One way to write an effective Hamiltonian for a Hamiltonian $H$ is via the equation

$$ \frac{1}{E-H_{eff}} = P\frac{1}{E-H}P $$

where P projects into the subspace we're interested in, and $H_{eff}$ is the effective Hamiltonian of this subspace. This equation is meant to hold for all $E$. I've heard two justifications for this idea; the first is that we want to match the eigenvalues of $H$ and $H_eff$, and both the left and the right hand sides of the above equation have singularities at the eigenvalues of $H$. The second is that Taylor expanding this equation gives

$$ \sum_i \frac{H_{eff}^i}{E^i}=\sum_i\frac{PH^iP}{E^i} $$ which, by matching powers of $E^i$, gives $H_{eff}^i=PH^iP$, so all powers of the Hamiltonians agree on the subspace.

In both of these justifications, it's not clear to me why we expect to be able to find an $H_{eff}$ that satisfies the equation. For example, if our subspace has only 2 basis elements, then $H_{eff}$ has four matrix elements to determine. In the first formulation, though, there are a potentially huge number of energies $E$ which we have to have poles at, and in the second formulation there are an infinite number of equations we need $H_{eff}$ to satisfy. Yet, we only have four unknowns to play with!

So, why do we think an $H_{eff}$ will exist? I would also be interested in better/more clear explanations of the effective Hamiltonian equation, since they might offer a better intuition for why we expect this equation to work.

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    $\begingroup$ Related 325701. $\endgroup$ Jul 11, 2017 at 0:45
  • $\begingroup$ Try to understand, how Dyson equation is working when you know the noninteracting, or a particular subspace, Green's function. The depicted approach to find the self-energies can be linked to how we should determine $H_{eff}$ in your language. $\endgroup$
    – Shasa
    Jul 11, 2017 at 9:32
  • $\begingroup$ I don't understand what this "Taylor expansion" is. Is $i$ an index for some basis set on the subspace? If so you can't expand the full resolvent $\frac{1}{E-H}$ in it, because that operator lives in full space and the subspace basis is not complete in full space. I think you should clarify what you mean by this expansion, it may well be something else that I don't know of. $\endgroup$ Jul 11, 2017 at 15:54
  • $\begingroup$ @Wolpertinger It's just the normal Taylor expansion in powers of $\frac{H}{E}$. Or equivalently, it's the power series that gives the inverse of $(E-H)$. $i$ is just an index, $i=0,1,2,...$ $\endgroup$ Jul 11, 2017 at 20:17
  • $\begingroup$ To be more explicit: $\frac{1}{E-H}=\frac{1}{E}\frac{1}{1-\frac{H}{E}}=\frac{1}{E}\sum_i (\frac{H}{E})^i$. $\endgroup$ Jul 11, 2017 at 20:19

1 Answer 1


To my knowledge effective Hamiltonians based on projection operators were first introduced by Feshbach (Feshbach1958, Feshbach1962, Feshbach1968). The works by Fano (Fano1961) on resonances in Helium and Bloch (notoriously hard to find reference, because the paper is in French...) on singular projection operators are related.

Firstly a comment on each of your "justifications":

  1. [...] we want to match the eigenvalues of $H$ and $H_{eff}$, and both the left and the right hand sides of the above equation have singularities at the eigenvalues of $H$.

Absolutely not. Instead the singularities of the left hand side (LHS) is a subset of the singularities of the full resolvent $\frac{1}{E-H}$. After all you are projecting onto the subspace.

  1. Taylor expanding this equation gives $$ \sum_i \frac{H_{eff}^i}{E^i}=\sum_i\frac{PH^iP}{E^i} $$ which, by matching powers of $E^i$, gives $H_{eff}^i=PH^iP$, so all powers of the Hamiltonians agree on the subspace.

See my comment to the question. I think this is just wrong, although I am not 100% sure what exactly this expansion is supposed to be.

To the actual question

So, why do we think an $H_{eff}$ will exist?

Because we can find it!

Feshbach derives it in Section II of his 1968 paper.

