John Preskills lecture notes (here) contain an equation for a general rotation operator (eqn 2.25, page 11): $$R(\hat{n},d\theta)=I-id\theta \hat{n}\cdot \vec{J},$$ where $\vec{J}$ is angular momentum, $\hat{n}$ is a unit vector along the axis of rotation and $d\theta$ is the infinitesimal rotation angle.

There is no motivation. This equation looks familiar from quantum mechanics courses but I cannot find it in any of my books. This operator has many interesting properties:

It is antiunitary $$i\frac{dR}{d\theta}=\hat{n}\cdot \vec{J}R, \qquad [R,\hat{n}\cdot\vec{J}]=0$$ It is invariant under rotations of $\hat{n}$ and $\vec{J}$.

But I am not sure what the physical significance of this object is. Does it act on the state vector? Under what circumstances?

I can almost see the necessity of this formula except perhaps for the sign of the second term.

  • $\begingroup$ the "physical significance" will depend on the physical context you are studying this in. Unless by "physical significance" you mean here the geometrical interpretation of this $\endgroup$
    – glS
    Nov 3, 2021 at 11:24

1 Answer 1


Quick answer to point you in the right direction.

  1. That's the expression for an infinitesimal rotation operator (near the identity). To get a general rotation operator you exponentiate it so that $R(\hat{n},{\theta}) = e^{-i \theta \vec n \cdot \vec J}$. This is eqn 2.26 of the lecture notes you linked.

  2. The note doesn't provide motivation for this equation because it is assumed you know about rotations (specifically here in 3d) being a Lie group, and correspondingly, about its Lie algebra. See https://en.wikipedia.org/wiki/3D_rotation_group#Lie_algebra [see also https://en.wikipedia.org/wiki/Angular_momentum_operator]. There, $\vec{L}$ (or in Preskill's notes $\vec{J}$) are elements of the Lie algebra. Their defining property is eq 2.27 $[J_k, J_l] = i \epsilon_{klm} J_m$.

  3. How to motivate / obtain the defining property of the Lie algebra. Just consider a 3d real vector, and think about rotating it very slightly away from its initial position (say pointing in the z-direction). There are three "different" rotations you can do: about the $x-y$, $y-z$ and $z-x$ planes; this corresponds to $R(\hat{z},d\theta_3) = 1 - i d\theta_3 \hat{z}\cdot J_z$; $R(\hat{x},d\theta_1) = 1 - i d\theta_1 \hat{x}\cdot J_x$; $R(\hat{y},\theta_2) = 1 - i d\theta_2 \hat{y}\cdot J_y$ respectively, i.e. "do nothing (identity matrix) minus a small action in the appropriate direction" . From there you can explicitly derive 3x3 matrices $J_x,J_y,J_z$ ('generators') which obey the commutation relations and which appear in the Wikipedia article. Note a more generic rotation is just summing over linear combinations of the generators as in the formula you are asking about. Note also the sign and imaginary factor in the formula is just a matter of convention.

  1. To go to quantum mechanics: the above gives the "defining" representation of so(3). We can ask, given the commutation relations are there different representations of the algebra, i.e. are there $n\times n$ matrices satisfying $[J_k,J_l] = i \epsilon_{klm} J_m$. Answer: yes, for any $n$. In particular there is one for $n = 2$ which are the spin-1/2 operators $J_x = \frac{1}{2} \sigma_x, J_y= \frac{1}{2} \sigma_y, J_z = \frac{1}{2} \sigma_z$ where $\vec\sigma$ are the standard Pauli matrices, which should be familiar to you. This is eqn 2.28 of Preskill's notes. Exponentiating an infinitesimal rotation gives a finite rotation matrix $R(\hat{n},\theta)$, but now acting on a 2-dimensional complex vector -- i.e. the state of a qubit. So that's how you implement rotations on a quantum system.
  • $\begingroup$ I can accept easily the commutation relations for angular momentum, and it's easy to see the transition from infinitesimal rotations to finite rotations. However, I don't understand the referenced Wikipedia article or Lie groups and Lie algebras. I'm studying elementary group theory in Lang's Undergraduate Algebra but have no idea which topics in that book would help me understand representations or Lie groups. What's the relationship between rotation operator and angular momentum operator? $\endgroup$
    – Anna Naden
    Nov 5, 2021 at 7:06
  • $\begingroup$ What is the motivation for the equations for R in 3, above? Could they have been written instead without the imaginary factor? $\endgroup$
    – Anna Naden
    Nov 5, 2021 at 8:28
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    $\begingroup$ I'm not familiar with Lang's book but I had a quick glance at its contents and it does not cover Lie groups or Lie algebras. So I think it is not useful in your context. Might I suggest a book written by a physicist for physicists: Howard Georgi's "Lie algebras in Particle Physics". Ignore the "particle physics" label, the relevant chapters are 2 and 3. $\endgroup$
    – nervxxx
    Nov 6, 2021 at 6:29
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    $\begingroup$ To answer your questions: just like how the momentum operator $\hat{p}$ is the generator of translations $T(x) = e^{-i x \hat{p}}$ (here I write for 1d), the angular momentum operator $\vec{\hat{J}}$ is the generator of rotations $R(\theta, \vec{n}) = e^{-i \theta \vec{n} \cdot \vec{\hat{J}}}$. As the Wikipedia article expounds, what the angular momentum refers to -- orbital or spin, or the sum -- is usually left to context. For case of quantum computing that you are probably interested in, we usually talk about some abstract qubit (2level system) that are realized in some physical manner $\endgroup$
    – nervxxx
    Nov 6, 2021 at 6:37
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    $\begingroup$ Re: imaginary factors or not. Let's not get hung up over a discussion on notation/convention, which is irrelevant. Physicists often define the Lie algebra with a factor of i. This endows the generators $\vec{J}$ with the property that it is hermitian. Mathematicians might not use the factor of i. This changes the commutation relations (trivially), and the generators become skew-hermitian. But no physics changes. $\endgroup$
    – nervxxx
    Nov 6, 2021 at 6:45

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