0. Cooper Pairing in “Exotic” Fermi Superfluids: An Alternative Approach
Lecture 2
Cooper Pairing in “Exotic” Fermi Superfluids: An Alternative Approach
Anthony J. Leggett Department of Physics University of Illinois at Urbana-Champaign
based largely on joint work with Yiruo Lin
supported in part by the National Science Foundation under grand no. DMR

1. Anisotropic Cooper pairing (textbook version)
Lecture 2
Anisotropic Cooper pairing (textbook version)
Can still use fundamental ansatz (l. 1)
same 𝜑!
Ψ 𝑁 𝒓 1 𝜎 1 … 𝒓 𝑁 𝜎 𝑁 =
𝑛∙𝒜∙𝜑 𝒓 1 𝜎 1 𝒓 2 𝜎 2 𝜑 𝒓 3 𝜎 3 𝒓 4 𝜎 4 … 𝜑 𝒓 𝑁−1 𝜎 𝑁−1 𝒓 𝑁 𝜎 𝑁
normalization
antisymmetrizer
and still require
𝜑 𝒓 1 𝜎 1 , 𝒓 2 𝜎 2 =𝜑 𝒓 1 − 𝒓 2 :𝜎 1 𝜎 2
i.e. COM at rest
but now allow 𝜑 to be nontrivial functions of 𝜎’s and 𝒓 12 .
However, because of 𝒜 (Pauli principle) 𝑆=0⇔𝑓 𝒓 1 − 𝒓 2 even, 𝑆=1⇔ 𝑓 𝒓 1 − 𝒓 2 odd.
Even-parity case is simple generalization of BCS:
Ψ 𝑁 =𝒩 𝑘 𝑐 𝑘 𝑎 𝑘↑ + 𝑎 −𝑘↓ + 𝑁/2 |𝑣𝑎𝑐
𝑐 𝒌 = 𝑐 −𝒌 =𝑓 𝒌 , 𝒌
⇒ generalized gap equation
Δ 𝑘 = 𝑘′ 𝑉 𝑘𝑘′ Δ 𝑘′ /2 𝐸 𝑘′
𝐸 𝑘′ ≡ 𝐸 𝑘 Δ 𝑘 2
in general (in 3D) gap has nodes at 2 or more points on F.S.

2. Odd-parity case: must now pair spins to form 𝑆=1
Odd-parity case: must now pair spins to form 𝑆=1. Description of general case (e.g. 3 𝐻𝑒−𝐵 ) complicated, but simplifies for ESP 3 𝐻𝑒−𝐴, 𝑆 𝑟 2 𝑅𝑢 𝑂 In
this case proper description is
equal spin pairing (↑↑ and ↓↓, no ↑↓+↓↑ )
𝜑 𝒓 1 − 𝒓 2 :𝜎 1 𝜎 2 = 𝑓 ↑↑ 𝒓 1 − 𝒓 2 |↑↑ + 𝑒 𝑖𝜑 𝑓 ↓↓ 𝒓 1 − 𝒓 2 |↓↓
⇒ Ψ 𝑁 =𝒩 𝑘 𝑐 𝑘↑ 𝑎 𝑘↑ + 𝑎 −𝑘 + + 𝑒 𝑖𝜑 𝑐 𝑘↓ 𝑎 𝑘↓ + 𝑎 −𝑘↓ + 𝑁/2 |𝑣𝑎𝑐
but in most TQC contexts usually adequate to replace by ↑!
Ψ 𝑁 2 ↑ = 𝒩 ↑ 𝑘 𝑐 𝑘 𝑎 𝑘↑ + 𝑎 −𝑘↑ + 𝑁/4 |𝑣𝑎𝑐
Ψ 𝑁 = Ψ 𝑁 2 ↑ Ψ 𝑁 2 ↓ ,
(etc.)
From now on, concentrate on single spin population (e.g. ↑) so omit ↑‘s and let 𝑁/2→𝑁 (i.e. 𝑁 is number of “relevant” particles) , Ψ 𝑁/2,↑ → Ψ 𝑁
Thus, at first sight,
Ψ 𝑁 =𝒩 𝑘 𝑐 𝑘 𝑎 𝒌 + 𝑎 −𝒌 + 𝑁/2 |𝑣𝑎𝑐
𝑐 𝒌 = −𝑐 −𝒌 (Pauli) [
↑: State 𝒌=0 has no partner! (contrast even-parity case, 𝑂↑𝑂↓ ), hence, strictly correct odd-parity GSWF for an odd number 2𝑁+1 of particles is
Ψ 𝑁 =𝒩 𝒌≠0 𝑐 𝑘 𝑎 𝑘 + 𝑎 −𝑘 + 𝑁/2 𝑎 𝑜 + |𝑣𝑎𝑐
(We will usually implicitly subtract off this odd particle in the accounting)

