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# 10: Quantum Physics

• • Johan Wevers

## Introduction to quantum physics

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:

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

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: $$\displaystyle n_k=\frac{g_k}{\exp((E_k-E_{\rm F})/kT)-1}$$.
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