Let $N\geq1$ be an integer and let us say that two matrices $U,V\in U(N)$ are related if $|U_{ij}|=|V_{ij}|$ for all indices $1\leq i,j\leq N$. When exactly are two unitary matrices related in this sense? Can we describe the symmetry group (of some kind of transformations of a matrix) that preserves norms of elements?

It is easy to verify that if $A,B$ are diagonal unitary matrices, then $U$ and $AUB$ are related. These two-sided multiplications by diagonal unitary matrices give a symmetry group of dimension $2N-1$. If $N\leq2$, one can show that if $U,V\in U(N)$ are related, then there are diagonal unitary matrices $A,B$ so that $V=AUB$. For $N\geq3$ this is no longer true; linearization and calculating some matrix ranks suggests that for a generic matrix $U\in U(3)$ the dimension of the set of matrices related to $U$ is 6 instead of 5. Where does this extra dimension come from and can we give a nice (= easy to calculate with) description of the corresponding symmetry?

If you need to make any genericity assumptions (such as all elements of the matrices being nonzero), do so. I'm not that interested in exceptional matrices. All dimensions $N\geq3$ are of interest to me, but if that is too complicated, I'm happy with the case $N=3$.

This problem turned up in something related to quantum mechanics. It sometimes happens that we can measure (easily or at all) only the norms of elements of a unitary matrix. We would like to know the full matrix, but we only know it up to some gauge — the problem is finding this gauge symmetry. This is related to finding the CKM and PMNS matrices.


I don't think that I have a full solution to this problem (and I don't think that it yet exists), however, for $N=3$ I think it is possible to make a sufficient comment.

First, one can note that each unitary matrix $U\in U(N)$ can be uniquely presented in the following form: \begin{align} U = \begin{bmatrix} e^{i\theta_1} & 0 & 0 & \cdots & 0\\ 0 & e^{i\theta_2} & 0 &\cdots &0 \\ \vdots & \vdots & \vdots &\ddots & \vdots\\ 0 & 0 & 0 & \cdots & e^{i\theta_n} \end{bmatrix} \tilde{U} \begin{bmatrix} 1 & 0 & 0 & \cdots & 0\\ 0 & e^{i\mu_2} & 0 &\cdots &0 \\ \vdots & \vdots & \vdots &\ddots & \vdots\\ 0 & 0 & 0 & \cdots & e^{i\mu_n} \end{bmatrix}, \end{align} where $\tilde{U}_{1j} = |U_{1j}|, \, \tilde{U}_{i1} = |U_{i1}|, \, i,j\in \{1,N\}$. Indeed, the first column and the first row of $U$ have $2N-1$ phases, which can be turned to identities by the $2N-1$ phases of $\theta,\, \mu$. One just need to solve a linear system to do the decomposition.

So now one may ask yourself, is it possible to reconstruct the unitary matrix $\tilde{U}$, knowing the $|U_{i,j}|$ and the fact that first row and first column are real and non-negative? For $N=3$ it looks possible.

So, for $N=3$ the unitary matrix $\tilde{U}$ is presented in the following form: \begin{equation} \tilde{U} = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ u_{21} & u_{22}e^{i\phi_{22}} & u_{23}e^{i\phi_{23}} \\ u_{31} & u_{32}e^{i\phi_{32}} & u_{33}e^{i\phi} \end{bmatrix}, \end{equation} where $\phi_{ij}$ are the phases we want to reconstruct from the knowledge of $u_{ij}$ in the formlua above. From the orthogonality relations for the columns of $\tilde{U}$ one can write down the the non-linear system of complex equations: \begin{equation} \begin{cases} u_{11}u_{12} + u_{21}u_{22}e^{i\phi_{22}} + u_{31}u_{32}e^{i\phi_{32}} = 0,\\ u_{11}u_{13} + u_{21}u_{23}e^{i\phi_{23}} + u_{31}u_{33}e^{i\phi_{33}} = 0,\\ u_{12}u_{13} + u_{22}u_{23}e^{i(-\phi_{22} + \phi_{23})} + u_{32}u_{33}e^{i(-\phi_{32} + \phi_{33})} = 0 \end{cases} \end{equation} It is important that the system is non-linear, so the linearization argument for phases $\phi_{ij}$ doesn't work here, because for $\phi_{ij}\rightarrow 0$ matrix $\tilde{U}$ immediately appears to be non-unitary. In fact, such system has a finite number of solutions (in fact no more than two). It can be proved just by school methods of solving this system, so I show it here.

I will just consider the first equation from the system, which gives me two real equations for real and imaginary parts, respectively: \begin{equation} \begin{cases} u_{21}u_{22}\sin(\phi_{22}) + u_{31}u_{32}\sin(\phi_{32}) = 0, \\ u_{11}u_{12} + u_{21}u_{22}\cos(\phi_{22}) + u_{31}u_{32}\cos(\phi_{32}) = 0 \end{cases}. \end{equation} Solving the above system we obtain: \begin{align} \cos(\phi_{22}) &= \dfrac{u^2_{31}u^2_{32} - u^2_{21}u^2_{22}-u_{11}^2u^2_{12}} {2u_{11}u_{12}u_{21}u_{22}}, \\ \cos(\phi_{33}) &= \dfrac{u^2_{21}u^2_{22} - u^2_{31}u^2_{32}-u_{11}^2u^2_{12}} {2u_{11}u_{12}u_{31}u_{32}}. \end{align} Note these exact solutions imply that number of solutions of $\phi_{ij}$ is no more than finite and the reconstruction formulas are direct. One can also note that the non-linear complex system remains valid under complex conjugation, so if $\phi_{ij}$ is a solution, than $-\phi_{ij}$ is also a solution.

Discussion So what does this result say about our problem? The above considerations are taken from the work:

Auberson, G., Andre Martin, and G. Mennessier. "On the reconstruction of a unitary matrix from its moduli." Communications in mathematical physics 140.3 (1991): 523-542.

In the introduction it is written, that for $N=3$ the number of solutions of the aforementioned system is exactly one, up to complex conjugation (exactly what we spoke about in the end).

From this, the following result follows:

Let $U,V$ be unitary matrices from $U(3)$, which have the same moduli of their elements. Then there exist $(\theta_1, \dots, \theta_3), \, (\mu_2, \mu_3)$ such that: \begin{equation} U = \mathrm{diag}(e^{i\theta_1},e^{i\theta},e^{i\theta_3})V \mathrm{diag}(1,e^{i\mu_1},e^{i\mu_2}) \end{equation} or \begin{equation} U = \mathrm{diag}(e^{i\theta_1},e^{i\theta},e^{i\theta_3})V^* \mathrm{diag}(1,e^{i\mu_1},e^{i\mu_2}), \end{equation} where * denotes the complex conjugation.

It implies that dimension of the group is 5, when $N=3$.

Further discussions: For $N >3$ the result of Auberson, G., claims that there are cases, when the number of the solutions of the related non-linear system is infinite (or even a continuum), which implies that action of the group cannot be represented by $2N-1$ phases and finite number of nice "sandwich formulas".

  • $\begingroup$ Also, for your interest I will give the following link: arxiv.org/pdf/math-ph/0603077.pdf. Authors consider more similair factorization, but for the case of unistochastic matrices. In section 2 they discuss the gauge invariance in this case. $\endgroup$ – Fedor Goncharov Jul 7 '17 at 15:13

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.