Can Anyon Statistics Explain High Temperature Superconductivity?

Home , Heike Kamerlingh Onnes

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2 The model

We follow the derivation of Hamiltonian model using the mean ﬁeld approxi- mation and Bogoliubov unitary transformation as in the reference [9], where we can write the following relation,

(ǫ(k) µ)2 + ∆2 1 1−p 1 p − +1 1 = ep kB T . (2) (1 p)f(Ek) pf(Ek) − − 3 BCS Theory ( p 1) → For p 1, we recover the usual BCS Theory. →

When p 1,we have the following usual form of the Fermi function f(EK ), → 1 f(EK )= , (3) (ǫ µ)2 + ∆2 − 1+ ep kBT so, we have,

1  1 1  1= V . | | Z 2 (ǫ(k) µ)2 + ∆2  (ǫ(k) µ)2 + ∆2 − (ǫ(k) µ)2 + ∆2  Xk −  − − −  p  p k T p k T  1+ e B 1+ e B  (4) Replacing the summation by integration over the energy states, we have,

µ+ǫD N(0)dǫ  1 1  1= V . 2 2 2 2 2 2 | | Zµ−ǫD 2 (ǫ µ) + ∆  (ǫ µ) + ∆ − (ǫ µ) + ∆  −  − − −  p  p k T p k T  1+ e B 1+ e B  (5)

2 Let ζ = ǫ µ, then we have the master equation for BCS theory as, −

1 ǫD dζ  1 1  = . (6) 2 2 2 V N(0) Z0 ζ + ∆2  ζ + ∆2 − ζ + ∆2  | |  −  p  pk T p k T  1+ e B 1+ e B 

Determine TC .

At ∆ = 0, we have T = TC, so we can write,

ǫ 1 D dζ  1 1  = . (7) V N(0) Z ζ ζ − ζ | | 0  −   k T k T  1+ e B C 1+ e B C  ζ Deﬁne = x, we can write, kB TC ǫD 1 dx = KBTC φ(x), (8) V N(0) Z x | | 0 1 1 where φ(x)= . 1+ e−x − 1+ ex Using integration by parts we have,

ǫD ǫD ǫD KBTC dx KBTC KBTC dφ(x) φ(x) = [φ(x) ln x]0 ln xdx. (9) Z0 x − Z0 dx

1 For a weak coupling limit, i.e. ⋗⋗1, kBTC ⋖⋖ǫD, ∆0 ⋖⋖kBTC ,we |V |N0 | | have, n o

ǫD KBTC ǫD [φ(x) ln x]0 = ln , (10) KBTC ǫD dφ(x) ∞ dφ(x) KBTC ln xdx ln xdx 0, 12563. (11) Z0 dx ≈ Z0 dx ≈− So, we have

1 ǫ = ln 1, 134 D . (12) V N(0) KBTC | | We can rewrite the last equation to obtain the well known formula for critical temperature for BCS Theory as,

3 1 − 1, 134ǫD V N(0) TC = e| | . (13) KB

Determine ∆0.

At T = 0, we have ∆ = ∆0, so from the equation we have,

1 ǫD dζ ǫ = = sinh−1 D , (14) V N(0) Z0 2 2 ∆0 | | ζ + ∆0 q so we have,

ǫD 1 1 = e |V |N(0) . (16) ∆0 2 From equations (13) and (16), we arrive at the universal BCS formula,

∆0 =1.76KBTC . (17) Using equations (16) and (17) in the master equation (6), we can plot the T ∆ universal curve y = y(x), where x = ,y = as follow, TC ∆0

1 sinh |V |N(0) 1 dz  1 1  = y . V N(0) Z √ 2 √ 2 − √ 2 0 z +1  1.76y z +1 1.76y z +1  | |  −  1+ e x 1+ e x  (18)

4 Anyon superconductivity

From equations (1) and (2), we can write the general form of the master equation for any value of p as follow,

ǫD 1 dζ 1 = φ( ) φ(r(ζ)) , (19) V N(0) Z ζ2 + ∆2 r(ζ) − | | 0 p

4 where, ζ2 + ∆2 pK T r(ζ)= e B , (20) and φ(r) determined from the following equation,

1 1−p 1 p +1 1 = r. (21) (1 p)φ(r) pφ(r) − −

4.1 Determine TC for any value of p

At ∆ = 0, we have T = TC and then from equations (19-21) we can write,

ǫD KBTC 1 dx 1 = φ( ) φ(r(x)) , (22) V N(0) Z x r(x) − | | 0 r(x) = ex, (23)

and again, φ(r) determined from the equation (21).

4.2 Determine ∆0 for any value of p

At T = 0, we have ∆ = ∆0 and from equation (20), we have r = , so equation (21) takes the form, ∞

1 1−p 1 p +1 1 = . (24) (1 p)φ(r) pφ(r) − ∞ − Then for any value of p,we have

φ(r)=0at T =0. (25) 1 Similarly, we can evaluate φ r by solving the following equation, 1−p p  1   1  1 +1 1 = . (26) 1 1 − r (1 p)φ  pφ   r   r   −    1 So at T =0, We have r = , = 0 and then equation (26) takes the form, ∞ r 1−p p  1   1  +1 1 = 0 (27) 1 1 − (1 p)φ  pφ   r   r   −   

5 1 The solution of (26) (for positive φ( r )) takes the form, 1 1 φ( )= at T =0. (28) r p Using (25) and (28) in the master equation (19), we have at T = 0 the following equation,

ǫD p dζ ǫ = = sinh−1 D , (29) V N(0) Z 2 2 ∆0 | | 0 ζ + ∆0 q or in the equivalent form, p ǫ = ∆ sinh . (30) D 0 V N(0) | | At weak coupling limit, from the equation (30), we have the following universal relation for ∆0 at any value of p as, p − V N(0) ∆0 =2ǫD e| | . (31) In the next section, we will study some special cases of Anyon superconductivity for some special values of p.

