Phys 7654 (Basic Training in CMP- Module III)/ Physics 7636 (Solid State II) Homework 1 Solutions

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1 Phys 7654 Basic Training in CMP- Modue III/ Physics 7636 Soid State II Homework 1 Soutions by Hitesh Changani adapted from soutions provided by Shivam Ghosh Apri 19, 011 Ex Phase sips in a wire I: variationa approach a We can put the G-L free energy density in the form 6.4. a,b by noting that, F L Ψ F L Ψ 0 = F L Ψ 0 1 F L Ψ/F L Ψ 0 1 where F L Ψ is given by, F L Ψ = α Ψ + β Ψ 4 For the case where Ψ = Ψ 0 i.e. the uniform case we have, F L Ψ 0 = α Ψ 0 + β Ψ = α β = α Pug in F L Ψ from into 1, and use 5 and = α/β, to get, F L Ψ F L Ψ 0 = F L Ψ 0 1 Ψ β α Ψ Emai: hjc55@corne.edu 1

2 = F L Ψ 0 1 Ψ + Ψ 4 Ψ where F L Ψ 0 = F cond = α /β is the condensation energy. Simiary the gradient part of the free energy density, can be recast into the form, F grad = = F cond 1 Ψ 8 F grad = m Ψ 9 Ψ =. α m β.. 1. β Ψ 10 m α α which has been written to make the famiiar identications of F cond, ξ and Ψ 0 in terms of G-L parameters α and β. Thus we get, F grad = F cond ξ Ψ 11 b We assume a variationa state, Ψ = Ψ 0 ϕ 1 + ϕ x x < = Ψ 0 x > 1 with ϕ = 1 ϕ 1, which satises the boundary condition Ψx = 0/Ψ 0 = ϕ 1. Pugging the expression for Ψ from 1 into 8 we nd F L = F cond 1 ϕ 1 + ϕ x, x < 13 Aso note that F L = 0 for x >. Thus we can compute the free energy change as, F L = ˆ F L dx = F cond ϕ C ϕ 14

3 where C ϕ is a second order poynomia given by ϕ C ϕ = 5 ϕ Now consider the contribution from the gradient term. F grad = F cond ξ ϕ for x < and F grad = 0 for x >. Integrating this part of the free energy density we get, ˆ F grad = F grad dx = 4 F cond ξ ϕ. 1 The tota change in free energy F tota is obtained by adding 14 and 16 F tota = F cond ϕ C ϕ + 4 ξ Minimizing this as a function of the variationa parameter so that F tota = opt = 0, we get the optima ength, opt = ξ 18 C ϕ Evauating the barrier U ϕ F tota opt, we get, U = 4 C ϕ F cond ϕ ξ 19 c OPTIONAL In the presence of a super current the ampitude part of G-L equation remains the same but the gradient part becomes F grad = F cond ξ ie c and aso the super current density J s is given by J s = n s [ θ e A m c ] AΨ 0 we can now express Ψ as an ampitude times a phase part Ψ = Ψ A e iθ and pug into the covariant derivative ie c ie c 1 AΨ = Ψ A e iθ + iψ θ ie AΨ c = Ψ A e iθ + i θ e A c Ψ 3 AΨ = Ψ A e iθ + i m J s Ψ 4 n s 3

4 So, F grad becomes and recognizing Ψ A = n s gives us [ F grad = F cond ξ Ψ Ψ A + 0 m [ Ψ A + ] Js Ψ A F grad Ψ = F cond ξ n s The change in the gradient free energy density can therefore be expressed as, F grad = F grad Ψ F grad Ψ 0 = F cond ξ n s m [ Ψ A + ] Js m J s n s J s n s0 where we have used the fact that the same constant current J s in the case of Ψx = Ψ 0, ensures Ψ A = 0 and θ = const. with n s0 = Note aso that we have not reay worried about A here and assume the eect is sma. Observe that the x > contributions vanish. We can then compute the tota change in free energy as before. The ony new contributions require the cacuation of the integra, which gives us, ˆ ˆ 1 Ψ A dx = 1 ϕ 1 + ϕ x [ F grad = F cond ξ ϕ The tota change can thus be written as, [ F tota = F cond + C ϕ ϕ + 4ξ m dx = 1 ϕ. 1 m We minimize the free energy to get an optima ca it opt,js opt.js = ] J s ϕ Ψ ϕ Ψ 4 0 ] ] Js ϕ + 4ξ ϕ 1 ϕ ϕ C ϕ ϕ + 4ξ m ξ 31 Js ϕ Ψ ϕ putting J s = 0 recovers for us our earier resut in 18. Aso observe that opt,js < opt. Note that U Js ϕ becomes m J U Js = 4 C ϕ ϕ + 4ξ s ϕ F cond ϕξ 3 1 ϕ which is more than U Js=0 in 19! Ψ 4 0 4

