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

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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?

\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

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