# Where does this expression for infinitesimal rotations come from?

In Chapter 2, page 11 of Preskill's quantum computing notes, he mentions without explaining that a counterclockwise inifinitesimal rotation by $$d\theta$$ about the axis $$\hat{n}$$ is given by

$$R(d\theta, \hat n) = I - id\theta \hat n \cdot \hat J$$ Where does $$\hat J$$ come from and how do we pick its components? Clearly, different choices of $$J_x, J_y, J_z$$ lead to different rotations. How does one know that the angular momentum will be relevant and this it appears in this form?

It just doesn't seem like we need $$\hat J$$ at all to describe a rotation in 3 dimensions. Given $$\hat n$$ and we can just look at the 2 dimensional space which is orthogonal to it and do a counterclockwise rotation of that space, so that we only need $$\hat n$$ and a 2 component object to describe our rotation.

Consider a qubit system. Rotations are implemented by operators $$U \in SU(2)^1$$. The Lie group $$SU(2)$$ is compact and connected. Hence, the exponential map $$\exp: \mathfrak{su}(2) \rightarrow SU(2)$$ is surjective.

This lets us write an arbitrary rotation in the form $$U = \exp(-i\theta X) \tag{1}$$ where $$X \in \mathfrak{su}(2)$$. The Pauli matrices$$^2$$ $$\vec{\sigma}$$ are a standard basis of the Lie algebra $$\mathfrak{su}(2)$$. We must divide them by $$2$$ to get the “right” structure constants. So, any $$X \in \mathfrak{su}(2)$$ can be written in the form $$X = \frac{\vec{\sigma}}{2}\cdot\vec{n}. \tag{2}$$ Combining equations (1) and (2), we can write an arbitrary rotation of a single qubit as $$U = \exp(-i\frac{\theta}{2}\vec{\sigma}\cdot\vec{n}).$$

Finally, let $$\theta = d\theta$$ be infinitesimal. Then, $$U = \exp(-i\frac{d\theta}{2}\vec{\sigma}\cdot\vec{n}) = \mathbb{I} -i\frac{d\theta}{2}\vec{\sigma}\cdot\vec{n} +\mathcal{O}(d\theta^2)$$ by definition of matrix exponential. If you truncate the series at first order, you arrive at the appropriate expression for an infinitesimal rotation.

1. For sake of clarity I do not explicitly mention that we are dealing with the image of the two dimensional irreducible representation of $$SU(2)$$. The two dimensional irrep because we are dealing with a single qubit system.
2. I personally take the definition of angular momentum to be the generators of rotation (equivalently, the quantity that is conserved when the Hamiltonian is invariant under rotations). Hence, by definition I identify $$\vec{\sigma}$$ with angular momentum. Choosing the basis of (the image of the two dimensional irreducible representation of) $$\mathfrak{su}(2)$$ to be the Pauli sigma matrices is indeed conventional. The convention falls out of the following. Non-relativistic physics takes place in Euclidean 3-space, $$\mathbb{R}^3$$. We fix an $$xyz$$ cartesian coordinate system so we can talk about the directions $$x$$, $$y$$, and $$z$$. We can then "align" the direction of the spin-z of our qubit with the direction of spatial $$z$$. This induces a choice of basis for $$\mathfrak{su}(2)$$, i.e., the Pauli sigma matrices.

In Chapter 2, page 11 of Preskill's quantum computing notes, he mentions without explaining that a counterclockwise infinitesimal rotation by $$d\theta$$ about the axis $$\hat{n}$$ is given by

$$R(d\theta, \hat n) = I - id\theta \hat n \cdot \hat J$$ Where does $$\hat J$$ come from and how do we pick its components?

$$\vec J$$ is the angular momentum of the system being rotated, in units of $$\hbar$$. (Or rather, $$\hbar \vec J$$ is the angular momentum.) (But I'll just set $$\hbar=1$$, to make life easier.)

For example, if the system is a single spin-1/2 particle, but we ignore spatial motion, then: $$\vec J^{(1/2)}_{\text{no space}} = \hat x \left(\begin{matrix}0 & 1/2 \\ 1/2 & 0\end{matrix}\right) +\hat y \left(\begin{matrix}0 & -i/2 \\ i/2 & 0\end{matrix}\right) +\hat z \left(\begin{matrix}1/2 & 0 \\ 0 & -1/2\end{matrix}\right)\;,$$ where $$\hat x$$, $$\hat y$$, and $$\hat z$$ are unit vectors along the usual axes.

