A wavefunction is a tool used in quantum mechanics to describe all "observables" of any physical system. "Observables" refer to quantities of a system's configuration which can be measured (i.e. position, momentum, angular momentum, energy, etc.). The wavefunction is a complex vector or complex vector function which is defined in a complex Hilbert space (named for the great mathematician David Hilbert). The wavefunction is postulated to provide all of the information necessary to determine the configuration of the system at a particular instant of time. The wavefunction (and hence the system) evolves according to the Schrödinger equation (or, equivalently, Heisenberg's equations of motion). Observables (the things we can measure in an experiment) are obtained from unitary operators acting on the wavefunction. Following the basic postulate of quantum mechanics, one can establish uncertainty relations between operators of conjugate variables. The most famous of these is the Heisenberg Uncertainty principle relating the position and momentum of a particle, which establishes a lower bound on the certainty of measuring these two variable simultaneously. One can state this quite simply as $\Delta x \Delta p \ge {\hbar \over 2}$ where $\Delta x$ and $\Delta p$ are the RMS position and momentum of a particle and $\hbar$ is Planck's reduced constant.
So, the wavefunction completely determines the current and time-evolution of the state of a system. But what is it exactly? In the early part of the 20th century this was a hotly debated topic. It is difficult to make physical sense of the wavefunction itself. It is like a sort of kernel for describing the current and time-evolved state of a quantum system. Current modern interpretation is that the wavefunction represents a probability amplitutde (as a purely complex vector) and a probability (defined in any number of bases) distribution of the system in that bases. For example, the wavefunction of a particle in one dimension $\psi(x)$ is a complex function defined over the real numbers. The positive definite function $|\psi(x)|^2$ (this can be shown easily assuming the most general form of $\psi(x)$ as a complex vector function) is interpreted as the probability density associated with the particle's position. More specifically, the probability of measuring the particle's position between $x_1$ and $x_2$ is:
(1)
\begin{align} P(x=x_1,x_2)=\int_{x_1}^{x_2}|\psi(x)|^2,dx \end{align}
In any choice of bases (position, momentum, angular momentum, etc.), the wave function provides a complete description of the physical system. An element of a (any) vector space can be expressed in any number of bases. The same applies to the wavefunction. However, the wavefunction (like any other vector) itself is not dependent on the basis chosen. Representing the wavefunction in a new bases does not alter the wavefunction, only its representation. One often finds calculations easier in one particular basis compared to another.
Since the interpretation of the wavefunction is that of a probability, the following normalization is required:
(2)
\begin{align} P(x=-\infty,\infty)=\int_{-\infty}^{\infty}|\psi(x)|^2dx = 1 \end{align}
Since the probability of the particle being anywhere must be 1. The same is true regardless of the chosen bases representation (the probability of the paricle having any momentum is 1, any energy is 1, ect.) Because of this the wavefunction is normalized so that the norm is 1.
Now that we have some working idea of what the wavefunction represents, let's consider a simple two particle system. Our wavefunction will now be a function of both particles positions $\psi(r_{1},r_{2})$ (let's assume the system is time-independent).
