Given two 1-qubit rotations $U=R_n (\theta)$ and $V=R_m(\phi)$ with $n$ and $m$ vectors defining a rotation and $\theta, \phi$ angles, define $D(U,V)=Tr(|U-V|)$ where $|U-V|=\sqrt{(U-V)^\dagger (U-V)}$ and $E(U,V)=max_{|\psi \rangle} ||(U-V)|\psi \rangle ||$ where $|| |\psi\rangle ||$ is the vector norm.

Im trying to prove that $D(U,V)=2E(U,V)$, which can be seen from taking $n$ and $m$ as the same vector. I also tried considering the case $U=R_z(\theta)$ and taking $m=\cos(\alpha)Z + \sin(\alpha)X$ but in this case I couldn't prove the result as I get a too complicated expression for the trace distance. Any help for proving this?


2 Answers 2


Let's start with expanding the calculation of $E$: $$ E(U,V)=\max_{|\psi\rangle}\sqrt{\langle\psi|(U-V)^\dagger(U-V)|\psi\rangle}. $$ Clearly, we want $|\psi\rangle$ to be the eigenvector with maximum eigenvalue of $$ 2I-V^\dagger U-U^\dagger V. $$ Let's note that

  • If $|\psi\rangle$ is an eigenvector of $U^\dagger V$, then the eigenvalue must be of the form $e^{i\theta}$ because $U^\dagger V$ is unitary

  • Furthermore, it is also an eigenvector of $V^\dagger U$, but with eigenvalue $e^{-i\theta}$ just by rearranging (multiplying by $V^\dagger U$) $U^\dagger V|\psi\rangle=e^{i\theta}|\psi\rangle$.

  • For a single-qubit rotation where $\text{det}(U)=\text{det}(V)=1$, there is a second eigenvector $|\psi^{\perp}\rangle$ whose eigenvalue is the conjugate (i.e. $e^{-i\theta}$ for $U^\dagger V$ and $e^{i\theta}$ for $V^\dagger U$) because the eigenvalues must have a product equal to 1.

So, $$ E(U,V)=\sqrt{2-e^{i\theta}-e^{-i\theta}}=2\cos\frac{\theta}{2} $$

Now let's consider $D(U,V)$. $$ D(U,V)=\text{Tr}\left(\sqrt{(U-V)^\dagger(U-V)}\right). $$ When taking the trace, we can use any basis we want. Let's use the eigenbasis of $2I-V^\dagger U-U^\dagger V$, $$ D(U,V)=\langle\psi|\sqrt{(U-V)^\dagger(U-V)}|\psi\rangle+\langle\psi^\perp|\sqrt{(U-V)^\dagger(U-V)}|\psi^\perp\rangle. $$ Square roots maintain the diagonalisation, which means we can do $$ D(U,V)=\sqrt{\langle\psi|(U-V)^\dagger(U-V)|\psi\rangle}+\sqrt{\langle\psi^\perp|(U-V)^\dagger(U-V)|\psi^\perp\rangle}, $$ and we can now make use of the eigenvector relations \begin{align*} D(U,V)&=\sqrt{2+e^{i\theta}+e^{-i\theta}}+\sqrt{2+e^{-i\theta}+e^{i\theta}} \\ &=4\cos\frac{\theta}{2} \\ &=2E(U,V). \end{align*}

For a completely different way to tackle the problem, let $U=e^{i\theta\underline{n}\cdot\underline{\sigma}}$ and $V=e^{i\phi\underline{m}\cdot\underline{\sigma}}$. Let's multiply out $U^\dagger V$. We get $$ U^\dagger V=\cos\theta\cos\phi I-i\underline{n}\cdot\underline{\sigma}\cos\phi\sin\theta+i\underline{m}\cdot\underline{\sigma}\sin\phi\cos\theta+(\underline{n}\cdot\underline{\sigma})\cdot(\underline{m}\cdot\underline{\sigma})\sin\theta\sin\phi. $$ Similarly, $$ V^\dagger U=\cos\theta\cos\phi I+i\underline{n}\cdot\underline{\sigma}\cos\phi\sin\theta-i\underline{m}\cdot\underline{\sigma}\sin\phi\cos\theta+(\underline{m}\cdot\underline{\sigma})\cdot(\underline{n}\cdot\underline{\sigma})\sin\theta\sin\phi. $$ So, when we add these two terms together, the middle two terms cancel immediately. The last term requires a little more thought. Note that $$ (\underline{n}\cdot\underline{\sigma})\cdot(\underline{m}\cdot\underline{\sigma})=(\underline{n}\cdot\underline{m})I+(\underline{n}\times\underline{m})\cdot\underline{\sigma}. $$ Also recal that the cross product is antisymmetric, so when we swap the order of terms, we get a negative sign. That means that the cross product term will also cancel. Hence $$ U^\dagger V+V^\dagger U=2I(\cos\theta\cos\phi+\underline{n}\cdot\underline{m})\sin\theta\sin\phi). $$ Overall, we have $$ 2I-U^\dagger V-V^\dagger=2I(1-\cos\theta\cos\phi-\underline{n}\cdot\underline{m}), $$ which makes both the maximum eigenvector and the trace very easy to deal with and, critically, the trace of $I$ is double the maximum eigenvector.


Use the fact that both distance measures are invariant under left- as well as right-multiplication (independently!) with an arbitrary unitary.

This way, you can map $U$ to $I$ and $V$ to $R_z(\phi)$. Now (i) the matrices are both diagonal, making the trace distance trivial to compute, and (ii) the problem is only characterized by a single angle.

This will simplify the problem tremendously.

  • $\begingroup$ Thank you! This really helps. Just to make sure I understand: Basically $Tr|U-V| = Tr|I-U^{\dagger} V|$. Letting $W=U^{\dagger}V$ we have to prove now that $D(I, W)=2E(I, W)$. Then we can multiply by both sides with some rotation that turns $W$ into $R_z$ $\endgroup$
    – Apo
    May 5, 2020 at 0:31
  • 1
    $\begingroup$ @Pam Precisely. Why do the full work if you can move to a convenient basis? $\endgroup$ May 6, 2020 at 14:19

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