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The second order correction to the energy of an homogeneous electron gas (jellium) is given rougly by $$E^{(2)}=\frac{Ne^2}{2a_0}(\epsilon_2^r+\epsilon_2^b)$$ with two terms, a direct term given by $$\epsilon_2^r=-\frac{3}{8\pi^5}\int\text{d}q\frac{1}{q^4}\int_{|\vec{p}+\vec{q}|>1}\text{d}k^3\int_{|\vec{p}+\vec{q}|>1}\text{d}^3p\frac{\theta(1-k)\theta(1-p)}{q^2+\vec{q}\cdot(\vec{k}+\vec{p})}$$ and an exchange term by $$\epsilon_2^b=\frac{3}{16\pi^5}\int\text{d}q\frac{1}{q^2}\int_{|\vec{p}+\vec{q}|>1}\text{d}k^3\int_{|\vec{p}+\vec{q}|>1}\text{d}^3p\frac{\theta(1-k)\theta(1-p)}{(\vec{q}+\vec{k}+\vec{p})^2[q^2+\vec{q}\cdot(\vec{k}+\vec{p})]}$$

An attempt to obtain such a result can be seen in How to calculate the second-order pertubation in an electron gas?

However I have a problem, if we are discussing second order corrections there is a factor 2 between direct and exchange terms that is bothering me.

Original derivation and issue

In the original paper by W. Macke, Z. Naturforsch. 5a. 192 (1950), Wacke writes in Eq. 17, that

$$V_{0n}=V_{n0}=\frac{e^2h^2}{\pi v}\left\{\frac{\delta_{s_1's_1}\delta_{s_2's_2}}{\mathfrak g^2}-\frac{\delta_{s_1's_2}\delta_{s_2's_1}}{(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)^2}\right\} $$ where $V_{n0}$ is the amplitude in the second order perturbation theory: $$E^{(2)}=\sum_{n\neq0} \frac{V_{n0}V_{0n}}{E_{0}-E_n}$$ where $|0\rangle$ indicates the ground state of a non-interacting electron gas.

As $V_{0n}V_{n0}=|V_{0n}|^2$, I was expecting the term in brackets to become $$\left\{\frac{\delta_{s_1's_1}\delta_{s_2's_2}}{\mathfrak g^2}-\frac{\delta_{s_1's_2}\delta_{s_2's_1}}{(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)^2}\right\}^2=\frac{\delta_{s_1's_1}\delta_{s_2's_2}}{\mathfrak g^4}+\frac{\delta_{s_1's_2}\delta_{s_2's_1}}{(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)^4}-\frac{2\delta_{s_1's_1}\delta_{s_2's_2}\delta_{s_1's_2}\delta_{s_2's_1}}{\mathfrak g^2(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)^2}$$

However Macke, in the equation just below equation 19, writes $$E^{(2)}=-\frac{m}{4}\left(\frac{v^2}{h^3}\right)^3\left(\frac{e^2h^2}{\pi v}\right)\sum_{ss'}\int\frac{d\mathfrak p_1 d\mathfrak p_2 d\mathfrak g}{(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)\mathfrak g}\\ \cdot\left\{\frac{\delta_{s_1's_1}\delta_{s_2's_2}}{\mathfrak g^4}+\frac{\delta_{s_2's_2}\delta_{s_1's_1}}{(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)^4}-\frac{2\delta_{s_1's_1}\delta_{s_2's_2}\delta_{s_1's_2}\delta_{s_2's_1}}{\mathfrak g^2(\mathfrak p_1-\mathfrak p_2 +\mathfrak g)^2}\right\}$$ take a close look at the symbols in the Dirac deltas of the second term. With Wacke's construction I get his result but I do not know how he makes that squaring of the amplitude. Did he made a mistake? I do not understand also how is he handling the spin sums. The paper is in German, so I am not completely sure if there is a key step there. My derivation can be seen below.

