# Black magic “Hartree” approximation

The question is about an unusual looking version of the Hartree or mean field approximation. The context is several papers I've been reading recently about the out of equilibrium dynamics of phase transitions in the early universe [1-3]. The procedure is to shift a (scalar) field $\Psi \to \psi + \langle \Psi \rangle$, where $\psi$ are fluctuations and $\langle \Psi \rangle$ is a background field. This is perfectly fine. Then you simplify the dynamics of the fluctuations by replacing cubic and quartic terms by quadratic terms like so:

$$\psi^4 \to 6 \langle \psi^2 \rangle \psi^2 - 8 \langle \psi \rangle^3 \psi + 6 \langle \psi \rangle^4 - 3 \langle \psi^2 \rangle^2,$$

$$\psi^3 \to 3 \langle \psi^2 \rangle \psi - 2 \langle \psi \rangle^3.$$

See footnote 11 of  for these full formulas; the other refs simplify these by using $\langle \psi \rangle = 0$. Also note this is supposed to work for bosonic fields.

In the papers these substitutions just come out of the blue.

I get that the idea is to make the problem solvable by reducing the order of the interaction terms (and then eventually determining a self-consistent $\langle\Psi\rangle$), but I'm very confused about the numerical coefficients. Where do they come from? Is there a systematic way to derive them, or consistency condition they must satisfy? Why the alternating signs? What is the relation of this "Hartree" approximation to the Hartree approximation which has to do with minimizing the energy for separable wavefunctions or summing up a particular class of Feynman diagrams?

I've tried the seemingly likely stories:

• It looks at first like maybe something to do with Wick contractions, but the signs are a problem even if all the numbers came out the right size - which they don't.
• Repeatedly applying the mean field theory rule $AB \to A\langle B\rangle + \langle A\rangle B - \langle A\rangle\langle B\rangle$ doesn't seem to work out either.
• Neither does writing $\psi^4 = (\psi - \langle\psi\rangle)^4 + \cdots$ and crossing off the $(\psi - \langle\psi\rangle)^4$ term, or same thing for $\psi^4 = (\psi^2 - \langle\psi^2\rangle)^2 + \cdots$
• It looks similar to the cumulant expansion but I can't get it out of that either.
• The online encyclopedia of integer sequences was completely useless!

A literature pointer would be acceptable if the derivation is long, but please give me an article that explains the result! So far I've only found things which quote it and act as if it's the most obvious thing in the world. Sorry if it is... I feel like I'm missing something very basic. :)