Let's then define an operator (like the ones mentioned above) that will exchange the particle at position $r_{1}$ for the particle at $r_{2}$, that is:
(3)
\begin{align} \hat P_{12} \psi(r_{1},r_{2}) = \psi(r_{2},r_{1}) \end{align}
This operator must obey certain requirements that satisfy the basic postulates of quantum mechanics, but we only need to worry about the implications of this transformation. First, we can imagine that the particles are distinguishible. How would we distinguish between 2 electrons, for instance? On the macroscopic scale, we might make a mark on one particle (small enough as to not affect the particle significantly). But at the scall of subatomic particles, this doesn't make any sense. The point of any fundamental constiuent of nature is that they are all indistinguishible; one electron's as good as the next (and so on for protons, neutrons, gluons, quarks, etc.). So, such a transformation defined by this exchange operator must result in only a change in the state of the two particle system by a (possibly) complex constant $c$:
(4)
\begin{align} \psi(r_{1},r_{2}) = c\psi(r_{2},r_{1}) \end{align}
At first glance, this may seem like a completely new wavefunction, and hence a new distinct state. But, remember that the wavefunction is not something "measurable", and a basic postulate of quantum mechanics is that the eignevalues (and eigenvectors) of the wavefunction (in whichever bases it is represented) are what's important to measurement (and hence, what is physically important). Vectors that differ by an arbitrary constant are not really different vectors at all (in the sense that the vector space that they span is linearly dependent). Performing the exchange operation again yields:
(5)
\begin{align} \hat P_{12} \psi(r_{2},r_{1}) = \psi(r_{1},r_{2}) = c^{2}\psi(r_{2},r_{1}) \end{align}
or
(6)
\begin{align} c^{2}\psi(r_{2},r_{1}) = \psi(r_{1},r_{2}) \end{align}
so $c = \pm 1$. This means that either the wavefunction is symmetric ($c=1$) or anti-symmetric ($c=-1$ under particle exchange. To determine which particles obey either transformation rule requires the complete relativistic theory of quantum mechanics (one of the greatest achievements of the 20th century). This analysis captures the flavor of the quantum rules of distinguishibility.
Let's consider one more aspect of this result. Remember that the wavefunction of a single particle may be interpreted as a probability. We should then interpret a wavefunction for two particles as a joint-probability density; that is:
(7)
\begin{align} \int_{-\infty}^{\infty}|\psi((r_{1},r_{2})|^2 dr_1dr_2 = 1 \end{align}
The probability of finding both particles anywhere. The symmetry properties above imply for anti-symmetric exchanges that:
(8)
\begin{align} \psi((r_{1},r_{2}=r_{1}) = -\psi((r_{2}=r_{1},r_{1}) \end{align}
Iff
(9)
\begin{align} \psi((r_{1},r_{2}=r_{1}) = 0 \end{align}
That is, the probability of finding both particles (which are anti-symmetric under exchange) at the same position is 0. This is Pauli's exclusion principle for particles obeying anti-symmetric exchanges. These particles cannot occupy the same quantum state simultaneously.
In relativistic quantum theory, every particle posses an intrinsic kind of quantized (in units of 1/2) angular momentum called spin (although it has no classic analogue like momentum). According to the spin-statistics theorem, particles with half-interger spin are anti-symmetric and particles with interger spin are symmetric. Anti-symmetric particles are called Fermions and symmetric particles Bosons. Both types of particle can be either fundamental or composite. For instance, the electron (fundamental) and the proton (3 quarks) are both fermions. The photon and the Meson (quark and anti-quark pairs) are both Bosons. Fermions cannot occupy the same quantum state simultaneously. There is no limit to the number of bosons that can occupy a given quantum state.
The Pauli-exclusion principle is one of the most important results of quantum theory. It helps explain a wide variety of physical phenomena. The way electrons are distributed in an atom's electron shell and the interaction between different atoms can be explained by this principle. If not for the "degeneracy pressure" (the exclusion of more than 2 electrons from occupying the same electron orbital) the atom would collapse in on itself. This also explains the behavior of the periodic table of elements.
A partciularly striking effect can be seen in neutron stars. Under such intense gravitational forces the usual atomic structure is disrupted, leaving the remaining particles supported by this "degeneracy pressure" solely. This exotic form of matter is known as degenerate matter {wiki}. In neutron stars, the electrons merge with the protons to form neutrons, which produce a larger degeneracy pressure. Neutron stars are the most "rigid" objects known - their Young modulus (or more accurately, bulk modulus) is 20 orders of magnitude larger than that of diamond! {wiki}
The statistical implications of their quantum state symmetries are explained in the next section.
If you didn't read the previous page, the Bose-Einstein distribution is derived assuming particles can occupy the same quantum state simulatenously (i.e. there is no limit to the number of particles in a given energy level). The Fermi-Dirac distribution is derived assuming that only one particle can occupy a given energy level. Using these constraints the two different distributions are derived.
