# 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,

$D (\mathbf{\hat n},\phi) = \exp \left( -i \phi \frac{\mathbf{\hat n} \cdot \mathbf J }{ \hbar} \right)$

$\mathbf{\hat n}$ is rotation axis, and $\mathbf{J}$ is angular momentum.

## The translation operator

The rotation operator $\,\mbox{R}(z, \theta)$, with the first argument $\,z$ indicating the rotation axis and the second $\,\theta$ the rotation angle, can operate through the translation operator $\,\mbox{T}(a)$ 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: $\mbox{T}(a)|x\rangle = |x + a\rangle$

Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):

$\,\mbox{T}(0) = 1$
$\,\mbox{T}(a) \mbox{T}(da)|x\rangle = \mbox{T}(a)|x + da\rangle = |x + a + da\rangle = \mbox{T}(a + da)|x\rangle \Rightarrow$
$\,\mbox{T}(a) \mbox{T}(da) = \mbox{T}(a + da)$

Taylor development gives:

$\,\mbox{T}(da) = \mbox{T}(0) + \frac{d\mbox{T}(0)}{da} da + ... = 1 - \frac{i}{h}\ p_x\ da$

with

$\,p_x = i h \frac{d\mbox{T}(0)}{da}$

From that follows:

$\,\mbox{T}(a + da) = \mbox{T}(a) \mbox{T}(da) = \mbox{T}(a)\left(1 - \frac{i}{h} p_x da\right) \Rightarrow$
$\,[\mbox{T}(a + da) - \mbox{T}(a)]/da = \frac{d\mbox{T}}{da} = - \frac{i}{h} p_x \mbox{T}(a)$

This is a differential equation with the solution $\,\mbox{T}(a) = \mbox{exp}\left(- \frac{i}{h} p_x a\right)$.

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 $\,[H,\mbox{T}(a)]=0$. 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 $\,l = r \times p$. This is the same in quantum mechanics considering $\,r$ and $\,p$ as operators. Classically, an infinitesimal rotation $\,dt$ of the vector r=(x,y,z) about the z-axis to r'=(x',y',z) leaving z unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

$\,x' = r \cos(t + dt) = x - y dt + ...$
$\,y' = r \sin(t + dt) = y + x dt + ...$

From that follows for states:

$\,\mbox{R}(z, dt)|r\rangle$$= \mbox{R}(z, dt)|x, y, z\rangle$$= |x - y dt, y + x dt, z\rangle$$= \mbox{T}_x(-y dt) \mbox{T}_y(x dt)|x, y, z\rangle$$= \mbox{T}_x(-y dt) \mbox{T}_y(x dt)|r\rangle$

And consequently:

$\,\mbox{R}(z, dt) = \mbox{T}_x (-y dt) \mbox{T}_y(x dt)$

Using $\,T_k(a) = \exp\left(- \frac{i}{h}\ p_k\ a\right)$ from above with $\,k = x,y$ and Taylor development we get:

$\,\mbox{R}(z, dt) = \exp\left[- \frac{i}{h}\ (x p_y - y p_x) dt\right]$$= \exp\left(- \frac{i}{h}\ l_z dt\right) = 1 - \frac{i}{h} l_z dt + ...$

with lz = x py - y px 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 $\mbox{R}(z, 0) = 1$:

$\,\mbox{R}(z, t + dt) = \mbox{R}(z, t) \mbox{R}(z, dt) \Rightarrow$
$\,[\mbox{R}(z, t + dt) - \mbox{R}(z, t)]/dt = d\mbox{R}/dt$$\,= \mbox{R}(z, t) [\mbox{R}(z, dt) - 1]/dt$$\,= - \frac{i}{h} l_z \mbox{R}(z, t) \Rightarrow$
$\,\mbox{R}(z, t) = \exp\left(- \frac{i}{h}\ t\ l_z\right)$

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

For the spin angular momentum about the y-axis we just replace $\,l_z$ with $\,S_y = \frac{h}{2} \sigma_y$ and we get the spin rotation operator $\,\mbox{D}(y, t) = \exp\left(- i \frac{t}{2} \sigma_y\right)$.

## 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:

$\,S_c = \mbox{D}(y, t) S_b \mbox{D}^{-1}(y, t)$

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:

$\,\frac{\hbar}{2} |c+\rangle = S_c |c+\rangle = \mbox{D}(y, t) S_b \mbox{D}^{-1}(y, t) |c+\rangle \Rightarrow$
$\,S_b \mbox{D}^{-1}(y, t) |c+\rangle = \frac{\hbar}{2} \mbox{D}^{-1}(y, t) |c+\rangle$

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. It is important, for instance, in Sakurai's Bell inequality.