3. Most interesting case (especially in 2D) is when 𝑐 𝑘 is (nontrivially) complex. Important example 3 𝐻𝑒−𝐴, 𝑆 𝑟 2 𝑅𝑢 𝑂 4 ? is 𝑝+𝑖𝑝 state:
𝑐 𝑘 =𝑓 𝒌 exp 𝑖 𝜑 𝑘  ∠ of 𝒌 on Fermi surface
In this case, pair wave function 𝐹 𝑘 is of form
𝐹 𝑘 = 𝑐 𝑘 𝑐 𝑘 2 = Δ 𝑘 2 𝐸 𝑘
Δ 𝑘 = Δ 𝑘 exp 𝑖 𝜑 𝑘
(so in 2D, no nodes). Thus, if we write
Δ 𝑘 𝑘 = 𝑘 𝐹 ≡ ∆ 𝑜 𝑘 𝐹 ,
then on and near F.S.,
Δ 𝑘 = Δ 𝑜 𝑘 𝑥 +𝑖 𝑘 𝑦 (hence, “𝑝+𝑖𝑝”)

4. However, from symmetry of gap equation (with 𝑉 𝑘𝑘′ = 𝑉 𝒌−𝒌′ )
Δ 𝑘 =− 𝑘′ 𝑉 𝒌−𝒌′ Δ 𝑘 ′ |2 𝐸 𝑘′
𝐸 𝑘 ≡ 𝜖 𝑘 Δ 𝑘 /2 =𝑓 𝒌
Δ 𝑘 must have same symmetry for all 𝑘 (down to 𝑘 →0). Does it also have same dependence on 𝑘 ? (i.e. is Δ 𝑘 ∝ 𝑘 ? If potential 𝑉 𝑟 can be Taylor-expended, yes – usually assumed in literature. Thus, usual assumption is
Δ 𝑘 = Δ 𝑜 𝑘 𝑥 +𝑖 𝑘 𝑦 , ∀𝒌 (including 𝑘 →0).
For future simplicity, write for 2D 𝑝+𝑖𝑝 state
Ω † ≡𝒩′ 𝑘 𝑐 𝑘 𝑒 𝑖 𝜑 𝑘 𝑎 𝑘 + 𝑎 −𝑘 +
so that
Ψ 𝑁 𝑝+𝑖𝑝 = Ω † 𝑁/2 |𝑣𝑎𝑐

5. The 2D 𝑝+𝑖𝑝 state has several intriguing properties:
1. Since by direct calculation 𝐿 𝑧 , Ω † =ℏ Ω †
we have
𝐿 𝑧 | Ψ 𝑁 = 𝑁ℏ 2 | Ψ 𝑁
2. Nevertheless, if we take the limit 𝑐 𝑘 →𝜃 𝑘 𝐹 −𝑘 , we find
Ψ 𝑁 = 𝑘< 𝑘 𝐹 𝑒 𝑖 𝜑 𝑘 𝑎 𝑘 + 𝑎 −𝑘 + 𝑁/2 |𝑣𝑎𝑐
= 𝑘< 𝑘 𝐹 𝑒 𝑖 𝜑 𝑘 𝑎 𝑘 + 𝑎 −𝑘 + |𝑣𝑎𝑐
≡ 𝑘∠ 𝑘 𝐹 𝑒 𝑖 𝜑 𝑘 𝑘∠ 𝑘 𝐹 𝑎 𝑘 + 𝑎 −𝑘 + | 𝑣𝑎𝑐 ≡ 𝑘< 𝑘 𝐹 𝑒 𝑖 𝜑 𝑘 ′2 |𝐹𝑆
So since an overall phase factor is physically irrelevant, 𝐿 𝑧 | Ψ 𝑁 =0!

