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 $\widehat(R) : H\to H$ that is the rule that assigns to each vector in the space $H$ the vector
$, \alpha \rangle_R = \widehat(R) , \alpha \rangle$
that is also in $H$. We will show that, in terms of the generators of rotation,
$\widehat (\mathbf,\phi) = \exp \left( -i \phi \frac \right),$
where $\mathbf$ is the rotation axis, $\widehat$ is angular momentum operator, and $\hbar$ is the reduced Planck constant.

The translation operator

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

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

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

Additionally, suppose a Hamiltonian $H$ is independent of the $x$ position. Because the translation operator can be written in terms of $p_x$, and _x,H= 0, we know that , \operatorname(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):
$\begin x' &= r \cos(t + dt) = x - y \, dt + \cdots \\ y' &= r \sin(t + dt) = y + x \, dt + \cdots \end$

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 \left(x p_y - y p_x\right) dt\right= \exp\left(-\frac L_z dt\right) = 1-\fracL_z dt + \cdots
with $L_z = x p_y - y p_x$ 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 $\operatorname(z, 0) = 1$:

\begin
&\operatorname(z, t + dt) = \operatorname(z, t) \operatorname(z, dt) \\ .1ex\Rightarrow & \frac = \frac = \operatorname(z, t) \frac = - \frac L_z \operatorname(z, t) \\ .1ex\Rightarrow & \operatorname(z, t) = \exp\left(- \frac\, t \, L_z\right)
\end

Similar to the translation operator, if we are given a Hamiltonian $H$ which rotationally symmetric about the $z$-axis, _z,H0 implies operatorname(z,t),H0. 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 $S_y = \frac \sigma_y$ (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 $A$ can be represented in another basis through the transformation
$A' = P A P^$
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:
$S_c = \operatorname(y, t) S_b \operatorname^(y, t)$

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 $, b+\rangle = D^(y, t) , c+\rangle$.

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.

See also

* Symmetry in quantum mechanics
* Spherical basis
* Optical phase space

References

*L.D. Landau and E.M. Lifshitz: ''Quantum Mechanics: Non-Relativistic Theory'', Pergamon Press, 1985
*P.A.M. Dirac: ''The Principles of Quantum Mechanics'', Oxford University Press, 1958
*R.P. Feynman, R.B. Leighton and M. Sands: ''The Feynman Lectures on Physics'', Addison-Wesley, 1965

{{DEFAULTSORT:Rotation Operator (Quantum Mechanics)
Rotational symmetry
Quantum operators
Unitary operators