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I am confused about the Feynman diagrams of spin Hamiltonian, for example, the Heisenberg model, the quartic terms like this $[1]$: \begin{align} V &= -\frac{z}{4}\frac{J}{N}\sum_{1,2,3,4}\delta_{G}(1+2-3-4)\left[ V^{(1)}_{(1234)}\alpha^{\dagger}_{1}\alpha^{\dagger}_{2}\alpha_{3}\alpha_{4} +V^{(2)}_{(1234)}\alpha^{\dagger}_{1}\beta_{2}\alpha_{3}\alpha_{4}+V^{(3)}_{(1234)}\alpha^{\dagger}_{1}\alpha^{\dagger}_{2}\beta_{3}^{\dagger}\alpha_{4} \right. \\ & \qquad\qquad\qquad \left. +V^{(4)}_{(1234)}\alpha^{\dagger}_{1}\alpha_{3}\beta^{\dagger}_{4}\beta_{2} +V^{(5)}_{(1234)}\beta_{4}^{\dagger}\alpha_{3}\beta_{2}\beta_{1}+V^{(6)}_{(1234)}\beta^{\dagger}_{4}\beta^{\dagger}_{3}\alpha_{2}^{\dagger}\beta_{1} +V^{(7)}_{(1234)}\alpha^{\dagger}_{1}\alpha^{\dagger}_{2}\beta^{\dagger}_{3}\beta^{\dagger}_{4} \right. \\ & \qquad\qquad\qquad \left. +V^{(8)}_{(1234)}\beta_{1}\beta_{2}\alpha_{3}\alpha_{4} +V^{(9)}_{(1234)}\beta^{\dagger}_{4}\beta^{\dagger}_{3}\beta_{2}\beta_{1} \right] \end{align}

$\alpha$,$\beta$ for magnon operators, the feynman diagrams for these vertices are$[1]$:enter image description here

single arrow for $\alpha$ magnon, double arrow for $\beta$ magnon.

Most of the vertices are represented by a solid circle, is this just a convention? Can I draw them all as $V^{(4)}$ ? In many body physics textbook , the two body interaction vertices are represented by a dashed line with 2-in and 2-out legs.

I have read A.L.Fetter and J.D.Walecka's, Quantum theory of many-particle systems and W.H.Dickhoff and D.van Neck's Many body theory exposed, but I can't find any many body textbook introduce Feynman diagrams of spin Hamiltonian. Although in many cases we rewrite the spin Hamiltonian as boson Hamiltonian, they are also different from boson perturbation introduced in many body textbook (BEC).


  1. Theory of Raman scattering in layered cuprate materials. C. M. Canali and S. M. Girvin. Phys. Rev. B 45, 7127 (1992).
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  • $\begingroup$ Do you have any other questions regarding my answer? $\endgroup$ – AlQuemist Apr 25 '18 at 8:05
  • $\begingroup$ @AlQuemist, yes, for magnetic case, the ground state is different from the conventional bosonic case in the standard many body textbook, there is no chance for BEC (Bose-Einstein condensate), so the diagram techniques for magnon seems similar with fermion diagram. I have a question: when you express some vertices by a solid circle, is the number of topological distinct diagrams the same as the original dashed line one? If there is any detailed reference that would be better. Thanks. $\endgroup$ – ZJX Apr 25 '18 at 11:21
  • $\begingroup$ "the diagram techniques for magnon seems similar with fermion diagram": No; the magnons are bosonic degrees of freedom, and their diagrammatic rules naturally follow those of bosons. $\endgroup$ – AlQuemist Apr 25 '18 at 15:46
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    $\begingroup$ "when you express some vertices by a solid circle, is the number of topological distinct diagrams the same as the original dashed line one": Depends on how one defines such diagrammes with the dashed intermediate lines removed. In the current case, the number of topologically distinct diagrammes remain the same as the explicit notation (with dashed lines), since, afaiu, the authors simply tried to make a "shorter" notation for diagrams. However, there exist also “Hugenholtz diagrammes” where some more profound reduction in notation is applied -- but they complicate one's life a lot. $\endgroup$ – AlQuemist Apr 25 '18 at 15:51
  • $\begingroup$ For a general detailed introduction to diagrammatics, see Bruus & Flensberg, “Many-body quantum theory in condensed matter physics” (2004) [ wcat ]. $\endgroup$ – AlQuemist Apr 25 '18 at 15:56
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The authors use the Dyson-Maleev bosonic representation of quantum spins to rewrite a 2d antiferromagnetic Heisenberg Hamiltonian in terms of two bosonic elementary excitations, $a$ and $b$, with an intricate interaction between them (Eq. 2.24) to include two-magnon scattering processes (see §II of the paper). The final Hamiltonian is therefore a bosonic Hamiltonian to which standard perturbation theory techniques can be applied.

