Category Archives: Quantum Physics

Matter and Waves

Since Huygens and Newton, physicists have known that light is described by (electromagnetic) waves and particles (photons). Such peculiar nature of light is called wave-particle duality. What about material particles such as electrons? de Broglie proposed a bold hypothesis that what is true for light is also true for material particles i.e. they will also exhibit wave nature. (This was indeed confirmed by experiments.) So how do we mathematically model such a wave? What physicists thought of using to study material particles was a complex plane wave called de Broglie wave. It looks like $$\psi(x,t)=Ae^{i(kx-\omega t)}$$ for 1-dimensional case. For 3-dimensional case, it would be \begin{equation}\label{eq:planewave}\psi({\bf r},t)=Ae^{i({\bf k}\cdot{\bf r}-\omega t)}\end{equation} Before we continue, one may wonder how physicists came up with this kind of wave. I can only speculate but such a complex plane wave was already well-known to physicists as it is a solution of Maxwell’s equation in electromagnetism. Hence, complex plane wave may describe electromagnetic wave, and naturally it became the first candidate for modeling material particles. In fact, it worked out well as we shall see and consequently complex numbers played a crucial role in building quantum mechanics.

Let us first study some properties of plane waves. The plane wave \eqref{eq:planewave} describes a free particle, more accurately a free particle in a state. In order for a plane wave to behave like a particle, we want it to be localized i.e. the wave is defined in a tiny region. (There is a more mathematically subtle reason why we require this.) We can achieve this by redefining $\psi({\bf r},t)$ as $$\psi({\bf r},t)=\left\{\begin{array}{ccc}Ae^{i({\bf k}\cdot{\bf r}-\omega t)} & \mbox{for} & {\bf r}\ \mbox{within a volume}\ V=L^3\\0 & \mbox{for} & {\bf r}\ \mbox{outside a volume}\ V=L^3\end{array}\right.$$ Physicists call this box renormalization. Physically the state of a particle must not depend on a particular location of the tiny box, so we require the periodicity condition $$\psi(x,y,z,t)=\psi(x+L,y,z,t)=\psi(x,y+L,z,t)=\psi(x,y,z+L,t)$$ Here, $L$ is called wave length. The periodicity condition implies that ${\bf k}$ is quantized as $${\bf k}=\frac{2\pi}{L}{\bf n}$$ where ${\bf k}=(k_x,k_y,k_z)$, ${\bf n}=(n_x,n_y,n_z)$, and $n_i=0,1,2,\cdots$, $i=x,y,z$. The vector ${\bf k}$ is called wave vector and for 1-dimensional case, $k$ is called wave number. If the wave is periodic in time, say $\psi(x,t)=\psi(x,t+T)$, then we obtain $e^{-i\omega T}=1$. The nonzero minimum value of $T$ is $T=\frac{2\pi}{\omega}$. $\omega=\frac{2\pi}{T}$ is called angular frequency. $kx-\omega t$ is called phase and if $kx-\omega t$ is constant, the wave moves at the speed $v_p=\frac{dx}{dt}=\frac{\omega}{k}$. This $v_p$ is called phase velocity. For 3-dimensional case, \begin{align*}\psi({\bf r},t)&=Ae^{i({\bf k}\cdot{\bf r}-\omega t)}\\&=Ae^{i{\bf k}\cdot\left({\bf r}-\frac{\omega t}{|{\bf k}|^2}{\bf k}\right)}\\&=Ae^{i{\bf k}\cdot\left({\bf r}-\frac{\omega t}{|{\bf k}|}\hat{\bf k}\right)}\end{align*}So, the phase velocity would be $${\bf v}_p=\frac{d{\bf r}}{dt}=\frac{\omega}{|{\bf k}|}\hat{\bf k}$$

The image of wave function $\psi(x,t)=Ae^{i(kx-\omega t)}$ is a circle. We are in fact quite familiar with this kind of waves. On a beautiful day, you go to a lake. You would be then tempted to throw a rock into the cam water. When you do, you would see circular water waves spreading out from the point of impact.

References:

[1] Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer, 2001

[2] Quantum Mechanics, H.-S. Song (in Korean)

The Klein-Gordon Equation

For the Schrödinger equation $$i\hbar\frac{\partial\psi}{\partial t}=\hat H\psi({\bf x},t),$$ the Hamiltonian $$\hat H=-\frac{\hbar^2}{2m_0}\nabla^2+V({\bf x})$$ corresponds to the nonrelativistic energy-momentum relation $$\hat E=\frac{\hat p^2}{2m_0}+V({\bf x})$$ where $$\hat E=i\hbar\frac{\partial}{\partial t},\ \hat p=-i\hbar\nabla$$ So, naturally considering the relativistic energy-momentum relation $$\frac{E^2}{c^2}-{\bf p}\cdot{\bf p}=m_0^2c^2$$ would be the starting point to obtain a relativistic generalization of the Schrödinger equation. Replacing $E$ and ${\bf p}\cdot{\bf p}$ by operators $$\hat E=i\hbar\frac{\partial}{\partial t}\ \mbox{and}\ \hat p\cdot\hat p=-\hbar^2\nabla^2$$ acting on a wave function $\psi$, we obtain the Klein-Gordon equation for a free particle \begin{equation}\label{eq:k-g}\left(\Box-\frac{m_0^2c^2}{\hbar^2}\right)\psi=0\end{equation} where $$\Box=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2$$

