Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced. For instance, the particles could be trapped in a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles.
(a) Describe the ground state of this system, for each of these three cases.
(b) Suppose that the system has one unit of energy (above the ground state). Describe the allowed states of the system, for each of the three cases. How many possible system states are there in each case?
(c) Repeat part (b) for two units of energy and for three units of energy.
(d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero). In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss.
(a) The ground state of this system is.
(b) The are five possible system states in each case.
(c) For two-unit of energy the graph is
and for three-unit of energy is
(d) The way that the behavior of the bosonic system differs from that of the system of distinguishable particles is discussed below.
We have to describe the ground state of the system, for each of these three cases.
As bosons do not follow the Pauli exclusion principle, particles in the ground state on the same level are distinguishable, but if they are fermions, each one will occupy a level starting from the lowest level, resulting in something like this:
We have to describe the allowed states of the system, for each of the three cases and find the possible system states in each case.
Consider a particle that has been promoted to the second lowest level, one energy unit above the ground state. There is only one way to do this for identifiable particles or bosons, which is to promote one particle from the lowest level to the second lowest level. There are five ways to promote indistinguishable particles or fermions, and the particle is promoted from the fifth level to the sixth level, as represented graphically:
We have to repeat part (b) for two units of energy and for three units of energy.
Consider a particle promoted to the second lowest level with two energy units above the ground state. We can promote one particle up to two energy levels, leaving four particles in the first level, or we can promote two particles to the second lowest level. There is only one way to do this for either, which is to promote one particle from the lowest level to the second lowest level. There are ten ways to do the first arrangement (where the two particles are promoted to the second and third lowest levels above the last filled level) and five ways to do the second arrangement (where the last particle is promoted to the third level above the last filled level) for indistinguishable particles or fermions, as shown graphically:
Consider a particle promoted to the second lowest level with three energy units above the ground state. We can promote one particle up to three energy levels for distinguishable particles or bosons, leaving four particles in the first level, or promote one particle to the second lowest level and one to the third lowest level, or promote three particles to the second lowest level, but there is only one way to do this for each. There are ten ways to make the first arrangement, twenty ways to do the second arrangement, and five ways to do the third arrangement for indistinguishable particles or fermions, and these arrangements are depicted graphically as:
We have to find the behavior of the bosonic system differ from that of the system of distinguishable particles.
The Boltzmann factor is proportional to the likelihood that a system with a temperature of , i.e.
The energy of any system is the same, therefore this factor is the same, but when we factor in degeneracy (from the previous section, the degeneracy of the bosons is 3 and the degeneracy of the fermions is 35), we can see that at low temperatures, the fermions are more likely to be found than the bosons. The bosons are also likely to be discovered in the ground state.
A star that is too heavy to stabilize as a white dwarf can collapse further to form a neutron star: a star made entirely of neutrons, supported against gravitational collapse by degenerate neutron pressure. Repeat the steps of the previous problem for a neutron star, to determine the following: the mass radius relation; the radius, density, Fermi energy, and Fermi temperature of a one-solar-mass neutron star; and the critical mass above which a neutron star becomes relativistic and hence unstable to further collapse.
An atomic nucleus can be crudely modeled as a gas of nucleons with a number density of (where ). Because nucleons come in two different types (protons and neutrons), each with spin , each spatial wavefunction can hold four nucleons. Calculate the Fermi energy of this system, in MeV. Also calculate the Fermi temperature, and comment on the result.
A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are atoms, the total magnetization is typically, where µa is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure . Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of and therefore reduces the magnetization by . However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin-wave is proportional to the square of .. (in the limit of long wavelengths). Therefore, since and .. for any "particle," the energy of a magnon is proportional
In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy because the difference in direction between neighboring dipoles is very small.
to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write *, where* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal , about times the mass of an electron. Another difference between magnons and phonons is that each magnon ( or spin-wave mode) has only one possible polarization.
(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by
Evaluate the integral numerically.
(b) Use the result of part () to find an expression for the fractional reduction in magnetization, Write your answer in the form , and estimate the constant for iron.
(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find , where differs from To only by a numerical constant. Estimate for iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is .)
(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverge in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section we will consider a different two-dimensional model in which magnetization does occur.)
In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occupied. To model this effect in a simple way, imagine that a hemoglobin molecule has just two sites, either or both of which can be occupied. This system has four possible states (with only oxygen present). Take the energy of the unoccupied state to be zero, the energies of the two singly occupied states to be , and the energy of the doubly occupied state to be (so the change in energy upon binding the second oxygen is ). As in the previous problem, calculate and plot the fraction of occupied sites as a function of the effective partial pressure of oxygen. Compare to the graph from the previous problem (for independent sites). Can you think of why this behavior is preferable for the function of hemoglobin?
In the text I claimed that the universe was filled with ionised gas until its temperature cooled to about 3000 K. To see why, assume that the universe contains only photons and hydrogen atoms, with a constant ratio of 109 photons per hydrogen atom. Calculate and plot the fraction of atoms that were ionised as a function of temperature, for temperatures between 0 and 6000 K. How does the result change if the ratio of photons to atoms is 108 or 1010? (Hint: Write everything in terms of dimensionless variables such as t = kT/I, where I is the ionisation energy of hydrogen.)
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