Rotation operator (quantum mechanics)

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

1 The translation operator
2 In relation to the orbital angular momentum
3 Effect upon the spin operator and upon states
4 See also

The translation operator

The rotation operator $\,\mbox{R}(z, t)$ , with the first argument $\,z$ indicating the rotation axis and the second $\,t = \theta$ the rotation angle, is based on the translation operator $\,\mbox{T}(a)$ , which is acting on the state $|x\rangle$ in the following manner:

$\mbox{T}(a)|x\rangle = |x + a\rangle$

We have:

$\,\mbox{T}(0) = 1$

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

$\,\mbox{T}(a) \mbox{T}(da) = \mbox{T}(a + da)$

Taylor development gives:

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

with

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

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 $\,l = r \times p$ . This is the same in quantum mechanics considering $\,r$ and $\,p$ as operators. An infinitesimal rotation $\,dt$ about the z-axis can be expressed by the following infinitesimal translations:

$\,x' = x - y dt$

$\,y' = y + x dt$

From that follows:

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

And consequently:

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

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

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

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

$\,\mbox{R}(z, t + dt) = \mbox{R}(z, t) \mbox{R}(z, dt) \Rightarrow$

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

$\,\mbox{R}(z, t) = \exp\left(- \frac{i}{h}\ t\ l_z\right)$

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 upon the spin operator and upon states

Main article: Spin (physics)#Spin and rotations

Operators can be represented by matrices. From linear algebra one knows that a certain matrix $\,A$ can be represented in another base through the basis transformation

$\,A' = P A P^{-1}$

where $\,P$ is the transformation matrix. If $\,b$ and $\,c$ are perpendicular to the y-axis and the angle $\,t$ lies between them, the spin operator $\,S_b$ can be transformed into the spin operator $S c$ 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 $\,S_c |c+\rangle = \frac{\hbar}{2} |c+\rangle$ . So we have: