Lecture 9: Superconductor DiamagnetismClean BCS Superconductor Diamagnetism at T = 0General ConsiderationsCalculation of RBCS Diamagnetism at Finite TemperaturesSuperconductors with Vanishing GapsCoherence FactorsUltrasonic AttenuationNuclear Spin Relaxation Rate (1/T1)MIT OpenCourseWarehttp://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 res ponse 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 field, current always flows in the direction of the applied field. As a result, the paramagnetic current response tensor Rµν = −i � 0 | jµp(�q, t), jνp(−�q, 0) | 0 � (9.1) is diagonal for an isotropic system. Within the context of line ar response theory, this definition of Rµν yields �jµp� = −Rµν Aν (� (9.2)q, ω) The total current also includes the diamagnetic piece �jµ� = �jµp� + �jµd� (9.3) 2 �jµd� = neAµ (9.4)2− mc arising from the A�A�term in the Hamiltonian. · Combining these terms, the total current to linear order in Aν is given by �jµ� = −Kµν (� q, ω)q, ω)Aν (� (9.5) with total current response tensor 2neKµν = Rµν + δµν (9.6) mc2 2For normal metals, the constant diamagnetic current − ne 2 Aµ is exactly cancelled by part mc of the paramagnetic current. In a superconductor, however, this piece of the current survives. We begin by calculating the (diagonal) paramagnetic current response Rµµ of the BCS ground state. � 1 1 Rµµ = |� n |jµp(�q)| 0 �|2 ω − (En − E0) + iη − ω + (En − E0) + iη (9.7) n 1sad +,# (e:9) is the mmti0r.1 (cistram) optmtor bm en dm witb mamh d and lpin fT - - - ?"he hx& of^^ (9.g) is ia& the free ehwbr~n* mww, r&%h t;o ther &emid p~t~flW p. The secad term fmmmmu geld ap~p*-~d the lam dmtr11&011 [erWn) apeti&m of the four agemtot bl&aW%ian fl, ~~ q, we am change fsom khe single pw%& mommtum~Wte bds to the BqpEuf4ov q&pti& basis atat- defined by the [unitary) &-ion� � � � � � Clean BCS Superconductor Diamagnetism at T = 0 3 with E� = ξ 2 + |Δ|2 (9.18)k �k 1 ξ� u� 2 = 1 + k (9.19)k 2 E� � k �1 ξ�v�k 2 =2 1 − Ek� (9.20) k Thus the BCS Hamiltonian is diagonalized by the Bogliubov states with spectrum {E�k}. 9.1.3 Calculation of Rµµ 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, jp(�q) = − ekµ + qµ c† c� (9.21)µm 2 �k+� k,σ q,σ �k,σ Expanding out the sum over spins σ and letting �k → −(�k + q�) for the σ = ↓ terms, we get �� �� � jµp(�q) = − ekµ + qµ c�† k+�qc�k↑ − c† k c −(�k+q�)(9.22) m �k 2 ↑ −� ↓↓ Now, we need to transform into the Bogoliubov basis by substituting relations (9.15) and (9.16) for c� and c† . To do so, we will also need to make use of the Hermitian conjugates of k↑−�k↓(9.15) and (9.16): c�† k = u�kγ�k † + v�k ∗γ −�k (9.23) ↑ ↑ ↓ c −� = − v� γk † + u� γ −� (9.24)k k� k k↓ ↑↓ Inserting these relations, we get � �� � c�† k+� c�k = u�k+�q γ�k† +� + v�k∗ +�q γ −(�k+�q)u�kγ�k + v�kγ† k(9.25)q↑↑ q↑ ↓ ↑−� ↓ = (u�k+q�u�k)γ�† γ�k + (v�kv� ∗ )γ k+�q)γ† k+q�↑↑ k+�q −(� ↓−�k↓ + (u� v� )γ† γ† + (u� v� ∗ )γ γ�k+q� k�k+q� −�kkk+�q −(�k+q�) k ↑ ↓ ↓ ↑ = (u� u )γ† γ� � v∗ )γ† γ + (v� v∗ , γ†k+q��k�k+q�k↑ − (vk �k+q� −�k −(�k+�q) k �k+q�){γ −(�k+q�)−�k }↑ ↓↓ ↓↓+ (u�k+q�v�k)γ�† γ† − (u�kv�k∗ +q�)γ�k γ −(�k+�q )k+q�↑ −�k↓ ↑ ↓ and � �� � c† c = + u� γ† γ† + u� γ (9.26)−�k↓−(�k+�q)↓ −v�k ∗γ�k↑ k −�k↓ −v�k+�q �k+�q↑ k+�q −(�k+q�)↓ = (v�k ∗v�k+q�)γ�k↑ γ�k† +q�↑ − (v�k ∗u�k+q�)γ�k↑ γ −(�k+�q)↓Substituting he back inta (9.22) and -cSq our &dm of {w6] G lfi* we get The ma- tz,~ d a,$+$ we Imam u w- &dw& N& that term in eqmtim (9.30) cme-vm total &pin projediim dmg tee & d b the effect ofiame&hg tMnet~dm~trmby~. At T = 0, the %a groundstats ha no qmdpmtide emiWions. Thus The only th~ stmi- is the tenn,eontsining the double neation qxratm 7&f17tD, YidW double exdation TkMqLn. lntheDCMt, w=Oand� � �� BCS Diamagnetism at Finite Temperatures 5 since the largest contribution comes from the smallest energy excitations, which have energy 2Δ. In the �q → 0 limit, however, p�k,�k+q� → 0, which gives 0 Rµµ(ω = 0, � q → 0) −→ 2Δ = 0 (9.34) Inserting this into the expression for the total current response tensor Kµν , we get 2 2ne ne Kµν (ω = 0, � q → 0) = Rµν (ω = 0, � q → 0) + mc2 δµν = 0 + mc2 δµν (9.35) Thus in a sup erconductor, the diamagnetic current survives in contrast to the cancellation that occurs for a normal metal As a result, in this limit at T = 0 2 ��j � ne A� (9.36)= − mc2 This is a curious result, as the specific form of A� is gauge dependent. In fact, the result is only true in the London Gauge in which � · A�= 0. How did this choice of gauge creep into our derivation? It is possible that the gap Δ depends on A�. In this case, we would have to solve a new self-consistent BCS equation in the presence of the altered form. According to rotational invariance, any correction to Δ must take the form Δ = Δ0 + �c A� (9.37)· where �c is some ve ctor relevant to the system. In the present case, the only relevant vector is �q. By choosing the London Gauge � · A�= 0 we ensure that q� A� = 0, thus guarantying the validity · of the result just derived. 9.2 BCS Diamagnetism at Finite Temperatures At finite temperatures T > 0, the quasiparticle state populations will in general