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The following identity

$(\boldsymbol a\cdot\boldsymbol J)(\boldsymbol b\cdot\boldsymbol J) = (\boldsymbol a\cdot\boldsymbol b) I + (\boldsymbol a×\boldsymbol b)\cdot\boldsymbol J$$\tag1$

is used to prove

$\exp(-\frac{i}{2} \theta\boldsymbol (e\cdot \boldsymbol J)) = I \cos(\frac{\theta}{2})-\mathrm i(\boldsymbol e\cdot \boldsymbol J) \sin(\frac{\theta}{2}) $$\tag2$

which is used in quantum mechanics for rotation operators,

where $\boldsymbol a,\boldsymbol b,\boldsymbol e \in \mathbb{R}^3, e$ is a unit vector, $I$ is the identity matrix and $ \boldsymbol J = (\boldsymbol J_1, \boldsymbol J_2, \boldsymbol J_3) $ where $\boldsymbol J_1, \boldsymbol J_2, \boldsymbol J_3 $ are 3×3 matrices.

But I didn't understand how identity 1 is true? Can it be verified using any other coordinate geometry relations? And how this identity helps to transform the the above exponential equation?

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    $\begingroup$ I got the equation and info that it can be verified using the given identity from a Caltech lecture note. hep.caltech.edu/~fcp/physics/quantumMechanics/angularMomentum/… $\endgroup$
    – Igris
    Sep 3, 2022 at 11:51
  • $\begingroup$ Is this about proving the equation in the title or proving the first equation in the post using the equation in the title? $\endgroup$
    – Mauricio
    Sep 3, 2022 at 12:14
  • $\begingroup$ The one in the title is just a property of Pauli matrices. $\endgroup$
    – Mauricio
    Sep 3, 2022 at 12:16
  • $\begingroup$ @Mauricio I want to prove Eq.(2), but also wanted to understand how we have Eq.(1). $\endgroup$
    – Igris
    Sep 3, 2022 at 12:31

1 Answer 1

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I'll continue with $\boldsymbol \sigma=(\sigma_1,\sigma_2,\sigma_3)$ instead of $\boldsymbol J$ to represent the Pauli matrices.


To prove Eq(1):

The commutators and anti-commutators between Pauli matrices are

$$ [\sigma_i,\sigma_j]=2\mathrm i\varepsilon_{ijk}\sigma_k,\quad \{\sigma_i,\sigma_j\}=2\delta_{ij}. $$ Adding up the two equations and dividing by 2 gives $$ \sigma_i\sigma_j=\delta_{ij}+\mathrm i\varepsilon_{ijk} \sigma_k. $$ The product of a vector $\boldsymbol a\in\mathbb C^3$ and the Pauli vector $\boldsymbol \sigma$ is defined as $$ \boldsymbol a\cdot\boldsymbol \sigma=a_i\sigma_i. $$ So $$ \begin{aligned} (\boldsymbol a\cdot\boldsymbol \sigma) (\boldsymbol b\cdot\boldsymbol \sigma) &=a_jb_k\sigma_j\sigma_k\\ &=a_jb_k(\delta_{jk}+\mathrm i\varepsilon_{jki}\sigma_i)\\ &=a_kb_k+\mathrm i\varepsilon_{ijk}a_jb_k\sigma_i\\ &=\boldsymbol a\cdot\boldsymbol b+\mathrm i(\boldsymbol a\times \boldsymbol b)\cdot\boldsymbol \sigma. \end{aligned} $$


To prove Eq(2):

As stated in the lecture, let $\boldsymbol a=\boldsymbol b=\boldsymbol e$, we get $$ (\boldsymbol e\cdot \boldsymbol \sigma)^2=I. $$ The exponential of a matrix is defined as the Tylor series $$ \begin{aligned} \exp\left(-\mathrm i\frac{\theta}{2}\boldsymbol e\cdot\boldsymbol \sigma\right) &=\sum_{n=0}^\infty \frac{(-\mathrm i\theta)^n}{2^nn!} \left(\boldsymbol e\cdot\boldsymbol \sigma\right)^n\\ &=\sum_{k=0}^\infty \frac{(-\mathrm i\theta)^{2k}}{2^{2k}(2k)!} \left(\boldsymbol e\cdot\boldsymbol \sigma\right)^{2k} - \mathrm i\left(\boldsymbol e\cdot\boldsymbol \sigma\right) \sum_{k=0}^\infty \frac{\mathrm i^{2k}\theta^{2k+1}}{2^{2k+1}(2k+1)!} \left(\boldsymbol e\cdot\boldsymbol \sigma\right)^{2k}\\ &=\sum_{k=0}^\infty \frac{(-)^k}{(2k)!}\left(\frac{\theta}{2}\right)^{2k} -\mathrm i \left(\boldsymbol e\cdot\boldsymbol \sigma\right) \sum_{k=0}^\infty \frac{(-)^k}{(2k+1)!}\left(\frac{\theta}{2}\right)^{2k+1}\\ &=\cos\frac\theta 2 -\mathrm i(\boldsymbol e\cdot\boldsymbol \sigma) \sin\frac{\theta}{2}. \end{aligned} $$ In the second line, the series is split into two parts, containing even and odd terms respectively.

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  • $\begingroup$ There is an identity matrix missing in the first term of final equation. $\endgroup$
    – Igris
    Sep 3, 2022 at 12:53
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    $\begingroup$ @Igris Yes, but when there is no ambiguity, a number is considered as the identity multiplied by a scalar. $\endgroup$
    – Luessiaw
    Sep 3, 2022 at 13:05
  • $\begingroup$ if we replace the Pauli spin matrix $\sigma$ by general angular momentum operator J, where there there is an additional $\frac{\hbar}{2}$ term in front of these Pauli spin matrices, we won't be able to take out that $\hbar$ terms from inside sine and cosine, right? $\endgroup$
    – Igris
    Sep 3, 2022 at 17:14
  • $\begingroup$ Substituting $\boldsymbol \sigma=\frac2\hbar\boldsymbol J$ into Eq.1, you'll get $(\boldsymbol e\cdot\boldsymbol J)^2= \frac{\hbar^2}4$, and the $\hbar$'s are eliminated in the series, except an extral $\frac2\hbar$ factor in front of sine in the final result. You can as well substitute $\boldsymbol \sigma=\frac2\hbar\boldsymbol J$ into the final line to get the result. $\endgroup$
    – Luessiaw
    Sep 3, 2022 at 17:38
  • $\begingroup$ how do we eliminate the $\frac{\hbar^2}{4}$. That term in the taylor expansion has a power to it. $\endgroup$
    – Igris
    Sep 3, 2022 at 17:43

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