5 Ex Phase sips in a wire II: exact soution a The tota Free energy density F tota = F L + F grad is given in 6.4. F tota = F cond ξ Ψ + F L 33 simiar to the Lagrangian of a physica system. We can get the 'equation of motion' Euer- Lagrange equation using Inserting 33 in 34 we get, F tota Ψ F tota Ψ = 0 34 F cond ξ d Ψ dx F LΨ = 0 35 where F L Ψ = df L/dΨ. Aso, it is convenient to choose a abe for the constants infront of the second derivative, c = F cond ξ 36 Mutipying 35 by dψ/dx and integrating we get, ˆ ˆ c dx d Ψ x dx.dψ dx dx df L x dψ.dψ dx = 0 37 c dψ F L Ψ dx H 38 where H is the constant of integration the constant of motion and the factor of 1/ in the rst term comes from integration by parts. We can compare this to Newton's equation of motion energy constraint for a partice in a one dimensiona potentia and make the identication V Ψ = F L Ψ. b We can x the constant of motion by noting that far away from the uctuation Ψ = Ψ 0 and 35 becomes F L Ψ 0 = H 39 using which we can express 38 as c dψ = F L Ψ F L Ψ 0 40 dx c dψ = F cond 1 Ψ 41 dx 5

6 x x Trying a soution of the form Ψx = Ψ 0 tanh 0 and using 1 tanh x = sech x we get, ξ c ξ = F cond 4 which is consistent with the denition of the constant c. x 0 determines the position where the order parameter woud vanish. To obtain x 0 use the fact that Ψx = 0 = Ψ 0 ϕ 1 x0 ϕ 1 = tanh 43 ξ Recovering the x x symmetry we arrive at the function, Ψx = Ψ 0 tanh x + ξ tanh 1 ϕ 1 44 c We can now pug in this soution for Ψx in 6.4. to get F tota, F tota = 3 F cond A ˆ dx sech 4 x x0 ξ 45 making a change of variabes u = x x 0 /ξ and evauating the integra, aows us to write F tota as F tota = 3ϕ 1 + ϕ 3 1 Fcond A ξ 46 Comparing the exact soution with the variationa soution we get, F tota,ex F tota,var = = 3 ϕ ϕ F cond A ξ 47 3 ϕ ϕ 5 ϕ For ϕ is very cose to zero this ratio is 0.9, for ϕ = 0.5 it is 0.94 and for ϕ = 1 it is 15/ indicating that the variationa approximation is rather good. Ex How sma to see phase sips? a We can get the diameter of the wire by equating U B = F cond A ξ to 10k B T c and using F cond = Hc /8π, 3 10 din m = 7.10k B in S.I.T c in K 49 Hc in Tesaξin m 6

7 For Pb, T c = 7.19K,H c = 8 10 Tesa, ξ = m. Using these parameters, we nd a diameter of.44 Angstrom. Beware of conversion units Let us aso compute the energy scaes in question for these parameters, to make sense our numbers and to ensure we have a reasonabe estimate Tc 10k B T c = 10 k B T room 50 T room ev 51 We now compute F cond = H c 8π F cond = H c µ 0 giving us, This demands that, Using the given vaue of ξ we get, which impies that the wire diameter is, This is in Gaussian units. When expressed in Tesa one may use F cond = π J/m ev/m 3 53 A ξ m 3 54 A m 55 d m 56 NOTE: Some of you got answers in miimeters because of wrong conversion factors! b To compute the number of phase sips happening per second we use, d dt θ θ 1 = e V 57 Now since θ θ 1 = πn where n is an integer, the number of phase sips per second dn/dt is given by dn dt = ev 58 h 7

8 where we have used e = e. Thus for V = 1µV the phase sip rate is s 1. 8