# When is many-body perturbation theory valid?

I'm calculating expectation values (thermal, time-independent) using many-body perturbation theory, but I'm unsure how to work out what values the parameter I'm expanding the perturbation series in can take.

I read that it's when the matrix elements, $\langle i | H_{pert} | j \rangle$ where $H_{pert}$ is the perturbative term of the Hamiltonian and $| i \rangle$ and $| j \rangle$ are the eigenvectors of the unperturbed Hamiltonian, are much smaller than the energy difference between $i$ and $j$. But I don't really understand what that means, or how it helps me to calculate what values my perturbation parameter can take - Is there a method I can use to figure it out?

Edit:

As requested, to make this concrete, I have a one dimensional fermionic Hubbard model with Hamiltonian

\begin{equation} H= -t \sum_{\langle l,m \rangle} (c^\dagger_l c_m + h.c.) + U \sum_l (n_{l\uparrow} - 1/2)(n_{l\downarrow}-1/2) \end{equation}

I have a special case where I know that $U$ is very small and I want to use many-body perturbation theory to see its effects on correlation functions. I calculate the correlation functions using the functional integral method (i.e. calculating a functional partition function). For this case, how would I go about finding out how small $U$ has to be in order for perturbation theory to be valid?

Secondly, (if this should be a separate question, please let me know!) if instead I have a random $U$, dependent on its position in the lattice,

\begin{equation} H= -t \sum_{\langle l,m \rangle} (c^\dagger_l c_m + h.c.) + \sum_l U_l (n_{l\uparrow} - 1/2)(n_{l\downarrow}-1/2) \end{equation}

I can then use a similar functional integral technique, but take an average over the functional partition function (e.g. over a Gaussian distribution). This average removes the $U_l$ and leaves $\Delta$, the variance of the distribution we've averaged over. In this case, it is $\Delta$ which the perturbation series is expanded in. How would I go about finding how small $\Delta$ has to be for the perturbation series to be valid?

I don't want an answer that's true for any system, I just want to understand how to go about finding it for any system. So if anyone knows of another system where it is shown how small the expansion term has to be, please let me know.

Thanks.

• Is many-body important in your question? For me it looks like careful reading about perturbation theory would help. – Misha Aug 30 '11 at 17:37
• @misha Yes many-body is important. I calculate the expectation values from a functional partition function. I've read about this kind of perturbation theory in lots of places, but none that I've found focus on this. If you know of any, that'd be great! – Calvin Aug 30 '11 at 19:48
• You should say that you are asking for a perturbative treatment of a spatially variable hopping in a Hubbard model, this makes the problem concrete. There is no general answer to the question, because it depends sensitively on things that are impossible to state in terms of crude matrix element bounds. – Ron Maimon Aug 31 '11 at 2:57
• @Calvin Trying to make the question general you lost some important details. No, I do not know many of Hubbard models. However, your idea that perturbation theory there should be different from perturbation theory elsewhere seems unnatural. Thus I assume that your question is not about perturbation theory but about how to apply perturbation to the particular problem which you (unfortunately) did not formulate in the question. – Misha Aug 31 '11 at 4:27
• @Misha I've updated my question. The reason I made it general is that I want to know know to find out when the perturbation is valid for any model, not just for this one. By that I don't mean something that's true for any system; I mean that I would like to understand how to go about finding out for a system. If you know of any examples (not just with the Hubbard model), please point me in the right direction. Thanks. – Calvin Aug 31 '11 at 12:42

Your "perturbation parameter" should just be something that sets the scale of $H_{pert}$ -- that is, the size of the matrix elements you specified compared to the bare energies of the unperturbed problem, which you know the exact solution of. To see why this is you just need to look at the explicit second order contribution from (nondegenerate!) perturbation theory, say for the ground state: $$E_0^{(2)}=-\sum_{j \neq 0}\frac{\langle 0 | H_{pert }| j \rangle \langle j | H_{pert }| 0 \rangle}{E_j - E_0}$$

You can read off that this is only going to be a small correction (and it has to be a small correction if we want to truncate the expansion here) if you can justify those matrix elements being smaller than the energy difference between the states. If you can't do that (say, you have no control over the strength of the perturbation) then you can't trust the perturbation expansion.

Edit 1: When we derive the perturbation expansion, the usual recipe is to do something along the lines $H=H_0+\alpha H_{pert}$, as you say, then expand in alpha -- but we must keep in mind that this guy $\alpha$ is a bookkeeping device that lets us make sure we group all the terms correctly by their order in alpha! After we're done bookkeeping and grouping terms, $\alpha \rightarrow 1$ always. It is not part of the physics and is not ours to play with. We expanded in it as a formal device. The real small parameter is going to be the ratio of the overall energy scale of $H_{pert}$ to $H_0$.

