10: Quantum Physics

Introduction to quantum physics

Black body radiation

Planck’s law for the energy distribution for the radiation of a black body is:

$$w(f)=\frac{8\pi hf^3}{c^3}\frac{1}{{\rm e}^{hf/kT}-1}~~,~~~ w(\lambda)=\frac{8\pi hc}{\lambda^5}\frac{1}{{\rm e}^{hc/\lambda kT}-1}$$

Stefan-Boltzmann’s law for the total power density can be derived from this: $P=A\sigma T^4$. Wien’s law for the maximum can also be derived from this: $T\lambda_{\rm max}=k_{\rm W}$.

The Compton effect

If light is considered to consist of particles, the wavelength of scattered light can be derived as:

$$\lambda'=\lambda+\frac{h}{mc}(1-\cos\theta)=\lambda+\lambda_{\rm C}(1-\cos\theta)$$

Electron diffraction

Diffraction of electrons at a crystal can be explained by assuming that particles have a wave character with wavelength $\lambda=h/p$. This wavelength is called the deBroglie-wavelength.

Wave functions

The wave character of particles is described by a wavefunction $\psi$. This wavefunction can be described in normal or momentum space. Both definitions are each others Fourier transform:

$$\Phi(k,t)=\frac{1}{\sqrt{h}}\int\Psi(x,t){\rm e}^{-ikx}dx~~~\mbox{and}~~~ \Psi(x,t)=\frac{1}{\sqrt{h}}\int\Phi(k,t){\rm e}^{ikx}dk$$

These waves define a particle with group velocity $v_{\rm g}=p/m$ and energy $E=\hbar\omega$.

The wavefunction can be interpreted as a measure with the probability $P$ of finding a particle somewhere (Born): $dP=|\psi|^2d^3V$. The expectation value $\left\langle f \right\rangle$ of a quantity $f$ of a system is given by:

$$\left\langle f(t) \right\rangle=\int\hspace{-1.5ex}\int\hspace{-1.5ex}\int\Psi^* f\Psi d^3V~~,~~\left\langle f_p(t) \right\rangle=\int\hspace{-1.5ex}\int\hspace{-1.5ex}\int\Phi^*f\Phi d^3V_p$$

This is also written as $\left\langle f(t) \right\rangle=\left\langle \Phi|f|\Phi \right\rangle$. The normalizing condition for wavefunctions follows from this: $\left\langle \Phi|\Phi \right\rangle=\left\langle \Psi|\Psi \right\rangle=1$.

Operators in quantum physics

In quantum mechanics, classical quantities are translated into operators. These operators are Hermitian because their eigenvalues must be real:

$$\int\psi_1^*A\psi_2d^3V=\int\psi_2(A\psi_1)^*d^3V$$

When $u_n$ is the eigenfunction of the eigenvalue equation $A\Psi=a\Psi$ for eigenvalue $a_n$, $\Psi$ can be expanded into a basis of eigenfunctions: $\Psi=\sum\limits_nc_nu_n$. If this basis is taken orthonormal, then it follows for the coefficients: $c_n=\left\langle u_n|\Psi \right\rangle$. If the system is in a state described by $\Psi$, the chance to find eigenvalue $a_n$ when measuring $A$ is given by $|c_n|^2$ in the discrete part of the spectrum and $|c_n|^2da$ in the continuous part of the spectrum between $a$ and $a+da$. The matrix element $A_{ij}$ is given by: $A_{ij}=\left\langle u_i|A|u_j \right\rangle$. Because $(AB)_{ij}=\left\langle u_i|AB|u_j \right\rangle=\langle u_i|A\sum\limits_n|u_n\rangle\left\langle u_n|B|u_j \right\rangle$ it then holds that: $\sum\limits_n|u_n\rangle\langle u_n|=1$.

The time-dependence of an operator is given by (Heisenberg):

$$\frac{dA}{dt}=\frac{\partial A}{\partial t}+\frac{[A,H]}{i\hbar}$$

with $[A,B]\equiv AB-BA$ the commutator of $A$ and $B$. For Hermitian operators the commutator is always complex. If $[A,B]=0$, the operators $A$ and $B$ have a common set of eigenfunctions. By applying this to $p_x$ and $x$ it follows that (Ehrenfest): $md^2\left\langle x \right\rangle_t/dt^2=-\left\langle dU(x)/dx \right\rangle$.

The first order approximation $\left\langle F(x) \right\rangle_t\approx F(\left\langle x \right\rangle)$, with $F=-dU/dx$ represents the classical equation.

Before using quantum mechanical operators which are a product of other operators, they should be made symmetrical: a classical product $AB$ becomes $\frac{1}{2} (AB+BA)$.

