 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.

$|\alpha \rangle _{R}=D(R)|\alpha \rangle$ In terms of the generators of rotation,

$D(\mathbf {\hat {n}} ,\phi )=\exp \left(-i\phi {\frac {\mathbf {\hat {n}} \cdot \mathbf {J} }{\hbar }}\right),$ where $\mathbf {\hat {n}}$ is rotation axis, and $\mathbf {J}$ is angular momentum.

## The translation operator

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

Translation of the particle at position x to position x + a: $\operatorname {T} (a)|x\rangle =|x+a\rangle$ Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing):

$\operatorname {T} (0)=1$ $\operatorname {T} (a)\operatorname {T} (da)|x\rangle =\operatorname {T} (a)|x+da\rangle =|x+a+da\rangle =\operatorname {T} (a+da)|x\rangle \Rightarrow \operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a+da)$ Taylor development gives:

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

$p_{x}=i\hbar {\frac {d\operatorname {T} (0)}{da}}$ From that follows:

$\operatorname {T} (a+da)=\operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a)\left(1-{\frac {i}{\hbar }}p_{x}da\right)\Rightarrow [\operatorname {T} (a+da)-\operatorname {T} (a)]/da={\frac {d\operatorname {T} }{da}}=-{\frac {i}{\hbar }}p_{x}\operatorname {T} (a)$ This is a differential equation with the solution

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

$x'=r\cos(t+dt)=x-ydt+\cdots$ $y'=r\sin(t+dt)=y+xdt+\cdots$ From that follows for states:

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

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

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

$\operatorname {R} (z,dt)=\exp \left[-{\frac {i}{h}}(xp_{y}-yp_{x})dt\right]=\exp \left(-{\frac {i}{h}}l_{z}dt\right)=1-{\frac {i}{h}}l_{z}dt+\cdots$ with $l_{z}=xp_{y}-yp_{x}$ the z-component of the angular momentum according to the classical cross product.

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

$\operatorname {R} (z,t+dt)=\operatorname {R} (z,t)\operatorname {R} (z,dt)\Rightarrow$ $[\operatorname {R} (z,t+dt)-\operatorname {R} (z,t)]/dt=d\operatorname {R} /dt=\operatorname {R} (z,t)[\operatorname {R} (z,dt)-1]/dt=-{\frac {i}{h}}l_{z}\operatorname {R} (z,t)\Rightarrow$ $\operatorname {R} (z,t)=\exp \left(-{\frac {i}{h}}\ t\ l_{z}\right)$ Similar to the translation operator, if we are given a Hamiltonian $H$ which rotationally symmetric about the z axis, $[l_{z},H]=0$ implies $[\operatorname {R} (z,t),H]=0$ . This result means that angular momentum is conserved.

For the spin angular momentum about the y-axis we just replace $l_{z}$ with $S_{y}={\frac {h}{2}}\sigma _{y}$ and we get the spin rotation operator

$\operatorname {D} (y,t)=\exp \left(-i{\frac {t}{2}}\sigma _{y}\right).$ ## Effect on the spin operator and quantum states

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

$A'=PAP^{-1}$ where $P$ is the basis transformation matrix. If the vectors $b$ respectively $c$ are the z-axis in one basis respectively another, they are perpendicular to the y-axis with a certain angle $t$ between them. The spin operator $S_{b}$ in the first basis can then be transformed into the spin operator $S_{c}$ of the other basis through the following transformation:

$S_{c}=\operatorname {D} (y,t)S_{b}\operatorname {D} ^{-1}(y,t)$ From standard quantum mechanics we have the known results $S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle$ and $S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle$ where $|b+\rangle$ and $|c+\rangle$ are the top spins in their corresponding bases. So we have:

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

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