With every physical rotation R, we postulate a quantum mechanical rotation operator D(R) which rotates quantum mechanical states.
| α ⟩ R = D ( R ) | α ⟩ In terms of the generators of rotation,
D ( n ^ , ϕ ) = exp ( − i ϕ n ^ ⋅ J ℏ ) n ^ is rotation axis, and J is angular momentum.
The rotation operator R ( z , θ ) , with the first argument z indicating the rotation axis and the second θ the rotation angle, can operate through the translation operator 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 ⟩ according to Quantum Mechanics).
Translation of the particle at position x to position x+a: T ( a ) | x ⟩ = | x + a ⟩
Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):
T ( 0 ) = 1 T ( a ) T ( d a ) | x ⟩ = T ( a ) | x + d a ⟩ = | x + a + d a ⟩ = T ( a + d a ) | x ⟩ ⇒ T ( a ) T ( d a ) = T ( a + d a ) Taylor development gives:
T ( d a ) = T ( 0 ) + d T ( 0 ) d a d a + . . . = 1 − i h p x d a with
p x = i h d T ( 0 ) d a From that follows:
T ( a + d a ) = T ( a ) T ( d a ) = T ( a ) ( 1 − i h p x d a ) ⇒ [ T ( a + d a ) − T ( a ) ] / d a = d T d a = − i h p x T ( a ) This is a differential equation with the solution T ( a ) = exp ( − i h p x a ) .
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 , T ( a ) ] = 0 . This result means that linear momentum for the system is conserved.
Classically we have for the angular momentum l = r × p . This is the same in quantum mechanics considering r and p as operators. Classically, an infinitesimal rotation d t of the vector r=(x,y,z) about the z-axis to r'=(x',y',z) leaving z unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):
x ′ = r cos ( t + d t ) = x − y d t + . . . y ′ = r sin ( t + d t ) = y + x d t + . . . From that follows for states:
R ( z , d t ) | r ⟩ = R ( z , d t ) | x , y , z ⟩ = | x − y d t , y + x d t , z ⟩ = T x ( − y d t ) T y ( x d t ) | x , y , z ⟩ = T x ( − y d t ) T y ( x d t ) | r ⟩ And consequently:
R ( z , d t ) = T x ( − y d t ) T y ( x d t ) Using T k ( a ) = exp ( − i h p k a ) from above with k = x , y and Taylor expansion we get:
R ( z , d t ) = exp [ − i h ( x p y − y p x ) d t ] = exp ( − i h l z d t ) = 1 − i h l z d t + . . . with lz = x py - y px 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 R ( z , 0 ) = 1 :
R ( z , t + d t ) = R ( z , t ) R ( z , d t ) ⇒ [ R ( z , t + d t ) − R ( z , t ) ] / d t = d R / d t = R ( z , t ) [ R ( z , d t ) − 1 ] / d t = − i h l z R ( z , t ) ⇒ R ( z , t ) = exp ( − i h t l z ) Similar to the translation operator, if we are given a Hamiltonian H which rotationally symmetric about the z axis, [ l z , H ] = 0 implies [ R ( z , t ) , H ] = 0 . This result means that angular momentum is conserved.
For the spin angular momentum about the y-axis we just replace l z with S y = h 2 σ y and we get the spin rotation operator D ( y , t ) = exp ( − i t 2 σ y ) .
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 − 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 = D ( y , t ) S b D − 1 ( y , t ) From standard quantum mechanics we have the known results S b | b + ⟩ = ℏ 2 | b + ⟩ and S c | c + ⟩ = ℏ 2 | c + ⟩ where | b + ⟩ and | c + ⟩ are the top spins in their corresponding bases. So we have:
ℏ 2 | c + ⟩ = S c | c + ⟩ = D ( y , t ) S b D − 1 ( y , t ) | c + ⟩ ⇒ S b D − 1 ( y , t ) | c + ⟩ = ℏ 2 D − 1 ( y , t ) | c + ⟩ Comparison with S b | b + ⟩ = ℏ 2 | b + ⟩ yields | b + ⟩ = D − 1 ( y , t ) | c + ⟩ .
This means that if the state | c + ⟩ is rotated about the y-axis by an angle t , it becomes the state | b + ⟩ , a result that can be generalized to arbitrary axes. It is important, for instance, in Sakurai's Bell inequality.
