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Any unitary matrix $U$ can be diagonalized by another unitary matrix $V$, $$U=VDV^\dagger,$$ where $D={\rm diag}(z_1,z_2,...,z_N)$ is diagonal.

If $U$ is taken at random uniformly with respect to Haar measure, then $V$ and $D$ are independent and $D$ has the Weyl distribution, $P(D)\propto \prod_{j<k}|z_k-z_j|^2$. I would like to know what is the space of all $V$'s. Which unitary matrices are eigenvectors of unitary matrices? What is their distribution?

On the one hand I would guess that $V$ is also uniformly distributed in the unitary group, but on the other hand this seems paradoxical. Because integration over $U$ can be decomposed as integration over $D$ and $V$ and then integration over $V$ would be the same as integration over $U$ again?

I have consulted many references about this subject, but they tend to focus on the eigenvalues.

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    $\begingroup$ $V$ and $D$ are only uniquely defined up to simultaneous multiplication by (resp conjugacy with) permutation matrices; how do you choose them to make them unique? $\endgroup$
    – David Handelman
    Commented Jul 24, 2020 at 14:45

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The invariance of the Haar measure implies that the probability to draw the matrix $U$ from the unitary group is unchanged if you replace $U$ by $U_0 U U_0^\dagger$, with $U_0$ an arbitrary unitary matrix. Since this conjugation changes the unitary matrix of eigenvectors from $V$ into $U_0V$, it means that $V$ and $U_0V$ are equally probably in the unitary group, which is another way to say that the matrix of eigenvectors is uniformly distributed in the unitary group.

If you wish to carry out the integration over $U$ by first integrating over the eigenvalues $D$ and then over the eigenvectors $V$, you will need to first specify a parameterization which uniquely fixes the eigenvectors. The uniformity of the eigenvector distribution holds for any such parameterization.

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  • $\begingroup$ But then how do you decompose the measure $dU$ as the product $dDdV$ if $dV=dU$? $\endgroup$
    – thedude
    Commented Jul 24, 2020 at 15:52
  • $\begingroup$ Sorry, but I still don't get it. So to integrate over $U$, I first integrate over its eigenvalues and then over $V$. To integrate over $V$, I first integrate over its eigenvalues and then over its eigenvectors, call that $V_2$. To integrate over $V_2$, I first integrate over its eigenvalues and then over its eigenvectors, call that $V_3$. And so on. You say I must choose a parametrization. Fine. Does that help me out of this conundrum? $\endgroup$
    – thedude
    Commented Jul 24, 2020 at 16:35
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    $\begingroup$ the correct statement is that $d\mu_{H}(U)=\prod_{n<m}|e^{i\phi_n}-e^{i\phi_m}|^2\prod_k d\phi_k d\mu_{H}(V)$, where $d\mu_{H}$ is the Haar measure and the $e^{i\phi_n}$'s are the eigenvalues of $U$; the fact that the Haar measure appears on both sides of the equation, but calculated for a different matrix, is by no means a paradox or a contradiction. $\endgroup$
    – Carlo Beenakker
    Commented Jul 24, 2020 at 16:39
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The crucial point is that the eigenvectors are not unique, as David Handelman commented. There is a gauge freedom since you can change their phase at will. If you want the map $U\to(D,V)$ to be bijective, you must fix the gauge. For example, you may require that the first element of every eigenvector is real and positive. In terms of real dimension, you then have $N^2$ for $U$, $N$ for $D$ and $N^2-N$ for $V$.

After fixing the gauge, you can decompose the integral over $U$ as integral over $D$ and $V$. But it is no longer true that $V$ is uniformly distributed in the unitary group.

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