My derivation

The Hamiltonian could be divided into $$H_0=\sum_{\mathbf{k},\lambda}\frac{\hbar^2 k^2}{2m}a^\dagger_{\mathbf{k},\sigma}a_{\mathbf{k},\sigma}$$ and the perturbation $$V=\frac{e^2}{2v}\sum_{\mathbf{k},\mathbf{k}',\mathbf{q}\neq0}\sum_{\sigma,\sigma'}\frac{4\pi}{q^2}a^\dagger_{\mathbf{k}-\mathbf{q},\sigma}a^\dagger_{\mathbf{p}+\mathbf{q},\sigma'}a_{\mathbf{p},\sigma'}a_{\mathbf{k},\sigma}$$ where $v$ is the volume of the electron gas.

Here is my take on the second order term, in units of $k_{\rm F}$, $$E^{(2)}\propto-\sum_{\mathbf p\mathbf p' \mathbf q'}\sum_{\lambda \lambda '}\frac{\Theta(|\mathbf p-\mathbf q'|-1)\Theta(|\mathbf p'+\mathbf q'|-1)\Theta(1- p)\Theta(1-p')}{(q')^2+\mathbf{q}'\cdot(\mathbf p'-\mathbf p)}\\ \times\left|\sum_{\sigma \sigma'}\sum_{q}\sum_{\mathbf k \mathbf k'}\frac{1}{q^2} \langle \mathrm 0|a_{\mathbf p\lambda}^\dagger a^\dagger_{\mathbf p'\lambda'} a_{\mathbf p'+\mathbf q',\lambda '} a_{\mathbf p-\mathbf q',\lambda} a^\dagger_{\mathbf k-\mathbf q,\sigma} a^\dagger_{\mathbf k'+\mathbf q,\sigma '} a_{\mathbf k'\sigma'} a_{\mathbf k\sigma}|0\rangle \right|^2$$ where $\mathbf p\mathbf p '\mathbf q'$ (spins $\lambda,\lambda'$) are the sum over $|n\rangle=a^\dagger_{\mathbf p-\mathbf q',\lambda} a^\dagger_{\mathbf p'+\mathbf q',\lambda '} a_{\mathbf p'\lambda'} a_{\mathbf p\lambda}|0\rangle$ and $\mathbf k,\mathbf k' \mathbf q$ (spins $\sigma\sigma'$) associated with the Coulomb interaction. The amplitude can be calculated to be $$\langle \mathrm 0|a_{\mathbf p\lambda}^\dagger a^\dagger_{\mathbf p'\lambda'} a_{\mathbf p'+\mathbf q',\lambda '} a_{\mathbf p-\mathbf q',\lambda} a^\dagger_{\mathbf k-\mathbf q,\sigma} a^\dagger_{\mathbf k'+\mathbf q,\sigma '} a_{\mathbf k'\sigma'} a_{\mathbf k\sigma}|0\rangle \\ =(\delta_{\mathbf p-\mathbf q',\mathbf k-\mathbf q}\delta_{\lambda \sigma}\delta_{\mathbf p'+\mathbf q',\mathbf k'+\mathbf q}\delta_{\lambda' \sigma'}-\delta_{\mathbf p-\mathbf q',\mathbf k'+\mathbf q}\delta_{\lambda \sigma'}\delta_{\mathbf p'+\mathbf q',\mathbf k-\mathbf q}\delta_{\lambda' \sigma})(\delta_{\mathbf p,\mathbf k}\delta_{\lambda \sigma}\delta_{\mathbf p',\mathbf k'}\delta_{\lambda' \sigma'}-\delta_{\mathbf p,\mathbf k'}\delta_{\lambda \sigma'}\delta_{\mathbf p',\mathbf k}\delta_{\lambda' \sigma})$$

