# Shouldn't the addition of angular momentum be commutative?

I have angular momenta $S=\frac{1}{2}$ for spin, and $I=\frac{1}{2}$ for nuclear angular momentum, which I want to add using the Clebsch-Gordan basis, so the conversion looks like:

\begin{align} \lvert 1,1\rangle &= \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2}\tfrac{1}{2}\rangle,\tag{4.21a} \\ \lvert 1,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2},-\tfrac{1}{2}\rangle + \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr),-\tfrac{1}{2}\tfrac{1}{2}\rangle\biggr),\tag{4.21b} \\ \lvert 1,-1\rangle &= \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr),-\tfrac{1}{2},-\tfrac{1}{2}\rangle,\tag{4.21c} \\ \lvert 0,0\rangle &= \frac{1}{\sqrt{2}}\biggl(\lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr)\tfrac{1}{2},-\tfrac{1}{2}\rangle - \lvert\bigl(\tfrac{1}{2}\tfrac{1}{2}\bigr),-\tfrac{1}{2}\tfrac{1}{2}\rangle\biggr),\tag{4.21d} \end{align}

where $F=I+S$, so this is the basis $\lvert F m_F \rangle = \sum_m \lvert\bigl(I S\bigr),m_I m_S\rangle$.

Now since adding angular momenta is commutative, the exchange between $I$ and $S$ shouldn't mathematically introduce any kind of difference.

In other words, in the basis described in those equations, it shouldn't matter whether I write it as $\lvert\bigl(I S\bigr),m_I m_S\rangle$ or $\lvert\bigl(S I\bigr),m_S m_I\rangle$, right?

Now the problem is the following: I have created the Hamiltonian matrix $H=-\vec{\mu}\cdot \vec{B} = -2 \mu B_z S_z/\hbar$ in the $\lvert F m_F \rangle$ representation, and actually the result depends on how you call those angular momenta, so the result could be

$$H = \begin{pmatrix} \mu_B B & 0 & 0 & 0 \\ 0 & - \mu_B B & 0 & 0 \\ 0 & 0 & 0 &\mu_B B \\ 0 & 0 & \mu_B B & 0 \end{pmatrix}$$

Or could be

$$H = \begin{pmatrix} \mu_B B & 0 & 0 & 0 \\ 0 & - \mu_B B & 0 & 0 \\ 0 & 0 & 0 &-\mu_B B \\ 0 & 0 & -\mu_B B & 0 \end{pmatrix}$$

Depending on how you "label" them, $I$ or $S$... which is very confusing!

This happens because the off-diagonal terms

$$\left\langle 1 0 \right| S_z \left| 0 0 \right\rangle = \frac{1}{2} \left( \left\langle (\frac{1}{2} \frac{1}{2}) \frac{1}{2} -\frac{1}{2} \right| + \left\langle (\frac{1}{2} \frac{1}{2}) -\frac{1}{2} \frac{1}{2} \right| \right) S_z \left( \left| (\frac{1}{2} \frac{1}{2}) \frac{1}{2} -\frac{1}{2} \right\rangle - \left| (\frac{1}{2} \frac{1}{2}) -\frac{1}{2} \frac{1}{2} \right\rangle \right)$$

will be either $\hbar/2$ or $-\hbar/2$ depending on your convention whether it's $\lvert\bigl(I S\bigr),m_I m_S\rangle$ or $\lvert\bigl(S I\bigr),m_S m_I\rangle$.

How can I understand this physically and mathematically? Shouldn't the addition be commutative and the process be blind to which labels I use?

• Maybe I'm just not getting it... but doesn't your Hamiltonian only depend on $S$? Why then would you expect it to be symmetric under the interchange $I \leftrightarrow S$? Also when you do your permutation you are free to redefine the phase of the states, which may be enough freedom if you really want the matrix to come out the same... Commented Oct 22, 2013 at 12:03
• @MichaelBrown Why expect it to be symmetric? because I=S=1/2, and the new basis depends directly on the commutativity of I+S=S+I=F, which is symmetric in that sense. So everything is symmetric up until now, but the Hamiltonian is not, while all what the Hamiltonian does it "pick" one of them! Why should it matter at all if everyyyyyyything is symmetric! Why should it matter if I take |SI mS mI> or |IS mI mS>? Does it really make sense to you? Please explain. Commented Oct 22, 2013 at 12:28

If you exchange $I$ and $S$, you change the last basis vector $e_4=|00\rangle$ into : $e'_4=-|00\rangle$. The new basis $e'$ is expressed from the old basis $e$ with the matrix $M= M^t = M^{-1} = (M^{-1})^t = Diag (1,1,1,-1)$, with $e' = M e$, and so it explains the new expression of the hamiltonian relatively to the new basis $e'$ , you have $H' = (M^{-1})^t H M^{-1}$.
• The physics does not change. The vector $|00\rangle$ has coordinates $(0,0,0,1)$ in the $e$-basis, and $(0,0,0,-1)$ in the new $e'$ basis. So, quantities like $\langle 10|H|00\rangle$ have the same value, whatever the basis you are using. Commented Oct 22, 2013 at 18:36
• Here I mean $|00\rangle = |0_I0_S\rangle$, that is the order $I, S$ is fixed and does not change. Commented Oct 22, 2013 at 18:40
• A physical quantity like $|\langle 1_I0_s|H|0_I0_S\rangle|^2$, does not change with a basis change. You are mixing the notion of elements of a basis like $e_i$, which change with your basis, with a FIXED state like $|0_I0_S\rangle$ . Consider a fixed vector $\vec v$. You may change the basis, so the coordinates of $\vec v$ (the $v^i$) change, but $\vec v$ itself (as a global entity) does not change. For a FIXED operator $H$, its matrix representation $M(H)$ is changing with a basis change, but $H$ (as a global entity) does not change , and quantities like $w^i(M(H))_{ij} v^j$ do not change. Commented Oct 23, 2013 at 8:06