AFAIU, the authors use a simplified notation for the Feynmann diagrams by removing the dashed intermediary lines, except for $ V^{(4)} $ (see Fig. 17 of the paper, included in the question).

The main need for such intermediary lines is to keep the ordering of the quartic interaction terms intact:

In the $n$-th order of a perturbative expansion, always appear expressions like $$ \int_0^\beta \mathrm{d}t_1 \cdots \int_0^\beta \mathrm{d}t_n \, \langle T_t \left[ W(t_1) \cdots W(t_n) \right] \rangle $$

where

$$ W(t) = \frac{1}{2} \sum_{\nu_1 \nu_2 \nu_3 \nu_4} V_{12,34} \, a_{\nu_1}^\dagger(t) \, a_{\nu_2}^\dagger(t) \, a_{\nu_4}(t) \, a_{\nu_3}(t) $$

is the Fock-space representation of a quartic interaction with $\nu_i$ as the quantum numbers, and $T_t$ is the time-ordering operator.

But one precaution must be taken regarding the ordering of the four operators in the two-particle interaction operator, $ W(t) $; namely, the two creation operators must always be to the left of the two annihilation operators. To make sure of that, one “splits” the time arguments of (1,3)- and (2,4)- operators, and adds an infinitesimal time, $ \eta = 0^+ $, to the time arguments of the creation operators, which gives the correct ordering when the time-ordering operator $T_t$ acts on $W$. In this way, the time-integral over $W$ becomes

$$ \int_0^\beta \mathrm{d}t \, W(t) = \int_0^\beta \mathrm{d}t \int_0^\beta \mathrm{d}t' \frac{1}{2} \sum_{\nu_1 \nu_2 \nu_3 \nu_4} V_{12,34}(t, t') \, a_{\nu_1}^\dagger(t^+) \, a_{\nu_2}^\dagger(t'^+) \, a_{\nu_4}(t') \, a_{\nu_3}(t) \, \delta(t - t') ~. $$

Diagrammatically, this time-splitting is denoted as a ‘link’ or intermediary line:

For further details, consult chp. 13 of Bruus and Flensberg, “Many-Body Quantum Theory in Condensed Matter Physics” (2004) [wcat].

Notice that the authors have a different ordering of the four operators in Eq. (2.24), $ a_1^\dagger \, a_2^\dagger \, a_3 \, a_4 $, which leads to a slightly different splitting,

$$ \int_0^\beta \mathrm{d}t \, W(t) = \int_0^\beta \mathrm{d}t \int_0^\beta \mathrm{d}t' \frac{1}{2} \sum_{\nu_1 \nu_2 \nu_3 \nu_4} V_{12,34}(t, t') \, a_{\nu_1}^\dagger(t^+) \, a_{\nu_2}^\dagger(t'^+) \, a_{\nu_3}(t') \, a_{\nu_4}(t) \, \delta(t - t') ~, $$

as apparent in the case of $V^{(4)}$. Upon such a splitting, we obtain, for example, the diagrams below for $ V^{(1)}, V^{(2)}, V^{(4)}, V^{(5)}, V^{(7)}$, and $ V^{(9)} $:

Remember that the authors use a singly-arrowed line for $a$ and a doubly-arrowed one for $b$; the time-directions of $a$ and $b$ are also different (see Fig. 1 of the paper):

Regardless of the splitting arrangement used, it is important only to be consistent throughout the perturbative expansion and diagrammatics.

Furthermore, sometimes for brevity of representation, the intermediate lines are neglected and collapsed into a single point, but such lines are very important to avoid miscalculations. This is indeed the reason for the explicit form of $ V^{(4)} $ in Fig. 17 of the paper.

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