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I first encountered the concept of renormalization in the context of statistical physics. Here, the renormalization "group" is a set of transformations of the system such that the Hamiltonian $H(J,\beta)$ is mapped to the same Hamiltonian with different parameters $H(J',\beta')$. This transformation has some fixed points, which correspond to critical points in the theory, where the system exhibits infinite correlation length and scale invariance.

Then I heard about the density matrix renormalization group (DMRG), which according to Wikipedia is a numerical variational method that allows to find the ground state of quantum many body systems.

I know ground states of quantum many body system exhibit quantum phase transitions when some (non thermal) parameters are varied. This confuses me, I don't understand the following: if DMRG is used to find ground states of quantum Hamiltonians, what does it have to do with renormalization and the theory of critical phenomena? Does DMRG have anything to do with quantum phase transitions? Why is it called "renormalization"?

If the word "renormalization" is not the same renormalization in DMRG as in the RG in statistical physics, can quantum phase transitions be regarded as fixed points of some renormalization procedure?

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  • $\begingroup$ I don't know anything about DMRG really, but the wikipedia page you link does summarise its relation to RG along the lines of "RG is about taking the low energy limit of a system, DMRG is about considering a smaller Hilbert space in a low energy limit". So I think the RG in DMRG is the Renormalisation Group, but somebody else will need to elaborate the connection. $\endgroup$ – jacob1729 Dec 8 '19 at 14:00
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It is instructive to come back first to NRG (Numerical Renormalization Group) proposed by Ken Wilson (Nobel prize laureate for his work on Renormalization Group in the context of critical phenomena). Consider a translation-invariant 1D quantum Hamiltonian (for example a quantum spin chain). Start with a sufficiently small system of size $\ell$ so that the Hamiltonian can be numerically diagonalized. Truncate the Hilbert space by keeping only the $M$ lowest eigenstates: ${\cal H}\longrightarrow {\cal H}_M$. Now, join two copies of this system by an interaction Hamiltonian $H_{\rm int}$ (for example an exchange coupling between nearest neighbors spins) to get a new system of size $2\ell$. The matrix associated to the full Hamiltonian $H=H_1\otimes\mathbb{I}+\mathbb{I}\otimes H_2+H_{\rm int}$ in the Hilbert space ${\cal H}_M\otimes{\cal H}_M$ can be computed numerically if $M$ is small enough. Because of the truncation, $H$ is not the true Hamiltonian of the system of size $2\ell$ but an effective one. $H$ is called the renormalized Hamiltonian because of the similarity with Wilson renormalization group where a renormalized action is obtained by integrating short-distance (i.e. high frequency) modes. In the NRG algorithm, high-energy states are not integrated out but simply thrown away. The procedure is iterated many times. For more details, I advice you to have a look to the review "The density-matrix renormalization group" (https://arxiv.org/abs/cond-mat/0409292) by U. Schollwöck.

NRG is not efficient for reasons discussed in the above-cited paper. DMRG was introduced by Steve White, former Ph.D. student of Kenneth Wilson, as an improvement of NRG. The truncated Hilbert space ${\cal H}_M$ is not constructed from the $M$ lowest eigenstates of the Hamiltonian but from the $M$ lowest eigenstates of the ground state reduced density matrix $\rho_1={\rm Tr}_2|\psi_0\rangle\langle\psi_0|$ of the full Hamiltonian. Therefore, the density matrix is said to be renormalized. With DMRG, the energy, as well as any local average, of the ground state can be estimated with a very high accuracy. Remember that a quantum phase transition occurs when the gap between the ground state and the first-excited state vanishes leading to a change of symmetry of the ground state.

To summarize, the term "Renormalization Group" is used for NRG and DMRG because an effective Hamiltonian is constructed at each step by throwing away higher-energy states that are assumed to be irrelevant. However, this renormalized Hamiltonian is used only to compute quantum averages. Critical exponents cannot be extracted from the law $J'={\cal R}(J)$ as would be done in Wilson Renormalization Group. With the DMRG algorithm, you have to run several simulations at different control parameters $\delta$ to estimate for example the magnetic critical exponent $\beta$ from the magnetization behavior close to the quantum critical point: $M\sim |\delta-\delta_c|^\beta$. You can also use Finite-Size Scaling $M\sim L^{-\beta/\nu}$ by performing several DMRG calculations for different lattice sizes.

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    $\begingroup$ You may also have a look to the related answer: physics.stackexchange.com/questions/451603/… $\endgroup$ – Christophe Dec 8 '19 at 15:05
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    $\begingroup$ Maybe one could add that the way DMRG chooses which states to keep is not based on energy (as in other renormalization schemes), but on entanglement, that is, the states which are most relevant for the entanglement are being kept. It thus works better where entanglement in the ground state plays an essential role. $\endgroup$ – Norbert Schuch Dec 8 '19 at 16:59
  • $\begingroup$ So do I understand it correctly that DMRG has nothing to do with quantum phase transitions? And NRG has nothing to do with phase transitions in general? They only share the name because they reduce the degrees of freedom of the system? $\endgroup$ – user2723984 Dec 9 '19 at 8:45
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    $\begingroup$ Yes, you understood correctly. $\endgroup$ – Christophe Dec 9 '19 at 9:02
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It's basically a misnomer that has nothing to do with the "real" renormalization group. (I've heard this confirmed directly from professional practitioners.) You can come up with some vague similarities if you really stretch for it, but I don't think there's any conceptual insight to be gained from doing so.

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  • $\begingroup$ How does this answer square with the one by Christophe which is in direct contradiction with you? $\endgroup$ – KF Gauss Dec 10 '19 at 4:52

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