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?

  • $\begingroup$ Related : Need help with solution of the Dirac equation $\endgroup$ – Frobenius Dec 12 '17 at 13:25
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    $\begingroup$ : ...and how do you imagine the outer product $\:a\times b\:$ of two matrices be??? ...and what are the dimensions of these matrices??? ...and what the product $\:a\cdot \boldsymbol{\sigma}\:$ ??? $\endgroup$ – Frobenius Dec 12 '17 at 13:30
  • $\begingroup$ Can you write out in full what $\vec{a}\cdot\vec{\sigma}$ would be: is it simply $a^i\,\sigma_i$ where the $a^i$ are now general $2\times2$ matrices rather than scalars? $\endgroup$ – WetSavannaAnimal Dec 12 '17 at 13:33
  • $\begingroup$ @Frobenius those operations can still make sense if $\vec a$ and $\vec b$ are vectors of operators. E.g. $(\vec a\times \vec b)_1= a_2 b_3 - a_3 b_2$, $\boldsymbol\sigma\times\boldsymbol\sigma=2i\boldsymbol\sigma$, $\boldsymbol\sigma\cdot\boldsymbol\sigma=3\mathbb 1$ $\endgroup$ – glS Dec 12 '17 at 13:34
  • $\begingroup$ 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 $\endgroup$ – Yair M Dec 12 '17 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 $\boldsymbol\sigma\cdot\boldsymbol\sigma=3\,\boldsymbol{1}$ you get for the LHS

$$(3\,\boldsymbol{1})(3\,\boldsymbol{1})=9\,\boldsymbol{1}$$ while for the RHS $$ 3\,\boldsymbol{1} + i(2i\boldsymbol\sigma)\cdot\boldsymbol\sigma=3\,\boldsymbol{1}-6\,\boldsymbol{1}=-3\,\boldsymbol{1}. $$ where $\:\boldsymbol{1}\:$ the identity.

  • $\begingroup$ thanks. A small question though: Isn't $\boldsymbol\sigma\times\boldsymbol\sigma=0$? $\endgroup$ – Yair M Dec 12 '17 at 13:44
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    $\begingroup$ For example, $(\boldsymbol\sigma\times\boldsymbol\sigma)_1=\boldsymbol\sigma_2 \boldsymbol\sigma_3 - \boldsymbol\sigma_3 \boldsymbol\sigma_2 = YZ - ZY = 2YZ = 2i X = 2i \boldsymbol\sigma_1$. $\endgroup$ – glS Dec 12 '17 at 13:48

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