In-Plane Zeeman Field and Rashba SOC

Ethan Lake(Dated: July 9, 2015)

We investigate unconventional superconductivity in the 2DEG with Rashba SOC and an appliedexternal magnetic field, where ER ⊥ B. When the Zeeman field is in-plane, pairing with finitecenter-of-mass momentum is possible, and so we develop a way to formulate the BCS theory withfinite momentum pairing.

Finite COMM Pairing

Since the introduction of an in-plane magnetic fieldwill generically result in pairing with finite COMM, it isworthwhile to develop the formulation of BCS theory inthis situation. We will work with two bands and supposethe existence of two momenta 2q1, 2q2 that optimize thepairing strength on each band. The Hamiltonian reads

H =∑kλ

ξkλa†kλakλ+

∑kk′λµ

Gkk′λµa†k+qλ

a†−k+qλak′+qµa−k′+qµ

We define the anomalous averages on each band as

Akλ = 〈a−k+qλak+qλ〉,

and the corresponding order parameters by

∆kλ =∑k′µ

Gkk′λµAk′µ

which allows us to write the interaction Hamiltonian inmean-field form as

Hint =∑kλ

(∆kλa

†k+qλ

a†−k+qλ + h.c.)−∑kk′λµ

Gkk′λµA∗kλAk′µ

The next step is, as usual, the diagonalization of theHamiltonian. As before, we introduce the spinor Ψkλ =(ak+qλ , a

†−k+qλ)T . However, our matrix h changes a bit.

First, introduce the combinations (P for +, M for −)

Pkλ =1

2(ξk+qλ + ξ−k+qλ), Mkλ =

1

2(ξk+qλ − ξ−k+qλ)

and then construct h by

hλ(k) =

(Pkλ +Mkλ 2∆kλ

2∆∗kλ −Pkλ +Mkλ

)which allows us to write

H =1

2

∑kλ

(Ψ†kλhλ(k)Ψkλ + ξkλ

)−∑kk′λµ

Gkk′λµA∗kλAk′µ

We didn’t have to do this before, since with q = 0 wehave ξk+q = ξ−k+q.

The next step is, of course, a Bogoliubov transforma-tion. By looking at the eigenvalues of h and recalling thatwe had v =

√(1− ξ/E)/2 in the zero COMM case, it’s

natural to choose (again, we look at 2∆ as the quantityof physical interest)

Ekλ =√P 2kλ + ∆2

kλ, v2kλ =

1

2(1− Pkλ/Ekλ) , u2kλ = 1−v2kλ

after which the standard calculation gives

H =∑kλ

[(Ekλ+Mkλ)γ†kλγkλ+

ξkλ − Ekλ2

−∑k′µ

Gkk′λµA∗kλAk′µ

]which looks like the usual Hamiltonian, except with anadded term in the first sum accounting for the energycreated by the finite pairing momentum.

At T = 0, the calculation goes through uneventfully.Varying over ∆, the self-consistent equations are

∆kλ = −∑k′µ

Gkk′λµ∆k′µ

2√P 2k′µ + ∆2

k′µ

If the dispersion ξ is quadratic and the interactionis momentum-independent, solving the self-consistentequations isn’t so bad. It’s interesting to see what qc wewould need to kill off superconductivity and send ∆→ 0.

For just one band, we have (assuming a momentum-independent interaction G and ξk+q = (k+ q)2/2m−µ):

1 = −Gρ∫ ωc

0

dξ√ξk + q2c/2m

We then find, making the usual assumptions, that

qc ≈√

2m∆

Of course, we’ve let q be arbitrary in our analysis. Inprinciple, we can obtain it by varying the thermodynamicpotential: ∂Ω/∂q = 0. However, this is usually easiersaid than done. With that, we turn to the application tothe Rashba + Zeeman problem.

Introducing an In-Plane Zeeman Field

When we put the Zeeman field in-plane and introducenon-vanishing SOC, the two Fermi surfaces become dis-torted away from perfect circular shapes. They becomeoblong in the direction normal to both the Rashba andZeeman fields. This means that finite momentum pairing

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is possible. Additionally, if the magnetic field dominatesover SOC, the spin vectors no longer wind completelyabout the out of plane direction.

There are two ways of approaching this problem. Thefirst is to orient the Zeeman field along the x directionand the Rashba field along the −z direction, with the2DEG in the x− y plane. The free Hamiltonian reads

H = εσ0 + α(kyσx − kxσy)−Bσx

It will again be helpful to transform to the chiral eigen-basis given by the two subbands in the problem. This isaccomplished by the transformation

c↑ =α(ikx + ky)−B√

2|αk −By|(a+ − a−)

c↓ =1√2

(a+ + a−)

which diagonalizes the free Hamiltonian as

H =∑kλ

ξkλa†kλakλ

with the dispersion given by

ξkλ =k2

2m±√α2k2x + (αky −B)2

where we choose the + sign for λ = 1 and the − forλ = 2.

