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: ( r ) = summationdisplay k a ( k ) e ik r (1) so the mean square radius is: ( r 2 ) = integraltext r 2 | | 2 dr integraltext | | 2 dr (2) The denominator is given by: integraldisplay | | 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 ( 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 ( r ) = i summationdisplay k e ik r k a ( k ) (5) and therefore integraldisplay r 2 | | 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 dN ( ) vextendsingle vextendsingle vextendsingle a k vextendsingle vextendsingle vextendsingle 2 integraltext dN ( ) | a | 2 N (0) parenleftBig k parenrightBig 2 integraltext d parenleftBig a parenrightBig 2 N (0) integraltext da 2 (7) 2 Using the approximation V k,k = V for k , k < D (8) we get a ( k ) = C E 2 k , and parenleftbigg a parenrightbigg 2 = 4 C 2 ( E 2 ) 4 , a 2 = C 2 ( E 2 ) 2 (9) Also, parenleftBig k parenrightBig 2 = 2 h 2 F m = h 2 v 2 F , and then ( r 2 ) = 4 h 2 v F integraltext d ( E 2 ) 4 integraltext d ( E 2 ) 2 = 4 h 2 v 2 F 1 6( E 2 ) 3 | 1 2( E 2 ) | = 4 3 h 2 v 2 F E 2 (10) and radicalBig ( r 2 ) = 2 3 hv F | E | = hv F 3 D bracketleftbigg e 2 N (0) V 1 bracketrightbigg hv F 3 D e 2 N (0) V for N (0) V 1 (11) So radicalBig ( r 2 ) hv F = 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 pair = ( E + 2 F ) pair (13) or [ E k + k cm / 2 + E k + k cm / 2 ] a k + summationdisplay k a k V kk = ( E + 2 F ) a k (14) where E = binding energy of the pair and E k = h 2 k 2 2 m . As in class lets define k = E k F . Then we obtain a k = Assuming V kk = V for 0 E h D we obtain the Cooper pair equation 1 = V summationdisplay k but ( k + k cm / 2 k + k cm / 2 ) = 2 k + hbar 2 4 m k 2 cm , so, to first order in k cm , the denominator is given by E 2 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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