### Time reversal operator has no eigenvectors - meaning what

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6 months ago by
The time reversal operator $\mathcal{T}$ is an antilinear operator, meaning

\begin{align}
\mathcal{T} (c_1 \phi_1 + \phi_2) &= c_1^* \mathcal{T}\phi_1 + \mathcal{T} \phi_2
\end{align}

If it has an eigenvector $\phi_n$ with $\mathcal{T} \phi_n = \lambda_n \phi_n$, then by applying it twice we find

\begin{align}
\mathcal{T}^2 \phi_n &= \mathcal{T} \lambda_n \phi_n = \lambda_n^* \mathcal{T} \phi_n\\
&= |\lambda_n|^2 \phi_n
\end{align}

We see that the linear operator $\mathcal{T}^2$ must have positive eigenvalues. The time reversal operator fulfills $\mathcal{T}^2 = (-\mathbb{1})^{2j}$, where $j$ is spin. For example

\begin{align}
\mathcal{T}^2 &= (K)^2 = 1  &  &\quad\text{spin-0}\\
\mathcal{T}^2 &= (-\mathrm{i} \sigma_y K)^2 = -\mathbb{1} & &\quad\text{spin-}\tfrac{1}{2}
\end{align}

where $K$ complex conjugates. We see that for spin-$\tfrac{1}{2}$ particles $\mathcal{T}^2 < 0$ contradicts the above statement. Hence $\mathcal{T}$ has no eigenvectors for particles with odd spin, while it may have some for particles with even spin.

Normally if a symmetry is present (i.e. $[H,\mathcal{T}] = 0$ ) it means that we can diagonalize both operators simultaneously and states with different eigenvalues of the symmetry operator would not mix, so that we get a conserved quantity and could split the Hilbert space in different subspaces. For $\mathcal{T}$ this does not work; What are the implications of this? How is this symmetry different from other symmetries? How is it different for particles of different spins?
why $K^2 \phi = - \phi$ when $\phi$ is an eigenvector of $K$, the time-reversal operator? make it explicit.
written 6 months ago by AlQuemist
That was a mistake of my part. I wanted to simplify things and showed the operator for spinless particles, but only particles of odd spin fulfill $\mathcal{T}^2 = -1$. I fixed that in my question
written 6 months ago by LidoDiCamaiore