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So, I was reading about the time reversal operator, and I came to know that it can be expressed as: $$T = KU$$ where, $U$ is an unitary operator and $K$ is the complex conjugation operator. Now, if I take the hermitian of the equation: $$T^\dagger = U^\dagger K^\dagger$$ Now, I am confused as how to interpret $K^\dagger$ in the above equation. Does it mean hermitian of the hermitian of the argument (would it not become just an identity if it were the case), or does it have a different interpretation?

I also encountered an example somewhere where $T$ has been taken as: $$T = K \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $$ $$\implies T^{\dagger} = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} K^\dagger$$ and then $T A T^{\dagger}$ was computed where $A$ was taken to be a general matrix $\begin{pmatrix} a & b\\ c & d\\ \end{pmatrix}$.

I would be thankful if somewhere could explain to me how $T^\dagger$ is interpreted, and how it is operated on some matrix, say the operation $TAT^\dagger$. It'd be more helpful if any examples could be provided, without it being too abstract.

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$\let\s=\sigma \let\pd=\partial \let\dn=\downarrow \let\up=\uparrow \let\dag=\dagger \def\hp{\hat p} \def\hs{\hat s} \def\hx{\hat x} \def\hy{\hat y} \def\hz{\hat z} \def\hI{{\hat I}} \def\hL{{\hat L}} \def\hP{{\skew{1}\hat P}} \def\hT{{\skew{1}\hat T}} \def\ket#1{|#1\rangle} \def\bra#1{\langle#1|} \def\braket#1#2{\langle#1|#2\rangle} \def\mxelm#1#2#3{\bra#1\,#2\,\ket#3} \def\half{{\textstyle{1 \over 2}}} \def\kd{\ket{\!\dn\,}} \def\ku{\ket{\!\up\,}}$ The answer is in keeping a clear distinction between abstract operators and their representations. Which implies a parallel distinction between abstract Hilbert spaces and their representations. The simple equation you wrote $$T = K\,U \tag1$$ doesn't and can't hold for abstract operators. In an abstract Hilbert space conjugation is meaningless (or, if you like better, can be defined in infinite different ways, so it's useless). Instead eq. (1) is meaningful if you're referring to a well defined representation, and the $U$ will be different in different representations of the same abstract space.

I'll show you some examples. First, a particle moving along a straight line. All observables are definable starting from position $\hx$ and momentum $\hp$. So to define time reversal it suffices to give its action on $\hx$ and $\hp$: $$\hT\,\hx\,\hT^\dag = \hx \qquad \hT\,\hp\,\hT^\dag = -\hp.\tag2$$ Note that $\hT$ preserves commutation relation $$[\hx,\hp] = i\,\hbar\,\hI.$$ In fact applying time reversal to LHS $$\hT\,[\hx,\hp]\, \hT^\dag = [\hx,-\hp] = -i\,\hbar\,\hI$$ and applying it to RHS $$\hT\,(i\,\hbar\,\hI)\,\hT^\dag = -i\,\hbar\,\hI$$ because $\hT\,i\,\hT^\dag=-i$ (antiunitarity). Note that for a antiunitary operator, as for a unitary one, $\hT^\dag=\hT^{-1}$.

The $\hat{}$ over operator symbols is there to remind you that these are abstract operators. I haven't yet defined any representation. For a particle on a line the most common representation is Schrödinger's: $$\hx \mapsto x \qquad \hp \mapsto -i\,\hbar{\pd \over \pd x}.$$ Then you can see that in order to satisfy (2) we just have to put $$\hT \mapsto K$$ where $K$ means complex conjugation of wavefunctions and of operator representations. In fact $Kx=x^*=x$, whereas momentum changes sign because of the "$i$" (BTW $K^\dag=K$).

Another useful representation for the Hilbert space of this physical system is momentum representation. It is defined by $$\hx \mapsto i\,\hbar{\pd \over \pd x} \qquad \hp \mapsto p.$$ If you tried to use again $\hT\mapsto K$ you'd get a wrong result: $\hp$ unchanged and $\hx$ reversed! The right solution is $$\hT \mapsto K\,P$$ where $P$ is the space inversion operator, which changes sign both to $x$ and to $p$: $$\hP\,\hx\,\hP^\dag = - \hx \qquad \hP\,\hp\,\hP^\dag = -\hp.$$ So you see that eq. (1) holds in both representations, but with different $U$'s: $U=I$ in Schr. rep, $U=P$ in momentum rep.


A second example is a spin-1/2 particle. If we are only concerned with spin states, the only observables are spin components. There are three linearly independent, say $\hs_x$, $\hs_y$, $\hs_z$. The system's Hilbert space actually is just a 2D complex vector space. In it a basis is customarily assumed consisting of eigenvectors of $\hs_z$: $\ku$, $\kd$ $$\hs_z \ku = \half\,\hbar\,\ku \qquad \hs_z \kd = -\half\,\hbar\,\kd.$$ Note that this doesn't completely define the basis, as arbitrary phase factors are still allowed for both vectors. In order to maximally define the basis the action of another observable is needed, e.g. $\hs_x$. The usual definition is $$\hs_x \ku = \half\,\hbar\,\kd \qquad \hs_x \kd = \half\,\hbar\,\ku$$

Once a basis is fixed, a representation of spin observables is also defined. It is $$\hs_x \mapsto \half\,\hbar\,\s_x \qquad \hs_y \mapsto \half\,\hbar\,\s_y \qquad \hs_z \mapsto \half\,\hbar\,\s_z$$ where $$\s_x = \pmatrix{0&1\cr1&0\cr} \qquad \s_y = \pmatrix{0&-i\cr i&0\cr} \qquad \s_z = \pmatrix{1&0\cr0&-1\cr}$$ are the usual Pauli matrices.

Now the action of $\hT$ on spin observables is to be defined. If we want to ensure time reversal invariance say for an atom, we must take for $\hs_x$, $\hs_y$, $\hs_z$, the same transformation law holding for orbital angular momentum. This is already fixed by transformation laws of $\hx$, $\hy$, $\hz$, $\hp_x$, $\hp_y$, $\hp_z$: the former are invariant, the latter change sign. Then $\hL_z$, being defined as $\hx\,\hp_y-\hy\,\hp_x$, changes sign too, and the same happens for the other components. We conclude that all spin components change sign.

Once again we see that the simple choice $\hT\mapsto K$ doesn't work: it's OK for $\hs_y$ but not for the other two. Remembering that 1/2-spin components anticommute with one another the solution is immediate: $$\hT \mapsto K \s_y.$$ I leave for you to check this is the right answer.

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