The uncertainty principle

If the uncertainty $\Delta A$ in $A$ is defined as: $(\Delta A)^2=\left\langle \psi|A_{\rm op}-\left\langle A \right\rangle|^2\psi \right\rangle=\left\langle A^2 \right\rangle-\left\langle A \right\rangle^2$ it follows that:

$$\Delta A\cdot\Delta B\geq \frac{1}{2} |\left\langle \psi|[A,B]|\psi \right\rangle|$$

From this follows: $\Delta E\cdot\Delta t\geq\hbar$, and because $[x,p_x]=i\hbar$: $\Delta p_x\cdot\Delta x\geq \frac{1}{2} \hbar$, and $\Delta L_x\cdot\Delta L_y\geq \frac{1}{2} \hbar L_z$.

The Schrödinger equation

The momentum operator is given by: $p_{\rm op}=-i\hbar\nabla$. The position operator is: $x_{\rm op}=i\hbar\nabla_p$. The energy operator is given by: $E_{\rm op}=i\hbar\partial/\partial t$. The Hamiltonian of a particle with mass $m$, potential energy $U$ and total energy $E$ is given by: $H=p^2/2m+U$. From $H\psi=E\psi$ the Schrödinger equation follows:

$$-\dfrac{\hbar^{2}}{2m}\bigtriangledown ^{2} \psi +U\psi=E\psi = i\hbar \dfrac{\partial \psi}{\partial t}$$

The linear combination of the solutions of this equation give the general solution. In one dimension it is:

$$\psi(x,t)=\left(\sum+\int dE\right)c(E)u_E(x)\exp\left(-\frac{iEt}{\hbar}\right)$$

The current density $J$ is given by: $\displaystyle J=\frac{\hbar}{2im}(\psi^*\nabla\psi-\psi\nabla\psi^*)$

The following conservation law holds: $\displaystyle\frac{\partial P(x,t)}{\partial t}=-\nabla J(x,t)$

Parity

The parity operator in one dimension is given by ${\cal P}\psi(x)=\psi(-x)$. If the wavefunction is split into even and odd functions, it can be expanded into eigenfunctions of $\cal P$:

$$\psi(x)= \underbrace{\frac{1}{2} (\psi(x)+\psi(-x))}_{\rm even:~\hbox{$\psi^+$}}+ \underbrace{\frac{1}{2} (\psi(x)-\psi(-x))}_{\rm odd:~\hbox{$\psi^-$}}$$ $[{\cal P},H]=0$.

The functions $\psi^+= \frac{1}{2} (1+{\cal P})\psi(x,t)$ and $\psi^-= \frac{1}{2} (1-{\cal P})\psi(x,t)$ both satisfy the Schrödinger equation. Hence, parity is a conserved quantity.

The tunnel effect

The wavefunction of a particle in an infinitely high potential well from $x=0$ to $x=a$ is given by $\psi(x)=a^{-1/2}\sin(kx)$. The energy levels are given by $E_n=n^2h^2/8a^2m$.

If the wavefunction with energy $W$ encounters a potential barrier where $W_0>W$ the wavefunction will, unlike the classical case, be non-zero within the potential barrier. If 1, 2 and 3 are the areas in front, within and behind the potential well, then:

$$\psi_1=A{\rm e}^{ikx}+B{\rm e}^{-ikx}~~,~~~ \psi_2=C{\rm e}^{ik'x}+D{\rm e}^{-ik'x}~~,~~~\psi_3=A'{\rm e}^{ikx}$$

with $k'^2=2m(W-W_0)/\hbar^2$ and $k^2=2mW$. Using the boundary conditions requiring continuity: $\psi=$continuous and $\partial\psi/\partial x=$continuous at $x=0$ and $x=a$ gives $B$, $C$ and $D$ and $A'$ expressed in $A$. The amplitude $T$ of the transmitted wave is defined by $T=|A'|^2/|A|^2$. If $W>W_0$ and $2a=n\lambda'=2\pi n/k'$ then: $T=1$ holds.

The harmonic oscillator

For a harmonic oscillator where: $U= \frac{1}{2} bx^2$ and $\omega_0^2=b/m$ the Hamiltonian $H$ is given by:

$$H=\frac{p^2}{2m}+\frac{1}{2} m\omega^2 x^2= \frac{1}{2} \hbar\omega+\omega A^\dagger A$$

with

$$A=\sqrt{\mbox{$\frac{1}{2}$}m\omega}x+\frac{ip}{\sqrt{2m\omega}}~~\mbox{and}~~ A^\dagger=\sqrt{\mbox{$\frac{1}{2}$}m\omega}x-\frac{ip}{\sqrt{2m\omega}}$$

$A\neq A^\dagger$ is non-Hermitian. $[A,A^\dagger]=\hbar$ and $[A,H]=\hbar\omega A$. $A$ is a so called raising ladder operator, $A^\dagger$ a lowering ladder operator. $HAu_E=(E-\hbar\omega)Au_E$. There is an eigenfunction $u_0$ for which $Au_0=0$ holds. The energy in this ground state is $\frac{1}{2} \hbar\omega$: the zero point energy. For the normalized eigenfunctions it follows that:

$$u_n=\frac{1}{\sqrt{n!}}\left(\frac{A^\dagger}{\sqrt{\hbar}}\right)^nu_0~~\mbox{with}~~ u_0=\sqrt[4]{\frac{m\omega}{\pi\hbar}}\exp\left(-\frac{m\omega x^2}{2\hbar}\right)$$

with $E_n=( \frac{1}{2} +n)\hbar\omega$.

