# Quantum Angular Momentum in $\mathbb{R}^{2+2}$ and $\mathfrak{su}(1,1)$ Representation

It can be shown that quantum angular momentum
\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*}
can be obtained purely mathematically by $\mathfrak{su}(2)$ Lie algebra representation as discussed here. Since $\mathfrak{su}(2)$ representation contains information on the symmetry of $\mathbb{R}^3$, one can speculate that the symmetry of the background space plays a crucial role in quantum mechanics.

For fun, let us consider quantum mechanics in $\mathbb{R}^{2+2}$, a 4-space with 2 time dimensions. (I said for fun, so let us not worry about whether it is physically meaniful or not for now.) In this case, can we also derive quantum angular momentum (or something like it) by a Lie algebra representation? If so, what is a relevant Lie algebra? To answer this question, we need to understand the symmetry of $\mathbb{R}^{2+2}$.

The rotations (actually Euclidean rotation and Lorentz boosts), in particular orthochronous Lorentz transformations i.e. time-orientation and parity preserving Lorentz transformations in  Minkowski 3-space $\mathbb{R}^{2+1}$ form the special pseudo orthogonal group $\mathrm{SO}^+(2,1)$, the identity component of the Lorentz group $\mathrm{O}(2,1)$. The $2+1$ dimensional spacetime $\mathbb{R}^{2+1}$ can be identified with the set of $2\times 2$ matrices of the form
$$\underline{X}=\begin{pmatrix} \eta & x+iy\\ -(x-iy) & -\eta \end{pmatrix}$$
with the inner product $\langle\ ,\ \rangle$ defined by
$$\langle\underline{X},\underline{Y}\rangle=-\frac{1}{2}\mathrm{tr}\left[\underline{X}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \underline{Y}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \right]$$
In particular
$$|\underline{X}|^2=\det\underline{X}$$
Here the matrix $\underline{X}$ is identified with the 3-vector $X=(\eta,x,y)\in\mathbb{R}^{2+1}$. The identification is an isometry. The indefinte special unitary group $\mathrm{SU}(1,1)$ acts isometrically on $\mathbb{R}^{2+1}$ by the action
$$\mathrm{SU}(1,1)\times\mathbb{R}^{2+1}\longrightarrow\mathbb{R}^{2+1};\ (U,X)\longmapsto UXU^{-1}$$
For a fixed $U\in\mathrm{SU}(1,1)$, the map
$$\mathbb{R}^{2+1}\longrightarrow\mathbb{R}^{2+1};\ X\longmapsto UXU^{-1}$$
is an orthochronous Lorentz transformation of $\mathbb{R}^3$. Thus the Lie group action induces a Lie group representation $\rho:\mathrm{SU}(1,1)\longrightarrow\mathrm{SO}^+(2,1)$. Since both $U$ and $-U$ result the same isometry, the representation $\rho$ is a 2:1 map. The kernal of $\rho$ is $\mathbb{Z}_2=\{\pm I\}$, so we have $\mathrm{SU}(1,1)/\mathbb{Z}_2=\mathrm{SO}^+(2,1)$. The quotient group $\mathrm{SU}(1,1)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(1,1)$ is called the projective indefinite special unitary group. The double cover $\mathrm{SU}(1,1)$ of $\mathrm{SO}^+(2,1)$ is connected (but not simply connected) and there exists a short exact sequence
$$1\rightarrow\mathbb{Z}_2\rightarrow\mathrm{SU}(1,1)\rightarrow\mathrm{SO}^+(2,1)\rightarrow 1$$
So $\mathrm{SU}(1,1)$ may be regarded as the spin group $\mathrm{Spin}(2,1)$. Since the spin group $\mathrm{Spin}(p,q)$ of a split signature is required to be connected (but not necessarily simply connected), it may not uniquely exist unlike the spin group $\mathrm{Spin}(n)$. For instance, the linear special group $\mathrm{SL}(2,\mathbb{R})$ may also be considered as the spin group $\mathrm{Spin}(2,1)$.