1. Boyanovsky, D., Cormier, D., de Vega, H., & Holman, R. (1997). Out of equilibrium dynamics of an inflationary phase transition. Physical Review D, 55(6), 3373–3388. doi:10.1103/PhysRevD.55.3373
2. Boyanovsky, D., & Holman, R. (1994). Nonequilibrium evolution of scalar fields in FRW cosmologies. Physical Review D, 49(6), 2769–2785. doi:10.1103/PhysRevD.49.2769
3. Chang, S.-J. (1975). Quantum fluctuations in a φ^{4} field theory. I. Stability of the vacuum. Physical Review D, 12(4), 1071–1088. doi:10.1103/PhysRevD.12.1071
• The second equation can be derived from the first.($\frac{d\psi^4}{d \psi} = 4 \psi^3$) – Trimok Jul 24 '13 at 12:00
• Some numerology, assuming $\langle \psi \rangle = 0$, gives : $3(\psi^4-(\psi^2-\langle \psi^2 \rangle)^2) = 6\langle \psi^2 \rangle\psi^2 - 3\langle \psi^2 \rangle^2$. We could see the first term as the possible Wick contractions between $\psi_1\psi_2\psi_3\psi_4$, and the second term (with a minus sign) as possible contractions between $(\psi_1\psi_2 - \langle\psi_1\psi_2 \rangle)$$(\psi_3\psi_4 - \langle\psi_3\psi_4 \rangle) – Trimok Jul 24 '13 at 13:00 • @Trimok Thanks. :) Re first comment: I'll take that as a consistency condition, but it doesn't help derive the first equation in the first place. Second comment: maybe I should have been clearer on what I tried. I could get that out (see the third attempt on my list), and it may have to satisfy me, but I was hoping to understand the full formulas for \langle\psi\rangle\neq 0. – Michael Brown Jul 25 '13 at 23:05 • Numerology 2 : Note, that in the full formulae, the sum of the coefficients is 1, while in the formulae with <\psi>=0, the sum of the coefficients is 3. In fact, adding to this the "consistency" rule \frac{d\psi^4}{d \psi} = 4 \psi^3 and suppose knowing the terms in the 1st formulae (but not the coefficients), and you get your coefficients (you will have 4 equations with 4 unknowns) – Trimok Jul 26 '13 at 11:29 ## 3 Answers I will derive here the second formula, since it is simpler. You can follow the same procedure to derive the first one (although the algebra will be a little bit more complicated). Assume that \psi is some stochastic variable. It is important to clarify that \psi does not need to be described by a Gaussian distribution. In fact, this derivation constitutes a Gaussian approximation. Consider the average \langle\psi^³\rangle. You may write: \psi=\langle\psi\rangle+\delta\psi, with \delta\psi=\psi-\langle\psi\rangle. Then, the average we are interested in, writes: \langle\psi^³\rangle=\langle(\langle\psi\rangle+\delta\psi)^3\rangle= \langle\psi\rangle^3+3\langle\psi\rangle^2\langle\delta\psi\rangle+3\langle\psi\rangle\langle\delta\psi^2\rangle+\langle\delta\psi^3\rangle. The previous formula is exact and no approximation has been used yet. Now, note that the second term in the r.h.s of the previous expansion vanishes since, by definition, \langle\delta\psi\rangle\equiv0. On the other hand, considering only small fluctuations, the leading term is given by the third one, and we can drop the term which is cubic in the fluctuations. With this, we have: \langle\psi^³\rangle\approx\langle\psi\rangle^3+3\langle\psi\rangle\langle\delta\psi^2\rangle=\langle\psi\rangle^3+3\langle\psi\rangle\langle(\psi-\langle\psi\rangle)^2\rangle=\langle\psi\rangle^3+3\langle\psi\rangle(\langle\psi^2\rangle-\langle\psi\rangle^2). Thus, we have \langle\psi^³\rangle\approx3\langle\psi\rangle\langle\psi^2\rangle-2\langle\psi\rangle^3. The second approximation comes around now. The idea is to write an "operational" equation for \psi leading to the previous result. Note that \psi is a random variable. If you consider a new stochastic variable \phi, defined as \phi\stackrel{\text{def}}{=}3\psi\langle\psi^2\rangle-2\langle\psi\rangle^3, you will have exactly that \langle\phi\rangle=3\langle\psi\rangle\langle\psi^2\rangle-2\langle\psi\rangle^3. Then, the true essence of the approximation is to approximate the random variable \psi^3 by the random variable \phi, \psi^3\approx3\psi\langle\psi^2\rangle-2\langle\psi\rangle^3, which lead to the approximated form of the average \langle\psi^3\rangle that we have derived previously. Repeat the same procedure for the first identity (the one for \psi^4) and you will get the correct coefficients. Well, I have no idea if this is the original derivation of the approximations, but I've been able to get these expression from a reasonable-ish, if ad hoc, procedure. Let's assume \psi is normally distributed with mean and variance \mu\equiv\langle\psi\rangle, \sigma^2 respectively. This is true for the mode amplitudes of a free field only, but what the hell, let's go with it. We're trying to approximate the interacting field by a "best-ish" free field anyway. Now we want to approximate \psi^4 with a homogenous polynomial of degree 4 in \psi,\psi^2,\langle\psi\rangle,\langle\psi^2\rangle only because these are the only terms that can go in the mean field Lagrangian. The most general form is$$ \text{approx} = (a_1 \langle\psi^2\rangle + a_2 \langle\psi\rangle^2) \psi^2 + (b_1 \langle\psi^2\rangle \langle\psi\rangle + b_2 \langle\psi\rangle^3) \psi + c_1 \langle\psi\rangle^4 + c_2 \langle\psi^2\rangle^2 + c_3 \langle\psi^2\rangle \langle\psi\rangle^2.$$There is a 7 dimensional space of approximations which we want to cut down to a unique "best" in the following sense: we want to match as many low order moments as we can:$$ \langle\psi^4 \psi^n\rangle = \langle\text{approx}\ \psi^n\rangle, $$for n=0 to as large as we can make it. It turns out that matching n=0,1,2 uniquely determine the solution a_1=6,a_2=0,b_1=0,b_2=-8,c_1=6,c_2=-3,c_3=0 (thank you Mathematica!) which matches the expression in the question. We can examine the higher moments. These cannot be matched with only the terms available, but it is straightforward to compute the errors. Surprisingly the errors get smaller at larger \mu, with relative errors going as \mu^{-6} at large \mu. I'm not sure what to make of this, because we mostly want to use the approximation at \mu=0. For \mu=0 the relative errors vanish for n=\text{odd} and either asymptote to O(1) or grow logarithmically for large n (not sure which is the behaviour from looking at the plots). Here are some distribution plots showing the obvious improvement at larger \mu (Monte Carlo - true dist on left, approx on right):  You can play the same game for \psi^3. There is a five dimensional space of approximations:$$ \text{approx} = a_1 \langle\psi\rangle \psi^2 + (b_1 \langle\psi^2\rangle + b_2 \langle\psi\rangle^2) \psi + c_1 \langle\psi\rangle^3 + c_2 \langle\psi\rangle \langle\psi^2\rangle.$$Matching$\langle\psi^3\psi^n\rangle$for$n=0,1,2$fixes the solution$a_1=3,b_1=3,b_2=-6,c_1=4,c_2=-3$. This doesn't match the given expression but notice that the approximation in the question is linear, since it is being used in the equation of motion rather than the Lagrangian (see original paper). So to get the best linear approximation we drop the$n=2$matching and set$a_1=0$. This gives the expression in the question. So if approximating$\psi^3$in the Lagrangian use the expression calculated here, but if approximating$\psi^3$in the equations of motion use the form given in the question. Whew! I only want to add a few details in the spirit of making the above answer of R. Garcia-Garcia more complete, and also clarifying one detail in the question. Using the questioner's notation, in fact what we've got is the rule$\Psi \rightarrow 3\langle\Psi^2\rangle\Psi - 2\langle\Psi\rangle^3$. By pure and simple algebraic manipulation, this rule can be demonstrated to be equivalent to the set of two rules$\psi^2 \rightarrow \langle\psi^2\rangle$and$\psi^3 \rightarrow 3\langle\psi^2\rangle\psi$. The rules involving only$\psi$, not$\Psi$, is what the question meant to ask. This equivalence should sound right, since the purpose of Hartree approximation, as mentioned above, is to approximate the high order of the provided stochastic variable by a new stochastic variable that has simple relations to the small orders of the provided one, and the two equivalent ways merely correspond to approximate either (background + fluctuation), i.e.$\Psi$, or (only fluctuation), i.e.$\psi$. Also, I think saying that some refs use$\langle\psi\rangle=0$to simplify the rules' expression, as in the question, is a bit misleading. It's not that we can choose to impose the condition$\langle\psi\rangle=0$; this condition (or rather, definition, since we want to interpret$\psi$as the fluctuation, separate from the background) is necessary to "derive" the (full) Hartree factorization rule$\Psi \rightarrow 3\langle\Psi^2\rangle\Psi - 2\langle\Psi\rangle^3\$, as demonstrated in R. Garcia-Garcia's answer. If you see some Hartree factorizations that involve non-zero average of first order of the field, then the field being averaged must be the background + fluctuation, not only fluctuation.

• +1 for the good points in clarifying both, the original question and my answer. – R. García-García Jul 30 '16 at 15:27