In order to find the effective Hamiltonian in a subspace which is energetically well separated from the rest of the Hilbert space people try to find a unitary transformation which makes the Hamiltonian block-diagonal in that subspace. Usually this procedure is done perturbatively and the corresponding formulae -usually to second order- is available. But I saw somewhere that the effective Hamiltonian satisfies the compact relation : $$ \frac{1}{E-H_{eff}}=P_s \frac{1}{E-H} P_s$$

Where $P_s$ is the projection operator into the subspace which we want its effective Hamiltonian.

So where does the above relation come from? Also it will be very helpful if you mention some references about about different ways of obtaining effective Hamiltonian systematically.

I saw the above formula in the book "Interacting Electrons and Quantum Magnetism" by Auerbach.

  • 4
    $\begingroup$ Just a side-note to make a connection with usual perturbation theory, where we are naturally interested in $P H^k P$ for all integer $k$ (i.e. how the Hamiltonian can connect different states in our subspace). This infinite list of operators can be captured by a single generating function $f(\epsilon) = P \frac{1}{1-\epsilon H} P$. Indeed, by the identity of geometric series we know that algebraically $f(\epsilon) = P \left( \sum_{n = 0}^\infty \epsilon^n H^n \right) P$ such that $PH^kP = \frac{\mathrm d^k f}{\mathrm d \epsilon^k} \big|_{\epsilon = 0}$. $\endgroup$ Commented Apr 15, 2017 at 0:49
  • 3
    $\begingroup$ [continuation] Note that $G_\textrm{eff}(E) = P \frac{1}{E-H} P = \frac{f\left( \frac{1}{E} \right)}{E}$, so on an algebraic level $G_\textrm{eff}(E)$ captures all the info we want for our conventional perturbation theory. $\endgroup$ Commented Apr 15, 2017 at 0:49

2 Answers 2


This approach is straightforward to understand if you realize that $1/(E-H)$ is nothing but the propagator (the Green's function) $G(E)=(E-H)^{-1}$. So this approach simply means that the effective propagator $G_\text{eff}(E)=(E-H_\text{eff})^{-1}$ is obtained by restricting the full propagator to the subspace of interest $G_\text{eff}(E)=P_sG(E)P_s$. One may wonder why not projecting the Hamiltonian directly to the subspace but projecting the propagator. The reason is that all physical observables are measured with respect to the density matrix $\rho(E)=-2\Im G(E+i0_+)$, which is the imaginary part of the propagator. For example, the expectation value of an operator $A$ evaluated on an eigenstate of the energy $E$ is given by

$$\bar{A}(E)=\text{Tr}\hat{A}\rho(E)=-2\text{Tr}\hat{A}\Im G(E+i0_+).$$

Now suppose we are only interested in the physical observables in the Hilbert subspace $\mathcal{H}_s$, then the information of the propagator $G(E)$ in this subspace will be sufficient to reproduce all measurement result, and hence an "effective" description of the subsystem. The Hamiltonian that will produce the effective propagator is therefore considered as the effective Hamiltonian for the subsystem. Of course, the effective Hamiltonian is typically only calculated perturbatively to some order, so approximations are introduced. But imagine if we could find the effective Hamiltonian to all orders, then it would agree with the full Hamiltonian on any physical measurements that take place in the subsystem (or the subspace).

Take a simple quantum mechanical problem for example. Consider a two-level system described by the Hamiltonian


where $t\ll U$ is treated as a perturbation. In the limit of $t\to 0$, we get two levels of the energies 0 and $U$ respectively. Now we are interested in the energy correction to the low-energy level (the level around energy 0). So we first calculate the propagator of the system

$$G=\frac{1}{E-H}=\left[\begin{matrix}\frac{E-U}{E^2-E U-t^2} & \frac{t}{E^2-E U-t^2} \\ \frac{t}{E^2-E U-t^2} & \frac{E}{E^2-E U-t^2} \end{matrix}\right].$$

The effective propagator for the low-energy level is obtained by restricting the propagator to the low-energy subspace, i.e. by taking the $\mathcal{P}_1 G(E) \mathcal{P}_1=G(E)_{11}$ component (at the first line and first column),

$$G_\text{eff}(E)=\frac{E-U}{E^2-E U-t^2}.$$

Now we wish to construct an effective Hamiltonian $H_\text{eff}$ such that the effective propagator can be produced by $G_\text{eff}(E)=1/(E-H_\text{eff})$. We find


