# Prove the Pauli matrices identity $(\vec a\cdot\vec \sigma)(\vec b\cdot\vec \sigma)=(\vec a\cdot \vec b)I+i(\vec a\times\vec b)\cdot\vec\sigma$

I wish to prove the following identity $$\Big(\vec{a}\cdot\vec{\sigma}\Big) \left(\vec{b}\cdot\vec{\sigma}\right) = \left(\vec{a}\cdot\vec{b}\right)I+i \left(\vec{a}\times\vec{b}\right) \cdot \vec{\sigma}$$

regarding pauli matrices and two arbitrary vector operators. The identity's proof is given in Wikipedia, and is very straightforward.

However, it relies on the assumption that the element (matrices) of the vector operator $\vec{b}$ commute with the Pauli matrices.

My question is: Is the identity only valid under this assumption or am I missing something?

• Commented Dec 12, 2017 at 13:25
• For instance $\vec{a}=\vec{J}$ might represent a vector operator representing some angular momentum in three directions ($x,y,z$). This would represent the interaction of electrons with a magnetic impurity in the Kondo model Commented Dec 12, 2017 at 13:35

The identity is not valid in general if $$a$$ and $$b$$ are vectors of operators. You can see it easily trying it with $$\vec a=\vec b=\boldsymbol\sigma$$.

Noting that $$\boldsymbol\sigma\times\boldsymbol\sigma=2i\boldsymbol\sigma$$ and $$\boldsymbol\sigma\cdot\boldsymbol\sigma=3 I$$ you get for the LHS

$$(3I)(3I)=9I$$ while for the RHS $$3I + i(2i\boldsymbol\sigma)\cdot\boldsymbol\sigma=3I-6 I=-3I.$$ where $$I$$ the identity.

• thanks. A small question though: Isn't $\boldsymbol\sigma\times\boldsymbol\sigma=0$? Commented Dec 12, 2017 at 13:44
• For example, $(\boldsymbol\sigma\times\boldsymbol\sigma)_1=\boldsymbol\sigma_2 \boldsymbol\sigma_3 - \boldsymbol\sigma_3 \boldsymbol\sigma_2 = YZ - ZY = 2YZ = 2i X = 2i \boldsymbol\sigma_1$.
– glS
Commented Dec 12, 2017 at 13:48