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.
Best Answer
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 $\:\theta$.