# How to prove $\exp(-\frac{i}{2} \theta (e.J)) = I \cos(\frac{\theta}{2})- i( e.J) \sin(\frac{\theta}{2})$ [closed]

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?

• 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/… Sep 3, 2022 at 11:51
• Is this about proving the equation in the title or proving the first equation in the post using the equation in the title? Sep 3, 2022 at 12:14
• The one in the title is just a property of Pauli matrices. Sep 3, 2022 at 12:16
• @Mauricio I want to prove Eq.(2), but also wanted to understand how we have Eq.(1). Sep 3, 2022 at 12:31

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.

• There is an identity matrix missing in the first term of final equation. Sep 3, 2022 at 12:53
• @Igris Yes, but when there is no ambiguity, a number is considered as the identity multiplied by a scalar. Sep 3, 2022 at 13:05
• 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? Sep 3, 2022 at 17:14
• 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. Sep 3, 2022 at 17:38
• how do we eliminate the $\frac{\hbar^2}{4}$. That term in the taylor expansion has a power to it. Sep 3, 2022 at 17:43