Rotation operator (quantum mechanics)

This article concerns the rotation operator, as it appears in quantum mechanics.

Quantum mechanical rotations
With every physical rotation $$R$$, we postulate a quantum mechanical rotation operator $$D(R)$$ which rotates quantum mechanical states. $$| \alpha \rangle_R = D(R) |\alpha \rangle$$

In terms of the generators of rotation, $$D (\mathbf{\hat n},\phi) = \exp \left( -i \phi \frac{\mathbf{\hat n} \cdot \mathbf J }{ \hbar} \right),$$ where $$\mathbf{\hat n}$$ is rotation axis, $$ \mathbf{J} $$ is angular momentum, and $$\hbar$$ is the reduced Planck constant.

The translation operator
The rotation operator $$\operatorname{R}(z, \theta)$$, with the first argument $$z$$ indicating the rotation axis and the second $$\theta$$ the rotation angle, can operate through the translation operator $$\operatorname{T}(a)$$ for infinitesimal rotations as explained below. This is why, it is first shown how the translation operator is acting on a particle at position x (the particle is then in the state $$|x\rangle$$ according to Quantum Mechanics).

Translation of the particle at position $$x$$ to position $$x + a$$: $$\operatorname{T}(a)|x\rangle = |x + a\rangle$$

Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing): $$\operatorname{T}(0) = 1$$ $$\operatorname{T}(a) \operatorname{T}(da)|x\rangle = \operatorname{T}(a)|x + da\rangle = |x + a + da\rangle = \operatorname{T}(a + da)|x\rangle \Rightarrow \operatorname{T}(a) \operatorname{T}(da) = \operatorname{T}(a + da)$$

Taylor development gives: $$\operatorname{T}(da) = \operatorname{T}(0) + \frac{d\operatorname{T}(0)}{da} da + \cdots = 1 - \frac{i}{\hbar} p_x da$$ with $$p_x = i \hbar \frac{d\operatorname{T}(0)}{da}$$

From that follows: $$\operatorname{T}(a + da) = \operatorname{T}(a) \operatorname{T}(da) = \operatorname{T}(a)\left(1 - \frac{i}{\hbar} p_x da\right) \Rightarrow \frac{\operatorname{T}(a + da) - \operatorname{T}(a)}{da} = \frac{d\operatorname{T}}{da} = - \frac{i}{\hbar} p_x \operatorname{T}(a)$$

This is a differential equation with the solution

$$\operatorname{T}(a) = \exp\left(- \frac{i}{\hbar} p_x a\right).$$

Additionally, suppose a Hamiltonian $$H$$ is independent of the $$x$$ position. Because the translation operator can be written in terms of $$p_x$$, and $$[p_x,H] = 0$$, we know that $$[H, \operatorname{T}(a)]=0.$$ This result means that linear momentum for the system is conserved.

In relation to the orbital angular momentum
Classically we have for the angular momentum $$\mathbf L = \mathbf r \times \mathbf p.$$ This is the same in quantum mechanics considering $$\mathbf r$$ and $$\mathbf p$$ as operators. Classically, an infinitesimal rotation $$dt$$ of the vector $$\mathbf r = (x,y,z)$$ about the $$z$$-axis to $$\mathbf r' = (x',y',z)$$ leaving $$z$$ unchanged can be expressed by the following infinitesimal translations (using Taylor approximation): $$\begin{align} x' &= r \cos(t + dt) = x - y \, dt + \cdots \\ y' &= r \sin(t + dt) = y + x \, dt + \cdots \end{align}$$

From that follows for states: $$\operatorname{R}(z, dt)|r\rangle = \operatorname{R}(z, dt)|x, y, z\rangle = |x - y \, dt, y + x \, dt, z\rangle = \operatorname{T}_x(-y \, dt) \operatorname{T}_y(x \, dt)|x, y, z\rangle = \operatorname{T}_x(-y \, dt) \operatorname{T}_y(x \, dt) |r\rangle$$

And consequently: $$\operatorname{R}(z, dt) = \operatorname{T}_x (-y \, dt) \operatorname{T}_y(x \, dt)$$

Using $$T_k(a) = \exp\left(- \frac{i}{\hbar} p_k a\right)$$ from above with $$k = x,y$$ and Taylor expansion we get: $$\operatorname{R}(z,dt)=\exp\left[-\frac{i}{\hbar} \left(x p_y - y p_x\right) dt\right] = \exp\left(-\frac{i}{\hbar} L_z dt\right) = 1-\frac{i}{\hbar}L_z dt + \cdots$$ with $$L_z = x p_y - y p_x$$ the $$z$$-component of the angular momentum according to the classical cross product.

To get a rotation for the angle $$t$$, we construct the following differential equation using the condition $$\operatorname{R}(z, 0) = 1 $$:

$$\begin{align} &\operatorname{R}(z, t + dt) = \operatorname{R}(z, t) \operatorname{R}(z, dt) \\[1.1ex] \Rightarrow {} & \frac{d\operatorname{R}}{dt} = \frac{\operatorname{R}(z, t + dt) - \operatorname{R}(z, t)}{dt} = \operatorname{R}(z, t) \frac{\operatorname{R}(z, dt) - 1}{dt} = - \frac{i}{\hbar} L_z \operatorname{R}(z, t) \\[1.1ex] \Rightarrow {}& \operatorname{R}(z, t) = \exp\left(- \frac{i}{\hbar}\, t \, L_z\right) \end{align}$$

Similar to the translation operator, if we are given a Hamiltonian $$H$$ which rotationally symmetric about the $$z$$-axis, $$[L_z,H]=0$$ implies $$[\operatorname{R}(z,t),H]=0$$. This result means that angular momentum is conserved.

For the spin angular momentum about for example the $$y$$-axis we just replace $$L_z$$ with $S_y = \frac{\hbar}{2} \sigma_y$ (where $$\sigma_y$$ is the Pauli Y matrix) and we get the spin rotation operator $$\operatorname{D}(y, t) = \exp\left(- i \frac{t}{2} \sigma_y\right).$$

Effect on the spin operator and quantum states
Operators can be represented by matrices. From linear algebra one knows that a certain matrix $$A$$ can be represented in another basis through the transformation $$A' = P A P^{-1}$$ where $$P$$ is the basis transformation matrix. If the vectors $$b$$ respectively $$c$$ are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle $$t$$ between them. The spin operator $$S_b$$ in the first basis can then be transformed into the spin operator $$S_c$$ of the other basis through the following transformation: $$S_c = \operatorname{D}(y, t) S_b \operatorname{D}^{-1}(y, t)$$

From standard quantum mechanics we have the known results $S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle$ and $S_c |c+\rangle = \frac{\hbar}{2} |c+\rangle$  where  $$|b+\rangle$$ and $$|c+\rangle$$ are the top spins in their corresponding bases. So we have: $$\frac{\hbar}{2} |c+\rangle = S_c |c+\rangle = \operatorname{D}(y, t) S_b \operatorname{D}^{-1}(y, t) |c+\rangle \Rightarrow$$ $$S_b \operatorname{D}^{-1}(y, t) |c+\rangle = \frac{\hbar}{2} \operatorname{D}^{-1}(y, t) |c+\rangle$$

Comparison with $S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle$ yields $$|b+\rangle = D^{-1}(y, t) |c+\rangle$$.

This means that if the state $$|c+\rangle$$ is rotated about the $$y$$-axis by an angle $$t$$, it becomes the state $$|b+\rangle$$, a result that can be generalized to arbitrary axes.