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}
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