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So I know that there are commutator relations for $L$ such as $[L_x,L_y] = i\hbar L_z$, but is there a relation for the anticommutator? For example, $L_xL_y + L_yL_x$?

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Strictly speaking anticommutators are not defined in a Lie algebra. The axioms only define the commutators. Clearly since we are used to represent the commutator as $$ [A,B] = AB - BA\,, $$ this might seem silly because, you know, why can't we just flip the sign? But the formal definition doesn't need $A$ and $B$ to be matrices and doesn't say what $AB$ is. It just defines an object that satisfies antisymmetry, bilinearity and the Jacoby identity.

Nevertheless it's possible to extend a Lie algebra to a Universal enveloping algebra which contains formal polynomials of the generators ($A\otimes B \otimes \cdots $ ) and in such an algebra the commutator is represented as one would expect ($A \otimes B - B \otimes A$). Needless to say, if you use matrices all this comes for free and $\otimes$ is just the row by column product.

In this universal enveloping algebra then the anticommutator would be obviously defined as $$ \{A,B\} \equiv A \otimes B + B \otimes A\,. $$ All this detour just to say that it doesn't come from the algebra but it depends on the representation (indeed one of the elements of the universal enveloping algebra in $\mathfrak{su}(2)$ is the Casimir $$L^2 \equiv L_x\otimes L_x + L_y\otimes L_y + L_z\otimes L_z$$ which is even used to label representations). In the fundamental representation of $\mathfrak{su}(2)$ the generators are Pauli matrices and we know $$ \{L_i,L_j\} = \frac{1}{2}\delta_{ij}\mathbb{1}\,. $$ More generally we might expect other tensors to appear in the right hand side. So the final answer would be $$ \{L_i,L_j\} = a(R)\delta_{ij}\mathbb{1} + M(R)_{ij}\,, $$ where $R$ denotes the representation considered, $a(R)$ is just a number and $M(R)$ satisfies $$ \mathrm{Tr}[M(R)_{ij}] = \sum_i M(R)_{ii} = \mathrm{Tr}[M(R)_{ij}L_k] = M_{ij}-M_{ji} = 0\,. $$ I took part of this from the answers to this question, you might want to check those! One thing that I should add if you are considering reading that thread is that in there they consider $\mathfrak{su}(N)$ for arbitrary $N$, the neat fact about $\mathfrak{su}(2)$ is that the so-called cubic casimir $d_{ijk} = \mathrm{Tr}[\{L_i,L_j\}L_k]$ actually vanishes, meaning that the extra piece is only the one orthogonal to the generators (that is the $M(R)$ piece)

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    $\begingroup$ But then this begs the question: why should angular momentum operators be related to a Lie algebraic structure? $\endgroup$ – ZeroTheHero Jan 17 at 2:08
  • $\begingroup$ Because rotations in 3 dimensions form a group and this group is a Lie group. Lie algebras emerge naturally from Lie groups (hence the name). $\endgroup$ – MannyC Jan 17 at 5:27
  • $\begingroup$ You should include (and maybe expand a bit) this in your answer as it’s quite essential to understanding why should Lie algebras be relevant to this question. $\endgroup$ – ZeroTheHero Jan 17 at 14:52
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Well, given any anticommutator, $$ \text{anti}(A, B) = AB + BA = AB - BA + 2BA = [A, B] + 2AB.$$

Thus any anticommutator relation can be written in terms of a commutator relation, with an extra term dangling off the back. But that extra term doesn't represent anything of value, so I don't see how it would be useful.

Personally, I've never come across anticommutators before, so I may be very wrong.

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  • $\begingroup$ what does “won’t contribute” mean in this context? contribute to what? $\endgroup$ – ZeroTheHero Jan 17 at 0:43
  • $\begingroup$ It means that they don't represent anything of value, or that they don't add some extra meaning. Please correct me if I'm wrong. $\endgroup$ – Hanting Zhang Jan 17 at 1:45
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The anticommutator does not appear much in elementary quantum mechanics: for instance, whereas commuting operators have common eigenvectors, anticommuting operators do not.

Anticommutators do occur quite naturally in the demonstration of Robertson's uncertainly relations. In fact, if one wishes to go from a lower bound to a strict equality to find states that satisfy $$ \Delta A\Delta B = \frac{1}{2}\vert\langle [\hat A,\hat B]\rangle\vert \tag{1} $$ rather than the more common inequality, one needs to set the average value of the anticommutator to $0$, i.e. $\langle \Delta A\Delta B+\Delta B\Delta A\rangle=0$ with $\Delta A=\hat A-\langle A\rangle$ etc. It is interesting to note here that, even if the right hand side of the uncertainty relation depends on the commutator, ones needs results on the anticommutator to get the "=" sign.

States that satisty the strict equality $\Delta A\Delta B = \frac{1}{2}\vert\langle [\hat A,\hat B]\rangle\vert$ are called intelligent and were studied in

  1. Aragone, C., et al. "Intelligent spin states." Journal of Physics A: Mathematical, Nuclear and General 7.15 (1974): L149,
  2. Aragone, C., E. Chalbaud, and S. Salamo. "On intelligent spin states." Journal of Mathematical Physics 17.11 (1976): 1963-1971.

Unfortunately, there doesn't seem to be open-access copies of these papers.

As alluded to in another answer, there is a deep relation between Lie algebras and commutators, but not anticommutators. In particular, the tools of representation theory can be used for various purposes when the observables close under commutation on a Lie algebra. The case of angular momentum follows because the operators $\hat L_x, \hat L_y, \hat L_z$ are infinitesimal generators of rotations, and the group of rotations is a Lie group.

Note that, in the special case of Pauli matrices, there is a neat relation for anticommutators: $\{\sigma_a,\sigma_b\}=2\delta_{ab}$ but this is quite specialized and such a clean relation does not hold for larger angular momentum matrices.

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