5 Special cases of Anyon superconductivity 1 5.1 p = 2 Anyon

From the uniﬁed statistics (equation(2)), we can ﬁnd the formula of f(Ek) as follow, 2 f(E )= . (32) k 1 2 (ǫ µ)2 + ∆2 2 − 1+ e p kBT      So, after some calculations, we arrive at the master equation for the case 1 p = 2 as follow,

6       ǫD   1 dζ  1 1  = . 2   2 V N(0) Z0 ζ 2  1 − 1  | | + ∆    2 2 2 2 2 2  p  2 ζ + ∆ 2 ζ + ∆   − p   p    1+ e kBT 1+ e kB T                   (33)

Determine TC

At ∆ = 0, we have T = TC, so from equations (32,33) we can write,

ǫD 1 dx  1 1  = KBTC . (34) 2 V N(0) Z x 1 − 1 | | 0    e−2x 2 e2x 2  (1 + ) (1 + ) 

1 1 Deﬁne φ(x)=  , we can calculate the last inte- 1 − 1    e−2x 2 e2x 2   (1 + ) (1 + )  gral as,

ǫD ǫD ∞ KBTC dx KBTC dφ(x) ǫD 1.535ǫD φ(x) [φ(x) ln x]0 dx ln x = ln +0.4228 = ln . Z0 x ≈ −Z0 x KBTC KBTC (35) So,we can write the equation(34) as, 1 1.535ǫ = ln D , (36) 2 V N(0) KBTC | | or on the equivalent form, 1 − 1.535ǫD 2 V N(0) TC = e | | . (37) KB Comparing equation (37), the critical temperature for the Anyon case (P = 1 2 ) and equation (13), the critical temperature for BCS theory (at the same value of Debye energy) we have,

7 1 TC p= 1 2 2 V N(0) =1.35e | | . (38) TCBCS 1 So, for some weak coupling limit, say = 5, we have V N(0) | | TC 16, 446TC . (39) p= 1 BCS 2 ≃ So, if the critical temperature in the standard BSC is equal to 1 Kelvin 1 (with = 5), then using a uniﬁed statistics with P = 1 , the critical V N(0) 2 temperature| | jumps to 16, 446 Kelvin. Notice that although the weak coupling is the reason of why we cannot explain some HTS using the standard BCS theory, in the Anyonic statistics the critical temperature is increased whenever the coupling limit is decreased!! 1 As an example, consider that we have more weak interaction (say = V N(0) 10), then we have, | |

TC 201TC !! (40) p= 1 BCS 2 ≃ So at the same condition, if the critical temperature in the standard BSC is 1 eual to 1 Kelvin then for a weak coupling = 5 we see that the critical V N(0) | | 1 temperature jumps to 16, 446 Kelvin and for more weak coupling = 10 V N(0) the critical temperature jumps to 201 Kelvin! | |

Determine ∆0

At T = 0, we have ∆ = ∆0, so from the equation (29) we have,

1 ǫD dζ ǫ = = sinh−1 D , (41) 2 V N(0) Z0 2 2 ∆0 | | ζ + ∆0 q so we have, 1 − 2 V N(0) ∆0 =2ǫD e | | .. (42) From (24) and (28), we have

∆0 =1.3KBTC. (43) T ∆ Finally, we can plot the universal curve y = y(x), where x = ,y = TC ∆0 1 for p = 2 as follow,

8   1 sinh 2|V |N(0)     1 y dz  1 1  =   . 2 V N(0) Z √ 2  1 − 1  0 z +1   | |  2.6y√z2 +1 2 2.6y√z2 +1 2   −         1+ e x 1+ e x                (44)

1 5.2 p = 3 Anyon superconductor 1 For p = 3 , we can determine φ(r) from equation (21) as,

3 φ(r)= . (45) 1 1 − (2r3 +2√r6 + r3 + 1)3 + (2r3 +2√r6 + r3 + 1) 3 1 − From equations (22-23) and after some calculations, we can ﬁnd the formula 1 of TC for the Anyon p = 3 as, 1 − 32.627 3 V N(0) TC = ǫD e | | . (46) KB Also, from the general equation (31) we have, 1 − 3 V N(0) ∆0 =2ǫD e | | . (47) Conclusion 1 In this paper, we calculated the critical temperature of some metals using the standard BCS theory but assuming that the particles inside this metals obey Anyonic statistics instead Fermi-Dirac statistics. We ﬁnd that the lower of the interaction we have, the greater of the critical temperature which may be a reasonable explanation of some high temperature superconductors which the standard BCS theory cannot explain it. Acknowledgement 2 I want to thank E. Ahmad for his advice and very useful discussion.

6 References

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