The average value of many measurements (expectation) of a physical quantity $A$ of a particle of wavefunction $\Psi$ is given by
$$ \left < \hat A \right > = \int \Psi^* \hat A \Psi\,\text d \vec r\quad \text{where} \; \text d\vec r=\text dx\,\text dy\,\text dz \; \text{in}\; 3\text{D} $$
Common operators
$$ \begin{aligned} \hat x &=x \qquad \qquad \;\;\hat{\underline r}=(x,y,z) \\ \hat p_x&=-i\hbar \frac{\partial}{\partial x} \qquad \hat{\underline p}=-i\hbar \left ( \frac{\partial }{\partial z}, \frac{\partial }{\partial y}\frac{\partial }{\partial z}\right ) \\ \hat L_x&=\hat y \hat p_z-\hat z \hat p_y \\ \hat H&=\hat T+\hat V=-\frac{\hbar^2}{2m}\nabla^2+V(\vec r) \\ &\quad \text{where}\;\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} \end{aligned} $$
Time independent Schrodinger equation
$$ \hat H \psi(\vec r)=E\psi (\vec r) $$
Simple wavefunction in the case of $V(\vec r,t)=V(\vec r)$
$$ \psi(\vec r,t)=\psi(\vec r)e^{-iEt/\hbar} $$
💼 Case: the hydrogen atom
For a hydrogen atom Schrodinger's time independent equation is of the following form:
$$ \begin{aligned} \hat H\psi(\vec r)&=E\psi(\vec r) \\ \left ( -\frac{\hbar^2}{2m_e}\nabla^2-\frac{Z e^2}{4\pi \epsilon_0 r} \right )\psi(r)&=E\psi(r) \end{aligned} $$
where $m_e$ is the electron mass, the potential is a Coulomb potential between the electron and $Z$ protons
We get the following eigen energies
$$ E_{n,l,m_l}=E_n=-\frac{Z^2 E_R}{n^2} $$
where $E_R$ is the Rydberg energy
$$ E_r=\frac{e^2}{8\pi \epsilon_0 a_0}=13.6\; \text{eV} \; \text{and}\; a_0=\frac{4\pi \epsilon_0 \hbar^2}{m_e e^2}=0.53 \AA $$
The eigenfunctions are products of two parts $\psi_{n,l,m_l}=R_{n,l}(r)Y_{l,m_l}(\theta,\phi)$
$$ \begin{aligned} R_{n,l}(r)&=\text{constant}\times \left ( \frac{Zr}{a_0} \right )^l e^{-Zr/n a_0} \sum^{n-l-1}{k=0}(-1)^k c_k \left ( \frac{Zr}{a_0}\right )\\ Y{l,m_l}(\theta,\phi)&=P_{l,m_l}(\theta)\Phi_{m_l}(\phi)\\ P_{l,m_l}(\theta)&=\text{Legendre polynomials} \\ \Phi_{m_l}(\phi)&=e^{im\phi} \end{aligned} $$
The angular eigenfunction are
$$ \begin{aligned} \hat L^2 Y_{l,m_l} &=l(l+1)\hbar^2 Y_{l,m_l} \\ \hat L_z Y_{l,m_l}&=m_l \hbar Y_{l,m_l}
\end{aligned} $$
In spherical coordinates the operator $\hat L_z$ is
$$ \hat L_z=-i\hbar \frac{\partial}{\partial \phi} $$
With allowed quantum numbers
$$ \begin{aligned} n&=1,2,\ldots \\ l&=0,1,2,\ldots,n-1 \\ m_l&=-l,-l+1,\ldots , l-1,l
\end{aligned} $$
🗒️Note: if we add the spin DOF, we get 4 quantum numbers $(n,l,m_l,m_s)$, with the spin quantum number $m_s=\pm 1/2$. The spin is usually denoted as $\chi_{m_s}$. The eigenequation for the spin is
$$ \hat S_z \chi_{m_s}=m_s\hbar \chi _{m_s} \qquad m_s=\pm \frac 12 $$
The complete wave function of the electron in a hydrogen atom is
$$ \Psi_{n,l,m_l,m_s}=\psi {n,l,m_l}(r\theta,\phi)\chi{m_s} $$
🗒️ Note: we usually denote spin up or spin down states as $\chi_{1/2}~=\chi_\uarr$ and $\chi_{-1/2}=\chi_\darr$
n | $l=0$ | $l=1$ | $l=2$ |
---|---|---|---|
1 | $e^{-r/a_0}$ | ||
2 | $\left ( 1-\frac{r}{2a_0} \right )e^{-r/2a_0}$ | $\frac{r}{a_0}e^{-r/2a_0}$ | |
3 | $\left ( 1-\frac{2r}{3a_0}+\frac{2r^2}{17a_0^2} \right ) e^{-r/3a_0}$ | $\frac{r}{a_0}\left ( 1-\frac{r}{6a_0} \right ) e^{-r/3a_0}$ | $\left ( \frac{r}{a_0} \right )^2 e^{-r/3a_0}$ |
Examples of $R_{n,l}(r)$
the ground state of hydrogen-like atom is described by the wavefunction $\psi_{1,0,0}=\psi_{1s}$ or the 1s-orbital, with the ground state energy
$$ E_1=E_{1s}=-13.6Z^2\; \mathrm{eV} $$
Other states are excited and have higher $E_n$.
We can extend the energy state to multi-electron hydrogen atoms (within the independent-particle approximation) by occupying the energy levels to build up the states of $N$-electron atoms by occupying the $N$ energy within the limitations of the Pauli exclusion principle and spin dof. (Aufbau principle)
💃 Example: $\psi_{1,0,0}$ as $1s$-orbital, $\psi_{2,1,m_l}$ with $m_l=0,\pm 1$ as $2$p-orbitals, $\psi_{3,2,m_l}$ with $m_l=0,\pm1,\pm 2$ as $3d$-orbital, etc. For these orbitals we can write:
$$ p_z=p_0 \qquad p_x=\frac{1}{\sqrt{2}} (p_1+p_{-1}), \quad p_y=\frac{1}{i\sqrt{2}}(p_1-p_{-1}) $$