Consider a gas of non-interacting, non-relativistic, identical bosons. Explain whether and why t

Consider a gas of non-interacting, non-relativistic,...

Key Concepts

Bose-Einstein Condensation (BEC)

Bose-Einstein condensation is a phase transition whereby a significant fraction of bosons occupies the ground state at temperatures below a critical threshold. This macroscopic occupation is a result of the quantum statistical properties of bosons, which allow multiple particles to exist in the same state, leading to collective quantum phenomena on a large scale.

Dimensionality and Phase Transitions

The spatial dimensionality of a system has a profound effect on the nature and existence of phase transitions. In three dimensions, thermal fluctuations are sufficiently suppressed allowing long?range order (such as a condensate) to form. In contrast, in lower dimensions such as two or one, enhanced fluctuations prevent the establishment of long-range order in an infinite, homogeneous system, thus typically inhibiting true Bose-Einstein condensation in the thermodynamic limit.

Density of States

The density of states determines how many quantum states are available at each energy level and varies with the dimensionality of the system. In three-dimensional systems, the energy dependence of the density of states is favorable for a macroscopic occupancy of the ground state, which is a key requirement for Bose-Einstein condensation. However, in one-dimensional and two-dimensional systems, the density of states does not support the required divergence at low energies, which is why BEC is not observed under the same conditions.

Transcript

00:01 So the first step is that we have three particles in the harmonic oscillator.

00:17 The total energy of the system is e is equal to 9h omega divided by 2.

00:24 And we need to find all the possible configurations of the particle if they are distinguishable, fermions or bosons.

00:31 The total energy of the system in the harmonic potential is, given by e is equal to age omega n1 plus n2 plus n3 plus 3 divided by 2 since e is equal to this so we are going to substitute it over here we get that 9h omega divided by 2 is equal to edge omega n1 plus n2 plus n3 plus 3 by 2 simplifying it a little bit we get 9 by 2 is equal to n1 plus n2 plus n3 plus 3 by 2 and when we'll solve it for n1 plus n2 plus n3 is equal to 3 we'll get the possible configurations would be n1 is equal to n2 is equal to n3 is equal to 1 so it means that we would have 1 -11 then the other would be n1 is equal to 0 n2 n2 is equal to 1 n3 is equal to 2 so the possible configurations would be 0 1 -2 -20 -20 -0 -2 -0 -21 -2 -1 -2 -1 -2 -1 -2 -1 -0 -0 -1 -0 -0 -0 -0 -0 -0 -0.

02:01 Then we would have 1 -2 -1 -2 and over here we would have 1 -n1 is equal to n -n2 is equal to n2 is equal.

02:18 Equal to 0 and 3 is equal to 3 and possible configurations would be 0 030 3 0 0 0 0.

02:35 Fine so now the part a is the distinguishable particles so we are interested in occupation number configurations so number one would be all three particles, three particles in state one, that is 0 ,3 ,000, 0 .00.

03:09 Find number two is that all three particles in different states, different states.

03:20 So that would be 1 ,1, 1 ,0, 0, 0, and number 3 would be 2, 2, 2, 2, 2.

03:29 Two particles in ground state and one in third.

03:37 So we would get two zero zero one zero.

03:48 So the second configuration, the second configuration is the most probable one because there are six possibilities.

04:01 And for it to happen, the first one has only one possibility and the third one has only two.

04:06 Three possibilities.

04:08 Fine, this first one has only one possibility and this third one has only three possibilities.

04:14 The second one has six possibilities.

04:16 So this is the most probable one...

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