Question

Determine the ground electronic state of the BeC molecule. Explain your answer.

The answer requires the molecular orbital diagram as well as an electron configuration.

Answer 1

Figure 1. Schematic showing two-color formation of molecules in their ground electronic state by Raman Photoassociation. In a BEC this process is initially coherent and leads to a wavefunction that is both atoms and moleculesWhen a system enters the BEC phase a sizable fraction of the total number of particles enters the ground state, until at some point almost all of your particles are in the ground state. In reality you almost always find that the particles prefer to go to the same state because of some tiny energy shifts in the system. For example, a ferromagnetic condensate can have many degenerate states, but there is an energy cost for particles that disagree. These systems will break symmetry by having all the particles choose the same (arbitrary) state. One exception is if the number of degenerate states is large, for instance larger than the number of particles. I have heard that claim that if the system has a sufficiently large degeneracy, for instance if all momentum states has the same energy, Bose-Einstein condensation might not occur. Other than that, if there really is no energy cost to arranging your particles however you want between two degenerate states, I would guess that the particles would on average be equally distributed. However, fluctuations should be enormous: the probability of any configuration (N atoms in state 1, M atoms in state 2) should all be equal. To be more specific about the questions in terms of the nature of the BEC state. Consider just a single atom. It is described by a wave function. As the atom is placed in a cooler environment it loses energy. This loss of energy changes the shape of the wave function. It makes the wave function bigger. The wavelength for the wave function of a gas atom is called the Thermal debroglie wavelength. The formula is: ?T=h2?mkT??????? where h is the Planck constant, k is the Boltzmann constant, T is the temperature, and m is the mass of the gas atom. The thing to note about the above formula is that as the temperature gets lower, the wavelength ?T gets bigger. When you reach the point that the wavelengths are larger than the distances between atoms, you have a BEC. The reference book I've got is "Bose-Einstein Condensation in Dilute Gases" by C. J. Pethick and H. Smith (2008) which has, on page 5. "An equivalent way of relating the transition [to BEC] temperature to the particle density is to compare the thermal de Broglie wavelength ?T with the mean interparticle spacing, which is of the ordern?1/3. ... At high temperatures, it is small and the gas behaves classically. Bose-Einstein condensation in an ideal gas sets in when the temperature is so low that ?T is comparable to n?1/3." When the wavelengths are longer than the inter-particle distances, the combined wave function for the atoms (which one must symmetrize by the rules of QM) becomes "coherent". That is, one can no longer treat the wave functions of different atoms as if they were independent. To illustrate the importance of this, let's discuss the combined wave function of two fermions that are widely separated with individual wave functions ?1(r1) and ?2(r2). For fermions, the combined symmetrized wave function is: ?(r1,r2)=(?1(r1)?2(r2)??1(r2)?2(r1))/2?. Now the Pauli exclusion principle says that exchanging two fermions causes the combined wave function to change sign. That's the purpose of the "-" in the above equation; swapping r1 for r2 gives you negative of the wave function before the change. Another, more immediate, way of stating the Pauli exclusion principle is that you can't find two fermions in the same position. Thus we must have that ?(ra,ra)=0 for any fermion wave function where ra can be any point in space. But for the case of the combined wave function of two waves that are very distant from one another there is no point ra where both of the wave functions are not zero (that is, the wave functions do not overlap). Thus the Pauli exclusion principle doesn't make a restriction on the combined wave function in the sense of excluding the possibility that both electrons could appear at the same point.That was already intrinsically required by the fact that the two wave functions were far apart. Now I used the above argument about "far apart fermions" because the Pauli exclusion principle has a more immediate meaning to a lot of people than the equivalent principle in bosons. The same argument applies to bosons, but in reverse. Bosons prefer to be found near one another, however if we write down the combined wave function for two widely separated bosons, there is still a zero probability of finding both bosons at the same location. Again the reason is the same as in the fermion case: ?1(rb) is only going to be nonzero in places where ?2(rb) is already zero. The bosons are too far apart to interact (in the sense of changing the probability of finding both at the same point from what you'd otherwise expect classically). Another way of saying all this is that in QM, there is no interaction between things except if their wave functions overlap in space. You use the thermal de Broglie wave function to determine how big a wave function has to be. If the atoms are closer than that, they're interacting in the sense of bose (or fermi) condensates.