I am wondering about the following mathematical question, motivated by a physics problem from topological band theory (a bit more on that at the very end). For now let me strip all the unimportant physics jargon.


Given a Hermitian matrix $H$ (i.e. $H = H^\dagger$), the question is whether there is a symmetric unitary matrix $U$ (i.e. $U = U^\top$ and $U\cdot U^\dagger = \mathbf{1}$) such that $H^\top = U\cdot H \cdot U^\dagger$ (Note: in my original question, the dagger at $U$ has been missing -- my mistake). The Hermitian matrix $H$ can be assumed to be positive definite, if it simplifies the problem.

There are in fact two questions:

(1) What is the condition on $H$ that guarantees that such symmetric unitary $U$ exists?

(2) Given that $U$ exists, is it possible to find it using some constructive algorithm?


Let me mention a few examples for $2\times 2$ matrices.

Using the Pauli matrices as basis, one easily finds that if $H$ is a symmetric matrix (i.e. linear combination of $\mathbf{1}$, $\sigma_x$ and $\sigma_z$), then the identity matrix $U = \mathbf{1}$ has this property. On the other hand, for $H=\sigma_y$, the choice $U = \sigma_x$ leads to $\sigma_x \cdot \sigma_y \cdot \sigma_x^\dagger = -\sigma_y = (\sigma_y)^\top$, which solves the problem.

However, for general matrices and especially of higher dimensions, I found it rather difficult to state anything specific.


Let me also mention a bit on the physics background:

One might wonder whether a model for spinless electrons hopping between a collection of sites respects time-reversal symmetry. The time-evolution of the electron states is modelled by a Hermitian Hamiltonian matrix $H$. It follows from more general considerations that spinless time-reversal symmetry is an ($i$) antiunitary symmetry, that ($ii$) commutes with the Hamiltonian, and that ($iii$) squares to $+\mathbf{1}$.

Since any antiunitary symmetry [condition ($i$)] can be expressed as a unitary followed by complex conjugation, it follows from ($iii$) that the unitary part is symmetric. Then condition ($ii$) then translates into the posed mathematical problem. I am thus trying to check whether the Hamiltonian $H$ exhibits spinless time-reversal symmetry.

While the symmetry is usually very apparent for a physical motivated basis of the Hilbert space, it could be that the basis in which I am given the Hamiltonian contains complicated phase factors, which make the time-reversal symmetry difficult to reveal.

  • $\begingroup$ I may be missing something, but doesn't $U=1$ work for any $H$? $\endgroup$
    – lisyarus
    Dec 4, 2019 at 17:45
  • $\begingroup$ Unfortunately it does not, because of the transpose condition. One must choose $U$ such that the imaginary part of the Hermitian $H$ is reversed by the transformation. $\endgroup$ Dec 4, 2019 at 17:48
  • $\begingroup$ Oh, didn't notice you make a specific distinction between hermitian and symmetric. $\endgroup$
    – lisyarus
    Dec 4, 2019 at 17:49

1 Answer 1


Edit. The OP has changed the equation to $H^\top=UHU^\ast$. In this case, the equation is always solvable and there is a simple solution. Since $H$ is Hermitian, we can always unitarily diagonalise it as $Q^\ast HQ=D$, where $Q$ is a unitary matrix whose columns form an orthonormal eigenbasis of $H$ and $D$ is a real diagonal matrix. It follows that $$ Q^\ast HQ=D=\overline{D}=\overline{Q^\ast HQ}=\overline{Q}^\ast H^\top\overline{Q}. $$ Hence $$ H^\top=\overline{Q}Q^\ast HQ\overline{Q}^\ast =\left(\overline{Q}Q^\ast\right)H\left(\overline{Q}Q^\ast\right)^\ast $$ and one may pick $U=\overline{Q}Q^\ast=\overline{Q}(\overline{Q})^\top$.

(Below was my old answer for the old question, in which the equation concerned was $H^\top=UHU$ for some symmetric unitary matrix $U$.)

This problem looks non-trivial. I haven't any answer but I can make some quick observations:

  • the desired $U$ exists if and only if there exists a unitary matrix $V$ such that $V^\top HV$ is real;
  • a necessary condition for the existence of $U$ is that $-iSK$ is unitarily similar to $iSK$, where $S$ and $K$ denote respectively the real and imaginary parts of $H$ (so that $S$ is real symmetric, $K$ is real skew symmetric and $H=S+iK$).

For the first bullet point, suppose that $H^\top = UHU$ for some symmetric unitary matrix $U$. Being symmetric, $U$ admits a Takagi factorisation $U=V\Sigma V^\top$, where $V$ is unitary and $\Sigma$ is a nonnegative diagonal matrix. As $U$ is unitary, $\Sigma$ is necessarily equal to $I$. Therefore $U=VV^\top$ and $H^\top = UHU$ implies that $V^\ast H^\top \overline{V}=V^\top HV$, i.e. $V^\top HV$ is real. Conversely, if $V^\top HV$ is real, then $H^\top=UHU$ for $U=VV^\top$.

When $n=2$, if $S=QDQ^\top$ is an orthogonal diagonalisation of the symmetric part $S$ of $H$ over $\mathbb R$, one can always pick $V=Q\operatorname{diag}(1,i)$ and $U=V^\top V=Q\operatorname{diag}(1,-1)Q^\top$.

For the second bullet point, suppose $V^\top HV$ is real. Then its symmetric part $V^\top SV$ and skew symmetric part $iV^\top KV$ are real, and so is $V^\top SV\left(\overline{iV^\top KV}\right)=-iV^\top SK\overline{V}$, meaning that $-iSK$ is unitarily similar to a real matrix. By Khakim Ikramov (2010), On complex matrices that are unitarily similar to real matrices, Mathematical Notes, 87(6): 821-827, a complex matrix $A$ is unitarily similar to a real matrix if and only if $A$ and $\overline{A}$ are unitarily similar. Hence $-iSK$ and $iSK$ are necessarily unitarily similar.

In particular, the nonzero eigenvalues of $-iSK$ must occur in sign pairs $(-\lambda,\lambda)$. In view of this, we see that when $n\ge3$, $H^\top=UHU$ more often than not is insolvable. E.g. when $S$ happens to be positive definite, $-iSK$ is similar to the Hermitian matrix $-iS^{1/2}KS^{1/2}$. While the nonzero eigenvalues of $-iK$ do occur in the form of $(-\lambda,\lambda)$ (because $K$ is real skew symmetric), this pairing structure is usually destroyed after a congruence via $S^{1/2}$.

  • $\begingroup$ Thank you for your valuable comments @user1551! The first part with the reality condition and the Takagi factorization I was able to work out myself (Sec. B of Supplemental Material here arxiv.org/abs/1907.10611), but the second part is entirely new to me. Unfortunately, I just noticed that one dagger $^\dagger$ has been missing in the postulated problem [see the Note in my edited version]. How does its inclusion modify your argument? $\endgroup$ Dec 7, 2019 at 13:36
  • $\begingroup$ @TomasBzdusek The equation $H^\top=UHU^\dagger$ is much easier to solve. Please see my new edit. $\endgroup$
    – user1551
    Dec 7, 2019 at 20:51

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