The interaction term is

Hint =∑

λ1,λ2,λ3,λ4

∫x,y,z

Gw,x,y,λ1...λ4U∗1λ1

(w)U∗2λ2(x)

× U2λ3(y)U1λ4

(z)a†λ1,wa†λ2,x

aλ3,yaλ4,z

where the matrix U that exhibits the diagonalization is

U =1√2

(α(ikx+ky)−B|αk−By|

−α(ikx+ky)−B|αk−By|

1 1

)

we shall also assume a constant repulsive interaction, andlet Gw,x,y,λ1...λ4

= U (bad notation!).As in the out-of-plane Zeeman field case, it’s kind of

a mess. We have the same sort of terms as before, andwill denote (1122) for a†k+q1a

†−k+q1a−k′+q2ak′+q2 , for ex-

ample. We have allowed for each band to have its ownpair momentum qλ, which will need to be determinedself-consistently later by minimizing the thermodynamicpotential. The intra-band and Josephson terms are (1 isthe smaller band)

U

4

(f∗k,1fk′,1(1111) + f∗k,2fk′,2(2222)− f∗k,1fk′2(1122)

− f∗k,2fk′,1(2211)

)

with the notation

fkλ =α(ikx + ky + qλ)−B√α2k2x + (αky + αqλ −B)2

where we have assumed that the vector q = qy is alongthe direction of the deformation of the Fermi surface.

While this isn’t so bad, the coefficient of the (1221)term (the one that we want to perform the unitary trans-form on) is a bit worse. Additionally, we still have manyterms with three identical band indices, like (1112). Un-like before, they are not any smaller than the other terms.If we can get rid of them, we need a better justification.

The Better Approach

The better approach is to set the spin quantizationaxis in-plane, aligned with the Zeeman field. This waywhen we take the B α limit, the spins will be nearlyparallel or antiparallel to the quantization axis, allowingmore simplification when the resulting terms are treatedperturbatively in α/B.

To do this, we re-define our coordinate system so thatthe electrons move in the x−z plane, with the Rashba Efield directed in the −y direction and the Zeeman field inthe +z direction. This means that the free Hamiltonianlooks like

Hf = εσ0 −Bσz + α(kxσz − kzσx)

The dispersion, as before, is

ξkλ =k2

2m±√α2k2z + (αkx −B)2

The diagonalization is a bit trickier in this case, butstill doable. The matrix that diagonalizes H is

U =1√2

(B−αkx−ξ1+εk

αkz

B−αkx−ξ2+εkαkz

1 1

)where ξ1 is the larger eigenvalue. Of course, the columnsof U aren’t normalized yet. The actual normalizationprocedure only yields useful expressions in the α Blimit, and involves a fair bit of algebra. We’ll skip writingthe full thing out, and will just quote the final result:

c↑ =

(1− α2k2z

8B2 − 4Bαkz

)a1 −

αkz2B − αkz

a2

c↓ =αkz

2B − αkza1 +

(1− α2k2z

8B2 − 4Bαkz

)a2

The above transformation is approximate, but remark-ably accurate. For αkz/B ∼ 0.1, the above expressionagrees with the exact result to about 1 part in 104. Aswe progress further, we can rest easy knowing that sim-plifying further will not sacrifice too much accuracy. We

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note that in the α → 0 limit, the transformation be-tween spin and band operators is the identity, which wasthe whole point of choosing the quantization axis to bein-plane in the first place.

Again, we write (1122) for a†k+q1a†−k+q1a−k′+q2ak′+q2

and compute the intra-band terms as (dropping terms oforder three and higher in αk/B):

(1111)

(Uα2kzk

′z

4B2 + 2Bα(kz + k′z) + α2kzk′z

)(2222)

(Uα2kzk

′z

4B2 − 2Bα(kz + k′z) + α2kzk′z

)The Josephson terms are

(1122)

(Uα2kzk

′z

4B2 + 2Bα(kz − k′z)− α2kzk′z

)(2211)

(Uα2kzk

′z

4B2 + 2Bα(k′z − kz)− α2kzk′z

)Finally, the (1221) term is

(1221)U

(1−

∑ki∈wxyz

α2k2i,z8B2 − 4Bαki,z

+α2wzxz

4B2 − 2Bα(wz + xz)2 + α2wzxz

+α2yzzz

4B2 − 2Bα(yz + zz)2 + α2yzzz

)where w, x, y, z are four different momenta.

As before, the idea is to perform a unitary transformon the (1221) term. We can see that after this is over,all terms except the 1 in the front will go as at leastU2α2k2z/B

2. The extra factor of U on the front will jus-tify dropping them, and so we can take the coefficient ofthe (1221) term to just be U for now.

The susceptibility then has the same form as usual:

χ∓(Q) =

∫d2p

(2π)2f(p)− f(p+Q)

ξ±(p)− ξ±(p+Q)

The dispersion has changed, which makes things a bitmore complicated. We use the approximation

ξ±(p) = p2(

1

2m± α2

2B

)±B ∓ αpx ∓

α2p2x2B

which is valid in the limit we’re working in. We write thesusceptibility as

χ∓(Q) =

∫d2p

(2π)2

f(p)

ξ±(p)− ξ±(p+Q)

+f(p)

ξ±(p)− ξ±(p−Q)

To actually do the integral, we need to also approxi-mate the Fermi surface as a circle. This of course isn’tstrictly true, but greatly aids in the calculation. Withthese assumptions, we can do the pz component of theintegral right away, which integrates to 2

√k2F − p2x.