6. (Moment of inertia is not extensive)
Relevant observation: how much energy does it cost to polarize Fermi sea to ang. momentum 𝑁ℏ/2?
Answer: 0 𝑁 1/3 !
area ∝ 𝑑 3/2  ℓ 𝑛 𝑑
ℓ 𝑚𝑎𝑥
(Moment of inertia is not extensive)
Possible resolution of apparent inconsistency of 1 and 2 (M. Stone):
In a finite system of dimension 𝑅 as Δ→0, Cooper-pair radius 𝜉~ℏ v 𝐹 /Δ eventually becomes >𝑅. So, for 𝜉≪𝑅 𝐿~𝑁ℏ/2, for 𝜉≫𝑅 𝐿∼0?
3. Behavior of “molecular” wave function φ at large 𝒓 1 − 𝒓 2 :

7. By inverting the general relation Ϝ 𝑘 = 𝐶 𝑘 1+| 𝐶 𝑘 | 2 and
substituting the equilibrium value of Ϝ 𝑘 , Δ 𝑘 /2 Ε 𝑘 ,
we obtain for 𝑘< 𝑘 Ϝ 𝜖 𝑘 <𝜇 the relation
𝑐 𝑘 = Δ 𝑘 /2 Ε 𝑘 1− | 𝜖 𝑘 −𝜇| Ε 𝑘
which in the limit 𝜖 𝑘 →0 reduces to 𝜇∕4 Δ 𝑘 ∗ +0( Δ 2 ∕ 𝜇 2 . Thus provided Δ 𝑘 ∝ 𝑘 for 𝑘 →0 as assumed,
𝑐 𝑘 = 𝑐𝑜𝑛𝑠𝑡 𝑘 𝑥 − 𝑖𝑘 𝑦 𝑎𝑠 𝑘→0
Thus the F. T., the “molecular wave function”
𝜑 𝒓 𝒓≡ 𝒓 1 − 𝒓 2 , behaves at large differences as
𝜑 𝑟 ~ 𝓏 − 𝓏≡𝓍+𝑖𝑦
and for these distances the many body GSWF has the “Pfaffian” form
Ψ Ν ~Ρ𝑓 1 𝑧 𝑖 − 𝑧 𝑗
←Moore – Read form for 𝜈= 5 2 QHE

8. ◦ ◦ 2. Anisotropic Cooper pairing: alternative approach
The ground state of a 𝑝+𝑖𝑝 Fermi superfluid looks “natural” when reached by path 1, much less so when matched by 2, since
(a) Δ𝐿 𝑎𝑡 𝑇 𝑐 ~Ν
(b) Cooper instability affects only states near Fermi surface, not far down in Fermi sea. So consider the ansatz 2 N
𝑇 𝑐 S ●
BCS
−1∕ 𝑘 Ϝ 𝑎 𝑠 1
BEC
Ψ Ν 𝑎𝑙𝑡 ~𝑛 𝑘> 𝑘 Ϝ 𝑐 𝑘 𝑎 𝑘 + 𝑎 −𝑘 + Μ 𝑘< 𝑘 Ϝ 𝑑 𝑘 𝑎 −𝑘 𝑎 𝑘 Μ | Ϝ𝑆
Μ≪Ν ◦
where 𝑑 𝑘 is related to the 𝑐 𝑘 of the textbook approach (call it 𝑐 𝑘 0 ) by ●
𝑘 Ϝ ●
𝑑 𝑘 = 𝑐 𝑘 −1 ◦ ●
This reproduces the “standard” value of 𝑛 𝑘 , since the number of holes in state 𝑘< 𝑘 Ϝ is 𝑑 𝑘 𝑑 𝑘 :
𝑛 𝑘 =1− 𝑑 𝑘 𝑑 𝑘 2 = 𝑐 𝑘 𝑐 𝑘
It also reproduces the value of 𝑎 𝑘 + 𝑎 −𝑘 + 𝑎 −𝑘′ 𝑎 𝑘 ′ for 𝒌 , 𝒌 ′ both above or both below 𝑘 Ϝ , but not for (e.g.) 𝑘> 𝑘 Ϝ , 𝑘 ′ < 𝑘 Ϝ . However this can be remedied by a small modification of Ψ Ν :

