Rotation operator (quantum mechanics)

Quantum mechanical rotations

With every physical rotation R, we postulate a quantum mechanical rotation operator \widehat{D}(R) : H\to H that is the rule that assigns to each vector in the space H the vector | \alpha \rangle_R = \widehat{D}(R) |\alpha \rangle that is also in H. We will show that, in terms of the generators of rotation, \widehat{D} (\mathbf{\hat n},\phi) = \exp \left( -i \phi \frac{\mathbf{\hat n} \cdot \widehat{\mathbf J }}{ \hbar} \right), where \mathbf{\hat n} is the rotation axis, \widehat{\mathbf{J}} is angular momentum operator, and \hbar is the reduced Planck constant.

The translation operator

Main article: Translation operator (quantum mechanics)

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

Main article: Spin (physics)#Rotations

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.

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