For example, if the system is a single spin-1/2 particle and we include spatial motion, then: $$\vec J^{(1/2)} = \left(\hat x \left(\begin{matrix}0 & 1/2 \\ 1/2 & 0\end{matrix}\right) +\hat y \left(\begin{matrix}0 & -i/2 \\ i/2 & 0\end{matrix}\right) +\hat z \left(\begin{matrix}1/2 & 0 \\ 0 & -1/2\end{matrix}\right) \right)\otimes \mathbf{I} +\mathbf{I}\otimes\vec{r} \times \vec{p}$$

Clearly, different choices of $$J_x, J_y, J_z$$ lead to different rotations. How does one know that the angular momentum will be relevant and this it appears in this form?

Yes, different systems have different angular momenta. But, we have already agreed on the conventions to use.

One can take the angular momentum operators as being defined as the generators of rotations.

If you are are looking for a little more detail and precision, I would recommend checking out M. Tinkham "Group Theory and Quantum Mechanics" Chapter 5 "Full Rotation Group and Angular Momentum."

For example, Tinkham writes: "...we may define the x component of the angular momentum as the operator whose conservation follows from the invariance of the Hamiltonian under rotations about the x axis... we then obtain a definition of angular momentum in terms of infinitesimal rotation..." $$J_x \psi = \lim_{\delta\theta_x\to 0}i\hbar\frac{R_{\delta \theta_x} - 1}{\delta \theta_x}\psi$$

Where does $$\hat J$$ come from

Your expression is a statement that $$\hat n\cdot \mathbf J$$ is the generator of infinitesimal rotations about $$\hat n$$. In quantum mechanics the total angular momentum operator is $$\mathbf J = \mathbf L + \mathbf S$$, the sum of the orbital angular momentum and spin angular momentum operators respectively, but to start we'll just look at $$\mathbf L$$ which is used for rotations of spatial coordinates:

From $$P = -i\hbar\frac{\mathrm d}{\mathrm d x}$$ we know that the momentum operator is the generator of spatial translations (up to a factor of $$\frac{-i}{\hbar}$$), i.e. $$T(\alpha)\psi(x) = \psi(x-\alpha)$$ where $$T(\alpha) = e^{\alpha\frac{-i}{\hbar}P}$$. This is simply due to the general relation $$e^{\alpha\frac{\mathrm d}{\mathrm d x}}f(x) = f(x-\alpha)$$ which can be verified by Taylor expanding both sides. More generally in 3 dimensions the translation operator in an arbitrary direction $$\hat n$$ is given by $$T(\alpha,\hat n) = e^{\alpha\frac{-i}{\hbar}\hat n \cdot \mathbf P}$$ such that $$T(\alpha,\hat n)\psi(\mathbf r) = \psi(\mathbf r - \alpha\hat n)$$, where $$\mathbf P = P_x\hat x + P_y\hat y + P_z\hat z$$.

Similarly, the angular momentum operators are the generators of angular translations. The simplest example to see this is with $$L_z$$ since the z axis is aligned with the azimuthal axis in spherical coordinates. The vector angular momentum operator is defined as the cross product $$\mathbf L = \mathbf R \times \mathbf P$$ where $$\mathbf R = X\hat x + Y\hat y + Z\hat z$$ which gives the z component $$L_z = XP_y - YP_x$$. Expressing this in the position basis, $$L_z = -i\hbar(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x})$$, and using the spherical-to-cartesian coordinate transformations $$x = r\sin\theta\cos\phi$$, $$y=r\sin\theta\sin\phi$$, and $$z = r\cos\theta$$ to compute the Jacobian, we find that $$L_z = -i\hbar \frac{\partial}{\partial \phi}$$. Therefore $$R_L(\alpha,\hat z) = e^{\alpha\frac{i}{\hbar}L_z}$$ such that $$R_L(\alpha,\hat z)\psi(r,\theta,\phi) = \psi(r,\theta,\phi-\alpha)$$. In other words $$L_z$$ is the generator of rotations about the z axis.