Angular momentum

For the angular momentum operators $L$: $[L_z,L^2]=[L_z,H]=[L^2,H]=0$. Also cyclically: $[L_x,L_y]=i\hbar L_z$. Not all components of $L$ can be known at the same time with arbitrary accuracy. For $L_z$:

$$L_z=-i\hbar\frac{\partial }{\partial \varphi}=-i\hbar\left(x\frac{\partial }{\partial y}-y\frac{\partial }{\partial x}\right)$$

The ladder operators $L_\pm$ are defined by: $L_\pm=L_x\pm iL_y$. Now $L^2=L_+L_-+L_z^2-\hbar L_z$. Further,

$$L_\pm=\hbar{\rm e}^{\pm i\varphi}\left(\pm\frac{\partial }{\partial \theta}+i\cot(\theta)\frac{\partial }{\partial \varphi}\right)$$

From $[L_+,L_z]=-\hbar L_+$ it follows that: $L_z(L_+Y_{lm})=(m+1)\hbar(L_+Y_{lm})$.

From $[L_-,L_z]=\hbar L_-$ it follows that: $L_z(L_-Y_{lm})=(m-1)\hbar(L_-Y_{lm})$.

From $[L^2,L_\pm]=0$ it follows that: $L^2(L_\pm Y_{lm})=l(l+1)\hbar^2(L_\pm Y_{lm})$.

Because $L_x$ and $L_y$ are Hermitian (this implies $L_\pm^\dagger=L_\mp$) and $|L_\pm Y_{lm}|^2>0$ it follows that: $l(l+1)-m^2-m\geq0\Rightarrow-l\leq m\leq l$. Further it follows that $l$ has to be integral or half-integral. Half-odd integral values give no unique solution for $\psi$ and are therefore dismissed.

Spin

The spin operators are defined by their commutation relations: $[S_x,S_y]=i\hbar S_z$. Because the spin operators do not act in the physical space $(x,y,z)$ the uniqueness of the wavefunction is not a criterium here: also half odd-integer values are allowed for the spin. Because $[L,S]=0$ spin and angular momentum operators do not have a common set of eigenfunctions. The spin operators are given by $\vec{\vec{S}}= \frac{1}{2} \hbar\vec{\vec{\sigma}}$, with

$$\vec{\vec{\sigma}}_x=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)~~,~~ \vec{\vec{\sigma}}_y=\left(\begin{array}{cc}0&-i\\i&0\end{array}\right)~~,~~ \vec{\vec{\sigma}}_z=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)$$

The eigenstates of $S_z$ are called spinors: $\chi=\alpha_+\chi_++\alpha_-\chi_-$, where $\chi_+=(1,0)$ represents the state with spin up ($S_z= \frac{1}{2} \hbar$) and $\chi_-=(0,1)$ represents the state with spin down ($S_z=- \frac{1}{2} \hbar$). Then the probability to find spin up after a measurement is given by $|\alpha_+|^2$ and the chance to find spin down is given by $|\alpha_-|^2$. Of course  $|\alpha_+|^2+|\alpha_-|^2=1$.

The electron will have an intrinsic magnetic dipole moment $\vec{M}$ due to its spin, given by $\vec{M}=-eg_S\vec{S}/2m$, with $g_S=2(1+\alpha/2\pi+\cdots)$ the gyromagnetic ratio. In the presence of an external magnetic field this gives a potential energy $U=-\vec{M}\cdot\vec{B}$. The Schrödinger equation then becomes (because $\partial\chi/\partial x_i\equiv0$):

$$i\hbar\frac{\partial \chi(t)}{\partial t}=\frac{eg_S\hbar}{4m}\vec{\sigma}\cdot\vec{B}\chi(t)$$

with $\vec{\sigma}=(\vec{\vec{\sigma}}_x,\vec{\vec{\sigma}}_y,\vec{\vec{\sigma}}_z)$. If $\vec{B}=B\vec{e}_z$ there are two eigenvalues for this problem: $\chi_\pm$ for $E=\pm eg_S\hbar B/4m=\pm\hbar\omega$. So the general solution is given by $\chi=(a{\rm e}^{-i\omega t},b{\rm e}^{i\omega t})$. From this it can be derived that: $\left\langle S_x \right\rangle= \frac{1}{2} \hbar\cos(2\omega t)$ and $\left\langle S_y \right\rangle= \frac{1}{2} \hbar\sin(2\omega t)$. Thus the spin precesses about the $z$-axis with frequency $2\omega$. This causes the normal Zeeman splitting of spectral lines.

