Rotation operator (quantum mechanics)

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

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 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.

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).}

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.

• Symmetry in quantum mechanics
• Spherical basis
• Optical phase space

• 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