Let $\hat{n}$ be a 3D unit vector and let $\vec{\sigma}$ be a vector of the Pauli matrices \begin{align} \vec{\sigma} = \left( \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right)\ ,\ \left(\begin{matrix}0 & - i \\ i & 0 \end{matrix}\right)\ ,\ \left(\begin{matrix} 1 & 0 \\ 0 & -1\end{matrix}\right)\right) \end{align} Let $\theta \in \mathbb{R}$ and $f$ be a function taking complex matrices to complex matrices. Show that \begin{align} f(\theta\ \hat{n} \cdot \vec{\sigma}) = \frac{f(\theta) + f(-\theta)}{2}I + \frac{f(\theta) - f(-\theta)}{2}\hat{n} \cdot \vec{\sigma} \end{align}

I have tried expressing $f$ as a Fourier Series in order to exploit the known fact that \begin{align} \exp(i\theta \hat{n} \cdot \vec{\sigma}) = \cos\theta I + i\sin\theta \hat{n} \cdot \vec{\sigma} \end{align} but I'm not sure if matrix-valued Fourier Series are even a thing. Could I please get some suggestions on how to prove this?

  • $\begingroup$ I may be wrong, but it seems $\hat n\cdot \vec \sigma$ isn't in $\mathbb C^2$ - it's instead a matrix acting on vectors in $\mathbb C^2$... so how do you apply $f$ to it? $\endgroup$
    – Ruslan
    Dec 3, 2014 at 7:54
  • $\begingroup$ Sorry, that was a typo. Basically what I mean is that $f$ acts on $\theta \hat{n} \cdot \vec{\sigma}$ and the output is something in $M(2,\mathbb{C})$. $\endgroup$
    – Paradox
    Dec 3, 2014 at 10:05
  • $\begingroup$ Wait, how do you then apply $f$ to $\theta$? Do you mean to apply it to $\theta I$ instead? $\endgroup$
    – Ruslan
    Dec 3, 2014 at 10:38
  • $\begingroup$ Yes that's a good point... I overlooked that before... I think you have to assume that $f$ can take both matrices or numbers as inputs (e.g. the exponential). The exact quote from "Quantum Computation and Quantum Information" - Nielsen & Chuang is "Let f(.) be any function from the complex numbers to the complex numbers". $\endgroup$
    – Paradox
    Dec 4, 2014 at 1:08

1 Answer 1


First thing to note is that, eigen values of $\hat{n}\cdot \vec{\sigma}$ is 1 and -1. Because,

$$\hat{n}\cdot \vec{\sigma} = \begin{bmatrix} n_3 & n_1-n_2i\\ n_1 + n_2i & -n_3 \end{bmatrix} $$ $$ \det(\hat{n}\cdot \vec{\sigma}-\lambda I) = \lambda^2 - (n_1^2 + n_2^2 + n_3^2) = \lambda^2 - 1 = 0 $$ By spectral decomposition theorem, we have two orthonormal eigen vectors and diagonalization as: $$\hat{n}\cdot \vec{\sigma} = |e_1\rangle\langle e_1| - |e_2\rangle\langle e_2|$$ By Sylvester's formula for functions that are analytic functions. We have $$f(\theta~\hat{n}\cdot \vec{\sigma}) = f(\theta) |e_1\rangle\langle e_1| + f(-\theta) |e_2\rangle\langle e_2|$$

Since the same orthonormal vectors form a basis, we have completeness relation that gives: $$ I = |e_1\rangle\langle e_1| + |e_2\rangle\langle e_2| $$ That implies, $$ |e_1\rangle\langle e_1| = \frac{I + \hat{n}\cdot \vec{\sigma}}{2}$$ $$ |e_2\rangle\langle e_2| = \frac{I - \hat{n}\cdot \vec{\sigma}}{2}$$

Substituting, we get: $$f(\theta~\hat{n}\cdot \vec{\sigma}) = f(\theta)\frac{I + \hat{n}\cdot \vec{\sigma}}{2} + f(-\theta)\frac{I - \hat{n}\cdot \vec{\sigma}}{2} $$

Rewriting: $$f(\theta~\hat{n}\cdot \vec{\sigma}) = \frac{f(\theta)+f(-\theta)}{2}I + \frac{f(\theta)-f(-\theta)}{2}\hat{n} \cdot \vec{\sigma} $$


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