View source for Rotation operator (quantum mechanics)

{{short description|Quantum operator}}
{{Other uses|Rotation operator (disambiguation)}}
{{Quantum mechanics}}

This article concerns the '''[[rotation]] [[Operator (physics)|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|angular momentum]] operator, and <math>\hbar</math> is the [[reduced Planck constant]].

==The translation operator==
{{Main|Translation operator (quantum mechanics)}}
The [[rotation]] [[Operator (physics)|operator]] <math>\operatorname{R}(z, \theta)</math>, with the first argument <math>z</math> indicating the rotation [[Cartesian coordinate system|axis]] and the second <math>\theta</math> the rotation angle, can operate through the [[Displacement operator|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 [[quantum state|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 function|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 series|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_(quantum_mechanics)|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==
{{Further|Bloch sphere#Rotations}} 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 series|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 matrices|Pauli Y matrix]]) and we get the [[Spin (physics)|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==
{{Main|Spin (physics)#Rotations}}
{{see also|Rotation group SO(3)#A note on Lie algebra|Change of basis#Endomorphisms}}
Operators can be represented by [[Matrix (mathematics)|matrices]]. From [[linear algebra]] one knows that a certain matrix <math>A</math> can be represented in another [[basis (linear algebra)|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

{{Physics operator}}

{{DEFAULTSORT:Rotation Operator (Quantum Mechanics)}}
[[Category:Rotational symmetry]]
[[Category:Quantum operators]]
[[Category:Unitary operators]]