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

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

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.


2 Answers 2


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).

  • $\begingroup$ 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! $\endgroup$
    – kandb
    Commented Jul 3, 2023 at 10:04

jd27 answers your question.

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

  • $\begingroup$ Thank you for your answer! Is it true then that Θ|𝑛,+⟩ is the eigenvector of 𝐧̂ ⋅𝝈 corresponding to the eigenvalue −1? It seems like the moral of the story is that Θ|𝑛,+⟩ and |𝑛,−⟩ lie on the same ray, which makes a lot of sense. Your proof is elegant and eminently lucid! $\endgroup$
    – kandb
    Commented Jul 3, 2023 at 10:18
  • 1
    $\begingroup$ No problem. Yes, that’s what I showed. To complete the proof, you just need to check that $\Theta|n,+\rangle\neq0$. $\endgroup$
    – LPZ
    Commented Jul 3, 2023 at 11:01

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