Past exam of the mathematical methods I course of the University of Oxford / 2015 / 12

Direct multiplication of the Pauli matrices gives ${\sigma }_j^2=I$, ${\sigma }_1{\sigma }_2=i{\sigma }_3$ and its cyclic permutations; reversing unequal factors changes the sign. Hence, using the Levi-Civita symbol,Repeated $\ell$ is summed, while the Kronecker delta handles the equal-index case.

For $v=(a,b{)}^T$, multiplying each Pauli matrix between $v^{†}$ and $v$ givesAll three components are real. Moreover $|x|=(|a{|}^2+|b{|}^2)/2=\Vert v{\Vert }^2/2$. Thus normalized two-component spinors map to the radius-$1/2$ version of the Bloch sphere; the usual unit-radius convention multiplies these coordinates by two.

Each Pauli matrix is Hermitian and the coefficients ${\alpha }_j$ are real, so $T^{†}=T$. Taking the conjugate transpose of the absolutely convergent matrix exponential series yieldsThe exponents $-iT/2$ and $iT/2$ commute. Therefore $U^{†}U=e^{-iT/2+iT/2}=I$, proving that $U$ is unitary.

If $Tv=tv$ with $v\ne 0$, induction gives $T^{k}v=t^{k}v$ for every nonnegative integer $k$. Substituting into the matrix exponential series givesThus each eigenvector of $T$ is an eigenvector of $U$, with eigenvalue $e^{it/2}$. Different eigenvalues of $T$ can produce the same exponential eigenvalue.

The anticommuting unequal Pauli matrices cancel the cross terms, leavingAlso $\mathrm{tr}T=0$, so its eigenvalues are $\pm |\alpha |$. ConsequentlyFor $|\alpha |>0$, the matrix exponential for a scalar square also gives $U=I\cos (|\alpha |/2)+iT\sin (|\alpha |/2)/|\alpha |$. At $\alpha =0$, its continuous limit is $U=I$. The two exponential eigenvalues coincide when $|\alpha |=2\pi k$, making $U=(-1{)}^{k}I$.

With only ${\alpha }_3$ nonzero, $T=\mathrm{diag}({\alpha }_3,-{\alpha }_3)$ andFor ${\alpha }_3\notin 2\pi \mathbb{Z}$, the eigenvectors are the nonzero multiples of $(1,0{)}^T$ and $(0,1{)}^T$. When ${\alpha }_3=2\pi k$, $U=(-1{)}^{k}I$ and every nonzero vector is an eigenvector.

The transformed components are $a^{\prime }=e^{i{\alpha }_3/2}a$, $b^{\prime }=e^{-i{\alpha }_3/2}b$. Hence $(a^{\prime }{)}^{\ast }{b}^{\prime }=e^{-i{\alpha }_3}{a}^{\ast }b$, while the two moduli are unchanged. Taking real and imaginary parts in the expression from part (ii) givesThis is a right-handed active rotation through $-{\alpha }_3$ about the third axis. The minus sign follows from the positive sign in $U=e^{+iT/2}$; changing that convention reverses the rotation. A $2\pi$ spatial rotation changes the spinor sign while leaving the three-vector unchanged.