The potential operator for two particles with spin $\pm \frac{1}{2} \hbar$ is given by:

$$V(r)=V_1(r)+\frac{1}{\hbar^2}(\vec{S}_1\cdot\vec{S_2})V_2(r)= V_1(r)+ \frac{1}{2} V_2(r)[S(S+1)-\frac{3}{2} ]$$

This makes it possible for two states to exist: $S=1$ (triplet) or $S=0$ (Singlet).

The Dirac formalism

If the operators for $p$ and $E$ are substituted in the relativistic equation $E^2=m_0^2c^4+p^2c^2$, the Klein-Gordon equation is found:

$$\displaystyle \left(\nabla^2-\frac{1}{c^2}\frac{\partial^2 }{\partial t^2}-\frac{m_0^2c^2}{\hbar^2}\right)\psi(\vec{x},t)=0$$

The operator $\Box-m_0^2c^2/\hbar^2$ can be separated:

$$\nabla^2-\frac{1}{c^2}\frac{\partial^2 }{\partial t^2}-\frac{m_0^2c^2}{\hbar^2}= \left\{\gamma_\lambda\frac{\partial }{\partial x_\lambda}-\frac{m_0c}{\hbar}\right\} \left\{\gamma_\mu\frac{\partial }{\partial x_\mu}+\frac{m_0c}{\hbar}\right\}$$

where the Dirac matrices $\gamma$ are given by: $\{\gamma_\lambda,\gamma_\mu\}= \gamma_\lambda\gamma_\mu+\gamma_\mu\gamma_\lambda=2\delta_{\lambda\mu}$ (In general relativity this becomes $2g_{\lambda\mu}$). From this it can be derived that the $\gamma$ are hermitian $4\times4$ matrices given by:

$$\gamma_k=\left(\begin{array}{cc}0&-i\sigma_k\\i\sigma_k&0\end{array}\right)~~,~~ \gamma_4=\left(\begin{array}{cc}I&0\\0&-I\end{array}\right)$$

With this, the Dirac equation becomes:

$$\left(\gamma_\lambda\frac{\partial }{\partial x_\lambda}+\frac{m_0c}{\hbar}\right) \psi(\vec{x},t)=0$$

where $\psi(x)=(\psi_1(x),\psi_2(x),\psi_3(x),\psi_4(x))$ is a spinor.

Atomic physics

Solutions

The solutions of the Schrödinger equation in spherical coordinates if the potential energy is a function of $r$ alone can be written as: $\psi(r,\theta,\varphi)=R_{nl}(r)Y_{l,m_l}(\theta,\varphi)\chi_{m_s}$, with

$$Y_{lm}=\frac{C_{lm}}{\sqrt{2\pi}}P_l^m(\cos\theta){\rm e}^{im\varphi}$$

For an atom or ion with one electron : $\displaystyle R_{lm}(\rho)=C_{lm}{\rm e}^{-\rho/2}\rho^l L_{n-l-1}^{2l+1}(\rho)$

with $\rho=2rZ/na_0$ and $a_0=\varepsilon_0 h^2/\pi m_{e}e^2$. The $L_i^j$ are the associated Laguere polynomials and the $P_l^m$ are the associated Legendre polynomials:

$$P_l^{|m|}(x)=(1-x^2)^{m/2}\frac{d^{|m|}}{dx^{|m|}}\left[(x^2-1)^l\right]~~,~~ L_n^m(x)=\frac{(-1)^mn!}{(n-m)!}{\rm e}^{-x}x^{-m}\frac{d^{n-m}}{dx^{n-m}}({\rm e}^{-x}x^n)$$

The parity of these solutions is $(-1)^l$. The functions are $2\sum\limits_{l=0}^{n-1}(2l+1)=2n^2$-fold degenerate.

