Bose 302

Bose 302

November 25, 2009 in Uncategorized | No comments

A Bose–Einstein condensate (BEC) is a state of matter of a dilute gas of weakly interacting bosons confined in an external potential and cooled to temperatures very near to absolute zero ( 0 K , −273.15 °C , or −459.67 °F ). Under such conditions, a large fraction of the bosons occupy the lowest quantum state of the external potential, and all wave functions overlap each other, at which point quantum effects become apparent on a macroscopic scale.

This state of matter was first predicted by Satyendra Nath Bose and Albert Einstein in 1924–25. Bose first sent a paper to Einstein on the quantum statistics of light quanta (now called photons). Einstein was impressed, translated the paper himself from English to German and submitted it for Bose to the Zeitschrift für Physik which published it. Einstein then extended Bose's ideas to material particles (or matter) in two other papers.

Seventy years later, the first gaseous condensate was produced by Eric Cornell and Carl Wieman in 1995 at the University of Colorado at Boulder NIST-JILA lab, using a gas of rubidium atoms cooled to 170 nanokelvin (nK) ( 1.7 × 10 ⁻⁷ K ). Cornell, Wieman, and Wolfgang Ketterle at MIT were awarded the 2001 Nobel Prize in Physics in Stockholm, Sweden for their achievements.

Theory

The slowing of atoms by use of cooling apparatus produces a singular quantum state known as a Bose condensate or Bose–Einstein condensate . This phenomenon was predicted in 1925 by generalizing Satyendra Nath Bose's work on the statistical mechanics of (massless) photons to (massive) atoms. (The Einstein manuscript, once believed to be lost, was found in a library at Leiden University in 2005.) The result of the efforts of Bose and Einstein is the concept of a Bose gas, governed by Bose–Einstein statistics, which describes the statistical distribution of identical particles with integer spin, now known as bosons. Bosonic particles, which include the photon as well as atoms such as helium-4, are allowed to share quantum states with each other. Einstein demonstrated that cooling bosonic atoms to a very low temperature would cause them to fall (or "condense") into the lowest accessible quantum state, resulting in a new form of matter.

This transition occurs below a critical temperature, which for a uniform three-dimensional gas consisting of non-interacting particles with no apparent internal degrees of freedom is given by:

where:

Einstein's argument

Consider a collection of N noninteracting particles which can each be in one of two quantum states, $\scriptstyle|0\rangle$ and $\scriptstyle|1\rangle$ . If the two states are equal in energy, each different configuration is equally likely.

If we can tell which particle is which, there are 2 ^N different configurations, since each particle can be in $\scriptstyle|0\rangle$ or $\scriptstyle|1\rangle$ independently. In almost all the configurations, about half the particles are in $\scriptstyle|0\rangle$ and the other half in $\scriptstyle|1\rangle$ . The balance is a statistical effect, the number of configurations is largest when the particles are divided equally.

If the particles are indistinguishable, however, there are only N+1 different configurations. If there are K particles in state $\scriptstyle|1\rangle$ , there are N-K particles in state $\scriptstyle|0\rangle$ . Whether any particular particle is in state $\scriptstyle|0\rangle$ or in state $\scriptstyle|1\rangle$ cannot be determined, so each value of K determines a unique quantum state for the whole system. If all these states are equally likely, there is no statistical spreading out; it is just as likely for all the particles to sit in $\scriptstyle|0\rangle$ as for the particles to be split half and half.

Suppose now that the energy of state $\scriptstyle|1\rangle$ is slightly greater than the energy of state $\scriptstyle|0\rangle$ by an amount E. At temperature T, a particle will have a lesser probability to be in state $\scriptstyle|1\rangle$ by exp(-E/T). In the distinguishable case, the particle distribution will be biased slightly towards state $\scriptstyle|0\rangle$ and the distribution will be slightly different from half and half. But in the indistinguishable case, since there is no statistical pressure toward equal numbers, the most likely outcome is that most of the particles will collapse into state $\scriptstyle|0\rangle$ .

In the distinguishable case, for large N, the fraction in state $\scriptstyle|0\rangle$ can be computed. It is the same as coin flipping with a coin which has probability p = exp(-E/T) to land tails. The fraction of heads is 1/(1+p), which is a smooth function of p, of the energy.

In the indistinguishable case, each value of K is a single state, which has its own separate Boltzmann probability. So the probability distribution is exponential:

For large N, the normalization constant C is (1-p). The expected total number of particles which are not in the lowest energy state, in the limit that $\scriptstyle N\rightarrow \infty$ , is equal to $\scriptstyle \sum_{n>0} C n p^n=p/(1-p)$ . It doesn't grow when N is large, it just approaches a constant. This will be a negligible fraction of the total number of particles. So a collection of enough bose particles in thermal equilibrium will mostly be in the ground state, with only a few in any excited state, no matter how small the energy difference.

Consider now a gas of particles, which can be in different momentum states labelled $\scriptstyle|k\rangle$ . If the number of particles is less than the number of thermally accessible states, for high temperatures and low densities, the particles will all be in different states. In this limit the gas is classical. As the density increases or the temperature decreases, the number of accessible states per particle becomes smaller, and at some point more particles will be forced into a single state than the maximum allowed for that state by statistical weighting. From this point on, any extra particle added will go into the ground state.

To calculate the transition temperature at any density, integrate over all momentum states the expression for maximum number of excited particles p/(1-p):

When the integral is evaluated with the factors of k _B and ℏ restored by dimensional analysis, it gives the critical temperature formula of the preceding section. Therefore, this integral defines the critical temperature and particle number corresponding to the conditions of zero chemical potential (μ = 0 in the Bose–Einstein statistics distribution).

Gross–Pitaevskii equation
Main article: Gross–Pitaevskii equation
The state of the BEC can be described by the wavefunction of the condensate $\psi(\vec{r})$ . For a system of this nature, $|\psi(\vec{r})|^2$ is interpreted as the particle density, so the total number of atoms is $N=\int d\vec{r}|\psi(\vec{r})|^2$

Provided essentially all atoms are in the condensate (that is, have condensed to the ground state), and treating the bosons using mean field theory, the energy (E) associated with the state $\psi(\vec{r})$ is:

Minimising this energy

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November 25, 2009 in News , things.hobby-site.com | Tags: things.hobby-site.com | No comments