Projecting the eigenvalue equation in full space $H|\Psi\rangle = E|\Psi\rangle$ onto the two subspaces with projection operator $P$ and $Q$ (where $P+Q$=1, $Q^2=Q$, $P^2=P$, $PQ=QP=0$) we get two coupled equations

$$\left[PH(P+Q)-EP\right]|\Psi\rangle = 0$$ $$\left[QH(P+Q)-EQ\right]|\Psi\rangle = 0$$

and hence

$$\left[E-H_{PP}\right]P|\Psi\rangle = H_{PQ}Q|\Psi\rangle$$ $$\left[E-H_{QQ}\right]Q|\Psi\rangle = H_{QP}P|\Psi\rangle.$$

We can write down a formal solution of the latter $$Q|\Psi\rangle = \frac{1}{E-H_{QQ}}H_{QP}P|\Psi\rangle.$$ and substitute into the former to get $$E P|\Psi\rangle = \left[H_{PP} + H_{PQ}\frac{1}{E-H_{QQ}}H_{QP}\right]P|\Psi\rangle.$$

This is a Schrödinger Equation in $P$-space where the subspace state is $P|\Psi\rangle$ and the effective Hamiltonian is $$H_{eff} = H_{PP} + H_{PQ}\frac{1}{E-H_{QQ}}H_{QP}.$$

That's it! Now you can go ahead and define your resolvents on the subspace and check that it is equal to the projected full space resolvent etc. For projection onto a continuous subspace $Q$ there will be issues with invertibility of $E-H_{QQ}$, but that's something else.

Two good resource I can recommend are Kukulin's textbook and Cohen-Tannoudji's book on Atom-Photon Interactions. The latter has an explicit derivation of why the resolvent in the subspace is the projected resolvent in full space.

Hope that helps, I might have missed the point of this question completely, but I like the topic.

  • $\begingroup$ Thank you for the answer, it's helpful. But I think I'm more confused now. The expression you wrote for $H_{eff}$ depends on a parameter $E$. What is the significance of this parameter? Do we expect $H_{eff}$ to be a good approximation for energies near $E$? How can we see that in the derivation you gave? Clearly $H_{eff}$ gives correct results for energies exactly at $E$, but that doesn't seem so useful. And if we're trying to find an effective H, how do we choose the parameter $E$ if we don't know the spectrum of $H$ to begin with? $\endgroup$ Jul 11, 2017 at 20:25
  • $\begingroup$ This does answer my question in the sense that if $H_{eff}$ is allowed to depend on $E$, obviously all the equations can be satisfied because we have a large number of parameters to play with. But then I am confused by the role of the specific $E$. $\endgroup$ Jul 11, 2017 at 20:26
  • $\begingroup$ @JahanClaes Effective Hamiltonians are always (or maybe usually, i'm sure you can find an exception) explicitly energy dependent. Maybe first a word on when you use them. Usually it would be some kind of scattering situation with some resonances in the continuum. So the spectrum of the Hamiltonian is continuous anyway, i.e. any $E$ will be valid. The method will still work for discrete systems. Another note: this is not a method to construct the eigenvalues for a Hamiltonian. It is a way to construct the dynamics of resonances as a subsystem. (to be cont.) $\endgroup$ Jul 11, 2017 at 21:54
  • $\begingroup$ Why is the effective Hamiltonian energy dependent? Mathematically: because of what you said. Physically: you are projecting on a subsystem. You will have some states there (choose a basis for the subspace), that couple to the continuum they sit in. Then of course their coupling strength will depend on which external state they will couple to. The argument applies vice versa (i.e. external to internal coupling) too. $\endgroup$ Jul 11, 2017 at 21:58
  • $\begingroup$ I still don't understand how you choose the energy $E$ for a given problem, or how the validity of the effective Hamiltonian depends on $E$. Perhaps context will help: I'm interested in things like the Heisenberg model as an effective Hamiltonian for the large-U Hubbard model. Scattering analogies are helpful, but not something I have good context for. $\endgroup$ Jul 11, 2017 at 22:02

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