Eigenvalue equations

The eigenvalue equations for an atom or ion with with one electron are:

Equation	Eigenvalue	Range
$H_{\rm op}\psi=E\psi$	$E_n=\mu e^4Z^2/8\varepsilon_0^2h^2n^2$	$n\geq1$
$L_{z\rm op}Y_{lm}=L_zY_{lm}$	$L_z=m_l\hbar$	$-l\leq m_l\leq l$
$L^2_{\rm op}Y_{lm}=L^2Y_{lm}$	$L^2=l(l+1)\hbar^2$	$l<n$
$S_{z\rm op}\chi=S_z\chi$	$S_z=m_s\hbar$	$m_s=\pm\frac{1}{2}$
$S^2_{\rm op}\chi=S^2\chi$	$S^2=s(s+1)\hbar^2$	$s=\frac{1}{2}$

Spin-orbit interaction

The total momentum is given by $\vec{J}=\vec{L}+\vec{M}$. The total magnetic dipole moment of an electron is then $\vec{M}=\vec{M}_L+\vec{M}_S=-(e/2m_{\rm e})(\vec{L}+g_S\vec{S})$ where $g_S=2.0023$ is the gyromagnetic ratio of the electron. Further: $J^2=L^2+S^2+2\vec{L}\cdot\vec{S}=L^2+S^2+2L_zS_z+L_+S_-+L_-S_+$. $J$ has quantum numbers $j$ with possible values $j=l\pm \frac{1}{2}$, with $2j+1$ possible $z$-components ($m_J\in\{-j,..,0,..,j\}$). If the interaction energy between $S$ and $L$ is small it can be stated that: $E=E_n+E_{SL}=E_n+a\vec{S}\cdot\vec{L}$. It can then be derived that:

$$a=\frac{|E_n|Z^2\alpha^2}{\hbar^2nl(l+1)(l+\frac{1}{2} )}$$

After a relativistic correction this becomes:

$$E=E_n+\frac{|E_n|Z^2\alpha^2}{n}\left(\frac{3}{4n}-\frac{1}{j+ \frac{1}{2} }\right)$$

The fine structure in atomic spectra arises from this. With $g_S=2$ follows for the average magnetic moment: $\vec{M}_{\rm av}=-(e/2m_{\rm e})g\hbar\vec{J}$, where $g$ is the Landé-g factor:

$$g=1+\frac{\vec{S}\cdot\vec{J}}{J^2}=1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)}$$

For atoms with more than one electron the following limiting situations occur:

1. $L-S$ coupling: for small atoms the electrostatic interaction is dominant and the state can be characterized by $L,S,J,m_J$. $J\in\{|L-S|,...,L+S-1,L+S\}$ and $m_J\in\{-J,...,J-1,J\}$. The spectroscopic notation for this interaction is: $^{2S+1}L_J$. $2S+1$ is the multiplicity of a multiplet.
2. $j-j$ coupling: for larger atoms the electrostatic interaction is smaller than the $L_i\cdot s_i$ interaction of an electron. The state is characterized by $j_i...j_n,J,m_J$ where only the $j_i$ of the not completely filled subshells are to be taken into account.

The energy splitting for larger atoms when placed in a magnetic field is: $\Delta E=g\mu_{\rm B}m_JB$ where $g$ is the Landé factor. For a transition between two singlet states the line splits in three parts, for $\Delta m_J=-1,0+1$. This results in the normal Zeeman effect. At higher $S$ the line splits up into additional parts: the anomalous Zeeman effect.

Interaction with the spin of the nucleus gives the hyperfine structure.

Selection rules

For the dipole transition matrix elements it follows that: $p_0\sim|\langle l_2m_2|\vec{E}\cdot\vec{r}\,|l_1m_1\rangle|$. Conservation of angular momentum demands that for the transition of an electron $\Delta l=\pm1$.

For an atom where $L-S$ coupling is dominant: $\Delta S=0$ holds (but not strictly), $\Delta L=0,\pm1$, $\Delta J=0,\pm1$ except for $J=0\rightarrow J=0$ transitions, $\Delta m_J=0,\pm1$, but $\Delta m_J=0$ is forbidden if $\Delta J=0$.

For an atom where $j-j$ coupling is dominant the following selection rules hold for the  electron involved in the transition  $\Delta l=\pm1$, and $\Delta j=0,\pm1$, For all other electrons: $\Delta j=0$. For all of the electrons in the atom taken together the net: $\Delta J=0,\pm1$ but no $J=0\rightarrow J=0$ transitions are allowed and $\Delta m_J=0,\pm1$, but $\Delta m_J=0$ is forbidden if $\Delta J=0$.

Interaction with electromagnetic fields

The Hamiltonian of an electron in an electromagnetic field is given by:

$$H=\frac{1}{2\mu}(\vec{p}+e\vec{A})^2-eV=-\frac{\hbar^2}{2\mu}\nabla^2+ \frac{e}{2\mu}\vec{B}\cdot\vec{L}+\frac{e^2}{2\mu}A^2-eV$$

where $\mu$ is the reduced mass of the system. The term $\sim A^2$ can usually be neglected, except for very strong fields or macroscopic motions. For $\vec{B}=B\vec{e}_z$ it is given by $e^2B^2(x^2+y^2)/8\mu$.