Let me give an example. The fermion Hubbard model has two terms -- (1) an easily diagonalized local interaction with a Coulomb energy scale called $U$; (2) a nearest neighbor hopping term that cannot be diagonalized in the same basis, with an energy scale (~the bandwidth) called $t$. Whether we can trust perturbation theory starting from this basis [that is, taking the hopping as a perturbation] depends only on the dimensionless scale $t/U$, and not the bookkeeping parameter we used to derive the formal terms of the expansion.

• Yes, by perturbation parameter I mean $\alpha$ with the term $\alpha H_{pert}$ in the Hamiltonian. But I consider the thermodynamic limit, so there's no way I could calculate whether it is small for all possible $i$ and $j$ anyway. Is there another way to determine how small $\alpha$ has to be (or have I misunderstood)? – Calvin Aug 30 '11 at 14:34
• Hm. It might help me understand your question if you tell me the explicit model you're using. I'm going to make an edit to my answer, and you tell me if I'm completely missing the point. :) – wsc Aug 30 '11 at 15:16
• For me, the parameter $\alpha$ is the $t$ in the fermion Hubbard model, and I want to expand in $\alpha$. The Hubbard model is fine as an example, but I can edit question to include more specifics of my own model if you like? – Calvin Aug 30 '11 at 15:43
• "the overall energy scale of $H_{\rm{pert}$" may involve the density of states and/or temperature of the system. – Slaviks Aug 30 '11 at 15:53
• @wsc Actually, more specifically, if I have a Hubbard model, but with $t$ dependent on where it is in the system (i.e. it depends on $l$, the position). I then average over $n$ replicas of the partition function to remove the $t_l$, and then I'm left with a parameter $\Delta$ which is the variance of the distribution of $t_l$. It is $\Delta$ I then expand in. This is a different question, but I'd like to know the answer to both this and the case without averaging using $t$. – Calvin Aug 30 '11 at 16:07

To assess whether perturbation theory is applicable in a particular case, you can apply variational perturbation theory:

Incorporate some parameters into your free comparison theory, with corresponding counterterms in the interaction, and do the perturbation theory as a function of these parameters. Typically, the best parameter choice is the one where the responses are least dependent on small changes of the parameters, and how much the responses change give you an idea of how much accuracy you can expect of your calculation.

In some cases, the improvement is drastic; see, e.g., http://en.wikipedia.org/wiki/Variational_perturbation_theory

The renormalizations in quantum field theory are a particular instantiation of variational perturbation theory where doing the variation is essential for getting finite results. See my paper ''Renormalization without infinities - a tutorial'' http://www.mat.univie.ac.at/~neum/ms/ren.pdf

I think we should first define exactly what we are talking about: the usual way to define finite-temperature many-body permutation theory is to split the Hamiltonian into two parts $H = H_0 + \lambda V$ and then expand thermal expectation values in the time-dependent perturbation series (a.k.a. Dyson series): $$\langle A\rangle = \frac 1Z \mathrm{Tr}(e^{-\beta H} A) = \frac 1Z \sum_{k=0}^\infty \frac{(-\lambda)^k}{k!} \int_0^\beta d^k\tau\ e^{-\beta H_0}\ T[A\ V(\tau_1)\cdots V(\tau_k)]$$

The question whether this is "valid" has actually several parts:

1. Does this series converge if we go to infinite order? For the Hubbard model, and indeed for any finite lattice and impurity model, the answer is yes, because the lattice regularises the ultraviolet divergence of the series. (We need to be a little bit more careful in the thermodynamic limit.)

2. Is every effect perturbative, i.e., can we truncate the perturbation series at some point to get it? Here the answer surprisingly is no, even for small values of $\lambda$. A famous counter-example is the Kondo effect, which is anomalous resistance minimum in metals with a small amount of magnetic impurities. This effect is non-perturbative.

3. Can we use the higher-order terms of the series to gauge the error of the expansion? Here the answer is, sadly, not in meaningful way. This is tricky in solid state systems: while, e.g., your total energy may converge quickly with the expansion order, many-body effects that are on a much smaller scale often re-order your energy levels, so you may still end up with the wrong ground state.

To make this more concrete, let us take the inter-electron interaction $U$ as expansion parameter. The first term of the energy expansion is given by Hartree-Fock: $$E^{(1)} = \sum_{ijkl} (U_{ijkl} - U_{ijlk}) \langle c^\dagger_i c_k \rangle \langle c^\dagger_j c_l \rangle$$ For basically any solid state system, Hartree-Fock is by far the largest contribution to the electronic energy: its characteristic scale is 10 eV or 110,000 K, whereas the energy scale for, e.g., the Kondo effect is typically of the order of 10 to 100 K. Thus, your criterion should work, and Hartree-Fock should give us a good enough answer for the properties of a solid or a molecule. For many materials, this is actually true, but there are systems, most famously strongly correlated systems, where Hartree-Fock fails catastrophically, yielding the wrong ground state, wrong excitation spectra, etc.