(10)
\begin{align} W = N \prod_i \frac{g_i^{n_i}}{n_i!} \end{align}
Bose-Einstein Statistics
First, we start with a system composed of bosons. Remember, there is no limit to the number of bosons which can occupy a given quantum state (energy level). Even though quantum mechanically there is no such thing as a distinguishibile particle, for each energy $\epsilon_i$ there are $f_i$ degeneracies associated with a given energy level. For instance, 2 particles may have the same total energy but different momenta, and hence can be distinguished. So, taking into account this degeneracy, the ways of distributing $n$ paticles among the energy levels with degeneracy $f_i$ is:
(11)
\begin{align} w(n_i,f_i) = \frac{(n_i+f_i-1)!}{n!(f_i-1)!} \end{align}
The exact derivation of this combinatoric has been omitted, but it’s derivation is similar to that of the Boltzmann distribution. Next, the number of ways that energy occupation levels can be realized is the product of the ways that each individual energy level can be populated. So, like the Boltzmann distribution, the function $W$ (which is directly related to the entropy) is:
(12)
\begin{align} W = \prod_i w(n_i,f_i) = \prod_i \frac{(n_i+f_i-1)!}{n_i!(f_i-1)!} \approx \prod_i \frac{(n_i+f_i)!}{n_i!f_i!} \end{align}
This approximation is valid if $g_i>>1$. Now we want to maximize W in the same way we did to find the Boltzmann distribution, subject to the contraints of particle number $N = \sum_i n_i$ and system energy $E = \sum_i \n_i \epsilon_i$. In the same way, it will be easier to maximize $ln(W)$:
(13)
\begin{align} g(n_i) = ln(W) + \alpha(N-\sum_i n_i) + \beta(E-\sum_i n_i \epsilon_i) \end{align}
Using Stirling's approximation ($ln(x!) \approx xln(x)-x$)
(14)
\begin{align} g( n_i ) = \sum_i ( n_i + f_i )ln( n_i + f_i ) - n_i ln( n_i )- f_i ln( f_i )+\alpha (N - \sum_i n_i) + \beta (E - \sum_i n_i \epsilon_i ) \end{align}
We now want to maximize this function (since this is equivalent to maximizing the entropy of the corresponding system’s macrostate) with respect to the number of microstates $n_i$:
(15)
\begin{align} \frac{dg}{ dn_i } = \sum_i ln( n_i + f_i ) ln ( n_i ) - \alpha - \beta \sum_i \epsilon_i = 0 \end{align}
Proving this is a maximum requires taking a second derivative (which won't be done here). Solving for $n_i$ yields:
(16)
\begin{align} n_i = \frac{ f_i }{ e^{\alpha + \beta \epsilon_i} - 1 } \end{align}
Where we can show (as for the Boltzmann distribution) that $\beta = \frac{1}{kT}$ and $\alpha = -\frac{ \mu }{kT}$ where $k$ is Boltzmann's constant, $T$ is the temperature, and $\mu$ is the chemical potential:
(17)
\begin{align} n_i = \frac{ f_i }{ e^{( \epsilon_i - \mu)kT} - 1 } \end{align}
Fermi-Dirac Statistics
Now, fermions obey the peculiar property of being forbidden to occupy the same quantum state. Statistically, this manifests itself as limiting the number of particles in a given energy sublevel to 1. It’s easy to see that in an energy level with $f_i$ sublevels, there are only $f_i$ ways of distributing 1 particle in that level. With 2 particles, we can put 1 in the first sublevel and put the other in the $f_i-1$ sublevels. The number of ways of distributing $n$ particles in $f$ sublevels is then:
(18)
\begin{align} w(n_i,f_i) = \frac{f_i!}{n_i!(f_i-n_i)!} \end{align}
And (like above) the number of ways of realizing the occupation of the various energy levels is:
(19)
\begin{align} W = \prod_i \frac{f_i!}{n_i!(f_i-n_i)!} \end{align}
Again, we would like to maximize this quantity (and maximizing $ln(W)$ instead will be more fruitful):
(20)
\begin{align} g(n_i) = ln(W) + \alpha(N-\sum_i n_i) + \beta(E-\sum_i n_i \epsilon_i) \end{align}
Skipping a bit of the (redundant) math above, the number of particles in a given energy level is given by:
(21)
\begin{align} n_i = \frac{ f_i }{ e^{( \epsilon_i - \mu)/kT} + 1 } \end{align}
Comparison to Maxwell-Boltzmann Distribution
Let's look at the comparison of the mean occupation numbers for the 3 different distribtuions:
(22)
\begin{align} \langle n_i \rangle =\frac{1}{ e^{( \epsilon _i - \mu )/kT}}$ \end{align}
(23)
\begin{align} \langle n_i \rangle = \frac{ 1 }{ e^{( \epsilon_i - \mu)kT} - 1 } \end{align}
(24)
\begin{align} \langle n_i \rangle = \frac{ 1 }{ e^{( \epsilon_i - \mu)/kT} + 1 } \end{align}
Or, for the general case, we could write:
(25)
\begin{align} \langle n_i \rangle = \frac{1}{ e^{( \epsilon_i - \mu)/kT} + a } \end{align}
With a=0 (Boltzmann), a=1 (Fermi-Dirac), and a=-1 (Bose-Einstein).
