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 $\hat{D}(R):H\to H$ that is the rule that assigns to each vector in the space $H$ the vector $$|\alpha {⟩}_R=\hat{D}(R)|\alpha ⟩$$ that is also in $H$. We will show that, in terms of the generators of rotation, $$\hat{D}(\widehat{𝐧},\varphi )=\exp (-i\varphi \frac{\widehat{𝐧}\cdot \widehat{𝐉}}{\hslash }),$$ where $\widehat{𝐧}$ is the rotation axis, $\widehat{𝐉}$ is angular momentum operator, and $\hslash$ is the reduced Planck constant.

The translation operator

Template:Main The rotation operator $\mathrm{R}(z,\theta )$, with the first argument $z$ indicating the rotation axis and the second $\theta$ the rotation angle, can operate through the translation operator $\mathrm{T}(a)$ 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 $x+a$: $\mathrm{T}(a)|x⟩=|x+a⟩$

Because a translation of 0 does not change the position of the particle, we have (with 1 meaning the identity operator, which does nothing): $$\mathrm{T}(0)=1$$ $$\mathrm{T}(a)\mathrm{T}(da)|x⟩=\mathrm{T}(a)|x+da⟩=|x+a+da⟩=\mathrm{T}(a+da)|x⟩\Rightarrow \mathrm{T}(a)\mathrm{T}(da)=\mathrm{T}(a+da)$$

Taylor development gives: $$\mathrm{T}(da)=\mathrm{T}(0)+\frac{d\mathrm{T}(0)}{da}da+\cdots =1-\frac{i}{\hslash }{p}_{x}da$$ with $$p_x=i\hslash \frac{d\mathrm{T}(0)}{da}$$

From that follows: $$\mathrm{T}(a+da)=\mathrm{T}(a)\mathrm{T}(da)=\mathrm{T}(a)(1-\frac{i}{\hslash }{p}_{x}da)\Rightarrow \frac{\mathrm{T}(a+da)-\mathrm{T}(a)}{da}=\frac{d\mathrm{T}}{da}=-\frac{i}{\hslash }{p}_x\mathrm{T}(a)$$

This is a differential equation with the solution

$$\mathrm{T}(a)=\exp (-\frac{i}{\hslash }{p}_{x}a).$$

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 $[H,\mathrm{T}(a)]=0.$ This result means that linear momentum for the system is conserved.

In relation to the orbital angular momentum

Template:Further 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 ${𝐫}^{\prime }=(x^{\prime },y^{\prime },z)$ leaving $z$ unchanged can be expressed by the following infinitesimal translations (using Taylor approximation): $$\begin{array}{cc}{x}^{\prime }& =r\cos (t+dt)=x-ydt+\cdots \\ y^{\prime }& =r\sin (t+dt)=y+xdt+\cdots \end{array}$$

From that follows for states: $$\mathrm{R}(z,dt)|r⟩=\mathrm{R}(z,dt)|x,y,z⟩=|x-ydt,y+xdt,z⟩={\mathrm{T}}_x(-ydt){\mathrm{T}}_y(xdt)|x,y,z⟩={\mathrm{T}}_x(-ydt){\mathrm{T}}_y(xdt)|r⟩$$

And consequently: $$\mathrm{R}(z,dt)={\mathrm{T}}_x(-ydt){\mathrm{T}}_y(xdt)$$

Using $$T_k(a)=\exp (-\frac{i}{\hslash }{p}_{k}a)$$ from above with $k=x,y$ and Taylor expansion we get: $$\mathrm{R}(z,dt)=\exp [-\frac{i}{\hslash }(x{p}_y-y{p}_x)dt]=\exp (-\frac{i}{\hslash }{L}_{z}dt)=1-\frac{i}{\hslash }{L}_{z}dt+\cdots$$ 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 $\mathrm{R}(z,0)=1$:

$$\begin{array}{cc}& \mathrm{R}(z,t+dt)=\mathrm{R}(z,t)\mathrm{R}(z,dt)\\ \Rightarrow & \frac{d\mathrm{R}}{dt}=\frac{\mathrm{R}(z,t+dt)-\mathrm{R}(z,t)}{dt}=\mathrm{R}(z,t)\frac{\mathrm{R}(z,dt)-1}{dt}=-\frac{i}{\hslash }{L}_z\mathrm{R}(z,t)\\ \Rightarrow & \mathrm{R}(z,t)=\exp (-\frac{i}{\hslash }t{L}_z)\end{array}$$

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

For the spin angular momentum about for example the $y$-axis we just replace $L_z$ with $S_y=\frac{\hslash }{2}{\sigma }_y$ (where ${\sigma }_y$ is the Pauli Y matrix) and we get the spin rotation operator $$\mathrm{D}(y,t)=\exp (-i\frac{t}{2}{\sigma }_y).$$

Effect on the spin operator and quantum states

Template:Main Template:See also 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: $$S_c=\mathrm{D}(y,t)S_b{\mathrm{D}}^{-1}(y,t)$$

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

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.

See also

• Symmetry in quantum mechanics
• Spherical basis
• Optical phase space

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