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Assume we have hamiltonian

$$ H = H_0 + \lambda V$$

where $ H_0 $ is unperturbed hamiltonian which we know the eigenstastes, and $ V$ is a perturbation.

In the effective hamiltonian approach using the canonical transformation, we transform the hamiltonian via

$$ H_{eff} = e^{S}He^{-S}$$

where $ S^\dagger = -S$, so $e^S$ is hermitian operator, so we are actually doing an unitary transformation of hamiltonian. Then expanding this term using the identity

$$ e^{S}He^{-S} = H + [S,H] + \frac{1}{2}[S,[S,H]] + \frac{1}{3!}[S,[S,[S,H]]] + ...$$

we get the effective hamiltonian as

$$ H_{eff} = H_0 + \lambda V + [S,H_0] + \lambda[S,V] + \frac{1}{2}[S,[S,H_0]] + \lambda \frac{1}{2}[S,[S,V]] + ...$$

But after the transformation, above hamiltonian looks more complicated than the original one. In this step, I don't see any reason why we do the canonical transformation to get the effective hamiltonian.

The book says that if we can find the operator $S$ which satisfies

$$\lambda V + [S,H_0] = 0$$

then we can eliminate the second and third terms of effective hamiltonian, but we still have the infinite series terms in the hamiltonian, which still looks complicated.

What is the essential point of this canonical transformion approach when getting the effective hamiltonian?

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  • $\begingroup$ There is another method, not involving a canonical transformation, the Feshbach projection operator method, leading to an effective Hamiltonian. It is discussed in en.wikipedia.org/wiki/Feshbach%E2%80%93Fano_partitioning for a specific situation but its range of applications is far more general. Basically a closed result is obtained on a subspace which is associated with a given projection operator $P$. $\endgroup$
    – Urgje
    Nov 9, 2015 at 10:31

2 Answers 2

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Canonical transformation

A canonical transformations of the Hamiltonian is given by \begin{equation} H' = e^{S} H e^{-S}, \end{equation} with $S$ anti-hermitian. The idea is to eliminate certain terms in the Hamiltonian by making a basis change. We trade old degrees of freedom by new ones in the hope that the new Hamiltonian becomes easier to solve. The specific choice of $S$ depends on the application. The canonical transformation with \begin{equation} [S,H_0] = -\lambda V, \end{equation} is called the Schrieffer-Wolff transformation. More applications can be found here.

Schrieffer-Wolff transformation

The idea behind this scheme is to facilitate the projection of the Hamiltonian into some low-energy subspace up to a specific order in $\lambda$. Consider the case where the first-order result is zero: \begin{equation} H_{eff} = P_g H_0 P_g + P_g V P_g = E_g P_g, \end{equation} where $P_g$ is the projection operator onto the (degenerate) ground state of the unperturbed system. An example is Anderson's superexchange mechanism for the Hubbard model in the limit of weak site hopping. In this model, tunnelling between sites only occurs via higher-order virtual processes.

So we want to eliminate $\lambda V$ from the Hamilonian, which will give a representation where the higher-order processes are manifest in the Hamiltonian. To better keep track of the order in $\lambda$, we take $H' = e^{\lambda S} H e^{-\lambda S}$. By choosing $S$ so that $[S,H_0] = V$, you can get rid of all terms at first order in $\lambda$ in the transformed Hamiltonian: \begin{equation} H' = H_0 + \frac{\lambda^2}{2} [ V, S ] + \mathcal O(\lambda^3). \end{equation} Every term in $H'$ except for $H_0$ represents a higher-order process. Up to second order, the effective Hamiltonian then becomes \begin{equation} H_{eff} = P_g H' P_g = E_g P_g + \frac{\lambda^2}{2} P_g [V, S] P_g. \end{equation}