We note that $H_\text{eff}$ is also a function of $E$, because the physics can change with respect to the energy scale. To find the eigen energy, one may solve the Schrodinger equation $H_\text{eff}(E)|\psi\rangle=E|\psi\rangle$. Because the subspace only contains a single state, in this case, the eigenstate is simply fixed with respect to the basis of the subspace (but actually implicitly varies with $E$ in the original basis of the full space), and the eigenenergy is given by $t^2/(E-U)=E$, whose solution is


One can see the effective Hamiltonian, if calculated exactly to all ordered, still contains the spectrum of the full system. But in general, we can only compute the effectively Hamiltonian perturbatively. In that case, it will only make sense to evaluate the effective Hamiltonian around the unperturbed energy level. So we evaluate $H_\text{eff}(E)$ at $E=0$ and find the second-order perturbation result $H_\text{eff}(E=0)=-t^2/U$. To obtain higher-order corrections, we feed back the second-order energy to the effective Hamiltonian and find the result that is accurate to the fourth-order in $t/U$, i.e. $H_\text{eff}(E=-t^2/U)=-t^2/U+t^4/U^3+\mathcal{O}[t^6]$. In this way, we can obtain the perturbative corrections order by order recursively.

  • $\begingroup$ Nice answer! Just two questions: 1) Is it clear that in principle $H_\textrm{eff}$ captures the full spectrum? I can see it in your example above, but I can't see whether or not that is an artifact of your simple model or something general. 2) Do you know how the definition you quote is related to the definition $e^{-iH_\textrm{eff} t} = P e^{-iHt} P$? Note that this $H_\textrm{eff}$ would depend on $t$, similar to how the definition you discuss depends on $E$. At first I thought they might be the same (after a natural substitution $E \leftrightarrow \frac{1}{t}$) but I can't quite see it. $\endgroup$ Commented Apr 15, 2017 at 0:52
  • $\begingroup$ @RubenVerresen 1) if $H_\text{eff}$ is evaluated to all orders then it should contain the full spectrum. 2) define the propagator in the time domain as $G(t)=e^{-iHt}$, then your definition is just $G_\text{eff}(t)=PG(t)P$, which is related to $G_\text{eff}(\omega)=PG(\omega)P$ by applying Fourier transform $G(\omega)=i\int \text{d} t G(t)e^{i\omega t}$ on both sides (note that the projection operator $P$ is time independent, so the can be move in or out of the integral freely). $\endgroup$ Commented Apr 15, 2017 at 17:13
  • 1
    $\begingroup$ Thank you! 2) Sounds good! 1) Hm but take for example the case $H = \left( \begin{array}{cc} A & 0 \\ 0 & B \end{array} \right)$ (where $A$ and $B$ are arbitrary Hermitian matrices) and $P$ is the projection onto the first-block. Then $P \frac{1}{E-H} P = (E-A)^{-1}$ which clearly contains no information about $B$. Am I overlooking something? $\endgroup$ Commented Apr 15, 2017 at 22:56
  • 2
    $\begingroup$ @RubenVerresen Well, except for the sick limit when the perturbation is completely turned off that the two subspaces don't talk to each other... $\endgroup$ Commented Apr 15, 2017 at 23:02
  • 3
    $\begingroup$ Generally, if $H=\left[\begin{matrix}H_{AA}&H_{AB}\\H_{BA}&H_{BB}\end{matrix}\right]$ and we would like to know the effective Hamiltonian in subspace A. $G=\frac{1}{E-H}=\left[\begin{matrix}E-H_{AA}&-H_{AB}\\-H_{BA}&E-H_{BB}\end{matrix}\right]^{-1}$, the effective propagator in subspace A is $G_{AA}=(E-H_{AA}-H_{AB}(E-H_{BB})^{-1}H_{BA})^{-1}=\frac{1}{E-H_{eff}}$. Then we know the effective Hamiltonian in subspace A is $H_{eff}=H_{AA}+H_{AB}(E-H_{BB})^{-1}H_{BA}$. Then, how to get the correction of effective Hamiltonian order by order? $\endgroup$
    – qfzklm
    Commented Apr 20, 2017 at 16:49

Everett explained the compact relation nicely. Regarding references, https://arxiv.org/abs/1005.2495 and https://arxiv.org/abs/1105.0675 (more mathematical) are nice. Also, the effective Hamiltonian is calculated to 6th order in appendix B of https://arxiv.org/abs/1704.03870 using the Schrieffer-Wolff formulation of degenerate perturbation theory. (I am an author of this one.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.