Negative sign in rotation operator again

Question

In Wikipedia's page on the rotation operator, section "In relation to the orbital angular momentum", they write $$ R(z,t) = exp((-i/h) \varphi L_z) $$
where $\varphi$ is the angle being rotated through My Schaum's textbook also has the negative sign.

However, this website does not have the negative sign. Its argument for deriving the rotation operator uses Taylor series. They say if you write

$ e^{iL_z \varphi / \hbar} \cdot f(\theta_0, \phi_0, r_0)$ (i.e. the operator acting on a particular point of the function $f$) and expand in Taylor series, you get:

\begin{align} e^{iL_z \varphi / \hbar} \cdot f(\theta_0, \phi_0, r_0)&=1 \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/1!) \varphi^1 (d/d\phi) \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/2!) \varphi^2 (d^2/d\phi^2) \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/3!) \varphi^3 (d^3/d\phi^3) \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/4!) \varphi^4 (d^4/d\phi^4) \cdot f(\theta_0, \phi_0, r_0) \\ &+ \cdots \end{align}

which is the Taylor expansion for $ f(\theta_0, \phi_0 + \varphi, r_0) $, so the expression $ e^{iL_z \varphi / \hbar} $ has successfully rotated the function's point through an angle $\varphi$.

That makes sense to me. But if you tried it with the negative sign as Wikipedia and Schaum do, the expression is instead:

\begin{align} e^{iL_z \varphi / \hbar} \cdot f(\theta_0, \phi_0, r_0)&=1 \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/1!) (-\varphi)^1 (d/d\phi) \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/2!) (-\varphi)^2 (d^2/d\phi^2) \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/3!) (-\varphi)^3 (d^3/d\phi^3) \cdot f(\theta_0, \phi_0, r_0) \\ &+ (1/4!) (-\varphi)^4 (d^4/d\phi^4) \cdot f(\theta_0, \phi_0, r_0) \\ &+ \cdots \end{align}

the Taylor expression for $ f(\theta_0, \phi_0 - \varphi, r_0) $. So that doesn't give you a rotation by $\varphi$ but by $-\varphi$, right?

So are the Wikipedia and Schaum formulation of the rotation operator wrong?

--

The linked question by Omry did not answer this question; the answer there suggested that it was okay to have the negative sign.

Answer 1

Let's make this concrete by using a $\text{spin-}\frac 1 2$ state in the $z$ direction, where we get to use the Pauli matrices, usually written as$$\sigma_x = \left[\begin{array}{cc}0&1\\1&0\end{array}\right]; ~~~\sigma_y = \left[\begin{array}{cc}0&-i\\i&0\end{array}\right]; ~~~\sigma_z = \left[\begin{array}{cc}1&0\\0&-1\end{array}\right]$$In this convention, the unit vector in the $+x$ direction is $\sqrt{\frac 1 2} \left[\begin{array}{c}1\\1\end{array}\right]$ while the unit vector in the $+y$ direction is $\sqrt{\frac 1 2} \left[\begin{array}{c}1\\i\end{array}\right]$, these having eigenvalue $+1$ for those operators. Note that the question of which one is "+y" comes down to which square root of negative 1 we choose to be "i", or, equivalently, what convention you take on the $\sigma_y$ matrix. The $+i$ bottom-left convention, however, is attested by both Wikipedia and MathWorld, and I think it's probably also in Griffiths' Introduction to Quantum Mechanics, so let's go with this convention.

Now we expect a rotation by $+\pi/2$ around the $z$-axis will turn the $+x$ direction into $+y$. So we form the rotation matrix for both signs to see which is "right". This is a bit tricky as $e^{i\pi} = e^{-i\pi} = -1$, and multiplying a wavefunction by $-1$ (or any phase factor $e^{i\phi}$), in quantum mechanics, doesn't change any observables: $$\langle\psi|\hat A|\psi\rangle = \langle\psi|e^{-i\phi} \hat A e^{i\phi}|\psi\rangle = \langle e^{i\phi}\psi| \hat A | e^{i\phi} \psi\rangle = \langle\psi'|\hat A|\psi'\rangle.$$ So we see that in the following expression, we should not use the Pauli matrix $\sigma_z$ willy-nilly but we should instead use $\frac 1 2 \sigma_z$ for our $\mathbf n \cdot \mathbf J / \hbar$ term, since after the exponent reaches $\pi$ all of our observables are the same as they used to be and a $2\pi$ rotation has been performed. So we write $$R^\pm_z(\theta) = \exp(\pm ~ i~\frac \theta 2~\sigma_z)=\left[\begin{array}{cc}e^{\pm~i~\theta/2}&0\\0&e^{\mp~i~\theta/2}\end{array}\right]$$and we find that for a rotation by $\pi/2$ this becomes$$R^\pm_z(\pi/2) = \sqrt{\frac 1 2} \left[\begin{array}{cc}1 \pm i&0\\0&1\mp i\end{array}\right].$$Operating on our $+x$ vector gives: $$R^\pm ~\hat x = \frac 1 2 \left[\begin{array}{c}1\pm i\\1\mp i\end{array}\right] = \frac {1\pm i} 2~\left[\begin{array}{c}1\\(1\mp i)^2/2\end{array}\right] = e^{\pm~i~\pi/4}~\left[\begin{array}{c}1\\\mp i\end{array}\right]$$Again, that phase prefactor does not change any observables so it is irrelevant. Therefore we find out that the correct way to rotate the angle forward by an amount $\theta$ is to use the negative exponent:$$R_z(\theta) = \exp\left(-i~\theta~\frac{\sigma_z}{2}\right).$$Wikipedia and Schaum are therefore right, consistent with their conventions. It is possible that the PhysicsPages result is also consistent with its own conventions, but it would probably require that for them, $L_z = -\sigma_z/2$ or their $f$ has a special form or something. It's certainly not 100% obvious that they're doing the right thing. Otherwise they have the "+ is clockwise" rotation convention, which is totally fine in normal physical terms but it is not the right-hand rule that you learned in your undergraduate coursework.