Rotation operator (quantum mechanics)

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

Quantum mechanical rotations

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

The translation operator

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} The rotation operator <math>\operatorname{R}(z, \theta)</math>, with the first argument <math>z</math> indicating the rotation axis and the second <math>\theta</math> the rotation angle, can operate through the translation operator <math>\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>|x\rangle</math> according to Quantum Mechanics).

Translation of the particle at position <math>x</math> to position <math>x + a</math>: <math>\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 display="block">\operatorname{T}(0) = 1</math> <math display="block">\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 display="block">\operatorname{T}(da) = \operatorname{T}(0) + \frac{d\operatorname{T}(0)}{da} da + \cdots = 1 - \frac{i}{\hbar} p_x da</math> with <math display="block">p_x = i \hbar \frac{d\operatorname{T}(0)}{da}</math>

From that follows: <math display="block">\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 display="block">\operatorname{T}(a) = \exp\left(- \frac{i}{\hbar} p_x a\right).</math>

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

In relation to the orbital angular momentum

Template:Further Classically we have for the angular momentum <math>\mathbf L = \mathbf r \times \mathbf p.</math> This is the same in quantum mechanics considering <math>\mathbf r</math> and <math>\mathbf p</math> as operators. Classically, an infinitesimal rotation <math>dt</math> of the vector <math>\mathbf r = (x,y,z)</math> about the <math>z</math>-axis to <math>\mathbf r' = (x',y',z)</math> leaving <math>z</math> unchanged can be expressed by the following infinitesimal translations (using Taylor approximation): <math display="block">\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 display="block">\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 display="block">\operatorname{R}(z, dt) = \operatorname{T}_x (-y \, dt) \operatorname{T}_y(x \, dt)</math>

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

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

<math display="block">\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>H</math> which rotationally symmetric about the <math>z</math>-axis, <math>[L_z,H]=0</math> implies <math>[\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>y</math>-axis we just replace <math>L_z</math> with <math display="inline">S_y = \frac{\hbar}{2} \sigma_y</math> (where <math>\sigma_y</math> is the Pauli Y matrix) and we get the spin rotation operator <math display="block">\operatorname{D}(y, t) = \exp\left(- i \frac{t}{2} \sigma_y\right).</math>

Effect on the spin operator and quantum states

{{#invoke:Labelled list hatnote|labelledList|Main article|Main articles|Main page|Main pages}} Template:See also Operators can be represented by matrices. From linear algebra one knows that a certain matrix <math>A</math> can be represented in another basis through the transformation <math display="block">A' = P A P^{-1}</math> where <math>P</math> is the basis transformation matrix. If the vectors <math>b</math> respectively <math>c</math> are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle <math>t</math> between them. The spin operator <math>S_b</math> in the first basis can then be transformed into the spin operator <math>S_c</math> of the other basis through the following transformation: <math display="block">S_c = \operatorname{D}(y, t) S_b \operatorname{D}^{-1}(y, t)</math>

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

Comparison with <math display="inline">S_b |b+\rangle = \frac{\hbar}{2} |b+\rangle</math> yields <math>|b+\rangle = D^{-1}(y, t) |c+\rangle</math>.

This means that if the state <math>|c+\rangle</math> is rotated about the <math>y</math>-axis by an angle <math>t</math>, it becomes the state <math>|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