1. May 26, 2017

### lichen1983312

I am trying to derive the DC electrical conductivity using the pertubation theory in Interaction picture and linear response theory. If working in a energy eigen basis and using the density matrix, the Fourier transform of the susceptibility can be written as

$\chi {(\omega )_{ij}} = i\int_0^\infty {dt{e^{i\omega t}}\sum\limits_n {\left\langle n \right|{e^{ - \beta {{\hat H}_0}}}[{{\hat J}_j}(0),{{\hat J}_i}(t)]\left| n \right\rangle } }$

Then use $I = \sum\limits_n {\left| n \right\rangle \left\langle n \right|}$ and ${{\hat J}_i}(t) = {e^{i{{\hat H}_0}t}}{{\hat J}_i}(0){e^{ - i{{\hat H}_0}t}}$ . We can rewrite ${\sum\limits_n {\left\langle n \right|{e^{ - \beta {{\hat H}_0}}}[{{\hat J}_j}(0),{{\hat J}_i}(t)]\left| n \right\rangle } }$ as

$\sum\limits_{m,n} {{e^{ - {E_m}\beta }}\left[ {\left\langle m \right|{{\hat J}_i}\left| n \right\rangle \left\langle n \right|{{\hat J}_j}\left| m \right\rangle {e^{i({E_m} - {E_n})t}} - \left\langle m \right|{{\hat J}_j}\left| n \right\rangle \left\langle n \right|{{\hat J}_i}\left| m \right\rangle {e^{i({E_n} - {E_m})t}}} \right]}$

so

${\chi _{ij}}(\omega ) = - i\int_0^\infty {dt{e^{i\omega t}}} \sum\limits_{m,n} {{e^{ - {E_m}\beta }}\left[ {\left\langle m \right|{{\hat J}_i}\left| n \right\rangle \left\langle n \right|{{\hat J}_j}\left| m \right\rangle {e^{i({E_m} - {E_n})t}} - \left\langle m \right|{{\hat J}_j}\left| n \right\rangle \left\langle n \right|{{\hat J}_i}\left| m \right\rangle {e^{i({E_n} - {E_m})t}}} \right]}$

I see in literature, to make the integral converge, a complex frequency $\omega + i\varepsilon$ is used to make the integrand vanish in $+ \infty$, therefore

$\begin{array}{l} {\chi _{ij}}(\omega + i\varepsilon ) = \sum\limits_{m,n} {{e^{ - {E_m}\beta }}\left[ {\frac{{\left\langle m \right|{{\hat J}_i}\left| n \right\rangle \left\langle n \right|{{\hat J}_j}\left| m \right\rangle }}{{\omega + i\varepsilon + {E_m} - {E_n}}} - \frac{{\left\langle m \right|{{\hat J}_j}\left| n \right\rangle \left\langle n \right|{{\hat J}_i}\left| m \right\rangle }}{{\omega + i\varepsilon + {E_n} - {E_m}}}{e^{i({E_n} - {E_m})t}}} \right]} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sum\limits_{m,n} {\frac{{\left\langle m \right|{{\hat J}_i}\left| n \right\rangle \left\langle n \right|{{\hat J}_j}\left| m \right\rangle }}{{\omega + i\varepsilon + {E_m} - {E_n}}}} ({e^{ - {E_m}\beta }} - {e^{ - {E_n}\beta }}) \end{array}$

My question is, since I am interested in the DC conductivity, I expect that finally I can remove ${i\varepsilon }$ in above expression when $\omega \to 0$. But I am not sure how to do this since the integral is not well defined if there is no imaginary part in frequency.

On the other hand, if I just simply set $\omega + i\varepsilon = 0$, I seem to be able to find the correct form of conductivity formula used in the quantum hall effect.

Can somebody help?

2. May 26, 2017

### Twigg

If I'm not mistaken, the inclusion of a complex part to the frequency is part of a limiting procedure, so the $\epsilon$ term vanishes anyways. In other words, you get the integral with no complex frequency by taking the limit of that expression as $\epsilon$ goes to 0. Then to get DC conductivity, you just plug in $\omega = 0$ and that should do it.