When a gauge transform $\vec{A}'=\vec{A}-\nabla f$, $V'=V+\partial f/\partial t$ is applied to the potentials the wavefunction is also transformed according to $\psi'=\psi{\rm e}^{iqef/\hbar}$ with $qe$ the charge of the particle. Because $f=f(x,t)$, this is called a local gauge transfor, in contrast with a global gauge transform which can always be applied.

Perturbation theory

Time-independent perturbation theory

To solve the equation $(H_0+\lambda H_1)\psi_n=E_n\psi_n$ one has to find the eigenfunctions of $H=H_0+\lambda H_1$. Suppose that $\phi_n$ is a complete set of eigenfunctions of the non-perturbed Hamiltonian $H_0$: $H_0\phi_n=E_n^0\phi_n$. Because $\phi_n$ is a complete set :

$$\psi_n=N(\lambda)\left\{\phi_n+\sum_{k\neq n}c_{nk}(\lambda)\phi_k\right\}$$

When $c_{nk}$ and $E_n$ are being expanded into $\lambda$: $\begin{array}{l} c_{nk}=\lambda c_{nk}^{(1)} + \lambda ^{2}  c_{nk}^{(2)} + .\; . \;.  \\ E_{n}=E_{n}^{0}+\lambda E_{n}^{(1)}+\lambda ^{2}  E_{n}^{(2)} + .\; .\; .  \end{array}$

and this is put into the Schrödinger equation the result is: $E_n^{(1)}=\left\langle \phi_n|H_1|\phi_n \right\rangle$ and
$\displaystyle c_{nm}^{(1)}=\frac{\left\langle \phi_m|H_1|\phi_n \right\rangle}{E_n^0-E_m^0}$ if $m\neq n$. The second-order correction of the energy is then given by:
$\displaystyle E^{(2)}_n=\sum_{k\neq n}\frac{|\left\langle \phi_k|H_1|\phi_n \right\rangle|^2}{E^0_n-E^0_k}$. So to first order: $\displaystyle\psi_n=\phi_n+\sum_{k\neq n} \frac{\left\langle \phi_k|\lambda H_1|\phi_n \right\rangle}{E_n^0-E_k^0}~\phi_k$.

In case the levels are degenerate the above does not hold. In that case an orthonormal set of eigenfunctions $\phi_{ni}$ is chosen for each level $n$, so that $\left\langle \phi_{mi}|\phi_{nj} \right\rangle=\delta_{mn}\delta_{ij}$. Now $\psi$ is expanded as:

$$\psi_n=N(\lambda)\left\{\sum_i\alpha_i\phi_{ni}+\lambda\sum_{k\neq n} c_{nk}^{(1)}\sum_i\beta_i\phi_{ki}+\cdots\right\}$$ $E_{ni}=E_{ni}^0+\lambda E_{ni}^{(1)}$ is approximated by $E_{ni}^0:=E_n^0$.

Substitution in the Schrödinger equation and taking scalar product with $\phi_{ni}$ gives: $\sum\limits_i\alpha_i\left\langle \phi_{nj}|H_1|\phi_{ni} \right\rangle=E_n^{(1)}\alpha_j$. Normalization requires that $\sum\limits_i|\alpha_i|^2=1$.

Time-dependent perturbation theory

From the Schrödinger equation $\displaystyle i\hbar\frac{\partial \psi(t)}{\partial t}=(H_0+\lambda V(t))\psi(t)$

and the expansion $\displaystyle\psi(t)=\sum_nc_n(t)\exp\left(\frac{-iE_n^0t}{\hbar}\right)\phi_n$ with $c_n(t)=\delta_{nk}+\lambda c_n^{(1)}(t)+\cdots$

follows: $\displaystyle c_n^{(1)}(t)=\frac{\lambda}{i\hbar}\int\limits_0^t\left\langle \phi_n|V(t')|\phi_k \right\rangle \exp\left(\frac{i(E_n^0-E_k^0)t'}{\hbar}\right)dt'$

N-particle systems

General

Identical particles are indistinguishable. For the total wavefunction of a system of identical indistinguishable particles:

1. Particles with a half-odd integer spin (Fermions): $\psi_{\rm total}$ must be antisymmetric w.r.t. interchange of the coordinates (spatial and spin) of each pair of particles. The Pauli principle results from this: two Fermions cannot exist in an identical state because then $\psi_{\rm total}=0$.
2. Particles with an integer spin (Bosons): $\psi_{\rm total}$ must be symmetric w.r.t. interchange of the coordinates (spatial and spin) of each pair of particles.