9. Ω + † ≡ 𝑘> 𝑘 𝐹 𝑐 𝑘 𝑎 𝑘 + 𝑎 −𝑘, + Ω − † ≡ 𝑘< 𝑘 𝐹 𝑑 𝑘 𝑎 −𝑘 𝑎 𝑘
Ω + † ≡ 𝑘> 𝑘 𝐹 𝑐 𝑘 𝑎 𝑘 + 𝑎 −𝑘, + Ω − † ≡ 𝑘< 𝑘 𝐹 𝑑 𝑘 𝑎 −𝑘 𝑎 𝑘
put
and write
Ψ 𝑁 𝑎𝑙𝑡 ≡ 𝑀 𝑘 𝑀 Ω + † Ω − † 𝑀 |FS , 𝑘 𝑀
slowly varying in phase as 𝑓(𝑀)
(one possible implementation:
Ψ 𝑁 𝑎𝑙𝑡 ~ 𝑑𝜑 exp 𝑖𝜑/2 ⋅Ω + † + exp −𝑖𝜑/2 ⋅Ω − † 𝑁/2 |𝐹𝑆 )
then both 𝑛 𝑘 and 𝑎 𝑘 + 𝑎 𝑘 + 𝑎 −𝑘′ 𝑎 𝑘 ′ same for all 𝒌, 𝒌 ′ (to order 𝑁 −1/3 )
as in textbook approach 𝐹 𝑘 𝑘< 𝑘 𝐹 = 𝑑 𝑘 ∗ 𝑑 𝑘 2 = 𝑐 𝑘 𝑜 𝑐 𝑘 ⟹ energy/particle of 2 states identical in thermodynamic limit.
At first sight Ψ 𝑁 𝑎𝑙𝑡 is just a rewriting of Ψ 𝑁 in different notation,
as in the s-wave case. However,
𝐿 𝑧 , Ω + † =ℏ Ω + † but 𝐿 2 , Ω − † =−ℏ Ω − †
⟹ 𝐿 𝑧 | Ψ 𝑁 𝑎𝑙𝑡 > =0 !
( +0 Δ/ 𝐸 𝐹 2 due to shift of 𝜇 in 𝑆 phase)
So … which is right, Ψ 𝑁 or Ψ 𝑁 𝑎𝑙𝑡 (or maybe a hybrid wave function)?
A possible answer: both, or neither! The MBGS may be degenerate within terms of relative order 𝑁 −1/3 in thermodynamic limit.
Note: Ψ 𝑁 𝑎𝑙𝑡 almost certainly corresponds to a quite different behavior of 𝜑 𝑟 in the limit 𝑟→∞. Plausibly (but not proved): the modification to 𝜑 𝑟 from its 𝑁−state value ∝ 𝑟 −3 .

10. 3. Fermionic quasiparticles: the textbook approach (BCS (isotropic) case)
Recall: in BCS formalism, even-no-parity (PNC) GS is written in form
Ψ BCS = 𝑘 𝑢 𝑘 + 𝑣 𝑘 𝑎 𝑘↑ + 𝑎 −𝑘↓ + |𝑣𝑎𝑐>.
≡ 𝑘 Φ 𝑘 ,Φ 𝑘 ≡ 𝑢 𝑘 |00 𝑘 + 𝑣 𝑘 |11 𝑘
𝑢 𝑘 = 𝑢 −𝑘 , 𝑣 𝑘 = 𝜐 −𝑘 , 𝑢 𝑘 real
We make standard Bogoliubov-Valatin transformation:
𝛼 𝑘𝜎 + = 𝑢 𝑘 𝑎 𝑘𝜎 + − 𝜎𝜈 𝑘 ∗ 𝑎 −𝑘, −𝜎
So (e.g.)
𝛼 𝑘↑ + = 𝑢 𝑘 𝑎 𝑘↑ + − 𝑣 𝑘 ∗ 𝑎 −𝑘↓
and so
𝑎 𝑘↑ + |𝛷 𝑘 ≡ 𝑢 𝑘 𝑎 𝑘↑ + − 𝑣 𝑘 ∗ 𝑎 −𝑘↓ 𝑢 𝑘 + 𝑣 𝑘 𝑎 𝑘↑ + 𝑎 −𝑘↓ + |GS
= 𝑢 𝑘 2 𝑎 𝑘↑ + |𝑣𝑎𝑐 + 𝑣 𝑘 2 𝑎 𝑘↑ + |GS
ACR’s!
≡𝛼 𝑘↑ + |𝐺𝑆 ≡ |1,0

