# Condition for a Hermitian matrix to fulfill $H^\top = U \cdot H\cdot U$ for some symmetric unitary $U$.

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.

$$\phantom{.}$$

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?

$$\phantom{.}$$

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.

$$\phantom{.}$$

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.

• I may be missing something, but doesn't $U=1$ work for any $H$? Commented Dec 4, 2019 at 17:45
• 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. Commented Dec 4, 2019 at 17:48
• Oh, didn't notice you make a specific distinction between hermitian and symmetric. Commented Dec 4, 2019 at 17:49

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 $$S^{1/2}$$.

• 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? Commented Dec 7, 2019 at 13:36
• @TomasBzdusek The equation $H^\top=UHU^\dagger$ is much easier to solve. Please see my new edit. Commented Dec 7, 2019 at 20:51