For a system of two electrons there are two possibilities for the spatial wavefunction. When $a$ and $b$ are the quantum numbers of electron 1 and 2 :

$$\psi_{\rm S}(1,2)=\psi_a(1)\psi_b(2)+\psi_a(2)\psi_b(1)~~,~~ \psi_{\rm A}(1,2)=\psi_a(1)\psi_b(2)-\psi_a(2)\psi_b(1)$$

Because the particles do not approach each other closely the repulsive energy at $\psi_{\rm A}$ in this state is smaller. The following spin wavefunctions are possible:

$$\chi_{\rm A}= \begin{array}{ll} \frac{1}{2} \sqrt{2}[\chi_+(1)\chi_-(2)-\chi_+(2)\chi_-(1)]&m_s=0 \end{array}$$

$$\chi_{\rm S}= \left\{\begin{array}{ll} \chi_+(1)\chi_+(2)&m_s=+1\\ \frac{1}{2} \sqrt{2}[\chi_+(1)\chi_-(2)+\chi_+(2)\chi_-(1)]&m_s=0\\ \chi_-(1)\chi_-(2)&m_s=-1 \end{array}\right.$$

Because the total wavefunction must be antisymmetric it follows that: $\psi_{\rm total}=\psi_{\rm S}\chi_{\rm A}$ or $\psi_{\rm total}=\psi_{\rm A}\chi_{\rm S}$.

For $N$ particles the symmetric spatial function is given by:

$$\psi_{\rm S}(1,...,N)=\sum\psi(\mbox{all permutations of }1..N)$$

The antisymmetric wavefunction is given by the determinant $\displaystyle\psi_{\rm A}(1,...,N)=\frac{1}{\sqrt{N!}}|u_{E_i}(j)|$

Molecules

The wavefunctions of atom $a$ and $b$ are $\phi_a$ and $\phi_b$. If the 2 atoms approach each other there are two possibilities: the total wavefunction approaches the bonding function with lower total energy $\psi_{\rm B}= \frac{1}{2}  \sqrt{2}(\phi_a+\phi_b)$ or approaches the anti-bonding function with higher energy $\psi_{\rm AB}= \frac{1}{2} \sqrt{2}(\phi_a-\phi_b)$. If a molecular-orbital is symmetric w.r.t. the connecting axis, like a combination of two s-orbitals it is called a $\sigma$-orbital, otherwise a $\pi$-orbital, like the combination of two p-orbitals along two axes.

The energy of a system is: $\displaystyle E=\frac{\left\langle \psi|H|\psi \right\rangle}{\left\langle \psi|\psi \right\rangle}$.

The energy calculated with this method is always higher than the real energy if $\psi$ is only an approximation for the solutions of $H\psi=E\psi$. Also, if there are more than one function to be chosen, the function which gives the lowest energy is the best approximation. Applying this to the function $\psi=\sum c_i\phi_i$ one finds: $(H_{ij}-ES_{ij})c_i=0$. This equation has only solutions if the secular determinant $|H_{ij}-ES_{ij}|=0$. Here, $H_{ij}=\left\langle \phi_i|H|\phi_j \right\rangle$ and $S_{ij}=\left\langle \phi_i|\phi_j \right\rangle$. $\alpha_i:=H_{ii}$ is the Coulomb integral and $\beta_{ij}:=H_{ij}$ the exchange integral. $S_{ii}=1$ and $S_{ij}$ is the overlap integral.

The first approximation in the molecular-orbital theory is to place both electrons of a chemical bond in the bonding orbital: $\psi(1,2)=\psi_{\rm B}(1)\psi_{\rm B}(2)$. This results in a large electron density between the nuclei and therefore an attraction which forms a bond. A better approximation is: $\psi(1,2)=C_1\psi_{\rm B}(1)\psi_{\rm B}(2)+C_2\psi_{\rm AB}(1)\psi_{\rm AB}(2)$, with $C_1=1$ and $C_2\approx0.6$.

In some atoms, such as C, it is energetically more suitable to form orbitals which are a linear combination of the s, p and d states. There are three ways of hybridization in C:

1. sp-hybridization: $\psi_{\rm sp}= \frac{1}{2} \sqrt{2}(\psi_{\rm 2s}\pm\psi_{{\rm 2p}_z})$. There are two hybrid orbitals which are placed along a line separated $180^\circ$. Further the 2p$_x$ and 2p$_y$ orbitals remain.
2. sp(^2\) hybridization: $\psi_{\rm sp^2}=\psi_{\rm 2s}/\sqrt{3}+c_1\psi_{{\rm 2p}_z}+c_2\psi_{{\rm 2p}_y}$, where $(c_1,c_2)\in\{(\sqrt{2/3},0),(-1/\sqrt{6},1/\sqrt{2})$ $,(-1/\sqrt{6},-1/\sqrt{2})\}$. The three sp$^2$ orbitals lay in one plane, with symmetry axes at an angle of $120^\circ$.
3. sp$^3$ hybridization: $\psi_{\rm sp^3}= \frac{1}{2} (\psi_{\rm 2s}\pm\psi_{{\rm 2p}_z}\pm\psi_{{\rm 2p}_y}\pm\psi_{{\rm 2p}_x})$. The four sp$^3$ orbitals form a tetrahedron with the symmetry axes at an angle of $109^\circ 28'$.

