# hw4.sol - Physics 240B Spring 2008 Homework 4 Solutions...

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Unformatted text preview: Physics 240B, Spring 2008 Homework 4 Solutions Jesse Noffsinger May 1, 2008 Solutions are in general not the original work of the author. Problem 1) a) Calculate the mean square radius of a Cooper pair: The Cooper pair wavefunction is given by: Ψ( r ) = summationdisplay k a ( k ) e ik · r (1) so the mean square radius is: ( r 2 ) = integraltext r 2 | Ψ | 2 dr integraltext | Ψ | 2 dr (2) The denominator is given by: integraldisplay | Ψ | 2 dr = summationdisplay k,k ′ a † ( k ′ ) a ( k ) integraldisplay e i ( k − k ′ ) · r dr = summationdisplay k | a ( k ) | 2 (3) For the numerator we use a trick: ∇ k Ψ( r ) = 0 = ∇ k parenleftBigg summationdisplay k ′ a ( k ′ ) e ik ′ · r parenrightBigg = e ik · r [ ∇ k a ( k ) + ira ( k )] (4) Upon summing over k we get r Ψ( r ) = i summationdisplay k e ik · r ∇ k a ( k ) (5) and therefore integraldisplay r 2 | Ψ | 2 dr = summationdisplay k,k ′ ∇ k ′ a † ( k ′ ) ∇ k a ( k ) integraldisplay e i ( k − k ′ ) · r dr = summationdisplay k |∇ k a ( k ) | 2 (6) and finally ( r 2 ) = ∑ k |∇ k a ( k ) | 2 ∑ k | a ( k ) | 2 ≈ integraltext ∞ dǫN ( ǫ ) vextendsingle vextendsingle vextendsingle ∂a ∂ǫ ∂ǫ ∂k vextendsingle vextendsingle vextendsingle 2 integraltext ∞ dǫN ( ǫ ) | a | 2 ≈ N (0) parenleftBig ∂ǫ ∂k parenrightBig 2 integraltext ∞ dǫ parenleftBig ∂a ∂ǫ parenrightBig 2 N (0) integraltext ∞ dǫa 2 (7) 2 Using the approximation V k,k ′ = − V Ω for ǫ k ,ǫ k ′ < ǫ D (8) we get a ( k ) = C E − 2 ǫ k , and parenleftbigg ∂a ∂ǫ parenrightbigg 2 = 4 C 2 ( E − 2 ǫ ) 4 , a 2 = C 2 ( E − 2 ǫ ) 2 (9) Also, parenleftBig ∂ǫ ∂k parenrightBig 2 = 2¯ h 2 ǫ F m = ¯ h 2 v 2 F , and then ( r 2 ) = 4¯ h 2 v F integraltext ∞ dǫ ( E − 2 ǫ ) 4 integraltext ∞ dǫ ( E − 2 ǫ ) 2 = 4¯ h 2 v 2 F 1 6( E − 2 ǫ ) 3 | ∞ 1 2( E − 2 ǫ ) | ∞ = 4 3 ¯ h 2 v 2 F E 2 (10) and radicalBig ( r 2 ) = 2 √ 3 ¯ hv F | E | = ¯ hv F √ 3 ǫ D bracketleftbigg e 2 N (0) V − 1 bracketrightbigg ≈ ¯ hv F √ 3 ǫ D e 2 N (0) V for N (0) V ≪ 1 (11) So radicalBig ( r 2 ) ∼ ¯ hv F Δ ∼ ξ = coherence length (12) b) Let ¯ hk cm denote the center of mass momentum. Then the Schrodinger equation gives: bracketleftBigg − ¯ h 2 2 m ( ∇ 2 1 + ∇ 2 2 ) + V bracketrightBigg Ψ pair = ( E + 2 ǫ F )Ψ pair (13) or [ E k + k cm / 2 + E − k + k cm / 2 ] a k + summationdisplay k ′ a k ′ V kk ′ = ( E + 2 ǫ F ) a k (14) where E = binding energy of the pair and E k = ¯ h 2 k 2 2 m . As in class let’s define ǫ k = E k − ǫ F . Then we obtain a k = Assuming V kk ′ = − V Ω for 0 ≤ E ≤ ¯ hω D we obtain the Cooper pair equation 1 = − V Ω summationdisplay k ′ but ( ǫ k + k cm / 2 − ǫ − k + k cm / 2 ) = 2 ǫ k + hbar 2 4 m k 2 cm , so, to first order in k cm , the denominator is given by E − 2 ǫ k , which is the same as that with k cm = 0....
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hw4.sol - Physics 240B Spring 2008 Homework 4 Solutions...

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