Rotation operator (quantum mechanics)

For other uses, see Rotation operator (disambiguation).

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

The translation operator

Main article: Translation operator (quantum mechanics)

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: $$\operatorname {T} (a+da)=\operatorname {T} (a)\operatorname {T} (da)=\operatorname {T} (a)\left(1-{\frac {i}{\hbar }}p_{x}da\right)\Rightarrow {\frac {\operatorname {T} (a+da)-\operatorname {T} (a)}{da}}={\frac {d\operatorname {T} }{da}}=-{\frac {i}{\hbar }}p_{x}\operatorname {T} (a)$$

This is a differential equation with the solution

$$\operatorname {T} (a)=\exp \left(-{\frac {i}{\hbar }}p_{x}a\right).$$

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

Further information: Bloch sphere § Rotations

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

$${\begin{aligned}x'&=r\cos(t+dt)=x-y\,dt+\cdots \\y'&=r\sin(t+dt)=y+x\,dt+\cdots \end{aligned}}$$

From that follows for states: $$\operatorname {R} (z,dt)|r\rangle =\operatorname {R} (z,dt)|x,y,z\rangle =|x-y\,dt,y+x\,dt,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|x,y,z\rangle =\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)|r\rangle$$

And consequently: $$\operatorname {R} (z,dt)=\operatorname {T} _{x}(-y\,dt)\operatorname {T} _{y}(x\,dt)$$

Using $$T_{k}(a)=\exp \left(-{\frac {i}{\hbar }}p_{k}a\right)$$ from above with $k=x,y$ and Taylor expansion we get: $$\operatorname {R} (z,dt)=\exp \left[-{\frac {i}{\hbar }}\left(xp_{y}-yp_{x}\right)dt\right]=\exp \left(-{\frac {i}{\hbar }}L_{z}dt\right)=1-{\frac {i}{\hbar }}L_{z}dt+\cdots$$ 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 $\operatorname {R} (z,0)=1$:

$${\begin{aligned}&\operatorname {R} (z,t+dt)=\operatorname {R} (z,t)\operatorname {R} (z,dt)\\[1.1ex]\Rightarrow {}&{\frac {d\operatorname {R} }{dt}}={\frac {\operatorname {R} (z,t+dt)-\operatorname {R} (z,t)}{dt}}=\operatorname {R} (z,t){\frac {\operatorname {R} (z,dt)-1}{dt}}=-{\frac {i}{\hbar }}L_{z}\operatorname {R} (z,t)\\[1.1ex]\Rightarrow {}&\operatorname {R} (z,t)=\exp \left(-{\frac {i}{\hbar }}\,t\,L_{z}\right)\end{aligned}}$$

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 for example the $y$-axis we just replace $L_{z}$ with ${\textstyle S_{y}={\frac {\hbar }{2}}\sigma _{y}}$ (where $\sigma _{y}$ is the Pauli Y matrix) 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

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 $A$ 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 ${\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }$ and ${\textstyle S_{c}|c+\rangle ={\frac {\hbar }{2}}|c+\rangle }$ 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 ${\textstyle S_{b}|b+\rangle ={\frac {\hbar }{2}}|b+\rangle }$ 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.

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