Bose Einstein

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:

\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:

\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:

\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 $c$:

\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:

\begin{align} \hat P_{12} \psi(r_{2},r_{1}) = \psi(r_{1},r_{2}) = c^{2}\psi(r_{2},r_{1}) \end{align}


\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:

\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:

\begin{align} \psi((r_{1},r_{2}=r_{1}) = -\psi((r_{2}=r_{1},r_{1}) \end{align}


\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.

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