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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,

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

Please see "rotations in quantum mechanics" for details.

The translation operator

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

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

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

$\,{\mbox{T}}(0)=1$

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

$\,{\mbox{T}}(a){\mbox{T}}(da)={\mbox{T}}(a+da)$

Taylor development gives:

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

with

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

From that follows:

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

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

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

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 $\,[H,{\mbox{T}}(a)]=0$. 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 $\,l=r\times p$. 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):

$\,x'=r\cos(t+dt)=x-ydt+...$

$\,y'=r\sin(t+dt)=y+xdt+...$

From that follows for states:

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

And consequently:

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

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

$\,{\mbox{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+...$

with l_z = x p_y - y p_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 ${\mbox{R}}(z,0)=1$:

$\,{\mbox{R}}(z,t+dt)={\mbox{R}}(z,t){\mbox{R}}(z,dt)\Rightarrow$

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

$\,{\mbox{R}}(z,t)=\exp \left(-{\frac {i}{h}}\ t\ l_{z}\right)$

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

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

Effect on the spin operator and quantum states

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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:

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

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:

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

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

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.

See also

• Symmetry in quantum mechanics
• Spherical basis

References

• 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
• Karin, Todd. Rotations in Quantum Mechanics. Unpublished Work.

See also

• Optical phase space

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