if we sum over $\mathbf k,\mathbf k'$ we get $$=\delta_{\mathbf q',\mathbf q}\delta_{\lambda\sigma}\delta_{\lambda'\sigma'}-\delta_{\mathbf q'-\mathbf p+\mathbf p',\mathbf q}\delta_{\lambda \sigma'}\delta_{\lambda \sigma}\delta_{\lambda '\sigma'}\delta_{\lambda' \sigma}-\delta_{\mathbf q,\mathbf p-\mathbf p'-\mathbf q'}\delta_{\lambda \lambda'}\delta_{\lambda \sigma'}\delta_{\lambda \sigma}\delta_{\lambda '\sigma'}\delta_{\lambda' \sigma}+\delta_{\mathbf q',-\mathbf q}\delta_{\lambda\sigma}\delta_{\lambda'\sigma'}$$ changing $\mathbf q\to-\mathbf q$ for some of the sums, we have $$=2\delta_{\mathbf q',\mathbf q}\delta_{\lambda\sigma}\delta_{\lambda'\sigma'}-2\delta_{\mathbf q'-\mathbf p+\mathbf p',\mathbf q}\delta_{\lambda \sigma'}\delta_{\lambda \sigma}\delta_{\lambda '\sigma'}\delta_{\lambda' \sigma}$$ similar to Eq. 17 in Macke, summing over $\sigma\sigma'$ we have $$=2\delta_{\mathbf q',\mathbf q}-2\delta_{\mathbf q'-\mathbf p+\mathbf p',\mathbf q}\delta_{\lambda \lambda'}$$

Putting it back,in the square and sums we have $$E^{(2)}\propto-\sum_{\mathbf p\mathbf p' \mathbf q'}\sum_{\lambda_1 \lambda_2}\frac{\Theta(|\mathbf p-\mathbf q'|-1)\Theta(|\mathbf p'+\mathbf q'|-1)\Theta(1- p)\Theta(1-p')}{(q')^2+\mathbf{q}'\cdot(\mathbf p'-\mathbf p)}\\ \times\left| \frac{1}{q'^2}-\frac{\delta_{\lambda_1\lambda_2}}{|\mathbf q'+\mathbf p'-\mathbf p|^2}\right|^2\tag{1}\label{this} $$ $$=-\sum_{\mathbf p\mathbf p' \mathbf q'}\sum_{\lambda_1 \lambda_2}\frac{\Theta(|\mathbf p-\mathbf q'|-1)\Theta(|\mathbf p'+\mathbf q'|-1)\Theta(1- p)\Theta(1-p')}{(q')^2+\mathbf{q}'\cdot(\mathbf p'-\mathbf p)}\\ \times\left[ \frac{1}{q'^4}-\frac{2\delta_{\lambda_1\lambda_2}}{q'^2|\mathbf q+\mathbf p'-\mathbf p|^2}+\frac{\delta_{\lambda_1\lambda_2}}{|\mathbf q'+\mathbf p'-\mathbf p|^4}\right] $$ $$=-2\sum_{\mathbf p\mathbf p' \mathbf q'}\frac{\Theta(|\mathbf p-\mathbf q'|-1)\Theta(|\mathbf p'+\mathbf q'|-1)\Theta(1- p)\Theta(1-p')}{(q')^2+\mathbf{q}'\cdot(\mathbf p'-\mathbf p)}\\ \times\left[ \frac{2}{q'^4}-\frac{2}{q'^2|\mathbf q+\mathbf p'-\mathbf p|^2}+\frac{1}{|\mathbf q'+\mathbf p'-\mathbf p|^4}\right] $$ $$=-2\sum_{\mathbf p\mathbf p' \mathbf q'}\frac{\Theta(|\mathbf p-\mathbf q'|-1)\Theta(|\mathbf p'+\mathbf q'|-1)\Theta(1- p)\Theta(1-p')}{(q')^2+\mathbf{q}'\cdot(\mathbf p'-\mathbf p)}\\ \times\left[ \frac{3}{q'^4}-\frac{2}{q'^2|\mathbf q'+\mathbf p'-\mathbf p|^2}\right] $$

