Rotation operator (quantum mechanics)

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

Quantum mechanical rotations

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

The translation operator

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

Translation of the particle at position x {\displaystyle x} to position x + a {\displaystyle x+a}: T ⁡ ( a ) | x ⟩ = | x + a ⟩ {\displaystyle \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): T ⁡ ( 0 ) = 1 {\displaystyle \operatorname {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 ) {\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)}

Taylor development gives: T ⁡ ( d a ) = T ⁡ ( 0 ) + d T ⁡ ( 0 ) d a d a + ⋯ = 1 − i ℏ p x d a {\displaystyle \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 ℏ d T ⁡ ( 0 ) d a {\displaystyle p_{x}=i\hbar {\frac {d\operatorname {T} (0)}{da}}}

From that follows: T ⁡ ( a + d a ) = T ⁡ ( a ) T ⁡ ( d a ) = T ⁡ ( a ) ( 1 − i ℏ p x d a ) ⇒ T ⁡ ( a + d a ) − T ⁡ ( a ) d a = d T d a = − i ℏ p x T ⁡ ( a ) {\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)}

This is a differential equation with the solution

T ⁡ ( a ) = exp ⁡ ( − i ℏ p x a ) . {\displaystyle \operatorname {T} (a)=\exp \left(-{\frac {i}{\hbar }}p_{x}a\right).}

Additionally, suppose a Hamiltonian H {\displaystyle H} is independent of the x {\displaystyle x} position. Because the translation operator can be written in terms of p x {\displaystyle p_{x}}, and [ p x , H ] = 0 {\displaystyle [p_{x},H]=0}, we know that [ H , T ⁡ ( a ) ] = 0. {\displaystyle [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 L = r × p . {\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} .} This is the same in quantum mechanics considering r {\displaystyle \mathbf {r} } and p {\displaystyle \mathbf {p} } as operators. Classically, an infinitesimal rotation d t {\displaystyle dt} of the vector r = ( x , y , z ) {\displaystyle \mathbf {r} =(x,y,z)} about the z {\displaystyle z}-axis to r ′ = ( x ′ , y ′ , z ) {\displaystyle \mathbf {r} '=(x',y',z)} leaving z {\displaystyle 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 + ⋯ {\displaystyle {\begin{aligned}x'&=r\cos(t+dt)=x-y\,dt+\cdots \\y'&=r\sin(t+dt)=y+x\,dt+\cdots \end{aligned}}}

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 ⟩ {\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 }

And consequently: R ⁡ ( z , d t ) = T x ⁡ ( − y d t ) T y ⁡ ( x d t ) {\displaystyle \operatorname {R} (z,dt)=\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)}

Using T k ( a ) = exp ⁡ ( − i ℏ p k a ) {\displaystyle T_{k}(a)=\exp \left(-{\frac {i}{\hbar }}p_{k}a\right)} from above with k = x , y {\displaystyle k=x,y} and Taylor expansion we get: R ⁡ ( z , d t ) = exp ⁡ [ − i ℏ ( x p y − y p x ) d t ] = exp ⁡ ( − i ℏ L z d t ) = 1 − i ℏ L z d t + ⋯ {\displaystyle \operatorname {R} (z,dt)=\exp \left[-{\frac {i}{\hbar }}\left(xp_{y}-yp_{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 {\displaystyle L_{z}=xp_{y}-yp_{x}} the z {\displaystyle z}-component of the angular momentum according to the classical cross product.

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

R ⁡ ( z , t + d t ) = R ⁡ ( z , t ) R ⁡ ( z , d t ) ⇒ d R d t = R ⁡ ( z , t + d t ) − R ⁡ ( z , t ) d t = R ⁡ ( z , t ) R ⁡ ( z , d t ) − 1 d t = − i ℏ L z R ⁡ ( z , t ) ⇒ R ⁡ ( z , t ) = exp ⁡ ( − i ℏ t L z ) {\displaystyle {\begin{aligned}&\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{aligned}}}

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

For the spin angular momentum about for example the y {\displaystyle y}-axis we just replace L z {\displaystyle L_{z}} with S y = ℏ 2 σ y {\textstyle S_{y}={\frac {\hbar }{2}}\sigma _{y}} (where σ y {\displaystyle \sigma _{y}} is the Pauli Y matrix) and we get the spin rotation operator D ⁡ ( y , t ) = exp ⁡ ( − i t 2 σ y ) . {\displaystyle \operatorname {D} (y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right).}

Effect on the spin operator and quantum states

Operators can be represented by matrices. From linear algebra one knows that a certain matrix A {\displaystyle A} can be represented in another basis through the transformation A ′ = P A P − 1 {\displaystyle A'=PAP^{-1}} where P {\displaystyle P} is the basis transformation matrix. If the vectors b {\displaystyle b} respectively c {\displaystyle c} are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle t {\displaystyle t} between them. The spin operator S b {\displaystyle S_{b}} in the first basis can then be transformed into the spin operator S c {\displaystyle S_{c}} of the other basis through the following transformation: S c = D ⁡ ( y , t ) S b D − 1 ⁡ ( y , t ) {\displaystyle 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 + ⟩ = ℏ 2 | b + ⟩ {\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle } and S c | c + ⟩ = ℏ 2 | c + ⟩ {\textstyle S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle } where | b + ⟩ {\displaystyle |b+\rangle } and | c + ⟩ {\displaystyle |c+\rangle } 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 + ⟩ ⇒ {\displaystyle {\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 D − 1 ⁡ ( y , t ) | c + ⟩ = ℏ 2 D − 1 ⁡ ( y , t ) | c + ⟩ {\displaystyle S_{b}\operatorname {D} ^{-1}(y,t)|c+\rangle ={\frac {\hbar }{2}}\operatorname {D} ^{-1}(y,t)|c+\rangle }

Comparison with S b | b + ⟩ = ℏ 2 | b + ⟩ {\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle } yields | b + ⟩ = D − 1 ( y , t ) | c + ⟩ {\displaystyle |b+\rangle =D^{-1}(y,t)|c+\rangle }.

This means that if the state | c + ⟩ {\displaystyle |c+\rangle } is rotated about the y {\displaystyle y}-axis by an angle t {\displaystyle t}, it becomes the state | b + ⟩ {\displaystyle |b+\rangle }, a result that can be generalized to arbitrary axes.

Rotation operator (quantum mechanics)

Quantum mechanical rotations

The translation operator

In relation to the orbital angular momentum

Effect on the spin operator and quantum states

See also