# Why do we still use perturbation theory, when we have advanced numerical methods and fast computers?

If my question sounds ignorant or even insulting, I apologise. I may be completely wrong, since I'm not a theoretical physicist.

So, I understand why perturbation theory was originally used in quantum mechanics and even earlier in other fields. There were just no better ways to solve some problems (at least approximately). And it's a great method for some problems with 'small' perturbations.

But now we have fast computers and advanced numerical methods. But not only do researchers still use PT for particular problems, much of theoretical work in quantum mechanics is still based on first-order (or second-order at best) PT. Take Fermi's golden rule for example.

But PT (first or second order, which are used) does not always give a good approximation, and I even heard that its convergence has not been proven in general.

So please, tell me, is there some work being done on moving beyond PT in the theoretical body of quantum mechanics, solid state physics and other fields? Or maybe, there is some advanced PT with high-order terms better suited to computer implementations?

• Computers do what you tell them to do (in a faster time than humans). Perturbation theory exists for all those Hamiltonians whose equations of motion cannot be explicitly solved (and thus numerically implemented); for those, the only computational methods invented so far are Taylor expansions. Other than that, nothing really forbids anybody from numerically implement anything of any sort, were they be able to do so (and people actually do, for example with numerical calculations of Feynman diagrams at almost any order). – gented Dec 24 '15 at 14:55
• In fields such a numerical relativity, perturbation theory (or a post-Newtonian approximation) is often used to check numerical simulations. I think this is an excellent reason to do perturbation theory. It can be a benchmark to make sure you are solving the problem correctly on a computer. – o0BlueBeast0o Dec 24 '15 at 16:02
• To add to the above comments: it is often easier to interpret equations than it is to interpret the results of numerical simulations. Perturbation theory allows us to generate simpler equations that can be used to interpret the numerical results. – march Dec 24 '15 at 17:17
• @GennaroTedesco, I can't imagine a problem for which the Taylor expansion is the only method. My point was exactly that for most problems, just taking first and second terms of Taylors is hardly the most effective way. Especially for development of the general QM theory. It's like linear or quadratic approximation which is still widely used in physics, despite its inaccuracy. – Yuriy S Dec 25 '15 at 7:13
• @YuriyS I agree it isn't the most effective: do you have examples where other methods are preferred (I personally don't) and easier to implement numerically? – gented Dec 25 '15 at 11:08

But now we have fast computers and advanced numerical methods. But not only do researchers still use PT for particular problems, much of theoretical work in quantum mechanics is still based on first-order (or second-order at best) PT.

The reason perturbation theory has not been abandoned is computational complexity. This field (CC) is the study of how the time required to obtain a solution grows as the size (N) of the problem increases (for example, increasing the number of grid points in a numerical integration). There are two primary categories of CC: 1) those for which the time increases as a polynomial of N (referred to as P-type), and 2) those for which time increases exponentially (referred to as EXPTIME).

I could write down the exact Schrodinger equation for an N-electron atom but unless N=1 (hydrogen, where an analytical solution is possible) the problem is of EXPTIME type and attempts to obtain numerical solutions to the exact equation will remain impractical no matter how fast computers become. Exponential growth in time soon overpowers any increase in resources.

This does not mean that simple first or second order perturbation theory is the only fallback. For the N-electron atom, for example, one usually assumes a set of hydrogen-like orbitals (generated via some central potential V(r)) and writes down the expectation value of the exact Hamiltonian for a Slater determinant of these orbitals. This is actually a first order perturbation theory result where the perturbing potential is the difference between the full Hamiltonian and the Hamiltonian that includes the central field V(r) but excludes the electron-electron interaction terms.

We don't stop at this first order result however. By equating the variational derivative of this expectation value with respect to each orbital to zero we obtain the Hartree-Fock (HF) equations (thereby converting an EXPTIME problem into a P problem). By iteratively solving the HF equations until the central potential V(r) obtained from the orbitals generates the exact same orbitals on the next iteration (a condition called self-consistency) we effectively sum a certain subset of perturbation theory terms to infinite order and obtain quite good approximations for the N-electron problem. Simple perturbation theory is the starting point but the final result goes far beyond.

It is true that there is no general proof of convergence of perturbation theory approaches and there are some examples where non-convergence can be explicitly demonstrated, yet many problems in physics simply cannot be approached any other way.

• After actually using numerical methods for real QM problems, I much better appreciate the truth of this answer. It's really not possible to give full numerical treatment to most problems with existing computational resources – Yuriy S Mar 7 at 14:46

For a lot of applications, first (sometimes second) order of perturbation theory gives adequate description in analytical form, which can be more suitable for further investigations than collection of data.

After all, the goal of such calculations is to test validity of specific physical model in question, by providing the values of certain physical observables. Sometimes seeking for numerical solution would be overkill, other time providing analytical treatment in terms of perturbation theory would emphasize limitations or other important features of theory.

The area of non-linear effects is not treated in the same way, because linear result would not describe the phenomena. Unfortunately, this doesn't mean that numerics will give the cure. Even if one gets numerical data, it is sometimes difficult to make conclusions just from it.

For the same reason we still use $\sqrt{1+x}\approx (1+\frac{1}{2}x)$: because in many circumstance one is not terribly interested by a long series of decimals but rather seeks to understand more qualitative aspects of the answer.

Perturbation theory for the eigenvalues/eigenvectors is a kind of sophisticated version of a series expansion, since one obtains eigenvalues and eigenvectors as a series in the powers of the perturbation.

Much in the way that, in a very large number of cases it’s enough to use $\sqrt{2}\approx 1.41$ or even $\sqrt{2}\approx 1.4$, there’s is rarely something fascinating with perturbation theory beyond the second order (there are exceptions). [As an aside, interesting problems often occur when perturbation theory is not applicable.]

If your Hilbert space is large - say $10^6$ for the purpose of argument although this is not particularly big - then the computer will spit out $10^6$ eigenvectors; each of these eigenvectors has $10^6$ components, most of which will be very small (else it’s not a perturbation). If a particular basis state enters in the exact eigenvector with probability $10^{-10}$, it’s likely this state is not terribly important to understand the properties of the eigenvector: your computer will have slaved to produce $10^6$ complicated eigenvectors, for which most of the components are of little interest. If you change the strength of the perturbation by just a bit, you have to redo the calculation to obtain a whole new set of eigenvectors, which will not differ much from the previous calculation.

With perturbation theory, on the other hand, the perturbative parameter is explicit so you can easily assess how things will change as you change this parameter. Basically, you don’t need to redo the calculation if you want to know $\sqrt{9.1}\approx 3.017,\sqrt{9.2}\approx 3.033$ or $\sqrt{9.3}\approx 3.050$: you can see how things change using $$\sqrt{9+\epsilon}\approx 3\left(1+\frac{\epsilon}{18}\right)\, .$$ If you need a numerically more accurate answer you can always include the term in $\epsilon^2$. It’s only if you really need accuracy that you would use a calculator.