# Why angular momentum about three independent axes?

The generic commutation relations for the angular momentum operator are $[J_x, J_y] = i \hbar J_z$, where the $J_i$, $i = x,y,z$ are the components of the angular momentum vector operator, $\mathbf J$. The spin components obey the same commutation relations. My question is, in all physical systems using a non-relativistic model for spin, the $i$ index above always runs from 1 to 3. What is special about three?

I understand that a $3 \times 3$ rotation matrix can be represented as an element of $SO(3)$ and that spin obeys the same commutation relations as angular momentum (The reps of $SO(3)$ are objects that are appropriate for acting on the kets in the space). So, in a three dimensional system, it makes sense that there are three independent directions in which we can define a spin component. However, even for the spin 1/2 electron system, which is a two dimensional Hilbert space, we still permit these three linearly independent rotations. Why is that?

Many thanks.

• It looks like you are confusing the dimension of space with the dimension of the Hilbert space. – fqq Jul 25 '14 at 11:50
• Hi fqq, could you please explain the difference? This would seem to mean the 'dimension of space' is always three? – CAF Jul 25 '14 at 12:35

I) Quantum mechanically, the Lie group associated with rotational symmetry is $$G~:=~Spin(3)~\cong~ SU(2),$$ which is a double cover of $SO(3)$, and has a 3-dimensional real Lie algebra $$L~:=~so(3)~\cong~ su(2)$$ with generators $J_i$ satisfying

$$[J_i,J_j]~=~i\hbar\sum_{k=1}^3\epsilon_{ijk} J_k.$$

II) In quantum mechanics, we are interested in Lie group representations $R: G \to GL(H)$ and Lie algebra representations $r: L \to {\rm End}(H)$, where $H$ is a complex Hilbert space. Here ${\rm End}(H)$ means the set of endomorphisms, which in this context means $\mathbb{C}$-linear maps: $H\to H$. In fact, we are interested in unitary representations.

Be aware that physicists refer to both the maps $R$ and $r$ and the vector space $H$ as "a representation", cf. this Phys.SE post.

III) Let us now take the Hilbert space $H$ to be 2 dimensional.

It seems OP is essentially asking:

How can the 3-dimensional Lie algebra $L$ be represented non-trivially in a 2-dimensional representation?

Answer: Well, remember that $r(J_i)$ should belong to ${\rm End}(H)$, the space of complex $2\times 2$ matrices, which has 4 complex dimensions. Even if we will only allow Hermitian $2\times 2$ matrices in order to have a unitary representation, we would still have 4 real dimensions at our disposal, one more than we need. In fact, it turns out that we can choose $r(J_i)$ to be proportional to the Pauli matrices $\sigma_i$.

In addition to Qmechanic's interesting answer, you might want to explicitly see that $\mathrm{SU(2)}$ only has three independent variables.

Let us start by writing the $U \in \mathrm{SU(2)}$ as: $$U=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ with $a,b,c,d \in \mathbb{C}$. Since we are looking at the special unitary group, we demand that: $$\mathrm{det}\left(U\right) = 1 \implies ad-bc=1 \tag{1}$$ and, of course, we also have that: $$U^{-1}=U^\dagger \tag{2}$$ Now, it follows from equation (1) that: $$U^{-1}= \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} =\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$ Furthermore: $$U^\dagger = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$$ Thus, from equation (2): $$\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} a^* & c^* \\ b^* & d^* \end{pmatrix}$$ which implies: $$\begin{array}{cc} d=a^* \; ,& b=-c^* \end{array}$$ and so we can write: $$U=\begin{pmatrix} a & b \\ -b^* & a^* \end{pmatrix}$$ with: $$a a^* + bb^* =1 \tag{3}$$ Since $a,b \in \mathbb{C}$, we see that $U$ has four real parameters. However, one of the four is fixed by equation (3) and so $\mathrm{SU(2)}$ only has three independent real parameters.