rotation operator (quantum mechanics)

With every physical rotation $R$, we postulate a quantum mechanical rotation operator $\backslash widehat\{D\}(R)\; :\; H\backslash to\; H$ that is the rule that assigns to each vector in the space $H$ the vector

$$|\; \backslash alpha\; \backslash rangle\_R\; =\; \backslash widehat\{D\}(R)\; |\backslash alpha\; \backslash rangle$$

that is also in $H$. We will show that, in terms of the generators of rotation,

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

where $\backslash mathbf\{\backslash hat\; n\}$ is the rotation axis, $\backslash widehat\{\backslash mathbf\{J\}\}$ is angular momentum operator, and $\backslash hbar$ is the reduced Planck constant.

{{Main|Translation operator (quantum mechanics)}}

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

Translation of the particle at position $x$ to position $x\; +\; a$: $\backslash operatorname\{T\}(a)|x\backslash rangle\; =\; |x\; +\; a\backslash rangle$

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

$$\backslash operatorname\{T\}(0)\; =\; 1$$

$$\backslash operatorname\{T\}(a)\; \backslash operatorname\{T\}(da)|x\backslash rangle\; =\; \backslash operatorname\{T\}(a)|x\; +\; da\backslash rangle\; =\; |x\; +\; a\; +\; da\backslash rangle\; =\; \backslash operatorname\{T\}(a\; +\; da)|x\backslash rangle\; \backslash Rightarrow\; \backslash operatorname\{T\}(a)\; \backslash operatorname\{T\}(da)\; =\; \backslash operatorname\{T\}(a\; +\; da)$$

Taylor development gives:

$$\backslash operatorname\{T\}(da)\; =\; \backslash operatorname\{T\}(0)\; +\; \backslash frac\{d\backslash operatorname\{T\}(0)\}\{da\}\; da\; +\; \backslash cdots\; =\; 1\; -\; \backslash frac\{i\}\{\backslash hbar\}\; p\_x\; da$$

with $$p\_x\; =\; i\; \backslash hbar\; \backslash frac\{d\backslash operatorname\{T\}(0)\}\{da\}$$

From that follows:

$$\backslash operatorname\{T\}(a\; +\; da)\; =\; \backslash operatorname\{T\}(a)\; \backslash operatorname\{T\}(da)\; =\; \backslash operatorname\{T\}(a)\backslash left(1\; -\; \backslash frac\{i\}\{\backslash hbar\}\; p\_x\; da\backslash right)\; \backslash Rightarrow\; \backslash frac\{\backslash operatorname\{T\}(a\; +\; da)\; -\; \backslash operatorname\{T\}(a)\}\{da\}\; =\; \backslash frac\{d\backslash operatorname\{T\}\}\{da\}\; =\; -\; \backslash frac\{i\}\{\backslash hbar\}\; p\_x\; \backslash operatorname\{T\}(a)$$

This is a differential equation with the solution

$$\backslash operatorname\{T\}(a)\; =\; \backslash exp\backslash left(-\; \backslash frac\{i\}\{\backslash hbar\}\; p\_x\; a\backslash right).$$

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

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

$$\backslash begin\{align\}$$

x' &= r \cos(t + dt) = x - y \, dt + \cdots \\

y' &= r \sin(t + dt) = y + x \, dt + \cdots

\end{align}

From that follows for states:

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

And consequently:

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

Using

$$T\_k(a)\; =\; \backslash exp\backslash left(-\; \backslash frac\{i\}\{\backslash hbar\}\; p\_k\; a\backslash right)$$

from above with $k\; =\; x,y$ and Taylor expansion we get:

$$\backslash operatorname\{R\}(z,dt)=\backslash exp\backslash left[-\backslash frac\{i\}\{\backslash hbar\}\; \backslash left(x\; p\_y\; -\; y\; p\_x\backslash right)\; dt\backslash right]\; =\; \backslash exp\backslash left(-\backslash frac\{i\}\{\backslash hbar\}\; L\_z\; dt\backslash right)\; =\; 1-\backslash frac\{i\}\{\backslash hbar\}L\_z\; dt\; +\; \backslash cdots$$

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 $\backslash operatorname\{R\}(z,\; 0)\; =\; 1$:

$$\backslash begin\{align\}$$

&\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{align}

Similar to the translation operator, if we are given a Hamiltonian $H$ which rotationally symmetric about the $z$-axis, $[L\_z,H]=0$ implies $[\backslash operatorname\{R\}(z,t),H]=0$. 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 $S\_y\; =\; \backslash frac\{\backslash hbar\}\{2\}\; \backslash sigma\_y$ (where $\backslash sigma\_y$ is the Pauli Y matrix) and we get the spin rotation operator

$$\backslash operatorname\{D\}(y,\; t)\; =\; \backslash exp\backslash left(-\; i\; \backslash frac\{t\}\{2\}\; \backslash sigma\_y\backslash right).$$

{{Main|Spin (physics)#Rotations}}

{{see also|Rotation group SO(3)#A note on Lie algebra|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

$$A\text{'}\; =\; P\; A\; P^\{-1\}$$

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:

$$S\_c\; =\; \backslash operatorname\{D\}(y,\; t)\; S\_b\; \backslash operatorname\{D\}^\{-1\}(y,\; t)$$

From standard quantum mechanics we have the known results $S\_b\; |b+\backslash rangle\; =\; \backslash frac\{\backslash hbar\}\{2\}\; |b+\backslash rangle$ and $S\_c\; |c+\backslash rangle\; =\; \backslash frac\{\backslash hbar\}\{2\}\; |c+\backslash rangle$ where $|b+\backslash rangle$ and $|c+\backslash rangle$ are the top spins in their corresponding bases. So we have:

$$\backslash frac\{\backslash hbar\}\{2\}\; |c+\backslash rangle\; =\; S\_c\; |c+\backslash rangle\; =\; \backslash operatorname\{D\}(y,\; t)\; S\_b\; \backslash operatorname\{D\}^\{-1\}(y,\; t)\; |c+\backslash rangle\; \backslash Rightarrow$$

$$S\_b\; \backslash operatorname\{D\}^\{-1\}(y,\; t)\; |c+\backslash rangle\; =\; \backslash frac\{\backslash hbar\}\{2\}\; \backslash operatorname\{D\}^\{-1\}(y,\; t)\; |c+\backslash rangle$$

Comparison with $S\_b\; |b+\backslash rangle\; =\; \backslash frac\{\backslash hbar\}\{2\}\; |b+\backslash rangle$ yields $|b+\backslash rangle\; =\; D^\{-1\}(y,\; t)\; |c+\backslash rangle$.

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

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

• 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

{{Physics operator}}

{{DEFAULTSORT:Rotation Operator (Quantum Mechanics)}}

Category:Rotational symmetry

Category:Quantum operators

Category:Unitary operators