# Rotation operator (quantum mechanics)

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

## Quantum mechanical rotations

With every physical rotation R, we postulate a quantum mechanical rotation operator D(R) which rotates quantum mechanical states.

${\displaystyle |\alpha \rangle _{R}=D(R)|\alpha \rangle }$

In terms of the generators of rotation,

${\displaystyle D(\mathbf {\hat {n}} ,\phi )=\exp \left(-i\phi {\frac {\mathbf {\hat {n}} \cdot \mathbf {J} }{\hbar }}\right)}$

${\displaystyle \mathbf {\hat {n}} }$ is rotation axis, and ${\displaystyle \mathbf {J} }$ is angular momentum.

## The translation operator

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

Translation of the particle at position x to position x+a: ${\displaystyle {\mbox{T}}(a)|x\rangle =|x+a\rangle }$

Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):

${\displaystyle \,{\mbox{T}}(0)=1}$
${\displaystyle \,{\mbox{T}}(a){\mbox{T}}(da)|x\rangle ={\mbox{T}}(a)|x+da\rangle =|x+a+da\rangle ={\mbox{T}}(a+da)|x\rangle \Rightarrow }$
${\displaystyle \,{\mbox{T}}(a){\mbox{T}}(da)={\mbox{T}}(a+da)}$

Taylor development gives:

${\displaystyle \,{\mbox{T}}(da)={\mbox{T}}(0)+{\frac {d{\mbox{T}}(0)}{da}}da+...=1-{\frac {i}{\hbar }}\ p_{x}\ da}$

with

${\displaystyle \,p_{x}=ih{\frac {d{\mbox{T}}(0)}{da}}}$

From that follows:

${\displaystyle \,{\mbox{T}}(a+da)={\mbox{T}}(a){\mbox{T}}(da)={\mbox{T}}(a)\left(1-{\frac {i}{h}}p_{x}da\right)\Rightarrow }$
${\displaystyle \,[{\mbox{T}}(a+da)-{\mbox{T}}(a)]/da={\frac {d{\mbox{T}}}{da}}=-{\frac {i}{h}}p_{x}{\mbox{T}}(a)}$

This is a differential equation with the solution ${\displaystyle \,{\mbox{T}}(a)={\mbox{exp}}\left(-{\frac {i}{h}}p_{x}a\right)}$.

Additionally, suppose a Hamiltonian ${\displaystyle \,H}$ is independent of the ${\displaystyle \,x}$ position. Because the translation operator can be written in terms of ${\displaystyle \,p_{x}}$, and ${\displaystyle \,[p_{x},H]=0}$, we know that ${\displaystyle \,[H,{\mbox{T}}(a)]=0}$. 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 ${\displaystyle \,l=r\times p}$. This is the same in quantum mechanics considering ${\displaystyle \,r}$ and ${\displaystyle \,p}$ as operators. Classically, an infinitesimal rotation ${\displaystyle \,dt}$ 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):

${\displaystyle \,x'=r\cos(t+dt)=x-ydt+...}$
${\displaystyle \,y'=r\sin(t+dt)=y+xdt+...}$

From that follows for states:

${\displaystyle \,{\mbox{R}}(z,dt)|r\rangle }$${\displaystyle ={\mbox{R}}(z,dt)|x,y,z\rangle }$${\displaystyle =|x-ydt,y+xdt,z\rangle }$${\displaystyle ={\mbox{T}}_{x}(-ydt){\mbox{T}}_{y}(xdt)|x,y,z\rangle }$${\displaystyle ={\mbox{T}}_{x}(-ydt){\mbox{T}}_{y}(xdt)|r\rangle }$

And consequently:

${\displaystyle \,{\mbox{R}}(z,dt)={\mbox{T}}_{x}(-ydt){\mbox{T}}_{y}(xdt)}$

Using ${\displaystyle \,T_{k}(a)=\exp \left(-{\frac {i}{h}}\ p_{k}\ a\right)}$ from above with ${\displaystyle \,k=x,y}$ and Taylor expansion we get:

${\displaystyle \,{\mbox{R}}(z,dt)=\exp \left[-{\frac {i}{h}}\ (xp_{y}-yp_{x})dt\right]}$${\displaystyle =\exp \left(-{\frac {i}{h}}\ l_{z}dt\right)=1-{\frac {i}{h}}l_{z}dt+...}$

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 ${\displaystyle \,t}$, we construct the following differential equation using the condition ${\displaystyle {\mbox{R}}(z,0)=1}$:

${\displaystyle \,{\mbox{R}}(z,t+dt)={\mbox{R}}(z,t){\mbox{R}}(z,dt)\Rightarrow }$
${\displaystyle \,[{\mbox{R}}(z,t+dt)-{\mbox{R}}(z,t)]/dt=d{\mbox{R}}/dt}$${\displaystyle \,={\mbox{R}}(z,t)[{\mbox{R}}(z,dt)-1]/dt}$${\displaystyle \,=-{\frac {i}{h}}l_{z}{\mbox{R}}(z,t)\Rightarrow }$
${\displaystyle \,{\mbox{R}}(z,t)=\exp \left(-{\frac {i}{h}}\ t\ l_{z}\right)}$

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

For the spin angular momentum about the y-axis we just replace ${\displaystyle \,l_{z}}$ with ${\displaystyle \,S_{y}={\frac {h}{2}}\sigma _{y}}$ and we get the spin rotation operator ${\displaystyle \,{\mbox{D}}(y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right)}$.

## Effect on the spin operator and quantum states

Operators can be represented by matrices. From linear algebra one knows that a certain matrix ${\displaystyle \,A}$ can be represented in another basis through the transformation

${\displaystyle \,A'=PAP^{-1}}$

where ${\displaystyle \,P}$ is the basis transformation matrix. If the vectors ${\displaystyle \,b}$ respectively ${\displaystyle \,c}$ are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle ${\displaystyle \,t}$ between them. The spin operator ${\displaystyle \,S_{b}}$ in the first basis can then be transformed into the spin operator ${\displaystyle \,S_{c}}$ of the other basis through the following transformation:

${\displaystyle \,S_{c}={\mbox{D}}(y,t)S_{b}{\mbox{D}}^{-1}(y,t)}$

From standard quantum mechanics we have the known results ${\displaystyle \,S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }$ and ${\displaystyle \,S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle }$ where ${\displaystyle \,|b+\rangle }$ and ${\displaystyle \,|c+\rangle }$ are the top spins in their corresponding bases. So we have:

${\displaystyle \,{\frac {\hbar }{2}}|c+\rangle =S_{c}|c+\rangle ={\mbox{D}}(y,t)S_{b}{\mbox{D}}^{-1}(y,t)|c+\rangle \Rightarrow }$
${\displaystyle \,S_{b}{\mbox{D}}^{-1}(y,t)|c+\rangle ={\frac {\hbar }{2}}{\mbox{D}}^{-1}(y,t)|c+\rangle }$

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

This means that if the state ${\displaystyle \,|c+\rangle }$ is rotated about the y-axis by an angle ${\displaystyle \,t}$, it becomes the state ${\displaystyle \,|b+\rangle }$, a result that can be generalized to arbitrary axes. It is important, for instance, in Sakurai's Bell inequality.