Quantum statistics

If a system exists in a state in which one lacks complete information about the system, it can be described by a density matrix $\rho$. If the probability that the system is in state $\psi_i$ is given by $a_i$, one can write for the expectation value $a$ of $A$: $\left\langle a \right\rangle=\sum\limits_ir_i\langle\psi_i|A|\psi_i\rangle$.

If $\psi$ is expanded into an orthonormal basis $\{\phi_k\}$ as: $\psi^{(i)}=\sum\limits_k c_k^{(i)}\phi_k$, then:

$$\left\langle A \right\rangle=\sum_k(A\rho)_{kk}={\rm Tr}(A\rho)$$

where $\rho_{lk}=c_k^*c_l$. $\rho$ is Hermitian, with Tr$(\rho)=1$. Further $\rho=\sum r_i|\psi_i\rangle\langle\psi_i|$. The probability to find eigenvalue $a_n$ when measuring $A$ is given by $\rho_{nn}$ if one uses a basis of eigenvectors of $A$ for $\{\phi_k\}$. For the time-dependence (in the Schrödinger picture operators are not explicitly time-dependent):

$$i\hbar\frac{d\rho}{dt}=[H,\rho]$$

For a macroscopic system in equilibrium $[H,\rho]=0$ holds. If all quantum states with the same energy are equally probable: $P_i=P(E_i)$, one can obtain the distribution:

$$P_n(E)=\rho_{nn}=\dfrac{e^{-E_n/kT}}{Z} \;\;\textrm{with the state sum}\;\;Z=\sum_n e^{-E_n/kT}$$    

The thermodynamic quantities are related to these definitions as follows: $F=-kT\ln(Z)$, $U=\left\langle H \right\rangle=\sum\limits_np_nE_n=\displaystyle-\frac{\partial }{\partial kT}\ln(Z)$, $S=-k\sum\limits_nP_n\ln(P_n)$. For a mixed state of $M$ orthonormal quantum states with probability $1/M$ it follows that: $S=k\ln(M)$.

The distribution function for the internal states for a system in thermal equilibrium is the most probable function. This function can be found by taking the maximum of the function which gives the number of states with Stirling’s equation: $\ln(n!)\approx n\ln(n)-n$, and the conditions $\sum\limits_kn_k=N$ and $\sum\limits_kn_kW_k=W$. For identical, indistinguishable particles which obey the Pauli exclusion principle the possible number of states is given by:

$$P=\prod_k\frac{g_k!}{n_k!(g_k-n_k)!}$$

This results in Fermi-Dirac statistics. For indistinguishable particles which do not obey the exclusion principle the possible number of states is given by:

$$P=N!\prod_k\frac{g_k^{n_k}}{n_k!}$$

This results in  Bose-Einstein statistics. So the distribution functions which explain how particles are distributed over the different one-particle states $k$ which are each $g_k$-fold degenerate depend on the spin of the particles. They are given by:

1. Fermi-Dirac statistics: integer spin. $n_k\in\{0,1\}$, $\displaystyle n_k=\frac{N}{Z_g}\frac{g_k}{\exp((E_k-\mu)/kT)+1}$
   with $\ln(Z_{\rm g})=\sum g_k\ln[1+\exp((E_i-\mu)/kT)]$.
2. Bose-Einstein statistics: half odd-integer spin. $n_k \in \mathbb{N}$, $\displaystyle n_k=\frac{N}{Z_g}\frac{g_k}{\exp((E_k-\mu)/kT)-1}$
   with $\ln(Z_{\rm g})=-\sum g_k\ln[1-\exp((E_i-\mu)/kT)]$.

Here, $Z_{\rm g}$ is the large-canonical state sum and $\mu$ the chemical potential. It is found by demanding that $\sum n_k=N$: $\displaystyle\lim\limits_{T\rightarrow0}\mu=E_{\rm F}$ the result of which is the Fermi-energy. $N$ is the total number of particles. The Maxwell-Boltzmann distribution can be derived from this in the limit $E_k-\mu\gg kT$: $$n_k=\frac{N}{Z}\exp\left(-\frac{E_k}{kT}\right)~~\mbox{with}~~ Z=\sum_kg_k\exp\left(-\frac{E_k}{kT}\right)$$ In terms of the Fermi-energy, Fermi-Dirac and Bose-Einstein statistics can be written as:

1. Fermi-Dirac statistics: $\displaystyle n_k=\frac{g_k}{\exp((E_k-E_{\rm F})/kT)+1}$.
2. Bose-Einstein statistics: $\displaystyle n_k=\frac{g_k}{\exp((E_k-E_{\rm F})/kT)-1}$.