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 D(R) which rotates quantum mechanical states.

$| \alpha \rangle_R = D(R) |\alpha \rangle$

In terms of the generators of rotation,

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

$\mathbf{\hat n}$ is rotation axis, and $\mathbf{J}$ is angular momentum.

The translation operator

Main article: Translation operator (quantum mechanics)

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

$\,p_x = i h \frac{d\mbox{T}(0)}{da}$

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):

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

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

From that follows for states:

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

And consequently:

$\,\mbox{R}(z, dt) = \mbox{T}_x (-y dt) \mbox{T}_y(x dt)$

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}\ (x p_y - y p_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

Main article: Spin (physics) § Rotations

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

$\,A' = P A P^{-1}$

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:

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

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
• 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