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.

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    $\begingroup$ With the minus sign, that corresponds to a positive (counter-clockwise) rotation, just like $x-a$ would correspond to a shift to the right for a function of Cartesian coordinates. In addition, note that a rotation of the coordinate system and a physical rotation of the object are in some sense opposite operations. $\endgroup$ – march Jul 8 '15 at 15:51
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    $\begingroup$ possible duplicate of Negative sign in rotation operator (quantum mechanics, angular momentum) $\endgroup$ – Omry Jul 8 '15 at 15:52
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    $\begingroup$ march, that makes much more sense. So Griffiths is defining the rotation operator to be rotating the WAVE FUNCTION, while Schaum and Wikipedia define it to be rotating the coordinate system containing the wave function. $\endgroup$ – a00 Jul 8 '15 at 16:03
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    $\begingroup$ I feel if anything the first question should be marked a duplicate of this more thorough one. $\endgroup$ – user10851 Jul 9 '15 at 21:35

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.

  • $\begingroup$ I think you're saying that to figure out which is the right way, try both formulations with the example that we know $\hat x $ rotated $\pi/2$ produces $\hat y$. I don't understand yet certain parts such as how you know $\sigma_z$ is involved, but I do see that the matrix you produced does rotate $\hat x$ correctly, assuming the assumption that these phase factors don't matter. I'm sure I will understand these points eventually. Thank you for the walk through! $\endgroup$ – a00 Jul 9 '15 at 15:36
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    $\begingroup$ The Pauli matrices give a great description of a spin-$\frac 1 2$ system where the ang. momentum in the $p$-direction of the state $|\psi\rangle$ is $\langle j_p\rangle=\frac\hbar 2\langle\psi|\sigma_p|\psi\rangle$. A nice check from orbital mechanics is that the $j^2$ "total spin" should be $\hbar^2s(s+1)$ which for $s=\frac 12$ is $\frac 34\hbar^2$, which is indeed what you get doing $\langle j_x^2+j_y^2+j_z^2\rangle$. Eigenvectors of those operators give us $x$, $y$, and $z$ states, but those eigenvectors aren't orthogonal because $\langle j^2\rangle>\hbar^2/4$, giving spatial "spread". $\endgroup$ – CR Drost Jul 10 '15 at 14:18
  • $\begingroup$ Your spatial spread statement is intriguing. I believe the fact that $\langle j^2 \rangle > \hbar^2/4$ is related to the fact that, for example, a supposedly "purely up spin" somehow has some character of "horizontal spin" that cannot be removed from it. (i.e., why we can measure x direction spin even when we are sure the spin is perfectly polarized in the positive z direction). It all ties into the bizarre nature of there being 6 dimensions of spin (+ and - z, + and - y, + and - x) rather than 3. $\endgroup$ – a00 Jul 14 '15 at 13:21
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    $\begingroup$ Well, I mean, among spin-$\frac 12$ particles there are only the 2 degrees of freedom of what's called the "Bloch sphere". But yes, that's exactly what I was alluding to: the length of $\vec j$ is $\sqrt 3$ longer than its component in any direction, even the one that it is most "pointed along", meaning that it must be "pointed along" a circle of size $\hbar \sqrt{1/2}$ which is necessarily "averaged out", and $\vec j$ refuses to be located on anything other than such a circle. $\endgroup$ – CR Drost Jul 14 '15 at 14:00
  • $\begingroup$ Huh. Very interesting. I will have to think on it further. Thank you for your additional comments. $\endgroup$ – a00 Jul 17 '15 at 15:18

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