1
$\begingroup$

I'm curious about for which $n$ there exists a $2\times 2$ unitary matrix $U(n)$ such that for $1 \le k \le n$ $$|U^k_{1,0}(n)|^2 = \frac kn,$$ where $U^k_{1,0}(n)$ is the lower left element of the $k$th power of $U(n)$.

The idea is that if such a matrix exists, then applying it $k$ times to the quantum state $|0\rangle$ will produce the quantum state $\sqrt{1-k/n}|0\rangle + \sqrt{k/n}|1\rangle$ modulo some phases, a "quantum coin" with probability $k/n$ of giving outcome 1.

It's easy to see that it exists for $n=2$, with an example being $$U(2) = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix},$$ but I couldn't find an example for any other $n$. A bit of brute force shows that no such matrices exist for $n=3$, and suggests that indeed no other $n$ works. Perhaps there's an elegant way to prove this? Or maybe there is a construction that I missed?


I originally posted this question on Mathematics Stack Exchange, but nobody was interested about it there.

$\endgroup$
1
$\begingroup$

Let's try a general construction. We know that, to satisfy the $k=1$ instance, we need $U_{1,0}=1/\sqrt{n}$, up to a possible phase. But we can always take one of the elements of $U$ to be real by removing a global phase.

We also know that the sum mod-square of the rows and columns for a unitary is 1, and that the rows must be mutually orthogonal. This gives us a lot of structure for $U$. $$ U=\frac{1}{\sqrt{n}}\left(\begin{array}{cc} \sqrt{n-1}e^{i\theta_1} & -e^{i(\theta_4+\theta_1)} \\ 1 & \sqrt{n-1}e^{i\theta_4} \end{array}\right). $$ Now we just start multiplying it together. We find that $$ |U_{1,0}^2(n)|^2=\frac{4(n-1)}{n^2}\cos^2((\theta_1-\theta_4)/2). $$ So, for all $n\geq 2$, there's a solution for $\theta_1-\theta_4$ that gives a working $k=2$. Next, $k=3$: $$ |U_{1,0}^3(n)|^2=\frac{(2 (n-1) \cos (\theta_1-\theta_4)+n-2)^2}{n^3} $$ Since we already know the value of $\cos^2((\theta_1-\theta_4)/2)=n/(2(n-1))$, we can use the double angle formula to find $\cos (\theta_1-\theta_4)=1/(n-1)$. Hence $$ |U_{1,0}^3(n)|^2=\frac{1}{n}\neq\frac{3}{n}. $$ Hence, this only works for $n=1,2$ where we do not require that the $k=3$ condition is satisfied.


For a more elegant version, note that if the relation holds for $k=n$, it must be that $$ U^n=e^{i\phi}\left(\begin{array}{cc} 0 & e^{-i\theta} \\ e^{i\theta} & 0 \end{array}\right)=e^{i\phi}(\cos\theta X+\sin\theta Y). $$ $U$ must be an $n^{th}$ root of this, so $$ U^k=e^{ik\phi/n}\left(I\cos(k\pi/(2n))+i\sin(k\pi/(2n))(\cos\theta X+\sin\theta Y)\right). $$ In other words, $$ |U^k_{1,0}|^2=\sin^2(k\pi/(2n))\neq \frac{k}{n}, $$ with the exception of $k=1,n=2$ which, as we already know, does work. (There are a few phase freedoms floating around here, but they don't make a difference to the final answer. I don't want to worry about them because it detracts from the simple concept of this answer.)

$\endgroup$
6
  • $\begingroup$ Your "more elegant" version is not correct. It excludes the case $k=1,n=2$, for which the construction does work. A possible solution for $|U^k_{1,0}|^2$ is $\sin^2(k\pi/(2n))$, but this is not the most general possible, as for $k=1,n=3$ it does not allow for $X^3 = X$. $\endgroup$ – Mateus Araújo Dec 7 '20 at 13:38
  • $\begingroup$ @MateusAraújo I was being lazy about the one case that we do know does work. As for your second comment, this was what I meant by "there are a few phase freedoms" One of those is that the rotation angle of $U$ can be any $\alpha$ such that $n\alpha\text{ mod }2\pi=\pi$. $\endgroup$ – DaftWullie Dec 7 '20 at 13:49
  • $\begingroup$ But it seems I did miss a factor of 1/2 in the angle of rotation. $\endgroup$ – DaftWullie Dec 7 '20 at 13:54
  • $\begingroup$ But the missing phases do matter. In general you have $|U^k_{1,0}|^2 = \sin^2(\frac{k\pi}{n}(\frac12 + l))$ for any integer $l$. $\endgroup$ – Mateus Araújo Dec 7 '20 at 14:05
  • $\begingroup$ I agree they give different numbers. I don't believe they give a different conclusion. $\endgroup$ – DaftWullie Dec 7 '20 at 14:13
1
$\begingroup$

This is a correct version of the "elegant" version of DaftWullie's answer.

If the relation holds for $k=n$, then $$ U^n = \begin{pmatrix} 0 & e^{-i\theta} \\ e^{i\theta} & 0 \end{pmatrix}, $$ modulo a global phase. Since the eigenvalues of $U^n$ are $\pm 1$, we have that $(I+U)/2$ and $(I-U)/2$ are orthogonal projectors, and thus $$ U = e^{i \frac{2\pi a}{n}}(I+U^n)/2 + e^{i(\frac\pi n+\frac{2\pi b}{n})}(I-U^n)/2$$ for some integers $a,b$. The $k$th power is then just $$ U^k = e^{i \frac{2\pi k a}{n}}(I+U^n)/2 + e^{i(\frac{\pi k} n+\frac{2\pi k b}{n})}(I-U^n)/2$$ Modulo a global phase, $U^k$ is then equal to $$ {U'}^k = I(1+e^{i(\frac{\pi k} n+\frac{2\pi k l}{n})})/2 + U^n(1-e^{i(\frac{\pi k} n+\frac{2\pi k l}{n})})/2,$$ for some integer $l$, which implies that $$|U^k_{1,0}|^2 = \sin^2\left(\frac{k\pi}{n}\left(\frac12+l\right)\right).$$ This will be equal to $k/n$ for $l=0$ and $n=2k$, for example. In general it doesn't work, though, because for $k=1$ we need that $$\frac{\pi}{n}\left(\frac12+l\right) = \pm\arcsin\frac{1}{\sqrt{n}}, $$ but $\arcsin\frac{1}{\sqrt{n}}$ is only a rational multiple of $\pi$ for $n=1,2,$ and $4$.

The cases $n=1,2$ do work, but for $n=4$ we would need $$\frac{\pi}{4}\left(\frac12+l\right) = \pm\frac{\pi}{6}, $$ which is not possible for integer $l$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.