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PS8Sol_f10

# PS8Sol_f10 - Physics 112 Fall 2010 Professor William...

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Physics 112 Fall 2010 Professor William Holzapfel Homework 8 Solutions Problem 1, Kittel 7.1: Density of Orbitals in one and two dimensions In one dimension the orbitals are of the form ψ n ( x ) = A sin( nπx/L ) where n is a positive integer. The energy is n = 2 2 m πn L 2 (1) and the density in n space is simply 2 dn , since there is one state per unit interval (the 2 comes from the spin degeneracy). We can invert (1) to get n ( ) = L π 2 m = dn = L 2 π 2 m d . Since the density of states D ( ) is defined by D ( ) d 2 dn , we have D ( ) = L π 2 m . (b) Now the orbitals are of the form ψ n ( x ) = A sin( n x πx/L ) sin( n y πy/L ), where n x , n y > 0 and the energy has the same form (1), but with n = n 2 x + n 2 y . The density of states in n -space is 1 4 · 2 · 2 πn dn = πn dn (2) where the 1/4 is because we’re only interested in the first quadrant, the 1/2 is for spin and the 2 πndn is just the area element in 2D n -space in polar coordinates. As above, we have D ( ) d πn dn = π L π 2 m · 1 2 L π 2 m d = mL 2 π 2 d = D ( ) = mL 2 π 2 . Problem 2, Kittel 7.2: Energy of a Relativistic Fermi Gas (a) For a relativistic gas in 3D the orbitals are still of the form ψ n ( x, y, z ) = A sin( n x πx/L ) sin( n y πy/L ) sin( n z πz/L ) but now the energy of the orbitals takes the form n = | p | c = π c L n (3) 1

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where n = n 2 x + n 2 y + n 2 z . We can still, however, use Kittel’s result that in n -space the occupied modes fill up a sphere of radius n F = 3 N π 1 / 3 (4) since this result is independent of the dispersion relation (3). The Fermi energy is then just F n F = π c L n F = πc 3 n π 1 / 3 . (b) Now using (3) and (4) the ground state energy is easily found: U 0 = 2 × 1 8 n F 0 4 πn 2 n dn = π 2 c L n F 0 n 3 dn = π 2 c 4 L 3 N π 4 / 3 = 3 4 N F . Problem 3, Kittel 7.5: Liquid 3 He as a Fermi Gas (a) The Fermi velocity v F is defined in terms of the Fermi energy: F 1 2 mv 2 F . To find the Fermi energy, we need to know the number density; we can find this from the given mass density of 3 He : n = ρ m M 3 He = 0 . 081 g / cm 3 3 amu = 0 . 081 g / cm 3 3 · 1 . 66 × 10 - 24 g = 1 . 63 × 10 22 1 cm 3 The Fermi velocity is then v F = 2 f m = v F = 2 2 2 m (3 π 2 n ) 2 / 3 m = m ( 3 π
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PS8Sol_f10 - Physics 112 Fall 2010 Professor William...

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