Rotation operator (quantum mechanics)

Short description: Quantum operator

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[math]\displaystyle{ i \hbar \frac{\partial}{\partial t} | \psi (t) \rangle = \hat{H} | \psi (t) \rangle }[/math]

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

## Contents

* 1 Quantum mechanical rotations
* 2 The translation operator
* 3 In relation to the orbital angular momentum
* 4 Effect on the spin operator and quantum states
* 5 See also
* 6 References

## Quantum mechanical rotations

With every physical rotation [math]\displaystyle{ R }[/math], we postulate a quantum mechanical rotation operator [math]\displaystyle{ D(R) }[/math] which rotates quantum mechanical states. [math]\displaystyle{ | \alpha \rangle_R = D(R) |\alpha \rangle }[/math]

In terms of the generators of rotation, [math]\displaystyle{ D (\mathbf{\hat n},\phi) = \exp \left( -i \phi \frac{\mathbf{\hat n} \cdot \mathbf J }{ \hbar} \right), }[/math] where [math]\displaystyle{ \mathbf{\hat n} }[/math] is rotation axis, [math]\displaystyle{ \mathbf{J} }[/math] is angular momentum, and [math]\displaystyle{ \hbar }[/math] is the reduced Planck constant.

## The translation operator

Main page: Physics:Translation operator (quantum mechanics)

The rotation operator [math]\displaystyle{ \operatorname{R}(z, \theta) }[/math], with the first argument [math]\displaystyle{ z }[/math] indicating the rotation axis and the second [math]\displaystyle{ \theta }[/math] the rotation angle, can operate through the translation operator [math]\displaystyle{ \operatorname{T}(a) }[/math] 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 [math]\displaystyle{ |x\rangle }[/math] according to Quantum Mechanics).

Translation of the particle at position [math]\displaystyle{ x }[/math] to position [math]\displaystyle{ x + a }[/math]: [math]\displaystyle{ \operatorname{T}(a)|x\rangle = |x + a\rangle }[/math]

Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing): [math]\displaystyle{ \operatorname{T}(0) = 1 }[/math] [math]\displaystyle{ \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) }[/math]

Taylor development gives: [math]\displaystyle{ \operatorname{T}(da) = \operatorname{T}(0) + \frac{d\operatorname{T}(0)}{da} da + \cdots = 1 - \frac{i}{\hbar} p_x da }[/math] with [math]\displaystyle{ p_x = i \hbar \frac{d\operatorname{T}(0)}{da} }[/math]

From that follows: [math]\displaystyle{ \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) }[/math]

This is a differential equation with the solution

[math]\displaystyle{ \operatorname{T}(a) = \exp\left(- \frac{i}{\hbar} p_x a\right). }[/math]

Additionally, suppose a Hamiltonian [math]\displaystyle{ H }[/math] is independent of the [math]\displaystyle{ x }[/math] position. Because the translation operator can be written in terms of [math]\displaystyle{ p_x }[/math], and [math]\displaystyle{ [p_x,H] = 0 }[/math], we know that [math]\displaystyle{ [H, \operatorname{T}(a)]=0. }[/math] This result means that linear momentum for the system is conserved.

## In relation to the orbital angular momentum

Further information: Bloch sphere § Rotations

Classically we have for the angular momentum [math]\displaystyle{ \mathbf L = \mathbf r \times \mathbf p. }[/math] This is the same in quantum mechanics considering [math]\displaystyle{ \mathbf r }[/math] and [math]\displaystyle{ \mathbf p }[/math] as operators. Classically, an infinitesimal rotation [math]\displaystyle{ dt }[/math] of the vector [math]\displaystyle{ \mathbf r = (x,y,z) }[/math] about the [math]\displaystyle{ z }[/math]-axis to [math]\displaystyle{ \mathbf r' = (x',y',z) }[/math] leaving [math]\displaystyle{ z }[/math] unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