Let $\mathcal{K}$ be the space of states $\psi$ as smooth functions on $\mathbb{R}^{2+1}$. It should be noted that $\mathcal{K}$ is not a Hilbert space but rather a Krein space. I will discuss about this some other time. Define a map $\Pi:\mathrm{SU(1,1)}\longrightarrow\mathrm{GL(\mathcal{K})}$ as follows: For each $U\in\mathrm{SU}(1,1)$, $\Pi(U):\mathcal{K}\longrightarrow\mathcal{K}$ is an isomorphism defined by
$$[\Pi(U)\psi](v)=\psi(\rho(U)^{-1}v),\ v\in\mathbb{R}^{2+1}$$
where $\rho$ is the covering map $\rho: \mathrm{SU}(1,1)\stackrel{2:1}{\longrightarrow}\mathrm{SO}^+(2,1)$. $\Pi$ is indeed a group homomorphism: For $U_1,U_2\in\mathrm{SU}(1,1)$,
\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 $\mathrm{SU}(1,1)$. The corresponding $\mathfrak{su}(1,1)$ Lie algebra representation $\pi$ 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 $\mathfrak{su}(1,1)$ has the canonical basis
\begin{align*}
X_1&=\frac{1}{2}\sigma_1=\frac{1}{2}\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}\\
X_2&=\frac{1}{2}\sigma_2=\frac{1}{2}\begin{pmatrix}
0 & i\\
-i & 0
\end{pmatrix}\\
X_3&=\frac{i}{2}\sigma_3=\frac{1}{2}\begin{pmatrix}
i & 0\\
0 & -i
\end{pmatrix}
\end{align*}
Let us calculate $\pi$ for the basis member $X_1$. $e^{\phi X_1}=\begin{pmatrix} \cosh\frac{\phi}{2} & \sinh\frac{\phi}{2}\\ \sinh\frac{\phi}{2} & \cosh\frac{\phi}{2} \end{pmatrix}$ and $\rho(e^{\phi X_1})=R_\phi^y$ where $R_\phi^y=\begin{pmatrix} \cosh\phi & -\sinh\phi & 0\\ -\sinh\phi & \cosh\phi & 0\\ 0 & 0 & 1\end{pmatrix}$ is rotation in $\mathbb{R}^{2+1}$ about the $y$-axis by a hyperbolic angle $\phi$. (Although I conveniently call this a rotation, note that this is not a Euclidean rotation but a rotation in spacetime. It is called a Lorentz boost in physics.) Let $v(\phi)$ be a curve in $\mathbb{R}^{2+1}$ defined by
$$v(\phi)=\rho(e^{\phi X_1})^{-1}v=(R_\phi^y)^{-1}v$$ so that $v(0)=v$. Write $v(\phi)=(\eta(\phi),x(\phi),y(\phi))$ and $v=(\eta,x,y)$. Then by the chain rule,
\begin{align*}
[\pi(X_1)\psi](v)&=\frac{\partial\psi}{\partial \eta}\frac{dx}{d\phi}|_{\phi=0}+\frac{\partial\psi}{\partial x}\frac{dx}{d\phi}|_{\phi=0}+\frac{\partial\psi}{\partial y}\frac{dy}{d\phi}|_{\phi=0}\\
&=x\frac{\partial\psi}{\partial \eta}+\eta\frac{\partial\psi}{\partial x}
\end{align*}
Hence,
$$\pi(X_1)=x\frac{\partial}{\partial \eta}+\eta\frac{\partial}{\partial x}$$
Using
\begin{align*}
e^{\phi X_2}&=\begin{pmatrix}
\cosh\frac{\phi}{2} & i\sinh\frac{\phi}{2}\\
-i\sinh\frac{\phi}{2} & \cosh\frac{\phi}{2}
\end{pmatrix},\ \rho(e^{\phi X_2})=R_\phi^x=\begin{pmatrix}
\cosh\phi & 0 & -\sinh\phi\\
0 & 1 & 0\\
-\sinh\phi & 0 & \cosh\phi
\end{pmatrix}\\
e^{\theta X_3}&=\begin{pmatrix}
e^{i\theta/2} & 0\\
0 & e^{-i\theta/2}
\end{pmatrix},\ \rho(e^{\sigma X_1})=R_\theta^\eta=\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\theta & -\sin\theta\\
0 & \sin\theta & \cos\theta
\end{pmatrix}
\end{align*}
one can find similar formulas for $\pi(X_2)$ and $\pi(X_3)$:
\begin{align*}
\pi(X_2)&=y\frac{\partial}{\partial \eta}+\eta\frac{\partial}{\partial y}\\
\pi(X_3)&=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}
\end{align*}
It turns out that
\begin{align*}
i\hbar\pi(X_1)&=L_y=i\hbar\left(\eta\frac{\partial}{\partial x}+x\frac{\partial}{\partial \eta}\right)\\
i\hbar\pi(X_2)&=L_x=i\hbar\left(y\frac{\partial}{\partial \eta}+\eta\frac{\partial}{\partial y}\right)\\
-i\hbar\pi(X_3)&=L_\eta=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}
are an analogue of the quantum angular momenta about the $y$-axis, $x$-axis and $\eta$-axis respectively. To see this, in $\mathbb{R}^{2+2}$ the momentum operator $p$ is given by

