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

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