[math]\displaystyle{ \begin{align} x' &= r \cos(t + dt) = x - y \, dt + \cdots \\\ y' &= r \sin(t + dt) = y + x \, dt + \cdots \end{align} }[/math]

From that follows for states: [math]\displaystyle{ \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 }[/math]

And consequently: [math]\displaystyle{ \operatorname{R}(z, dt) = \operatorname{T}_x (-y \, dt) \operatorname{T}_y(x \, dt) }[/math]

Using [math]\displaystyle{ T_k(a) = \exp\left(- \frac{i}{\hbar} p_k a\right) }[/math] from above with [math]\displaystyle{ k = x,y }[/math] and Taylor expansion we get: [math]\displaystyle{ \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 }[/math] with [math]\displaystyle{ L_z = x p_y - y p_x }[/math] the [math]\displaystyle{ z }[/math]-component of the angular momentum according to the classical cross product.

To get a rotation for the angle [math]\displaystyle{ t }[/math], we construct the following differential equation using the condition [math]\displaystyle{ \operatorname{R}(z, 0) = 1 }[/math]:

[math]\displaystyle{ \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} }[/math]

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

For the spin angular momentum about for example the [math]\displaystyle{ y }[/math]-axis we just replace [math]\displaystyle{ L_z }[/math] with [math]\displaystyle{ S_y = \frac{\hbar}{2} \sigma_y }[/math] (where [math]\displaystyle{ \sigma_y }[/math] is the Pauli Y matrix) and we get the spin rotation operator [math]\displaystyle{ \operatorname{D}(y, t) = \exp\left(- i \frac{t}{2} \sigma_y\right). }[/math]

## Effect on the spin operator and quantum states

Operators can be represented by matrices. From linear algebra one knows that a certain matrix [math]\displaystyle{ A }[/math] can be represented in another basis through the transformation [math]\displaystyle{ A' = P A P^{-1} }[/math] where [math]\displaystyle{ P }[/math] is the basis transformation matrix. If the vectors [math]\displaystyle{ b }[/math] respectively [math]\displaystyle{ c }[/math] are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle [math]\displaystyle{ t }[/math] between them. The spin operator [math]\displaystyle{ S_b }[/math] in the first basis can then be transformed into the spin operator [math]\displaystyle{ S_c }[/math] of the other basis through the following transformation: [math]\displaystyle{ S_c = \operatorname{D}(y, t) S_b \operatorname{D}^{-1}(y, t) }[/math]

From standard quantum mechanics we have the known results [math]\displaystyle{ S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle }[/math] and [math]\displaystyle{ S_c |c+\rangle = \frac{\hbar}{2} |c+\rangle }[/math] where [math]\displaystyle{ |b+\rangle }[/math] and [math]\displaystyle{ |c+\rangle }[/math] are the top spins in their corresponding bases. So we have: [math]\displaystyle{ \frac{\hbar}{2} |c+\rangle = S_c |c+\rangle = \operatorname{D}(y, t) S_b \operatorname{D}^{-1}(y, t) |c+\rangle \Rightarrow }[/math] [math]\displaystyle{ S_b \operatorname{D}^{-1}(y, t) |c+\rangle = \frac{\hbar}{2} \operatorname{D}^{-1}(y, t) |c+\rangle }[/math]

Comparison with [math]\displaystyle{ S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle }[/math] yields [math]\displaystyle{ |b+\rangle = D^{-1}(y, t) |c+\rangle }[/math].

This means that if the state [math]\displaystyle{ |c+\rangle }[/math] is rotated about the [math]\displaystyle{ y }[/math]-axis by an angle [math]\displaystyle{ t }[/math], it becomes the state [math]\displaystyle{ |b+\rangle }[/math], 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

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Operators in physics

General|

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Particles|

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Operators for operators|

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Quantum|

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Energy|

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* Transition dipole moment

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Particle physics|

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