$$p^\mu =i\hbar \nabla^\mu =i\hbar\left(\frac{\partial}{\partial\eta},-\nabla\right)=i\hbar\left(\frac{\partial}{\partial\eta},-\frac{\partial}{\partial x},-\frac{\partial}{\partial y}\right) \ \ \ \ \ \ (1)$$
What may be called quantum angular momentum in $\mathbb{R}^{2+2}$ may be foramlly found by $L=r_\mu\times p^\mu$ with $p^\mu$ in (1). The cross product $v\times w$ in $\mathbb{R}^{2+1}$is given by
$$v\times w:=\left|\begin{array}{ccc} e_0 & e_1 & -e_2\\ v^1 & v^2 & v^3\\ w^1 & w^2 & w^3 \end{array}\right|$$
where $e_0=(1,0,0)$, $e_1(0,1,0)$, and $e_2=(0,0,1)$. Let $r_\mu=(\eta,x,y)$. Then
\begin{align*}
L&=r_\mu\times(i\hbar\nabla^\mu)\\
&=\left|\begin{array}{ccc}
e_0 & e_1 & -e_2\\
\eta & x & y\\
\frac{\partial}{\partial\eta} & -\frac{\partial}{\partial x} & -\frac{\partial}{\partial y}
\end{array}\right|\\
&=i\hbar\left[\left(-x\frac{\partial}{\partial y}+y\frac{\partial}{\partial x}\right)e_0+\left(y\frac{\partial}{\partial\eta}+\eta\frac{\partial}{\partial y}\right)e_1+\left(\eta\frac{\partial}{\partial x}+x\frac{\partial}{\partial\eta}\right)e_2\right]
\end{align*}
If we write $L=(L_\eta,L_x,L_y)$, then
\begin{align*}
L_\eta&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\\
L_x&=i\hbar\left(y\frac{\partial}{\partial\eta}+\eta\frac{\partial}{\partial y}\right)\\
L_y&=i\hbar\left(\eta\frac{\partial}{\partial x}+x\frac{\partial}{\partial\eta}\right)
\end{align*}
In quantum mechanics in $\mathbb{R}^{2+2}$, the conservation of the above quantum angular momentum $L$ is expected.

References:

[1] Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer-Verlag 2000

[2] F. Reese Harvey, Spinors and Calibrations, Academic Press 1990

[3] Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

# Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation

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 $\mathfrak{su}(2)$. This hints us that the symmetry of the background space plays a crucial role in quantum mechanics.