11. Similarly the operator
𝛼 −𝑘↓ + ≡ 𝑢 𝑘 𝑎 −𝑘↓ + + 𝜐 𝑘 ∗ 𝑎 𝑘↑
when applied to the BCS groundstate, generates the state |01 𝑘 .
However, from the four operators 𝑎 𝑘↑ + , 𝑎 𝑘↑ ∗ , 𝑎 −𝑘↓ + , 𝑎 −𝑘↓ one can generate two other linearly independent (orthogonal) combinations:
𝛽 𝑘↑ ≡ 𝜐 𝑘 𝑎 𝑘↑ + + 𝑢 𝑘 𝑎 −𝑘↓
𝛽 −𝑘↓ ≡ 𝜐 𝑘 𝑎 −𝑘↓ + − 𝑢 𝑘 𝑎 𝑘↑
Appling these to the BCS groundstate gives, e.g.,
𝛽 𝑘↑ |Φ 𝑘 ≡ 𝜐 𝑘 𝑎 𝑘↑ + + 𝑢 𝑘 𝑎 −𝑘↓ 𝑢 𝑘 + 𝜐 𝑘 𝑎 𝑘↑ + 𝑎 −𝑘↓ + | GS
= 𝜐 𝑘 𝑢 𝑘 𝑎 𝑘↑ + | 𝑣𝑎𝑐 − 𝑢 𝑘 𝜐 𝑘 𝑎 𝑘↑ + | GS ≡0
and similarly for 𝛽 𝑘2 . So 𝛽 𝑘1 and 𝛽 𝑘2 are pure annihilators. At first sight this looks trivial, since 𝛽 𝑘↑ is just the H.C. of 𝛼 −𝑘↓ + , i.e. 𝛼 −𝑘↓ , and we are all used to the fact that the 𝛼 ′ 𝑠 annihilate the ground state. But it’s worth noting for future reference…

12. To summarize, in simple BCS theory the simplest MB energy eigenstates can be expressed in the form of a tensor product of occupation states Φ 𝑘 referring to the pair of plane-wave states 𝒌↑,−𝒌↓ . The (even-numbered-parity) “ground pair” state is
Φ 𝑘 𝐺𝑃 = 𝑢 𝑘 |00 + 𝑣 𝑘 |11
There is a second “completely-paired” (even-numbered parity) state orthogonal to Φ 𝑘 𝐺𝑃 ,:
Φ 𝑘 𝐸𝑃 =− 𝑣 𝑘 ∗ |00 + 𝑢 𝑘 |11
which can in fact be generalized by successive application of 𝛼 𝑘↑ + and 𝛼 −𝑘↓ +
The odd-number-parity energy eigenstates are
|10 ≡ 𝛼 𝑘↑ + | Φ 𝑘 𝐺𝑃 and |01 ≡ 𝛼 −𝑘↓ + | Φ 𝑘 𝐺𝑃
while the operators 𝛽 𝑘↑ ≡ 𝛼 −𝑘↓ and 𝛽 −𝑘↓ ≡ 𝛼 𝑘↑ annihilate Φ 𝑘 𝐺𝑃
All the above analysis generalizes straightforwardly to the anisotropic case, including the 𝑝+𝑖𝑝 state. (irrespective of whether we use the textbook or alternative approach*).
*In the latter, Φ 𝑘 is simply 𝑢 𝑘 𝑒 −𝑖 𝜑 𝑘 |00 + 𝑣 𝑘 |11 .