(26)
\begin{align} \langle n_i \rangle = \frac{1}{ e^{( \epsilon_i - \mu)/kT} } \end{align}
In general, the partition function is defined as:
(27)
\begin{align} Z = \prod_i Z_i \end{align}
with
(28)
\begin{align} Z_{i} = \sum_{n_i=0}^{\infty} e^{- \beta n_i ( \epsilon_i - \mu)} \end{align}
Defined in this way, the partition function is directly related to the probability of the system occupying a given microstate.
So, the partition functions for the particular distributions are:
Boltzmann:
(29)
\begin{align} Z_{i} = e^{- \beta ( \epsilon_i - \mu)}} \end{align}
Bosons:
(30)
\begin{align} Z_{i} = \frac{1}{1 – e^{- \beta ( \epsilon_i - \mu)}} \end{align}
Fermions:
(31)
\begin{align} Z_{i} = \frac{1}{1 + e^{- \beta ( \epsilon_i - \mu)}} \end{align}
With these partition functions we can derive all of the thermodynamic quantities of interest (number of particles, internal energy, pressure, etc.) in the same way we did for the Boltzmann distribution:
Number of Particles
(32)
\begin{align} N = \frac{1}{\beta} \left(\frac{\partial ln (Z)}{\partial \mu } \right)_{\beta,V} \end{align}
Internal Energy
(33)
\begin{align} U = \left(\frac{\partial ln (Z)}{\partial \beta } \right)_{\mu,V} + \mu N \end{align}
Pressure
(34)
\begin{align} P = \frac{1}{\beta} \left(\frac{\partial ln (Z)}{\partial V } \right)_{\mu,\beta} \end{align}
These results can also be used to construct a Bose of Fermi ideal gas equations of state.
Let's call $\alpha = (\epsilon_i - \mu)/kT$ and plot the three different average occupation numbers versus $\alpha$:
For the Boltzmann and Bose distributions the particles tend to accumulate in the lower energy levels as $\alpha$ (temperature) is decreased.
Clearly, in the limit as $T->\infty$ these distributions all converge to the same distribution (the Boltzmann). It is only at low temperatures (or high density, equivalently) that the quantum effects of the exclusion principle can be seen. This can be examined by considering the average interparticle distance in the system. The average interparticle distance in the system can be written as:
(35)
\begin{align} \frac{ V}{N}^{1/3} \end{align}
where:
$V=volume$
$N=number of particles$
The Thermal deBroglie wavelength for an ideal gas is:
(36)
\begin{align} \Lambda = \frac{ h }{ (2 \pi mkT)^{1/2} } \end{align}
If the average interparticle spacing is on the order of the Thermal deBroglie wavelength, quantum effects will begin to become important and the substance must be treated as a Fermi or Bose gas. This condition is:
(37)
\begin{align} ( \frac{ V }{ N } )^{1/3} \le \Lambda \end{align}
As discussed at the end of the previous article, the quantum effects of a system of Bosons or Fermions are only seen when the energy of the system is low (or equivalently, the particle density is high). This article outlines some of these effects.
Bose-Einstein Condensation
A Bose–Einstein condensate (BEC) is a state of a system of bosons confined by an external potential and cooled to temperatures near to absolute zero. This state of matter was predicted in 1924 by Satyendra Nath Bose and Albert Einstein. It wasn’t until 1995 that the effect was confirmed experimentally.