The rotations in $\mathbb{R}^3$ form the special orthogonal group $\mathrm{SO}(3)$. $\mathrm{SO}(3)$ is not simply connected (see [3] for detalis) and the special unitary group $\mathrm{SU}(2)$ is the universal covering group of $\mathrm{SO}(3)$. ($\mathrm{SU}(2)=S^3$ so it is simply connected.) The covering map $\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a $\mathrm{SU}(2)$ representation. To see this, note that Euclidean 3-space $\mathbb{R}^3$ can be identified with the set of $2\times 2$ hermitian matrices of the form
$$\underline{X}=\begin{pmatrix} z & x-iy\\ x+iy & -z \end{pmatrix}$$
with the inner product $\langle\ ,\ \rangle$ defined by
$$\langle\underline{X},\underline{Y}\rangle=\frac{1}{2}\mathrm{tr}(\underline{X}\cdot\underline{Y})$$
In particular
$$|\underline{X}|^2=\frac{1}{2}\mathrm{tr}\underline{X}^2=-\det\underline{X}$$
Here the hermitian matrix $\underline{X}$ is identified with the vector $(x,y,z)\in\mathbb{R}^3$. Since $|X|^2=-\det\underline{X}$, the identification is an isometry. $\mathrm{SU}(2)$ acts on $\mathbb{R}^3$ isometrically by the action
$$\mathrm{SU}(2)\times\mathbb{R}^3\longrightarrow\mathbb{R}^3;\ (U,X)\longmapsto U^{-1}XU$$
For a fixed $U\in\mathrm{SU}(2)$, the map
$$\mathbb{R}^3\longrightarrow\mathbb{R}^3;\ X\longmapsto U^{-1}XU$$
is an orientation preserving isometry of $\mathbb{R}^3$. Thus the Lie group action induces a Lie group representation $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$. Since both $U$ and $-U$ result the same isometry, the representation $\rho$ is a 2:1 map. The kernal of $\rho$ is $\mathbb{Z}_2=\{\pm I\}$, so we have $\mathrm{SU}(2)/\mathbb{Z}_2=\mathrm{SO}(3)$. The quotient group $\mathrm{SU}(2)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(2)$ is called the projective special unitary group. The double cover of the special orthogonal group $\mathrm{SO}(n)$ is called the spin group and is denoted by $\mathrm{Spin}(n)$. Hence, the double cover $\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is the spin group $\mathrm{Spin}(3)$.

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. 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 $X_3$. $e^{\theta X_3}=\begin{pmatrix} e^{i\theta/2} & 0\\ 0 & e^{-i\theta/2} \end{pmatrix}$ and $\rho(e^{\theta X_3})=R_\theta^z$ 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 by
$$v(\theta)=\rho(e^{\theta X_3})^{-1}v=(R_\theta^z)^{-1}v$$ so that $v(0)=v$. Write $v(\theta)=(x(\theta),y(\theta),z(\theta))$ and $v=(x,y,z)$. Then by the chain rule,
\begin{align*}
[\pi(X_3)\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,
$$\pi(X_3)=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$$
Using
\begin{align*}
e^{\phi X_2}&=\begin{pmatrix}
\cos\frac{\phi}{2} & -\sin\frac{\phi}{2}\\
\sin\frac{\phi}{2} & \cos\frac{\phi}{2}
\end{pmatrix},\ \rho(e^{\phi X_2})=R_\phi^y=\begin{pmatrix}
\cos\phi & 0 & -\sin\phi\\
0 & 1 & 0\\
\sin\phi & 0 & \cos\phi
\end{pmatrix}\\
e^{\sigma X_1}&=\begin{pmatrix}
\cos\frac{\sigma}{2} & i\sin\frac{\sigma}{2}\\
i\sin\frac{\sigma}{2} & \cos\frac{\sigma}{2}
\end{pmatrix},\ \rho(e^{\sigma X_1})=R_\sigma^x=\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\sigma & -\sin\sigma\\
0 & \sin\sigma & \cos\sigma
\end{pmatrix}
\end{align*}
one can find similar formulas for $\pi(X_2)$ and $\pi(X_1)$:
\begin{align*}
\pi(X_2)&=z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\\
\pi(X_1)&=z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}
\end{align*}
Note that
\begin{align*}
i\hbar\pi(X_1)&=L_x=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\
-i\hbar\pi(X_2)&=L_y=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\
i\hbar\pi(X_3)&=L_z=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}
i.e. the angular momenta about the $x$-axis, $y$-axis and $z$-axis respectively.

References:

[1] Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer-Verlag 2000

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

[3] Shlomo Sternberg, Group Theory and Physics, Cambridge University Press 1994