I'm having trouble fully wrapping my head around unitary matrices. I'm working on them in relation to quantum mechanics. The question specifically I am working on is:

Given the Pauli matrices $\sigma_x$, $\sigma_y$ and $\sigma_z$, write down explicitly the operators $e^{it\sigma_x}$, $e^{it\sigma_y}$ and $e^{it\sigma_z}$. Verify that these are unitary.

So the first part, I think I have, using Jordan blocks to apply the function $e$ to the operators. But how to verify that they are unitary? This is what I know about unitary operators:

  • $VV^\dagger$=$V^\dagger V$=$I$ - we can use this if all the matrices are diagonalisable, they are commutative and we can 'cancel' the 2 unitary operators.
  • $U$ is diagonalisable and hence unitary iff $U=VDV^\dagger$, where $V$ is unitary, and $D$ is diagonal and unitary, and we know from linear algebra that $D$ and $U$ are similar matrices.

But this is where I get confused! Is this not just an endless loop of showing what is unitary, because don't we have to show that $V$ is also unitary? And so where does it stop?

I just really don't know how to find the unitary matrix for the given questions. Do I have to find the eigenvectors of $e^{it\sigma_x}$, and this would make up the basis of $U$, then I can easily find $U^\dagger$? because I did this for the $\sigma_x$ matrix, and got eigenvectors of $0$, which can't make up the matrix.

  • $\begingroup$ you just need to show that $VV^\dagger=V^\dagger V =1$, where $V^\dagger$ is the transpose conjugate. If this is true then $V$ is unitary and hence (unitarily) diagonalizable $\endgroup$
    – glS
    Commented May 14, 2018 at 11:45
  • $\begingroup$ @Frobenius thanks for the hint! im struggling to see where (01) came from though, could you add a bit more detail to that? I can see the use of Eulers formula, but I cant see where identity and sigma come into it in the way they do. $\endgroup$
    – Learn4life
    Commented May 15, 2018 at 9:56

3 Answers 3


Although a hint to reach equation (01) of my comment is given by ohneVal in her/his answer, I'll give a proof found in many textbooks.

So, let $\:\sigma\:$ any finite square complex matrix (not necessarily $\:2\times 2\:$ hermitian like the Pauli ones) with property \begin{equation} \sigma^{2}=I=\text{identity} \tag{A-01} \end{equation} In general \begin{equation} e^{it\sigma}=\cos\left(t\sigma\right)+i\sin\left(t\sigma\right) \tag{A-02} \end{equation} But here based on the series expressions of the trigonometric functions \begin{align} \cos z & = 1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots\ =\ \sum^{\infty}_{k=0}\frac{(-1)^kz^{2k}}{(2k)!} \tag{A-03a}\\ \sin z & = z-\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\ =\ \sum^{\infty}_{k=0}\frac{(-1)^kz^{2k+1}}{(2k+1)!} \tag{A-03b} \end{align} replacing $\:z \longrightarrow t\sigma\:$ in (A-03) we have(1) \begin{align} \cos\left(t\sigma\right)=& \sum^{\infty}_{k=0}\frac{(-1)^k(t\sigma)^{2k}}{(2k)!} =I\cos t \tag{A-04a}\\ \sin\left(t\sigma\right) =&\sum^{\infty}_{k=0}\frac{(-1)^k(t\sigma)^{2k+1}}{(2k+1)!}=\sigma\sin t \tag{A-04b} \end{align} and (A-02) is written \begin{equation} e^{i t \sigma}=I\cos t+i\sigma\sin t \tag{A-05} \end{equation} For your information : \begin{equation} H = \text{hermitian} \Longrightarrow U= e^{i\mathrm H}=\text{unitary and } \det U=e^{i\cdot tr(H)} \tag{A-06} \end{equation} where $\:tr(H)\:$ the trace of $\:H$, a real number.

Note that the Pauli matrices are hermitian and traceless.

(1) From (A-01) \begin{align} I & = \sigma^{0} = \sigma^{2}= \sigma^{4} = \sigma^{6}=\cdots=\sigma^{2k} \tag{A-07a}\\ \sigma & =\sigma^{3}= \sigma^{5} = \sigma^{7}=\cdots=\sigma^{2k+1} \tag{A-07b} \end{align}

(2) If $\:\mathbf{n}=(n_{1},n_{2},n_{3}) \in \mathbb{R}^{3}$ , $\Vert\mathbf{n}\Vert^{2}=n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1 $, is a real unit 3-vector, and $\:\boldsymbol{\sigma}=(\sigma_{1},\sigma_{2},\sigma_{3})\:$ are the three Pauli matrices then the matrix \begin{equation} \left(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\right) = n_{1}\sigma_{1}+n_{2}\sigma_{2}+n_{3}\sigma_{3}= \begin{bmatrix} n_3 & n_1-in_2 \\ n_1+in_2 &-n_3 \end{bmatrix} \tag{A-08} \end{equation} is hermitian and traceless, as the Pauli matrices are, and moreover \begin{equation} \left(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)^{2} = I \tag{A-09} \end{equation} So if $\:t \in \mathbb{R}\:$ then from (A-05) \begin{equation} U \equiv e^{i t \left(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)}=I\cos t+i\left(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\sin t \tag{A-10} \end{equation} is a unitary matrix. Replacing (not accidentally) $\:t \longrightarrow -\theta/2\:$ \begin{equation} U\left(\mathbf{n},\theta\right) \equiv e^{-i\tfrac{\theta}{2} \left(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)}=I\cos \left(\theta/2\right)-i\left(\mathbf{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\right)\sin \left(\theta/2\right) \tag{A-11} \end{equation} which is the (special) unitary matrix representation of a rotation around $\:\mathbf{n}\:$ through an angle $\:\theta$.


By definition the only thing you need to check is that $V V^\dagger = V^\dagger V = \mathbb{I}$. You can use the Taylor expansion of the exponential, along with the property $\sigma_i^2=\mathbb{I}$ for $i=x,y,z$ to write concrete expressions for your operators ($e^{it\sigma_j}$).

Alternatively take the adjoint (dagger) on both sides and use its linearity along with the properties of the Pauli matrices to simplify the terms. You should be able to prove that $$\left(e^{it\sigma_j}\right)^{\dagger} = e^{-it\sigma_j}$$ from which unitarity follows.


Here's a slightly more abstract, but very general proof.

First prove to yourself that $\exp(A)^† = \exp(A^†)$. You can do this with the definition of the matrix exponential.

Next prove that $\exp(a_1 A)\exp(a_2 A) = \exp((a_1+a_2)A)$. This should be similar to proving that $e^a e^b = e^{a+b}$ for normal exponentiation. Also note that $\exp(A)\exp(B) \ne \exp(A+B)$ for matrices in general.

Finally, show that $(i\sigma)^† = -i\sigma$, and use that together with the previous two steps to show that $\exp(i\sigma) \exp(i\sigma)^† = \exp(0) = I$.


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