Free solutions of the Schrödinger equation with $V({\bf x})=0$ are of the form $$\psi=\exp\left[\frac{i}{\hbar}(-Et+{\bf p}\cdot{\bf x})\right]$$ They are also free solutions of the Klein-Gordon equation \eqref{eq:k-g} with the energy condition $$E=\pm c\sqrt{m_0^2c^2+p^2}$$ The solutions yielding negative energies appear to be unphysical and initially considered so by physicists, but later they were interpreted as antiparticles. Antiparticles are indeed seen in nature. In reality, antiparticles also have positive energies. Antiparticles as wave functions with negative energies is merely an interpretation of the mathematical representation of the energy condition. If antiparticles weren’t discovered, the negative energy condition would have been still thought to be unphysical.

Other than allowing solutions with negative energies, there was another issue with the Klein-Gordon equation noted by physicists. The conservation of four-current density $$j_\mu=\frac{i\hbar}{2m_0}(\psi^\ast\nabla_\mu\psi-\psi\nabla_\mu\psi^\ast),$$ where $\psi^\ast$ denotes the complex conjugate of $\psi$ and $\nabla_\mu=\left(-\frac{1}{c^2}\frac{\partial}{\partial t},\nabla\right)$, implies that the quantity $$\rho=\frac{i\hbar}{2m_0c^2}\left(\psi^\ast\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^\ast}{\partial t}\right)$$ can be considered as a probability density. However, the problem is that $\rho$ can be negative. This is due to the appearance of first-order partial derivative $\frac{\partial\psi}{\partial t}$, which is the consequence of the Klein-Gordon equation being of second-order in time. Because of this, the Klein-Gordon equation was not regarded as a physically viable relativistic generalization of the Schrödinger equation and physicists were instead looking for a relativistic generalization of first-order in time like the Schrödinger equation. Such an equation was finally discovered by P. A. M. Dirac and is called the Dirac equation. On the other hand, the Klein-Gordon equation drew attention of physicists again after they realized that $\rho$ can be interpreted as charge density, and indeed charged pions $\pi^+$ and $\pi^-$ were discovered. Today, the Klein-Gordon equation is an important relativistic equation that describe charged spin-0 particles.

References:

[1] Walter Greiner, Relativistic Quantum Mechanics, 3rd Edition, Springer-Verlag, 2000

Heat Equation and Schrödinger Equation

There is an intriguing relationship between Schrödinger equation for a free particle and homogeneous heat equation.

1-dimentional Schrödinger equation for a free particle is
\begin{equation}
\label{eq:se}
i\hbar\frac{\partial\psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\psi(x,t)}{\partial x^2}.
\end{equation}
Take the Wick rotation $t\mapsto \tau=it$. Then the Schrödinger equation \eqref{eq:se} turns into
\begin{equation}
\label{eq:hheq}
\frac{\partial\phi(x,\tau)}{\partial\tau}=\frac{\hbar}{2m}\frac{\partial^2\phi(x,\tau)}{\partial x^2},
\end{equation}
where $\phi(x,\tau)=\psi\left(x,\frac{\tau}{i}\right)$. \eqref{eq:hheq} is a homogeneous heat equation with diffusion coefficient $\alpha^2=\frac{\hbar}{2m}$. Conversely, apply the Wick rotation $t\mapsto\tau=-it$ to the 1-dimensional homogeneous heat equation
\begin{equation}
\label{eq:hhe2}
\frac{\partial u(x,t)}{\partial t}=\alpha^2\frac{\partial^2 u(x,t)}{\partial x^2}.
\end{equation}
Then the resulting equation is
\begin{equation}
\label{eq:se2}
i\hbar\frac{\partial w(x,\tau)}{\partial t}=-\alpha^2\hbar\frac{\partial^2 w(x,\tau)}{\partial x^2},
\end{equation}
where $w(x,\tau)=u\left(x,-\frac{\tau}{i}\right)$. \eqref{eq:se2} is a Schrödinger equation for a free particle with $m=\frac{1}{2\alpha^2}$. The solution of \eqref{eq:hhe2} with homogeneous boundary conditions takes the form
$$u(x,t)=\sum_{n=0}^\infty A_ne^{-\lambda_n^2\alpha^2 t}X_n(x).$$
Its Wick rotated solution is
\begin{align*}
w(x,\tau)&=\sum_{n=0}^\infty w_n(x,\tau)\\
&=\sum_{n=0}^\infty A_ne^{-i\lambda_n^2\alpha^2\tau}X_n(x).
\end{align*}
$i\hbar\frac{\partial w(x,\tau)}{\partial\tau}=\lambda_n^2\alpha^2\hbar w(x,\tau)$. Thus for each $n=0,1,2,\cdots$, $E_n=\lambda_n^2\alpha^2\hbar$ is the energy and $\omega_n=\lambda_n^2\alpha^2$ is the frequency of the wave $w_n(x,\tau)$.

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