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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.

$|\alpha {⟩}_R=D(R)|\alpha ⟩$

In terms of the generators of rotation,

$D(\hat{\mathbf{n}},\varphi )=\exp (-i\varphi \frac{\hat{\mathbf{n}}\cdot \mathbf{J}}{\hslash })$

Please see "rotations in quantum mechanics" for details.

The translation operator

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

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

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

$\text{T}(0)=1$

$\text{T}(a)\text{T}(da)|x⟩=\text{T}(a)|x+da⟩=|x+a+da⟩=\text{T}(a+da)|x⟩\Rightarrow$

$\text{T}(a)\text{T}(da)=\text{T}(a+da)$

Taylor development gives:

$\text{T}(da)=\text{T}(0)+\frac{d\text{T}(0)}{da}da+...=1-\frac{i}{h}{p}_{x}da$

with

$p_x=ih\frac{d\text{T}(0)}{da}$

From that follows:

$\text{T}(a+da)=\text{T}(a)\text{T}(da)=\text{T}(a)(1-\frac{i}{h}{p}_{x}da)\Rightarrow$

$[\text{T}(a+da)-\text{T}(a)]/da=\frac{d\text{T}}{da}=-\frac{i}{h}{p}_x\text{T}(a)$

This is a differential equation with the solution $\text{T}(a)=\text{exp}(-\frac{i}{h}{p}_{x}a)$.

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 $[H,\text{T}(a)]=0$. 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):

$x^{\prime }=r\cos (t+dt)=x-ydt+...$

$y^{\prime }=r\sin (t+dt)=y+xdt+...$

From that follows for states:

$\text{R}(z,dt)|r⟩$$=\text{R}(z,dt)|x,y,z⟩$$=|x-ydt,y+xdt,z⟩$$={\text{T}}_x(-ydt){\text{T}}_y(xdt)|x,y,z⟩$$={\text{T}}_x(-ydt){\text{T}}_y(xdt)|r⟩$

And consequently:

$\text{R}(z,dt)={\text{T}}_x(-ydt){\text{T}}_y(xdt)$

Using $T_k(a)=\exp (-\frac{i}{h}{p}_{k}a)$ from above with $k=x,y$ and Taylor development we get:

$\text{R}(z,dt)=\exp [-\frac{i}{h}(x{p}_y-y{p}_x)dt]$$=\exp (-\frac{i}{h}{l}_{z}dt)=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 $\text{R}(z,0)=1$:

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

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

$\text{R}(z,t)=\exp (-\frac{i}{h}t{l}_z)$

Similar to the translation operator, if we are given a Hamiltonian $H$ which rotationally symmetric about the z axis, $[l_z,H]=0$ implies $[\text{R}(z,t),H]=0$. 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 $\text{D}(y,t)=\exp (-i\frac{t}{2}{\sigma }_y)$.

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

$A^{\prime }=PA{P}^{-1}$

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:

$S_c=\text{D}(y,t)S_b{\text{D}}^{-1}(y,t)$

From standard quantum mechanics we have the known results $S_b|b+⟩=\frac{\hslash }{2}|b+⟩$ and $S_c|c+⟩=\frac{\hslash }{2}|c+⟩$ where $|b+⟩$ and $|c+⟩$ are the top spins in their corresponding bases. So we have:

$\frac{\hslash }{2}|c+⟩=S_c|c+⟩=\text{D}(y,t)S_b{\text{D}}^{-1}(y,t)|c+⟩\Rightarrow$

$S_b{\text{D}}^{-1}(y,t)|c+⟩=\frac{\hslash }{2}{\text{D}}^{-1}(y,t)|c+⟩$

Comparison with $S_b|b+⟩=\frac{\hslash }{2}|b+⟩$ yields $|b+⟩=D^{-1}(y,t)|c+⟩$.

This means that if the state $|c+⟩$ is rotated about the y-axis by an angle $t$, it becomes the state $|b+⟩$, 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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