# Derivation of Time-Reversed Ket for a Spin-1/2 Particle

I'm self-learning quantum mechanics from the 3rd edition of Sakurai and Napolitano's Modern Quantum Mechanics, and I've hit a bump in a derivation. The authors note that the eigenket of $$\mathbf{S}\cdot\hat{\mathbf{n}}$$ with eigenvalue $$\hbar/2$$ can be written as

$$\begin{equation*} |\hat{\mathbf{n}};+\rangle = e^{-iS_{z}\alpha/\hbar}e^{-iS_{y}\beta/\hbar}|+\rangle \end{equation*}$$

where $$\mathbf{S}$$ is the spin angular momentum operator; $$\alpha$$ and $$\beta$$ are, respectively, the polar and azimuthal angles characterizing the unit vector $$\hat{\mathbf{ n}}$$; and $$|+\rangle$$ is the eigenvector of $$S_{z}$$ with eigenvalue $$\hbar/2$$.

I understand the derivation up to this point. They then say that

$$$$\Theta|\hat{\mathbf{n}};+\rangle = e^{-iS_{z}\alpha/\hbar}e^{-iS_{y}\beta/\hbar}\Theta|+\rangle = \eta|\hat{\mathbf{n}};-\rangle$$$$

where $$|\hat{\mathbf{n}};-\rangle$$ is the eigenket of $$\mathbf{S}\cdot\hat{\mathbf{n}}$$ with eigenvalue $$-\hbar/2$$, and $$\Theta$$ is the time-reversal operator which, for an angular momentum operator $$\mathbf{J}$$, satisfies

$$\begin{equation*} \Theta \mathbf{J}\Theta^{-1} = -\mathbf{J}. \end{equation*}$$

I'm not sure how they use the previous equation to get to the one before that. Is it true that because $$\Theta \mathbf{J}\Theta^{-1} = -\mathbf{J}$$, then a similar rule holds for the exponential of the components of $$\mathbf{J}$$? Also, by writing $$\eta$$, it seems like the authors are indicating a phase factor rather than an operator, but I'm not sure. I think I could figure it out once I understand why

$$\begin{equation*} \Theta e^{-iS_{z}\alpha/\hbar}e^{-iS_{y}\beta/\hbar}|+\rangle = e^{-iS_{z}\alpha/\hbar}e^{-iS_{y}\beta/\hbar}\Theta|+\rangle. \end{equation*}$$

Frankly, I'm not even sure this forum is the right place to ask this question, so I apologize in advance if it's inappropriate.

Let $$J$$ be any component of $$S$$.

From $$\Theta J \Theta^{-1} = -J$$ we obtain $$\Theta J = - J \Theta$$. Since $$\Theta$$ is anti-linear we have $$\Theta iJ = -i \Theta J = iJ \Theta$$.

Further $$\Theta J^2 \Theta^{-1} = \Theta J \Theta^{-1} \Theta J \Theta^{-1} = J^2$$. So that $$\Theta J^2 = J^2 \Theta$$.

By induction we obtain $$\Theta (iJ)^n = (iJ)^n \Theta$$ for any $$n \in \mathbb{N}$$.

And therefore $$\Theta \exp (iJ) = \Theta \sum_{n=0}^\infty \frac{1}{n!} (iJ)^n = \exp(iJ) \Theta ,$$ which gives us the identity you have asked about (putting the $$-$$, $$\hbar$$, $$\alpha$$, $$\beta$$ in front of $$J$$ does not change anything because they are real numbers).

• Thank you for your answer! So, then, I guess we have that for any anti-linear operator $\Theta$ and any self-adjoint operator (what other requirements are needed?) $J$ that is odd under time reversal, we have that $\Theta$ commutes with the exponential of $iJ$. That's interesting! Commented Jul 3, 2023 at 10:04

I just wanted to add that it is not obvious that $$\Theta |+\rangle\propto |-\rangle$$ once you have your last identity.
This is why I prefer proving things in one go. Noting that: $$(n\cdot \sigma)|n,+\rangle =|n,+\rangle$$ using the conjugation relation: $$\Theta (n\cdot \sigma) \Theta^{-1}=-n\cdot \sigma$$ so: $$(n\cdot \sigma) \Theta |n,+\rangle =-\Theta|n,+\rangle$$ i.e. $$\Theta |n,+\rangle$$ is a $$-1$$ eigenstate of $$n\cdot \sigma$$ so: $$\Theta |n,+\rangle \propto |n,-\rangle$$ Hope this helps
• No problem. Yes, that’s what I showed. To complete the proof, you just need to check that $\Theta|n,+\rangle\neq0$.