SouthernMiss Math Forum » Topic: Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation

SouthernMiss Math Forum » Topic: Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation http://www.math.usm.edu/mathforum/ SouthernMiss Math Forum » Topic: Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation en Thu, 23 May 2013 21:32:27 +0000 lee on "Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation" http://www.math.usm.edu/mathforum/topic.php?id=164#post-493 Fri, 08 Jun 2012 07:43:37 +0000 lee 493@http://www.math.usm.edu/mathforum/ A much more detailed discussion of the above posting can be found <a href="http://www.math.usm.edu/lee/matharchives/?p=1239">here</a>. I also discussed quantum angular momentum in $\mathbb{R}^{2+2}$ and $\mathfrak{su}(1,1)$ representation <a href="http://www.math.usm.edu/lee/matharchives/?p=1253">here.</a> lee on "Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation" http://www.math.usm.edu/mathforum/topic.php?id=164#post-329 Wed, 04 Jan 2012 15:03:37 +0000 lee 329@http://www.math.usm.edu/mathforum/ In classical mechanics, the angular momentum of a body is given by $$L=r\times p$$ where $r$ and $p$ denote radius arm and linear momentum respectively. In quantum mechanics, the angular momentum of a spinning particle can be obtained by replacing linear momentum $p$ by momentum operator $-i\hbar\nabla$. As a result, the components of quantum mechanical angular momentum $L$ is given by \begin{align*} L_x&=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\ L_y&=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\ L_z&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right) \end{align*} It is interesting to see that the quantum mechanical angular momentum can be obtained purely from algebra, more specifically from representation theory. The relevant representations are the representations of the special unitary group $\mathrm{SU}(2)$ and its Lie algebra $\mathfrak{su}(2)$. Let $\mathcal{H}$ be the Hilbert space of states $\psi$ as smooth functions on $\mathbb{R}^3$. Define a map $\Pi:\mathrm{SU(2)}\longrightarrow\mathrm{GL(\mathcal{H})}$ as follows: For each $U\in\mathrm{SU}(2)$, $\Pi(U):\mathcal{H}\longrightarrow\mathcal{H}$ is an isomorphism defined by $$[\Pi(U)\psi](v)=\psi(\rho(U)^{-1}v),\ v\in\mathbb{R}^3$$ where $\rho$ is the universal covering map $\rho: \mathrm{SU}(2)\stackrel{2:1}{\longrightarrow}\mathrm{SO}(3)$. $\Pi$ is indeed a group homomorphism: For $U_1,U_2\in\mathrm{SU}(2)$, \begin{align*} \Pi(U_1)[\Pi(U_2)\psi](v)&=\Pi(U_2\psi)(\rho(U_1)^{-1}v)\\ &=\psi(\rho(U_2)^{-1}\rho(U_1)^{-1}v)\\ &=\psi((\rho(U_1)\rho(U_2))^{-1}v)\\ &=\psi(\rho(U_1U_2)^{-1}v)\\ &=[\Pi(U_1U_2)\psi](v) \end{align*} Hence, $\Pi$ is an infinite dimensional real representation of $\mathrm{SU}(2)$. Here the fact that $\rho$ is a group homomorphism is used. For details, see <a href="http://www.math.usm.edu/mathforum/topic.php?id=162#post-327">here</a>. We can also obtain the corresponding representation $\pi$ of the Lie algebra $\mathfrak{su}(2)$. $\pi$ can be computed as $$\pi(X)=\frac{d}{dt}\Pi(e^{tX})|_{t=0}$$ So, \begin{align*} [\pi(X)\psi](v)&=\frac{d}{dt}[\Pi(e^{tX}\psi](v)|_{t=0}\\ &=\frac{d}{dt}\psi(\rho(e^{tX})^{-1}v)|_{t=0} \end{align*} The Lie algebra $\mathfrak{su}(2)$ has the canonical basis \begin{align*} X_1&=\frac{i}{2}\sigma_1=\frac{i}{2}\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}\\ X_2&=\frac{i}{2}\sigma_2=\frac{i}{2}\begin{pmatrix} 0 & i\\ -i & 0 \end{pmatrix}\\ X_3&=\frac{i}{2}\sigma_3=\frac{i}{2}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \end{align*} Let us calculate $\pi$ for the basis member $X_1$. $e^{\theta X_1}=\begin{pmatrix} e^{i\theta/2} & 0\\ 0 & e^{-i\theta/2} \end{pmatrix}$ and $\rho(e^{\theta X_1})=R_\theta^z$ where $R_\theta^z=\begin{pmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}$ is rotation in $\mathbb{R}^3$ about the $z$-axis by angle $\theta$. Let $v(\theta)$ be a curve in $\mathbb{R}^3$ defined as $v(\theta)=R_\theta^zv$ so that $v(0)=v$. Write $v(\theta)=(x(\theta),y(\theta),z(\theta))$. Then by the chain rule, \begin{align*} [\pi(X_1)\psi](v)&=\frac{\partial\psi}{\partial x}\frac{dx}{d\theta}|_{\theta=0}+\frac{\partial\psi}{\partial y}\frac{dy}{d\theta}|_{\theta=0}+\frac{\partial\psi}{\partial z}\frac{dz}{d\theta}|_{\theta=0}\\ &=y\frac{\partial\psi}{\partial x}-x\frac{\partial\psi}{\partial y} \end{align*} Hence, $$\pi(X_1)=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$$ One can find similar formulas for $\pi(X_2)$ and $\pi(X_3)$. Note that $L_z=i\hbar\pi(X_1)$, the angular momentum about the $z$-axis. This hints us that the symmetry of the space plays a crucial role in quantum mechanics.