Rotation operator (quantum mechanics)

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https://en.wikipedia.org/wiki/Rotation_operator_(quantum_mechanics)

This article concerns the rotation operator, as it appears in quantum mechanics.

Quantum mechanical rotations

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

The translation operator

Main article: Translation operator (quantum mechanics)

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

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

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

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

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

This is a differential equation with the solution

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

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

In relation to the orbital angular momentum

Further information: Bloch sphere § Rotations

Classically we have for the angular momentum ${\displaystyle \mathbf{L}=\mathbf{r}\times \mathbf{p}.}$ This is the same in quantum mechanics considering ${\displaystyle \mathbf{r}}$ and ${\displaystyle \mathbf{p}}$ as operators. Classically, an infinitesimal rotation ${\displaystyle dt}$ of the vector ${\displaystyle \mathbf{r}=(x,y,z)}$ about the ${\displaystyle z}$-axis to ${\displaystyle {\mathbf{r}}^{\prime }=(x^{\prime },y^{\prime },z)}$ leaving ${\displaystyle z}$ unchanged can be expressed by the following infinitesimal translations (using Taylor approximation):

$${\displaystyle \begin{array}{rl}{x}^{\prime }& =r\cos (t+dt)=x-ydt+\cdots \\ y^{\prime }& =r\sin (t+dt)=y+xdt+\cdots \end{array}}$$

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

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

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

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

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

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

Effect on the spin operator and quantum states

Main article: Spin (physics) § Rotations

See also: Rotation group SO(3) § A note on Lie algebra, and Change of basis § Endomorphisms

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 $${\displaystyle A^{\prime }=PA{P}^{-1}}$$ 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: $${\displaystyle S_c=\mathrm{D}(y,t)S_b{\mathrm{D}}^{-1}(y,t)}$$

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 ${\displaystyle |b+⟩}$ and ${\displaystyle |c+⟩}$ are the top spins in their corresponding bases. So we have: $${\displaystyle \frac{\hslash }{2}|c+⟩=S_c|c+⟩=\mathrm{D}(y,t)S_b{\mathrm{D}}^{-1}(y,t)|c+⟩\Rightarrow }$$ $${\displaystyle S_b{\mathrm{D}}^{-1}(y,t)|c+⟩=\frac{\hslash }{2}{\mathrm{D}}^{-1}(y,t)|c+⟩}$$

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

This means that if the state ${\displaystyle |c+⟩}$ is rotated about the ${\displaystyle y}$-axis by an angle ${\displaystyle t}$, it becomes the state ${\displaystyle |b+⟩}$, a result that can be generalized to arbitrary axes.