Rotation Operator (quantum Mechanics)
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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,

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

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

## The translation operator

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

Translation of the particle at position x to position 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):

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

Taylor development gives:

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

with

${\displaystyle p_{x}=i\hbar {\frac {d\operatorname {T} (0)}{da}}}$

From that follows:

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

This is a differential equation with the solution

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

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 ${\displaystyle [H,\operatorname {T} (a)]=0.}$ 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 ${\displaystyle l=r\times p.}$ This is the same in quantum mechanics considering ${\displaystyle r}$ and ${\displaystyle p}$ as operators. Classically, an infinitesimal rotation ${\displaystyle dt}$ of the vector ${\displaystyle r=(x,y,z)}$ about the z-axis to ${\displaystyle r'=(x',y',z)}$ leaving z unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

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

From that follows for states:

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

And consequently:

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

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:

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

with ${\displaystyle l_{z}=xp_{y}-yp_{x}}$ 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 ${\displaystyle \operatorname {R} (z,0)=1}$:

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

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 ${\displaystyle [\operatorname {R} (z,t),H]=0}$. 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

${\displaystyle \operatorname {D} (y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right).}$

## Effect on the spin operator and quantum states

Operators can be represented by matrices. From linear algebra one knows that a certain matrix ${\displaystyle A}$ 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:

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

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:

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

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.