When is many-body perturbation theory valid?

Question

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.

Answer 1

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

Answer 2

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.