Rotation operator (quantum mechanics) explained

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

Quantum mechanical rotations

With every physical rotation

, we postulate a quantum mechanical rotation operator

\widehat{D}(R):H\toH

that is the rule that assigns to each vector in the space

the vector

| \alpha \rangle_R = \widehat(R) |\alpha \rangle

that is also in

. We will show that, in terms of the generators of rotation,

\widehat (\mathbf,\phi) = \exp \left(-i \phi \frac \right),

where

\hatn

is the rotation axis,

\widehat{J

} is angular momentum operator, and

\hbar

is the reduced Planck constant.

The translation operator

\operatorname{R}(z,\theta)

, with the first argument

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

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): $\operatorname(0) = 1$ $\operatorname(a) \operatorname(da)|x\rangle = \operatorname(a)|x + da\rangle = |x + a + da\rangle = \operatorname(a + da)|x\rangle \Rightarrow \operatorname(a) \operatorname(da) = \operatorname(a + da)$

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

From that follows: $\operatorname(a + da) = \operatorname(a) \operatorname(da) = \operatorname(a)\left(1 - \frac p_x da\right) \Rightarrow \frac = \frac = - \frac p_x \operatorname(a)$

This is a differential equation with the solution

$\operatorname(a) = \exp\left(- \frac p_x a\right).$

is independent of the

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

L=r x p.

This is the same in quantum mechanics considering

and

as operators. Classically, an infinitesimal rotation

of the vector

r=(x,y,z)

about the

-axis to

r'=(x',y',z)

leaving

unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

\beginx' &= r \cos(t + dt) = x - y \, dt + \cdots \\y' &= r \sin(t + dt) = y + x \, dt + \cdots\end

From that follows for states: $\operatorname(z, dt)|r\rangle = \operatorname(z, dt)|x, y, z\rangle = |x - y \, dt, y + x \, dt, z\rangle = \operatorname_x(-y \, dt) \operatorname_y(x \, dt)|x, y, z\rangle = \operatorname_x(-y \, dt) \operatorname_y(x \, dt) |r\rangle$

And consequently: $\operatorname(z, dt) = \operatorname_x (-y \, dt) \operatorname_y(x \, dt)$

Using $T_k(a) = \exp\left(- \frac p_k a\right)$ from above with

k=x,y

and Taylor expansion we get:

\operatorname(z,dt)=\exp\left[-\frac{i}{\hbar} \left(x p_y - y p_x\right) dt\right] = \exp\left(-\frac L_z dt\right) = 1-\fracL_z dt + \cdots

with

L_z=xp_y-yp_x

the

-component of the angular momentum according to the classical cross product.

To get a rotation for the angle

, we construct the following differential equation using the condition

\operatorname{R}(z,0)=1

$\begin&\operatorname(z, t + dt) = \operatorname(z, t) \operatorname(z, dt) \\[1.1ex]\Rightarrow & \frac = \frac = \operatorname(z, t) \frac = - \frac L_z \operatorname(z, t) \\[1.1ex]\Rightarrow & \operatorname(z, t) = \exp\left(- \frac\, t \, L_z\right)\end$

Similar to the translation operator, if we are given a Hamiltonian

which rotationally symmetric about the

-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

-axis we just replace

L_z

with

S_y = \frac \sigma_y

(where

\sigma_y

is the Pauli Y matrix) and we get the spin rotation operator

\operatorname(y, t) = \exp\left(- i \frac \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

can be represented in another basis through the transformation

A' = P A P^

where

is the basis transformation matrix. If the vectors

respectively

are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle

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(y, t) S_b \operatorname^(y, t)

From standard quantum mechanics we have the known results $S_b |b+\rangle = \frac |b+\rangle$ and $S_c |c+\rangle = \frac |c+\rangle$ where

|b+\rangle

and

|c+\rangle

are the top spins in their corresponding bases. So we have:

\frac |c+\rangle = S_c |c+\rangle = \operatorname(y, t) S_b \operatorname^(y, t) |c+\rangle \Rightarrow

S_b \operatorname^(y, t) |c+\rangle = \frac \operatorname^(y, t) |c+\rangle

Comparison with $S_b |b+\rangle = \frac |b+\rangle$ yields

|b+\rangle=D^-1(y,t)|c+\rangle

This means that if the state

|c+\rangle

is rotated about the

-axis by an angle

, it becomes the state