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Let's say I have two unitary operations $U_1$ and $U_2$, which together give a rotation of the following form:

$$ U_1\cdot U_2 = \begin{pmatrix} e^{i\varphi} & 0 \\ 0&e^{-i\varphi} \end{pmatrix} $$

Now I want to insert a 3rd unitary matrix of the form $U_3=\begin{pmatrix} e^{i\varphi_1} & 0 \\ 0&e^{-i\varphi_1} \end{pmatrix}$ between them and chose $\varphi_1$, such that:

$$U_1 U_3 U_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$$

I think this should not be possible, because just between these two matrices, the reference frame in which $\varphi_1$ acts might be completely different than whatever is the reference frame behind the two unitaries. What do you think?

1 Answers1

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Let unitary matrices $U_1,U_2$ with $U_1U_2=\operatorname{diag}(e^{i\varphi},e^{-i\varphi})$ be given and let there exist $\varphi_1\in[0,2\pi)$ such that $U_3=\operatorname{diag}(e^{i\varphi_1},e^{-i\varphi_1})$ satisfies $U_1U_3U_2=\operatorname{id}_2$. This is anything but impossible, even beyond the trivial case of $U_1,U_2$ both being diagonal from the start.

The first equation implies $U_1=\operatorname{diag}(e^{i\varphi},e^{-i\varphi})U_2^\dagger$ which combined with the second identity yields $$ \begin{pmatrix}e^{i\varphi}&0\\0&e^{-i\varphi}\end{pmatrix}U_2^\dagger U_3U_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}\quad\Rightarrow\quad U_2^\dagger \begin{pmatrix}e^{i\varphi_1}&0\\0&e^{-i\varphi_1}\end{pmatrix}U_2=\begin{pmatrix}e^{-i\varphi}&0\\0&e^{i\varphi}\end{pmatrix}\,.\tag{1} $$ Now it is well known that a unitary transformation ($A\to U^\dagger AU$) leaves the eigenvalues invariant which forces $\boxed{\varphi_1=\varphi\text{ or }\varphi_1=-\varphi}$. Now this makes for a handful of different cases depending on the original $\varphi$:

  • If $\varphi=0$ then $U_2$ can be anything, $U_1=U_2^\dagger=U_2^{-1}$ and $U_3=\operatorname{id}_2$.
  • The more interesting case is $\varphi\in (0,2\pi)$. If $\varphi=-\varphi_1$ then (1) forces $[\operatorname{diag}(e^{-i\varphi},e^{i\varphi}),U_2]=0$ so $U_2$ has to be diagonal, the same then for $U_1$ (as $U_1$ then is a product of two diagonal matrices.
  • Similarly, if $\varphi\in (0,2\pi)$ with $\varphi=\varphi_1$ then $U_2=FD$ where $\scriptsize F=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is the flip matrix and $D$ is any diagonal unitary matrix.
Frederik vom Ende
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