The behavior of a system of bosons at very low temperatures can be explained in terms of the behavior of the distributions discussed previously. As mentioned, there is no limit to the number of bosons that can occupy a given quantum state. As a consequence, as the temperature is lowered, a large fraction (with no limit) of bosons collapse (condense) into the lowest energy state (since the degeneracy associated with any level is limitless). This type of condensation is a condensation in state space (particles occupying the same energy level) and so is not a condensation in the typical sense. This type of quantum effect produces striking macroscopic effects. As a consequence of quantum mechanics, systems only acquire energy in discrete steps (in the thermodynamic limit these steps can be regarded as continuous). If a system is at such a low temperature that it is in the lowest energy state, it is no longer possible for it to reduce its energy discretely, even by friction. This is the so-called effect of superfluidity. The effect of viscosity is effectively eliminated.
The transition to superfluid behavior occurs below a critical temperature, which for a uniform three-dimensional boson gas consisting of non-interacting particles is:
(38)
\begin{align} T_c = \frac{ N }{V \zeta (3/2) }^{2/3} \frac{ h^2 }{ 2 \pi m k_b } \end{align}
where:
$T_c$ is the critical temperature.
$n = N/V$ is particle density
$m$ is the mass of the boson
$h$ is Planck’s constant
$k_b$ is Boltzmann’s constant
$\zeta$ is the Riemann zeta function ~ 2.6
The derivation is quite involved and won’t be presented here (it can be found in any Stat. Mech. Textbook).
If the number of particles and the volume are held fixed, then the above condition reduces to:
(39)
\begin{align} T < T_c = \frac{ h^2 }{2 \pi m k_b}^{2/3} \frac{ N }{ V \zeta (3/2) }^{2/3} \end{align}
For $T<T_c$ the system consists of two phases:
i) a "normal" phase consisting of $N_e=N \frac{T}{T_c}^{3/2}$ particles distributed over the excited states ($\epsilon > 0$).
ii) a condensed phase $N_0 = N-N_e$ of particles accumulated in the ground state.
Fermi Electron Gas in Metals
The application of the Fermi-Dirac statistical ensemble has removed a number of inconsistencies and descrepencies in conduction of electrons in metals. One such descrepency is the observed specific heats in metals appeared to be almost completely accoutable for by lattice vibrations alone, with practically no contribution from the electron gas. Theories requried that each electron contribute a mean thermal energy $3/2kT$ (much like that of an ideal gas), contributing an additional $3/2k$ to the specific heat.
If we consider the gas an ideal Fermi gas, we can estimate the Fermi energy (for say sodium):
(40)
\begin{align} \epsilon_f = (\frac{3N}{8 \pi V})^{2/3} \frac{h^2}{2m^'} \end{align}
where $m^'$ is the effective mass of an electron gas (difference in actual mass and mass of electrons in sodium structure, $m^'/m = 0.98$. The electron density can be written as:
(41)
\begin{align} \frac{N}{V} = \frac{n_en_a}{a^3} \end{align}
where $n_e$ is the number of conduction electrons per atom, $n_a$ is the number of atoms per unit cell and a is the lattice constant (or cell length). For sodium:
(42)
\begin{align} \epsilon_f = 3.14 eV \end{align}
The Fermi temperature is then:
(43)
\begin{align} T_f = \frac{\epsilon_F}{k} = 3.64 x 10^4 K \end{align}
Thus, at room temperature ($~3 x 10^2 K$), the conduction electrons in sodium are highly degenerate. Most metals Fermi temperature are $10^4 - 10^5 K$. So, the specific heat is no longer given by the temperature independent classical formula,
(44)
\begin{align} C_V = \frac{\pi^2}{2}Nk(\frac{kT}{\epsilon_F}) \end{align}
which is much smaller than the ideal specific heat, since at ordinary temperatures $\frac{kT}{\epsilon_F} = \frac{T}{T_F} = O(10^{-2})$. Thus, the almost complete dependance of the specific heat on lattice vibrations is explained.