3. May 26, 2017

### lichen1983312

I mean we need to deal with a integral like
$\int_0^\infty {dt{e^{i\omega t}}} {e^{i({E_m} - {E_n})t}} = \int_0^\infty {dt{e^{i({E_m} - {E_n} + \omega )t}} = } \left. {\frac{{{e^{i({E_m} - {E_n} + \omega )t}}}}{{i({E_m} - {E_n} + \omega )}}} \right|_0^\infty$
if $\omega$ have no imaginary part, ${{e^{i({E_m} - {E_n} + \omega )t}}}$ is not well defined in infinity. we cannot simply take a limiting process that the imagenary part of $\omega$ approching zero to evaluate the integral, right?

4. May 26, 2017

### Twigg

Can you tell us what reference you are looking at?

5. May 26, 2017

### lichen1983312

6. May 29, 2017

### Twigg

Sorry for the late reply. Notice for any nonzero value of the imaginary part of frequency, the integral will converge. The shift $\epsilon$ can be made as small as you like, until the error vanishes. Expanding, $$\frac{1}{stuff + i\epsilon} = \frac{1}{stuff} - \frac{i\epsilon}{stuff^{2}} + \mathcal{O}(\epsilon^2)$$ No matter how small your tolerance for error is, you can always find a satisfactory value of $\epsilon$. In short, you never actually set $\epsilon$ to 0, but you set it to an arbitrary small number so that it doesn't matter anymore.

If you are concerned whether this procedure gives a physical result, then consider what the physical implications of shifting the frequency are. The infinitesimal complex part guarantees causality. To set $\epsilon$ to 0 would violate that. So, to get the result you are looking for, you just use an arbitrarily small value of $\epsilon$. You end up dropping it out of the formula because it's unnecessary clutter and has no experimental meaning.

Does this help?

7. May 29, 2017

### lichen1983312

Thanks very much for your help. I can vaguely feel adding a small $\epsilon$ here has something to do with the causality, which is the part I tried to skip when I was deriving the conductivity. But, like you said, we are studying physics, adding something non-physical can always make trouble for understanding. Plus, your argument
$\frac{1}{stuff + i\epsilon} = \frac{1}{stuff} - \frac{i\epsilon}{stuff^{2}} + \mathcal{O}(\epsilon^2)$
seem doesn't apply here, since we already know when $\varepsilon = 0$ the integral doesn't converge at all. So mathematically, this is really a case no matter how small $\varepsilon$ is, it matters.

8. May 29, 2017

### Twigg

I'm not saying that the expansion I gave approximates the integral when $\epsilon = 0$. I'm saying $\epsilon$ can't be 0 for physical reasons, but it can be chosen to be arbitrarily small. To summarize, the causality requirement implies that the Fourier transform of the susceptibility, taken as a function of complex frequencies, can't have any poles in the upper half plane. That means if you try to evaluate the integral for $\epsilon = 0$, you get an un-causal result that has poles on the real axis (regardless of whether or not it converges). To prove this, you can show that the imaginary part of the exact integral when $\epsilon = 0$ gives you a factor of $\delta(\omega - E_m - E_n)$ in the imaginary part (I may have mixed up my n's and m's but you get the point). If you leave in some non-zero value of $\epsilon$, you get a causal result that has only poles in the lower half plane. The reason we can get rid of the $\epsilon$ is that we can make it any positive number greater than 0 and still keep causality happy. If you can make that value of $\epsilon$ as small as you like, then taking the zero-th order term in the series expansion is an exact result.

9. May 29, 2017

### lichen1983312

Thanks very much. I think it makes great sense, and I will also learn a little bit more about the casualty.

10. May 29, 2017

### Twigg

Here's a pdf that explains it pretty well. If you have Arfken, IIRC I think he goes over it in his section on dispersion relations in his chapter of complex analysis, but I can't find my copy at the moment to verify.

11. May 29, 2017

### lichen1983312

I got it, thanks very much!

12. May 30, 2017

### DrDu

You can also motivate the epsilon adiabatically switching on the perturbation, i.e., $H'\sim Aj \exp \epsilon t$, so that it vanishes in the far past.