13. same! So far, so good… But what if our Hamiltonian is more general:
𝐻 = 𝜎𝜎′ 𝑑𝑟 − ℏ 2 2𝑚 𝛿 𝜎 𝜎 ′ 𝜓 𝜎 † 𝒓 𝛻 2 𝜓 𝜎′ 𝒓 + 𝑈 𝜎𝜎′ 𝒓 𝜓 𝜎 † 𝒓 𝜓 𝜎′ 𝒓
𝛼𝛽𝛾𝛿 𝑑𝒓 𝑑 𝒓 ′ 𝑉 𝛼𝛽𝛾𝛿 𝒓,𝒓′ 𝜓 𝛼 † 𝒓 𝜓 𝛽 † 𝒓′ 𝜓 𝛾 𝒓′ 𝜓 𝛿 𝒓 ?
We can still use our general definition of a “completely paired” 𝑁(=even) – particle state:
same!
Ψ 𝑁 =𝒩∙𝒜∙ 𝜑 𝒓 1 𝜎 1 , 𝒓 2 𝜎 2 𝜑 𝒓 3 𝜎 3 , 𝒓 4 𝜎 4 … 𝜑 𝒓 𝑁−1 𝜎 𝑁−1, 𝒓 𝑁 𝜎 𝑁
≡𝒩′ 𝜎𝜎′ 𝑑𝒓 𝑑 𝒓 ′ 𝜑 𝒓𝝈,𝒓′𝜎′ 𝜓 𝜎 † 𝒓 𝜓 𝜎′ † 𝒓′ 𝑁/2 |𝑣𝑎𝑐
Theorem: can always find complete orthonormal set 𝑛, 𝑛 (i.e. 𝑛, 𝑛 ′ =0 , 𝑛|𝑛′ = 𝛿 𝑛𝑛′ , 𝑛 , 𝑛 ′ = 𝛿 𝑛 𝑛 ′ ) such that
𝜑 𝒓𝝈, 𝒓 ′ 𝜎 ′ = 𝑛 𝑐 𝑛 𝜒 𝑛 𝑟𝜎 𝜒 𝑛 𝑟′𝜎′
Thus, can write any completely paired state in the form
Ψ 𝑁 =𝒩" 𝑚 𝑐 𝑚 𝑎 𝑚 + 𝑎 𝑚 𝑁/2 |𝑣𝑎𝑐

14. As long as we deal only with the even-number-parity states, we can continue the analogy to BCS:
Ψ BCS = 𝑚 Φ 𝑚 𝐺𝑃 , Φ 𝑚 𝐺𝑃 ≡ 𝑢 𝑚 |00 + 𝑣 𝑚 |11 , |00 ≡ |00 𝑚 𝑚 , etc.
𝑐 𝑚 ≡ 𝑣 𝑚 / 𝑢 𝑚 , 𝑢 𝑚 𝑣 𝑚 2 =1
and the “excited-pair” state is given by
Φ 𝑚 𝐸𝑃 =− 𝑣 𝑚 ∗ |00 + 𝑢 𝑚 | 𝑐 𝑚 →− 𝑐 𝑚 ∗−1
This is an energy eigenstate if Φ 𝑚 𝐺𝑃 is, since 𝑛 𝑚 →1− 𝑛 𝑚 and 𝐹 𝑚 → −𝐹 𝑚 .
In the BCS case we found the values of the 𝑐 𝑘 ’s (or equivalently of the quantities 𝑛 𝑘 and 𝐹 𝑘 ) by minimizing the sum of the single – particle and pairing terms in 𝐻 . Can we do the same here? Since the functions 𝜒 𝑛 , 𝜒 𝑛 are unknown a priori, our calculation should also find them. Suppose we write
𝑈 𝑚 𝑟 ≡ 𝑈 𝑚 𝜒 𝑚 𝒓𝜎 , 𝑉 𝑚 𝒓 ≡ 𝑉 𝑚 𝜒 𝑚 𝒓𝜎
and minimize the s.p. + pairing terms with respect to the functions 𝑈 𝑚 𝒓𝜎 , 𝑉 𝑚 𝒓𝜎 , subject to the orthogonality constraints. 𝑈 𝑚 , 𝑈 𝑚′ ~ 𝛿 𝑚 𝑚 ′ , etc. . The resulting equations are, formally, exactly the standard BdG equations (see below) but with much stronger orthogonality constraints. So we can obtain an explicit solution this way only in special cases (e.g. 𝑚 =TR of 𝑚).

