# Second-order perturbation jellium revisited

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".

• Isn't it just a typo?
Apr 3, 2023 at 10:36
• @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 Apr 3, 2023 at 14:12
• 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?
Apr 4, 2023 at 9:31
• @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. Apr 4, 2023 at 9:40
• 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)
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).