Does any operator $\mathbf{T} = (T_1,T_2,T_3)$ that satisfies the commutation relations $[T_i, T_j] = i\hbar\epsilon_{ijk}T_k$ represent an angular momentum operator?
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asked Apr 9, 2015 at 23:19
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$\begingroup$ I believe the answer is yes, and that this may even be taken to be the definition of the angular momentum operators. $\endgroup$– Blake HartleyCommented Apr 9, 2015 at 23:35
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2 Answers
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No.
The commutation relation merely means that the $T_i$ form the Lie algebra $\mathfrak{su}(2)$. There are $\mathrm{SU}(2)$s (and consequently $\mathfrak{su}(2)$s) which have nothing to do with angular momentum, e.g. the $\mathrm{SU}(2)$ in the electroweak symmetry group $\mathrm{U}(1)\times\mathrm{SU}(2)$.
answered Apr 10, 2015 at 0:13
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$\begingroup$ sorry what's the difference between $\mathfrak{su}(2)$ and $SU(2)$ ? $\endgroup$ Commented Apr 10, 2015 at 0:15
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$\begingroup$ @SuperCiocia: That between a Lie algebra and a Lie group. $\endgroup$– ACuriousMind ♦Commented Apr 10, 2015 at 0:17
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$\begingroup$ @SuperCiocia Isospin generators also satisfy that Lie algebra, but have nothing to do with angular momentum. $\endgroup$ Commented Apr 10, 2015 at 0:23
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$\begingroup$ So if I had an operator that obeyed those commutation relations, is there a test (or some other way) to know whether or not it is angular momentum? $\endgroup$ Commented Apr 10, 2015 at 13:49
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$\begingroup$ @SuperCiocia: No, the information what physical quantity a given group/operator represents is extrinsic to the group/operator. You can only say that anything has not these relations isn't angular momentum. $\endgroup$– ACuriousMind ♦Commented Apr 10, 2015 at 15:01
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Not necessarily. Some operators representing other physical quantities can be transformed so that they have the same algebraic structure of the angular momentum operators. For example, the inverse of "Jordan-Wigner transformation". Of course, you can think of them as effective angular momentum.
