Which of $\{J_0,J_1,J_2\}$ and $\{J_0,J_+,J_-\}$ is a generator for the Lie algebra $\mathfrak{su}(2)$?

Question

I have just started learning about Lie algebra in the context of quantum mechanics and got confused with this:

Some sources say the generators are $J_0,J_1$ and $J_2$ and some use $J_0,J_+$ and $J_−$. Which set is correct? Or if both are correct what key concept am I missing here?

My understanding is that if have certain commutation relations then we know that Lie Algebra is such and such. But if we have two such choices then this understanding falls apart? How do I figure out what is $\mathfrak{su}(2)$ Lie Algebra in general?

Answer 1

Something which none of the other answers mentioned is that, assuming OP means $J_0=\sigma_1,J_1=\sigma_2,J_2=\sigma_3$, and $J_\pm=J_1\pm iJ_2$ or something like that, neither is actually a basis of $\mathfrak{su}(2)$.

My personal opinion is that none areas of mathematics are as butchered by the physicist's usual lack of precision as representation theory is. It is not always a problem, but it's nontheless good to get it right sometimes.

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So first of all, the Lie algebra $\mathfrak{su}(2)$ consists of traceless, antihermitian matrices. The Pauli matrices are hermitian. But for example, let's define $$ T_i=-\frac{i}{2}\sigma_i, $$ then $$ [T_i,T_j]=-\frac{1}{4}[\sigma_i,\sigma_j]=-\frac{1}{4}2i\epsilon_{ijk}\sigma_k=-\frac{i}{2}\epsilon_{ijk}\sigma_k=\epsilon_{ijk}T_k. $$ Then the system $T_1,T_2,T_3$ does provide a basis for $\mathfrak{su}(2)$.

Secondly, despite the involvement of matrices with complex entries, $\mathfrak{su}(2)$ is a real Lie algebra, because the antihermiticity condition is not invariant under multiplication with $i$.

If we allow multiplication of elements with $i$, we get the set of all traceless matrices, which is $\mathfrak{sl}(2,\mathbb C)$, which I'll be considering as a complex Lie algebra (taken this way, $\mathfrak{sl}(2,\mathbb C)$ is the "complexification" of $\mathfrak{su}(2)$).

Thus, if complex linear combinations are allowed, then $(T_1,T_2,T_3)$, $(J_0,J_1,J_2)$, $(J_0,J_ +,J_-)$ etc. are all valid generators of $\mathfrak{sl}(2,\mathbb C)$.

I am noting here that $\mathfrak{sl}(2,\mathbb C)$ can also be "decomplexified" to obtain a real Lie algebra of dimension 6. For example if $T_1,T_2,T_3$ are the three antihermitian matrices I have written above, then $\mathfrak{sl}(2,\mathbb C)_\mathbb R$ is a real Lie algebra of dimension 6 whose generators can be taken to be say $T_1,T_2,T_3,iT_1,iT_2,iT_3$.

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In the physics literature, for unitary Lie algebras, the generators are often taken to be hermitian rather than antihermitian because quantum mechanics prefer hermitian operators, and in many cases, complexifications and decomplexifications are left implicit and unmentioned and people will just happily multiply by $i$ without giving a second thought. But it should be noted that most matrices that are called the generators of $\mathfrak{su}(2)$ cannot actually be taken to be generators of $\mathfrak{su}(2)$, but of its complexification $\mathfrak{sl}(2,\mathbb C)=\mathfrak{su}(2)_\mathbb C$. This is especially true for the ladder opeators $J_\pm$, as those involve complex linear combinations.

Answer 2

Now that this question has been migrated to Mathematics.SE, I want to clarify that the following answer is from a physics perspective, using terms the way that they are conventionally defined in physics literature.

Both are correct. You can choose many different sets of generators for any given Lie algebra. The generators span a vector space, so switching generators just corresponds to changing basis vectors in that vector space. The commutation relations may change depending on the choice of basis. However, since any new set of generators can be written as linear combinations of any other set of generators, specifying the commutation relations for a given set of generators is sufficient to fix the commutation relations for all other sets of generators. Thus the Lie algebra is determined by the commutation relations of any one set of generators.

If you have a set of generators and their commutation relations, and you are concerned about determining which Lie algebra they belong to, you can enforce a normalization condition. This is analogous to choosing an orthonormal basis in a Euclidean vector space. In this situation, you could enforce a normalization condition like $\text{tr}(t_it_j)\propto\delta_{ij}$. This does not hold for the set $J_0$, $J_+$, and $J_-$ because $J_+J_-$ and $J_-J_+$ are not traceless.

Answer 3

The generators of a Lie group are just a basis for the corresponding Lie algebra. The standard way to find the Lie algebra of a matrix Lie group like $SU(2)$ is to start with the defining expressions of the group:

$$g^\dagger g = \mathbb I,\qquad \operatorname{det}(g) = 1$$

From there, you say that $g = e^{iA} \approx \mathbb I + i A$, and try to determine the properties of $A$. One finds that $$g^\dagger g \simeq \mathbb I + i(A-A^\dagger) = \mathbb I \implies A=A^\dagger$$ and $$\operatorname{det}(g) \simeq \operatorname{det}(\mathbb I + iA) \simeq 1 + i\operatorname{Tr}(A) = 1 \implies \operatorname{Tr}(A) = 0$$

so the matrix Lie algebra $\frak{su}(2)$ is the space of $2\times 2$ Hermitian, trace-free matrices. You can choose whatever basis you'd like for this space, but the Pauli matrices

$$\sigma_1 = \pmatrix{0 & 1 \\ 1 & 0} \qquad \sigma_2 = \pmatrix{0 & -i \\ i & 0} \qquad \sigma_3 = \pmatrix{1 & 0 \\ 0 & -1}$$

are a conveniently available choice, with commutation relations $$[\sigma_i,\sigma_j] = 2i\epsilon_{ijk} \sigma_k$$

Answer 4

Just to add a little bit to J. Murray's already excellent answer: The designation of the generators with a subscript $+$ and $-$ comes from the sum and difference of the non-diagonal generators. [This can be generalized to arbitrary SU(n).] The reason for doing this is to treat these generators as some kind raising and lower operators. If you want to know more you can read about Cartan sub-algebras.