# Prove that for one-qubit unitaries $\text{Tr}|U-V|=2\max_\psi\|(U-V)|\psi\rangle\|$

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?

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.

• 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 $R_z$
– Apo
May 5, 2020 at 0:31
• @Pam Precisely. Why do the full work if you can move to a convenient basis? May 6, 2020 at 14:19