MIT OpenCourseWare http ocw mit edu 8 512 Theory of Solids II Spring 2009 For information about citing these materials or our Terms of Use visit http ocw mit edu terms Lecture 9 Superconductor Diamagnetism In this lecture we will apply linear response theory to the diamagnetism of a clean BCS super conductor 9 1 Clean BCS Superconductor Diamagnetism at T 0 9 1 1 General Considerations Based on symmetry arguments it is easy to see that when an isotropic system is placed in an external eld current always ows in the direction of the applied eld As a result the paramagnetic current response tensor 9 1 R i 0 j p q t j p q 0 0 is diagonal for an isotropic system Within the context of linear response theory this de nition of R yields j p R A q 9 2 The total current also includes the diamagnetic piece j j p j d 9 3 2 j d ne A mc2 9 4 A term in the Hamiltonian arising from the A Combining these terms the total current to linear order in A is given by j K q A q 9 5 with total current response tensor K R ne2 mc2 9 6 2 ne For normal metals the constant diamagnetic current mc 2 A is exactly cancelled by part of the paramagnetic current In a superconductor however this piece of the current survives We begin by calculating the diagonal paramagnetic current response R of the BCS ground state 1 1 p 2 n j q 0 R 9 7 En E0 i En E0 i n 1 sad e 9 is the mmti0r 1 cistram optmtor bm en d m witb m a m h d and lpin fT of 9 g is ia the freeehwbr n mww r ht o ther emid he hx fmmmmugeld ap p d the lam dmtr11 011 erWn p t flW p The secad term apeti m of the four agemtot bl aW ian fl q we am change fsom khe single pw m o m m t u m W t b e d s to basis atat defined by the unitary the BqpEuf4ov q p t i o in Clean BCS Superconductor Diamagnetism at T 0 3 with u k2 v k2 2 2 k k 1 1 2 E k k 1 1 2 E k E k 9 18 9 19 9 20 Thus the BCS Hamiltonian is diagonalized by the Bogliubov states with spectrum E k 9 1 3 Calculation of R To evaluate the matrix element in equation 9 7 we need the explicit form of the paramagnetic current operator In second quantized notation in terms of the basis of single particle momentum eigenstates e q j p q k c ck 9 21 q k m 2 k Expanding out the sum over spins and letting k k q for the terms we get e q j p q k c c k c c k q 9 22 k q k m 2 k Now we need to transform into the Bogoliubov basis by substituting relations 9 15 and 9 16 for c k and c To do so we will also need to make use of the Hermitian conjugates of k 9 15 and 9 16 c u k k c k k v k k v k k 9 23 u k k 9 24 Inserting these relations we get c c k u k q v k q k q u k k v k k q 9 25 k q k u k q u k k v k v k q k q k k q u k q v k u k v k q k q k k k q u k q u k k v k v k q k q v k v k q k q k k k q u k q v k u k v k q k k q k q k and c c k q k v k k u k v k q k v k v k q k k q k q u k q k q v k u k q k k q 9 26 Substituting h e back inta 9 22 and c S q our dm of w6 G lfi we get t z d a we Imam u w dw N that term in eqmtim 9 30 cme vmtotal pinprojediim dmg tee d b the effect ofiame hg The ma t M n e t d m t r m b y At T 0 the agroundstats ha no qmdpmtide emiWions Thus The only t h stmi YidW TkMqLn double exdation lntheDCMt w O a n d is the tenn eontsiningthe double neation qxratm 7 f17tD BCS Diamagnetism at Finite Temperatures 5 since the largest contribution comes from the smallest energy excitations which have energy 2 In the q 0 limit however p k k q 0 which gives R 0 q 0 0 0 2 9 34 Inserting this into the expression for the total current response tensor K we get K 0 q 0 R 0 q 0 ne2 ne2 0 mc2 mc2 9 35 Thus in a superconductor the diamagnetic current survives in contrast to the cancellation that occurs for a normal metal As a result in this limit at T 0 j ne2 A mc2 9 36 is gauge dependent In fact the result is This is a curious result as the speci c form of A only true in the London Gauge in which A 0 How did this choice of gauge creep into our derivation In this case we would have to solve a new self It is possible that the gap depends on A consistent BCS equation in the presence of the altered form According to rotational invariance any correction to must take the form 0 c A 9 37 where c is some vector relevant to the system In the present case the only relevant vector is q 0 we ensure that q A 0 thus guarantying the validity By choosing the London Gauge A of the result just derived 9 2 BCS Diamagnetism at Finite Temperatures At nite temperatures T 0 the quasiparticle state populations will in general be nonzero Thus we expect all four terms of equation 9 30 to contribute to the matrix element in 9 7 Pulling all of these terms together we get f E k q f E k e2 q 2 R q 2 k 2 2k k q 9 38 m 2 E k E k q i k f E f E 1 f E f E 1 k q k k q k 2 p k k q E E E k E k q i k k q i If we now let 0 and take the limit q 0 p k k q 0 as before but k k q 1 since u k 2 v k 2 1 This yields k 2 f R q 0 0 2 e 0 m E k …