There is for sure a more rigorous derivation than what I will write here, but I think it suffices to say that since the choice of "the z axis" for a particular spherical coordinate system is totally arbitrary and can be chosen to point in any direction, then the result that $$L_z = \hat z\cdot \mathbf L$$ is the generator of rotations about $$\hat z$$ should generalize to state that $$\hat n \cdot \mathbf L$$ is the generator of rotations about $$\hat n$$.

$$R_L(\alpha,\hat n) = e^{\alpha\frac{-i}{\hbar}\hat n\cdot\mathbf L}$$

Now for $$\mathbf S$$. Although there is no associated translations from which we can derive that $$\mathbf S$$ generates rotations in a similar way to how we derived that for $$\mathbf L$$, we can understand that the spin state of a particle represents a direction in 3D space which can be described by angular coordinates, so the spin angular momentum operators should analogously generate rotation for those coordinates of "spin space", so to speak.

$$R_{S}(\alpha, \hat n) = e^{\alpha\frac{-i}{\hbar}\hat n\cdot \mathbf S}$$

If we rotate the system as a whole then we must rotate both the positional coordinates (using $$\mathbf L$$) and the "spin coordinates" (using $$\mathbf S$$). More formally, the quantum state is the tensor product of the spatial state and the spin state, $$|\psi\rangle = |\psi_L\rangle\otimes|\psi_S\rangle$$, and the different angular momentum operators act on their respective states, i.e. $$\mathbf L(|\psi_L\rangle\otimes|\psi_S\rangle) = (\mathbf L|\psi_L\rangle)\otimes|\psi_S\rangle$$ and $$\mathbf S(|\psi_L\rangle\otimes|\psi_S\rangle) = |\psi_L\rangle\otimes(\mathbf S|\psi_S\rangle)$$. Therefore the total angular momentum operator $$\mathbf J = \mathbf L + \mathbf S$$ is needed to generate rotations on whole state:

$$R(\alpha,\hat n) = e^{\alpha\frac{-i}{\hbar}\hat n \cdot \mathbf J}$$

To arrive at your original relation we can just expand this expression to first order for small $$\alpha$$:

$$R(\delta\alpha,\hat n) = e^{\delta\alpha\frac{-i}{\hbar}\hat n\cdot\mathbf J} = \hat I - \delta\alpha\frac{i}{\hbar}\hat n\cdot\mathbf J + \mathcal O(\delta\alpha^2)$$

how do we pick its components? Clearly, different choices of $$J_x,J_y,J_z$$ lead to different rotations

$$\mathbf J$$ is a vector and so the choice of components corresponds to a choice of coordinate system (although certain coordinate systems won't have corresponding operators). However vectors are defined independently of which basis they are expanded in and furthermore the operator $$\hat n \cdot \mathbf J$$ is the result of a dot product and is a coordinate independent quantity. Therefore the expression for the rotation operator doesn't depend on your choice of components for expressing $$\mathbf J$$.

It just doesn't seem like we need $$\hat J$$ at all to describe a rotation in 3 dimensions. Given $$\hat n$$ and we can just look at the 2 dimensional space which is orthogonal to it and do a counterclockwise rotation of that space, so that we only need $$\hat n$$ and a 2 component object to describe our rotation.

It is true that to describe a rotation by an angle all we need is an oriented plane (which is a 2D object like you said), which in 3D space is equivalent to specifying a vector to represent the plane which it is normal to and to represent the angle of rotation with its magnitude. However in 3D space there are 3 linearly independent planes of rotation ($$xy,yz$$, and $$zx$$) so we need 3 linearly independent angular momentum operators to generate rotations in each.

That being said, a general rotation of a 3D object from one orientation to another is uniquely specified by a sequence of 3 rotations and in fact only requires 2 independent axes to perform the rotations around. Common rotation conventions include ZYZ, ZXZ, XYX, etc. which specify a sequence with 3 angles $$(\alpha,\beta,\gamma)$$ to rotate by, in order, about those respective axes. In the case of ZYZ for example, the rotation operator would be

$$R(\alpha,\beta,\gamma) = e^{-\gamma\frac{iJ_z}{\hbar}}e^{-\beta\frac{iJ_y}{\hbar}}e^{-\gamma\frac{iJ_z}{\hbar}}$$

which only uses $$J_y$$ and $$J_z$$.