15. Now let’s turn to the odd-number-parity states
Now let’s turn to the odd-number-parity states. By exact analogy with the BCS case, the combinations
𝛽 𝑚 ≡ 𝜐 𝑚 𝑎 𝑚 + + 𝑢 𝑚 𝑎 𝑚
𝛽 𝑚 ≡ 𝜐 𝑚 𝑎 𝑚 + − 𝑢 𝑚 𝑎 𝑚
are pure annihilators: 𝛽 𝑚 | Ψ 𝑁 = 𝛽 𝑚 | Ψ 𝑁 =0 Note that any linear combination of PA’s is itself a PA! Moreover the states
𝛼 𝑚 + ≡ 𝑢 𝑚 𝑎 𝑚 + − 𝜐 𝑚 ∗ 𝑎 𝑚 ≡ 𝛽 𝑚 †
𝛼 𝑚 + ≡ 𝑢 𝑚 𝑎 𝑚 + + 𝜐 𝑚 ∗ 𝛼 𝑚 ≡ 𝛽 𝑚 †
create the states |1,0 𝑚 and |0,1 𝑚 respectively |1,0 𝑚 ≡ occupied, 𝑚 empty, etc.) and 𝛼 𝑚 + 𝛼 𝑚 + generates the “excited pair” state Φ 𝑚 𝐸𝑃 . However, these two states are not in general energy eigenstates. This is fairly obvious, since when written in the (basis of the 𝜒 𝑚 and 𝜒 𝑚 even the single-particle term 𝐻 is nondiagonal. So the odd-parity energy eigenstates must be linear combinations of the states |1,0 𝑚 and |0,1 𝑚 :
𝛾 𝑖 † = 𝑚 𝜆 𝑖𝑚 𝛼 𝑚 + + 𝜇 𝑖𝑚 𝛼 𝑚 +
How to find these linear combinations? Standard method is mean-field (BdG) approach: in spirit of BCS, factorize potential term in Hamiltonian:
𝑉 𝜓 𝛼 † 𝑟 𝜓 𝐵 † 𝑟′ 𝜓 𝛾 𝑟′ 𝜓 𝛿 𝑟 → 𝜓 𝛼 † 𝑟 𝜓 𝛽 † 𝑟′ 𝜓 𝛾 𝑟′ 𝜓 𝛿 𝑟
+ 𝜓 𝛾 𝑟′ 𝜓 𝛿 𝑟 𝜓 𝛼 † 𝑟 𝜓 𝛽 † 𝑟′
and define
Δ 𝛼𝛽 𝑟,𝑟′ ≡ 𝛾𝛿 𝑉 𝛼𝛽𝛾𝛿 𝒓,𝒓′ 𝜓 𝛾 𝑟′ 𝜓 𝛿 𝒓

16. Then the effective mean-field Hamiltonian is bilinear in 𝜓 and 𝜓 † :
𝐻 𝑚𝑓 = 𝛼𝛽 𝑑𝑟 𝜓 𝛼 † 𝑟 𝐻 𝑜 𝜓 𝛽 𝑟
𝛼𝛽 𝑑𝑟𝑑𝑟′ Δ 𝛼𝛽 𝒓,𝒓′ 𝜓 𝛼 † 𝑟 𝜓 𝛽 † 𝑟′ +𝐻.𝐶.
where the quantity 𝜓 𝛾 𝑟′ 𝜓 𝛿 𝑟 occuring in Δ 𝛼𝛽 must eventually be determined self-consistently. We now seek the creation operators of odd-parity energy eigenstates in the form
𝛾 𝑖 † = 𝑑𝑟 𝑢 𝑟 𝜓 † 𝑟 +𝜐 𝑟 𝜓 𝑟
with the normalization constraint
𝑢 𝑟 𝜐 𝑟 2 𝑑𝑟=1
in words, we create an extra particle with wave function 𝑢 𝑟 and an extra hole with wave function𝜐 𝑟 . Demanding that
𝐻 𝑚𝑓 , 𝛾 𝑖 † = 𝐸 𝑖 𝛾 𝑖 †
yields the famous Bogoliubov-de Gennes (BdG) equations, which in general, because of the spin degree of freedom, are 4×4 : here I give for simplicity the version for a single spin species (but keep the spatial “nonlocality”):

17. 𝐻 𝑜 𝑢 𝑖 𝒓 + Δ 𝒓,𝒓′ 𝜐 𝑖 𝒓′ 𝑑 𝑟 ′ = 𝐸 𝑖 𝑢 𝑖 𝒓
Δ ∗ 𝒓,𝒓′ 𝑢 𝑖 𝒓′ 𝑑 𝑟 ′ − 𝐻 𝑜 ∗ 𝜐 𝑖 𝒓 =𝐸 𝑖 𝜐 𝑖 𝒓
or schematically
𝐻 𝑜 Δ Δ − 𝐻 𝑜 ∗ 𝑢 𝜐 =𝐸 𝑢 𝜐
While the 𝑢’s do not in general by themselves form an orthonormal set, the spinors 𝑢,𝜐 do, i.e.
𝑢 𝑖 , 𝑢 𝑗 + 𝜐 𝑖 , 𝜐 𝑗 = 𝛿 𝑖𝑗 (generated by equations themselves). Hence the set of 𝛾 𝑖 † form a complete set of anticommuting Fermi operators, with 𝛾 𝑖 | Ψ BCS =0.
It is often pointed out in the literature that if 𝑢 𝑖 𝜐 𝑖 solves the BdG equations with energy eigenvalue 𝐸 𝑖 , then −𝜐 𝑖 ∗ 𝑢 𝑖 ∗ solves them with eigenvalue −𝐸 𝑖 , and so one says that this combination creates a “negative-energy state”. But this is illusory: the corresponding operator (call it 𝛾 𝑖 † ) is actually a combination of pure annihilators and hence itself simply a pure annihilator (in fact, with a  rotation of 𝜐 relation to 𝑢, it is just 𝛾 𝑖 ).

