Anti-unitary operator and hamiltonian

Question

For a symmetry represented by a unitary operator $U$ to be a dynamical symmetry, we require the condition that $Ue^{(-iHt/\hbar)}=e^{(-iHt/\hbar)}U$ which implies $UHU^*=H$.

If instead $U$ is an anti-unitary opertor, show that the above equation would imply that $UHU^*=-H$.

I'm not too sure how to do this question. I don't really understand how the first implication is derived from the condition, and secondly I don't see how this changes for an anti-unitary operator. $H$ is the Hamiltonian, and the definitions of unitary operator and anti-unitary operators are as follows:

A unitary operator $U$ on a Hilbert space is a linear map $U :\mathcal{H} \rightarrow \mathcal{H}$ that obeys $UU^*=U^*U=1_{\mathcal{H}}$ ($U^*$ being the adjoint).

An anti-unitary operator on a hilbert space is a surjective linear map $A :\mathcal{H} \rightarrow \mathcal{H}$ obeying $\langle A\phi |A\psi \rangle = \overline {\langle \phi | \psi \rangle} = \langle \psi | \phi \rangle$

Answer 1

A unitary operator is a linear surjective operator $U : {\cal H} \to {\cal H}$ that preserves the norm. It is equivalent to $U^*=U^{-1}$, namely $UU^*=U^*U=I$, where $U^*$ henceforth denotes the adjoint of $U$.

An antiunitary operator is an antilinear surjective operator $U : H \to H$ that preserves the norm. It is equivalent to $U$ bijective such that $$\langle U\psi|\phi\rangle = \overline{\langle \psi| U\phi\rangle}\:,\quad \forall \psi, \phi \in {\cal H}\:.$$ Now suppose that, in either cases, for all $t\in \mathbb{R}$ $$U e^{-itH} = e^{-itH}U\:.$$ By applying $U^{-1}$ on the right, we get the equivalent condition $$Ue^{-itH} U^{-1}= e^{-itH}\:.\tag{1}$$ From spectral calculus or other more elementary procedures, e.g., expanding the exponential as a series if $H$ is bounded and paying attention to $U i H = -iUH$ in view if antilinearity of $U$ if it is the case, (1) entails $$ e^{\mp itUHU^{-1}} = e^{-itH}\:.$$ Computing the derivative at $t=0$ (Stone's theorem) of both sides (on the relevant dense domain of $H$ which turns out to be invariant under $U^{-1}$, directly form the uniqueness part of Stone's theorem): $$\pm UHU^{-1} = H\:,$$ that is $$UHU^{-1} = \pm H\:,\tag{2}$$ where the sign $-$ is reserved to the antiunitary case. In case of a unitary oparator, we have also found that $$UHU^{*} = H$$ because $U^*=U^{-1}$. In case of an antiunitary $U$, with a suitable definition ($\dagger$) of adjoint operator for antilinear operators, we can equivalently rewrite (2) as $$UHU^{*} = -H\:.$$

However the definition of adjoint of an antiunitary operator is usually delicate and, to my personal experience, it is a source of mistakes. Dealing with symmetries it is much better to use $U^{-1}$ in both cases in place of $U^*$.

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$(\dagger)$ $\langle \psi|A \phi\rangle = \overline{\langle A^*\psi| \phi\rangle}$ for all $\psi,\phi\in {\cal H}$ assuming $A$ everywhere defined and antilinear.

Answer 2

There are a couple of confusing (or even wrong?) points in the post. First, I assume $U^*$ means $U^\dagger$, the adjoint of $U$. A unitary symmetry means $UHU^\dagger=H$.

An anti-unitary operator is first of all an anti-linear operator instead of a linear one. If $U$ is anti-unitary symmetry, then one still has $UHU^\dagger=H$, there should not be an extra minus sign. However, the definition of adjoint for anti-linear operator is different from that of a linear operator.

Edit: the other answer is correct. Usually for a time-reversal symmetry (which is the most common way one gets anti-unitary symmetry) we also take $t$ to $-t$ so $UHU^\dagger=H$. But if $U$ is just anti-unitary without $t$ going to $-t$, then because $Ui=-iU$ we have the extra minus sign.