where the factors of 2 come from the sum over the spins and the fraction of $1/|\mathbf q'+\mathbf p-\mathbf p'|^4$ disappears by a change of variables and is added to the $1/q^4$ term, however as it has $\delta_{\lambda_1\lambda_2}$ term it only contributes a factor of 1 and not 2. Eq.\eqref{this} is similar to Macke's Eq. 17, however we do not agree on the sums. For me part of the sum is inside the $V_{n0}$ while Macke has the sum outside (sums over 4 spin numbers). Thus I end up with a direct term that is 3/2 times larger than the exchange term, which is not what is expected by Macke (2:1).

I have checked G. Giuliani's book Quantum Theory of the Electron Liquid but there is no hint of what happens with that square. Just a simple "keeping a careful record of the multiplicities of the various terms".

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  • $\begingroup$ Isn't it just a typo? $\endgroup$
    – Adam
    Commented Apr 3, 2023 at 10:36
  • $\begingroup$ @adam I do not think it is a typo, if not there would be an extra term $(1/(q+k+p)'^4)$. To be clear there would be a square term $|1/q^2 +\delta_{\lambda_1\lambda_2}/(|q+k+p|'^2)|^2$ thus even with a change of variable the direct term would contribute 3 times the exchange, not 2 $\endgroup$
    – Mauricio
    Commented Apr 3, 2023 at 14:12
  • $\begingroup$ It is hard to fill the gap without more details in the derivations of the various expressions. But it feels like $|\sum_{p',k',q'}\frac1{q'^2}...|^2$ should become $\sum_{p',k',q'}\frac1{q'^2}...\sum_{p'',k'',q''}\frac1{q''^2}(...)^*$, right? $\endgroup$
    – Adam
    Commented Apr 4, 2023 at 9:31
  • $\begingroup$ @Adam I am hoping to avoid the "check my results" situation, however I cannot seem to get it right. Which step would you like me to add? Some are already in the linked answer. I agree with your statement , for me $|\sum 1/q'^2|^2=\sum' 1/q'\sum'' 1/q''^2$ so in the end there are three sums (and not two) one for variables $q$ coming from the sum over state $|n\rangle$, one for $q'$ and one for $q''$. This does not coincide with the solution in the link. $\endgroup$
    – Mauricio
    Commented Apr 4, 2023 at 9:40
  • $\begingroup$ Worse than "check my results" is "redo the calculation on your own to see where the problem is" ;-) If at least we have what $|n\rangle$, $E_n$ and $\langle 0|H_1|n\rangle$ are, it is simpler to see where the problem is (e.g. I don't have the book with me) $\endgroup$
    – Adam
    Commented Apr 4, 2023 at 11:17

1 Answer 1

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With @Adam permission's, here I copy the solution to this problem:

you assume that your excited state is of the form (in his notation) $|n\rangle=|p_1 s_1,p_2 s_2; p_1' s_1, p_2' s_2\rangle$, that is, you restric the holes and particles to have the same spins (no primes for the last two spin states $s_1$ and $s_2$) whereas he allows different spin states ($s_1'$ and $s_2'$)

This point resolves the issue. In my calculation I assumed, $|n\rangle=a^\dagger_{\mathbf p-\mathbf q',\lambda} a^\dagger_{\mathbf p'+\mathbf q',\lambda '} a_{\mathbf p'\lambda'} a_{\mathbf p\lambda}|0\rangle$ (2 spin sums) while I should instead unrestrict the spin variables as in the correct $|n\rangle=a^\dagger_{\mathbf p-\mathbf q',s_1} a^\dagger_{\mathbf p'+\mathbf q',s_2} a_{\mathbf p's_1'} a_{\mathbf p s_2'}|0\rangle$ (sum over 4 spins variables).

I recover the expected result now, without the right factors of 2, which might be related to double counting. But that is a question for another moment.

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