Example

As an example, consider a degenerate ground state with $E_g=0$ and let $P_e$ be the projection onto an excited subspace with unperturbed energy $E_e$. Furthermore, assume that $V$ only connects off-diagonal elements between the ground state and this excited state: \begin{equation} V = P_g V P_e + P_e V P_g. \end{equation} Now take \begin{equation} S = \frac{P_g V P_e - P_e V P_g}{E_e}. \end{equation} and note that $S^\dagger = -S$ and that \begin{equation} [S,H_0] = \frac{1}{E_e} [P_g V P_e - P_e V P_g,H_0] = P_g V P_e + P_e V P_g = V. \end{equation} The transformed Hamiltonian becomes \begin{align} H' & = H_0 + \frac{\lambda^2}{2E_e} [ V, P_g V P_e - P_e V P_g ] \\ & = H_0 + \frac{\lambda^2}{E_e} \left( P_e V P_g V P_e - P_g V P_e V P_g \right). \end{align} Note the interpretation of both second-order terms. The term $P_g V P_e V P_g$ represents a two-step tunnelling process from the ground state to an excited state and back. The effective low-energy Hamiltonian becomes \begin{equation} H_{eff} = P_g H' P_g = -\frac{\lambda^2}{E_e} P_g V P_e V P_g. \end{equation} In the superexchange model of Anderson I mentioned before, this effective Hamiltonian represents an antiferromagnetic exchange interaction.

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  • $\begingroup$ What is the difference between standard perturbation theory (calculating the first ,second order energy and wavefunctions, treated in standard QM book) and this effective perturbation approach? If we throw away the higher order terms in effective hamiltonian, isn't it same procedure as calculating first or second order energy in perturbation theory? Actually, I was reading books about frusted magnetism link. This chapter introduces effective hamiltonian approach to solve the highly degenerate ground state. $\endgroup$
    – user42298
    Nov 10, 2015 at 5:39
  • $\begingroup$ @user42298 There is no difference imho. It is just a convenient trick. Since you speak of effective Hamiltonian I assumed some sort of perturbative approach. Also, the transformed Hamiltonian $H'$ is equivalent to the original one, so I don't understand why you would call it effective. It becomes effective only when you project it on the ground state or restrict it to the ground state. If there is a first-order correction, the effective Hamiltonian is just $E_0 P_g + \lambda P_g V P_g$ which is the result of first-order degenerate perturbation theory. $\endgroup$
    – Praan
    Nov 10, 2015 at 13:50
  • $\begingroup$ @user42298 I have found a non-perturbative application of the canonical transformation. The idea is to get rid of a specific term and afterwards bring the Hamiltonian back to a closed form. $\endgroup$
    – Praan
    Nov 10, 2015 at 19:06
  • $\begingroup$ So you are saying that canonical transformation itself is the same as the original hamiltonian, since it is just an unitary transformation. If we project it onto ground state subspace or cut off higher order terms, then it becomes effective hamiltonian. Generally, the transformed hamiltonian involves the off-diagonal components, so describing ground state using only effective hamiltonian is not exact. But if a transformation makes block diagonal form or off-diagonal components for ground state subspace turns out to be small, we can easily describe ground state using fewer basis. $\endgroup$
    – user42298
    Nov 11, 2015 at 6:04
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As it stands the effective Hamiltonian contains the same information as the original one. But note that, given the condition $λV+[S,H₀]=0$, $S$ is proportional to $λ$ so the remaining terms are of order $λ{^2}$ and higher. Thus truncation by omitting these terms gives a result correct to order $λ$. I suppose this is the motivation behind the procedure. Does the book say nothing about this?

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  • $\begingroup$ I think your answer is right. The introduction of this chapter says "transforming away some degrees of freedom, and generating an effective interaction among the remaining ones". To solve the hamiltonian exactly, we have to search the whole Hilbert space. But by this kind of transformation, we can at least remove the first order term, so we don't need to search that space. Plus, if we truncate the higher order terms of effective hamiltonian, we can remarkably reduce the hilbert space again. I think this type of reduction is the essence of canonical transformation. $\endgroup$
    – user42298
    Nov 6, 2015 at 11:30
  • $\begingroup$ @user42298 Why would you want to get rid of the most significant term in the perturbation if it is nonzero? To me, this makes sense only when the first-order result vanishes. Is this procedure even possible if your unperturbed subspace has a first-order perturbative correction? $\endgroup$
    – Praan
    Nov 7, 2015 at 14:32

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