18. Problems with the MF-BdG approach:
(a) Galilean invariance: (simple BCS problem)
𝑚 =−𝒌
𝑚=𝒌
𝑣 𝑠 =0
−𝒌+ 𝑚𝒗 𝑠 ∕ℏ
𝒌+ 𝑚𝒗 𝑠 ∕ℏ
𝑣 𝑠 ≠0
Rest frame of condensate
Frame moving with velocity v 𝑠 w.r.t. condensate:
𝛼 𝑚 + = 𝑢 𝑘 𝑎 𝑘↑ + − 𝜐 𝑘 ∗ 𝑎 −𝑘↓
Ρ=ℏ 𝑢 𝑘 2 𝒌− 𝜐 𝑘 2 −𝒌
=ℏ𝒌 √
𝛼 𝑚 + ′ = 𝑢 𝑘 𝑎 𝑘+ 𝑚𝑣 𝑠 /ℏ + − 𝜐 𝑘 ∗ 𝑎 −𝑘+ 𝑚𝑣 𝑠 /ℏ
≡𝜬− 𝜬 0
𝜬− 𝜬 0 ′ =ℏ 𝑢 𝑘 2 𝑘+𝑚 𝑣 𝑠 /ℏ
− 𝜐 𝑘 2 −𝑘+ 𝑚𝑣 𝑠 /ℏ)
=ℏ𝑘+ 𝑢 𝑘 2 − 𝜐 𝑘 𝑚v 𝑠
Total momentum of groundstate =0
But: Galilean invariance requires simply
𝛲− 𝛲 0 ′ = 𝛲− 𝛲 0 + 𝑚v 𝑠 !!
note discrepancy depends on 𝑢 𝑘 , 𝜐 𝑘 , hence cannot be fixed simply by involving “spontaneously broken 𝑈 1 symmetry”.

19. Solution: when creating “hole” component of odd-parity state,
MUST ADD A COOPER PAIR!
thus, correct “Bog 𝑄P” creation operator is
𝐶 † ≡𝒩 𝑚 𝑐 𝑚 𝑎 𝑚 + 𝑎 𝑚 +
𝛼 𝑚 + = 𝑢 𝑚 𝑎 𝑚 + − 𝜐 𝑚 ∗ 𝑎 𝑚 𝐶 †
For condensate at rest, changes nothing (still have 𝑷− 𝑷 𝑜 =ℏk). But for condensate moving, adds an extra 2 𝜐 𝑘 𝑚𝑣 𝑠 , so that
𝑷− 𝑷 𝑜 ′=ℏ 𝑢 𝑘 𝒌+𝑚 𝒗 𝑠 /ℏ − 𝑣 𝑘 −𝒌+𝑚 𝒗 −𝑠 /ℏ 𝑣 𝑘 2 𝑚 𝒗 𝑠/ℏ
as required.
=ℏ𝒌+ m𝒗 𝑠 = 𝑃− 𝑃 𝑜 + m𝒗 𝑠
(6)
NMR of Majorana fermions (M.A. Silaev, PRB (2011)): consistent calculation based on MF Hamiltonian  spectral weight in longitudinal resonance absorption above Larmor frequency. (independent of dipole coupling constant 𝑔 𝐷 )
Problem: violates sum rule! (for 𝑔 𝐷 = 0, 𝜔𝜒 "(ω)𝑑ω=0, and for nonzero 𝑔 𝐷 , ∝ 𝑔 𝐷 )
Solution: need to consider response also of added Cooper pair (E. Taylor et al., arXiv: )
Moral: In any situation where Cooper pairs are behaving “nontrivially,”
MUST ENFORCE PARTICLE NUMBER CONSERVATION!