We make the following definition:

m∗ =

(1

m± α2

B

)−1And introduce the parameters

A+ =Q2 ± 2αQxm∗

2QkF, A− =

Q2 ∓ 2αQxm∗2Q

kF

which means that we can write, after a bit of manipula-tion,

χ∓(Q) = −kFm∗2π2Q

∫ 1

−1dx

√1− x2x+A+

+

√1− x2

−x+A−

We’ll omit the calculation of the integrals for brevity,

which are standard but fairly involved. When the smokeclears, we find that

χ∓(Q) = −m∗2π

1− 1

2

√1−

(2kFQ

)2

+4m∗Qxα

Q2+

4m2∗Q

2xα

2

Q4

− 1

2

√1−

(2kFQ

)2

− 4m∗Qxα

Q2+

4m2∗Q

2xα

2

Q4

which reduces to our previous expression for χ when α→0, as it should. Plots of χ for two different SOC strengthsare shown in the figures. Of interest here is that χ is nota constant for the smaller Fermi surface, as it was before.

The Self-Consistent Equations

Our task now is to solve the self consistent equations.By looking at the coefficients in front of each interactionterm, we are led to make the following definition:

gλ(k) =k sinφ

2B/α± k sinφ

where we use the + sign for λ = 1 and the − sign for λ =2. This means that the interaction potential is (excludingthe susceptibility)

Gkk′λµ = Ugλ(k)gµ(k′)

To begin, let us consider U to be very small and ignorethe intra-band susceptibility term χλ for each band. Theself-consistent equations are then

∆λ(k) = −U2

∫d2k′

(2π)2

gλ(k)gλ(k′)∆λ(k′)

Ek′λ

+gλ(k)gµ(k′)∆µ(k′)

Ek′µ

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with

Ekλ =√P 2kλ + ∆2

λ(k)

where the function P is as defined in the first section andµ is the opposite index as λ.

From this, we see that ∆λ(k) must be of the formgλ(k)C(λ), where C is a constant function that only de-pends on the band index λ. That is, the only momentumdependence in ∆ must be of the form g(k). If this werenot the case, we could vary k on one side of the equationwithout changing the other side, and the self-consistentequations would not hold. We are thus lead to the fol-lowing ansatz for ∆:

∆λ(k) = ∆λeiθλgλ(k)

where ∆λ is a real positive constant. If we plug this intothe self-consistent equations and cancel out the k depen-dent terms on either side, we obtain the two equations

1 = −U∫

d2k′

(2π)2g21(k′)

2Ek′,1− U

∫d2k′

(2π)2∆2g

22(k′)ei(θ2−θ1)

2∆1Ek′,2

1 = −U∫

d2k′

(2π)2g22(k′)

2Ek′,2− U

∫d2k′

(2π)2∆1g

21(k′)e−i(θ2−θ1)

2∆2Ek′,1

since g2λ(k′) and Ek′λ are both positive-definite, we mustrequire that θ2 − θ1 = π. However in this case, we seethat we also require∫

d2k′

(2π)2g21(k′)

Ek′,1<

∫d2k′

(2π)2g22(k′)

Ek′,2

∆2

∆1

and ∫d2k′

(2π)2g22(k′)

Ek′,2<

∫d2k′

(2π)2g21(k′)

Ek′,1

∆1

∆2

which can not be simultaneously satisfied. Evidently, theonly solution to the self-consistent equations is the trivialone.

We could have predicted this earlier, based on someresults from my last set of notes. There, we saw thata superconducting ground state was only possible if theproduct of the two Josephson coefficients was larger thanthe product of the two intra-band coefficients. Since thetwo products are equal in this case, a nontrivial solutioncannot be found.

The main point here is that including the intra-bandsusceptibility χλ is essential to realize a superconductingstate. If χ is negative the criterion mentioned above willbe satisfied, and a superconducting state can be realized.However, introducing χ into the self-consistent equations

makes finding a nice ansatz for the order parameter a bittrickier, especially since χ doesn’t factorize into functionsof each momentum.

FIG. 1: A plot of the susceptibility as a function of Q, themomentum transfer. The color scale is normalized to thevalue at Q = (0, 0). Although m∗ is different for each band,for reasonable values of α2/B this affect is small. In this plotα is taken to be rather small, and so we see a mostly circularappearance as in the α = 0 case.

FIG. 2: The same as in the last figure, but with a larger valueof α.

If there was no ∆λ(k) in Ekλ, we could solve the self-consistent equations (in principle) by using a Neumannintegral series, although the actual procedure wouldlikely be a bit messy. Barring an expansion into angularharmonics like last time, we’ll need to come up with abetter way of approaching the problem.