Spin-statistics theorem

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Spin-statistics theorem As we discussed in P301, all sub-atomic particles with which we have experience have an internal degree of freedom known as intrinsic spin, which comes in integral multiples of hbar/2 (i.e. h/4p, so it has dimensions of angular momentum). The value of this spin has remarkably powerful consequences for the behavior of many-body systems: FERMIONS (odd-integer multiple of hbar/2=s=hbar/2; 3hbar/2; 5hbar/2 etc.) Y F (x 1 ,x 2 ) = - Y F (x 2 ,x 1 ) BOSONS (even-integert multiple of hbar/2=s=0, hbar, 2*hbar, 3*hbar etc. Y B (x 1 ,x 2 ) = Y B (x 2 ,x 1 ) This connection between the intrinsic spin of the

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Page 1: Spin-statistics theorem

Spin-statistics theoremAs we discussed in P301, all sub-atomic particles with which we have experience have an internal degree of freedom known as intrinsic spin, which comes in integral multiples of hbar/2 (i.e. h/4p, so it has dimensions of angular momentum). The value of this spin has remarkably powerful consequences for the behavior of many-body systems:

FERMIONS (odd-integer multiple of hbar/2=s=hbar/2; 3hbar/2; 5hbar/2 etc.)

YF(x1,x2) = - YF(x2,x1)

BOSONS (even-integert multiple of hbar/2=s=0, hbar, 2*hbar, 3*hbar etc.

YB(x1,x2) = YB(x2,x1)

This connection between the intrinsic spin of the particle and the “exchange symmetry” of the many-body wavefunction is known as the spin-statistics theorem. We won’t try to prove it (it comes out of relativistic quantum field theory), but over the next couple of weeks we will look at some of its important consequences.

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Spin-statistics theoremAs we discussed in P301, all sub-atomic particles with which we have experience have an internal degree of freedom known as intrinsic spin, which comes in integral multiples of hbar/2 (i.e. h/4p, so it has dimensions of angular momentum). The value of this spin has remarkably powerful consequences for the behavior of many-body systems:

FERMIONS (odd-integer multiple of hbar/2=s=hbar/2; 3hbar/2; 5hbar/2 etc.)

YF(x1,x2) = - YF(x2,x1)

Pauli Exclusion Principle: NOTE That an important consequence of the above symmetry exchange requirement on fermion many-body wave functions is that you cannot have two fermions or more in the same one-body state. Thus fermion occupation numbers are limited to the values 0 or 1.

In the limit of very low temperature, essentially all states up to some value of the energy will be occupied (<na>=1), and those above this energy will be empty. We call the energy level at which the occupation number changes from 1 to zero the Fermi Energy (EF).

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Degenerate Fermi Gas/liquid

<n>

EEf

1

T=0

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Degenerate Fermi Gas/liquid

<n>

Em

1

T=0T > 0

At extremely low temperatures, the chemical potential is essentially equal to the Fermi Energy, but as T increases from zero, the two differ slightly.

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CALM: Lithium metal• Consider a small piece of isotopically pure liquid 7Li metal. Both the

7Li nuclei, and the conduction electrons in this material are fermions with net spin hbar/2, and let's assume we can treat them as ideal gases rather than liquids. Do either the nuclei or the electrons form a degenerate Fermi gas at 500K? Explain any difference between the two gases in this regard.• Yes for both (6 responses)• Yes for Nuclei but no for electrons (1 response)• Yes for Electrons but not for Nuclei (3 responses)

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CALM: Lithium metal• Consider a small piece of isotopically pure liquid 7Li metal. Both the

7Li nuclei, and the conduction electrons in this material are fermions with net spin hbar/2, and let's assume we can treat them as ideal gases rather than liquids. Do either the nuclei or the electrons form a degenerate Fermi gas at 500K? Explain any difference between the two gases in this regard.• Yes for both (6 responses)• Yes for Nuclei but no for electrons (1 response)• Yes for Electrons but not for Nuclei (3 responses)

• The Fermi temperature quoted for metals (e.g. in table 9.1) is that of the electrons in those metals. Always keep in mind that the Fermi Energy depends on the mass of the Fermion (why?), and much more massive particles have much smaller Fermi energies for a given density.

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CALM: Lithium metal• Consider a small piece of isotopically pure liquid 7Li metal. Both the

7Li nuclei, and the conduction electrons in this material are fermions with net spin hbar/2, and let's assume we can treat them as ideal gases rather than liquids. Do either the nuclei or the electrons form a degenerate Fermi gas at 500K? Explain any difference between the two gases in this regard.• Yes for both (6 responses)• Yes for Nuclei but no for electrons (1 response)• Yes for Electrons but not for Nuclei (3 responses)

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Example• Estimate the ratio of the lattice (phonon) and electronic contributions to

the specific heat of Copper at 300K and 1K. For copper: qD=343K and TF=81600K.

• You may wish to recall the behaviour of the lattice contribution as shown in the figure below

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White Dwarfs The figure below comes from Wikipedia and shows the relationship between the mass and radius of a White Dwarf star. Determine the Fermi temperature for the electrons in such a star with a mass of 0.2 MSun, assuming its composition to be Carbon and Oxygen in equal measure, and confirm that the electrons in these stars are degenerate and that relativity is marginally important in this case.

http://en.wikipedia.org/wiki/White_dwarf

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White Dwarfs Examine the mechanical equilibrium of a White Dwarf by considering how the gravitational and hydrodynamic potential energies vary with the radius of the star. For the hydrodynamic energy, assume that this is dominated by the degeneracy pressure from the electrons. At equilibrium how should the mechanical energy of the system vary with a virtual change in radius dR? Use this result to determine how the radius of such a star must vary with its mass (assuming a non-relativistic model is adequate).

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Laser Cooling/Trapping of atoms

We will discuss this in more detail toward the end of the semester, but it is possible to slow-down (cool) atoms by passing them through a region with counter-oriented laser beams tuned to just below an optical transition [“optical molassses”; so that atoms moving toward the laser will see photons Doppler shifted onto the resonance and absorb the photon (along with its momentum), and other atoms will interact minimally with the photons]. Using this technique, along with simple evaporative cooling, you can get VERY COLD gases of atoms in an optical/magnetic trap

http://www.ptb.de/en/org/4/44/443/melcol_e.jpg

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U.C. Boulder group’s BEC

This is a map of the momentum distribution in the gas cloud (measured by looking at the gas after the trap has been turned off) for various temperatures (yes, it is O(10-7 K). The sharp white peak in the middle is the BEC showing up at ~ 200 nK

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Bose-Einstein Condensation

There is a great applet at this web sit:http://www.colorado.edu/physics/2000/bec/evap_cool.html

You can slowly cool a model gas down and get BEC, or cool it down too quickly and just have all the gas leave your trap (you can even stop the gas in some state and then have it reestablish equilibrium in another smaller trap). The site shows the temperature of the gas, you can watch it cool as some molecules (the most energetic) leave the trap, and it shows the temperature at which BEC would be expected for the density in the trap.See also the movie at:

http://www.colorado.edu/physics/2000/bec/images/evap2.gif

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Another interesting applet (more for chapter 14 when we talk about cooling)Is at the related website:

http://www.colorado.edu/physics/2000/bec/lascool4.html