Which magnetic fields support a zero mode?

Rupert L. Frank; Michael Loss

doi:10.1515/crelle-2022-0015

Article

Which magnetic fields support a zero mode?

Rupert L. Frank and Michael Loss

Published/Copyright: April 28, 2022

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From the journal Journal für die reine und angewandte Mathematik (Crelles Journal) Volume 2022 Issue 788

Abstract

This paper presents some results concerning the size of magnetic fields that support zero modes for the three-dimensional Dirac equation and related problems for spinor equations. It is a well-known fact that for the Schrödinger equation in three dimensions to have a negative energy bound state, the 3/2 norm of the potential has to be greater than the Sobolev constant. We prove an analogous result for the existence of zero modes, namely that the 3/2 norm of the magnetic field has to greater than twice the Sobolev constant. The novel point here is that the spinorial nature of the wave function is crucial. It leads to an improved diamagnetic inequality from which the bound is derived. While the results are probably not sharp, other equations are analyzed where the results are indeed optimal.

Funding statement: Partial support through U.S. National Science Foundation grants DMS-1363432 and DMS-1954995 (Rupert L. Frank) and DMS-1856645 (Michael Loss) and through the Deutsche Forschungsgemeinschaft (German Research Foundation) through Germany’s Excellence Strategy EXC-2111-390814868 (Rupert L. Frank) is acknowledged.

A Some computations involving the Dirac matrices

The construction of zero modes in higher dimensions is more complicated and, as mentioned before, was accomplished by Dunne and Min [11] using information about the Dirac equation on the sphere. The advantage of their construction is that it delivers automatically the dimension of the zero mode space. If one is satisfied with less information, then there is, we believe, a simpler way to construct the Dunne–Min zero modes. Moreover, it gives the opportunity to get acquainted with some of the properties of the Dirac matrices. The basic idea is due to Adolf Hurwitz in his posthumously published paper “Über die Komposition der quadratischen Formen” [25]. In this paper he gave a complete classification of matrices γj, j=1,…,d, satisfying the relations

γj⁢γk+γk⁢γj=2⁢δi⁢j.

For our purpose we shall assume the matrices γj to be self-adjoint in the space ℂN with the usual inner product.

Theorem A.1.

Let d=2⁢ν+1 or d=2⁢ν, and consider the N×N hermitian matrices γj, j=1,…,d, satisfying

γi⁢γj+γj⁢γi=2⁢δi⁢j.

Then N=2ν and, if γj′ is another set of 2ν×2ν hermitian matrices satisfying the same relations, then there exists a 2ν×2ν unitary matrix A such that γj′=A*⁢γj⁢A for j=1,…,d.

The proof proceeds by reducing the γ matrices to a unitarily equivalent, but canonical set of matrices using an inductive procedure. We omit the proof and refer to [25].

Corollary A.2.

Let R be a d×d orthogonal matrix and define

γj′=∑k=1dRj⁢k⁢γk.

Then there exists a unitary matrix A such that for all j=1,…,d one has γj′=A*⁢γj⁢A.

The computation with γ matrices can be sometimes tedious and the following framework called “second quantization” is quite helpful.

In the remainder of this section, we assume that d=2⁢ν+1 is odd.

We single out the matrix γ1 and define the “annihilation” and “creation” operators

cj:=12⁢(γ2⁢j+i⁢γ2⁢j+1),cj*=12⁢(γ2⁢j-i⁢γ2⁢j+1),j=1,2,…,ν,

so that

γ2⁢j=cj+cj*,γ2⁢j+1=1i⁢(cj-cj*).

One easily checks that

cj⁢cj*+cj*⁢cj=I,cj2=cj*2=0

and, for k≠ℓ,

ck⁢cℓ+cℓ⁢ck=0,ck*⁢cℓ*+cℓ*⁢ck*=0,ck⁢cℓ*+cℓ*⁢ck=0.

Note that the matrix γ1 is not involved in these definitions.

Lemma A.3.

There exists a vector ϕ∈C2ν, a vacuum, such that ∥ϕ∥=1 and

cj⁢ϕ=0,j=1,…,ν.

Proof.

Since c12=0, it is clear that there exists ϕ≠0 such that c1⁢ϕ=0. Let k be the first index such that ck⁢ϕ≠0. Setting ψ=ck⁢ϕ, we see because of the commutation relations that cj⁢ψ=0, j=1,…,k-1, and ck⁢ψ=ck2⁢ϕ=0. Thus, replacing ϕ by ψ, we have that cj⁢ψ=0, j=1,…,k. Continuing in this fashion, we have a vector ϕ such that ck⁢ϕ=0 for all k=1,…,ν. ∎

Lemma A.4.

Let β→=(β1,…,βν) be a sequence with βj∈{0,1}, j=1,…,ν. Then the vector

|β→〉=ck1*β1⁢⋯⁢ckν*βν⁢ϕ

is non-zero if and only if the indices k1,…,kν are all distinct. In this case the vector is normalized. Moreover, the vectors |β→〉 form an orthonormal basis in C2ν.

In view of this lemma, we will sometimes denote ϕ=|0〉.

Proof.

If one or more of the indices are not distinct, then by commuting the various operators results in a square of one of the ci*, which is zero. Hence we may assume that the indices k1,…,kν are all distinct. We also may assume that β1=1 because otherwise ck1β1=I and we may move on to the next index. We have

∥ck1*β1⁢⋯⁢ckν*βν⁢ϕ∥2=(|0〉,ckνβν⁢⋯⁢ck1β1⁢ck1*β1⁢⋯⁢ckν*βν⁢ϕ)

and using ck1β1⁢ck1*β1=I-ck1*β1⁢ck1β1, we find

(|0〉,ckνβν⁢⋯⁢ck1β1⁢ck1*β1⁢⋯⁢ckν*βν⁢ϕ)=(|0〉,ckνβν⁢⋯⁢ck2β2⁢ck2*β2⁢⋯⁢ckν*βν⁢ϕ)

-(|0〉,ckνβν⁢⋯⁢ck1*β1⁢ck1β1⁢⋯⁢ckν*βν⁢ϕ).

The second term on the right side vanishes because the indices are distinct and thus ck1β1 either commutes or anti-commutes with all the matrices on the right and once it hits ϕ it yields zero. In this fashion we may move the annihilation matrices to the right and obtain that this state is normalized. Incidentally this also makes it clear that the state vanishes if two indices are the same on account of the fact that cj2=cj*2=0. From this argument it also follows that for β→≠β→′,

(|β→〉,|β→′〉)=0

and hence we have 2ν orthonormal vectors which constitute an orthonormal basis. ∎

Lemma A.5.

The vacuum is unique (up to a constant phase).

Proof.

Suppose that v is another vacuum, i.e., ∥v∥=1 and for all α=1,…,ν,

cα⁢v=0.

We may assume that 〈0|v〉=0. Then

(v,ck1*β1⁢⋯⁢ckν*βν⁢ϕ)

is always zero and therefore, by Lemma A.4, v=0, which is a contradiction. ∎

We note that γ1⁢ϕ satisfies the same properties as ϕ, namely, ∥γ1⁢ϕ∥=1 and

cα⁢γ1⁢ϕ=-γ1⁢cα⁢ϕ=0

for all α. By the uniqueness result of Lemma A.5 there is a θ∈ℝ such that γ1⁢ϕ=ei⁢θ⁢ϕ. Since γ1 is self-adjoint, we have ei⁢θ=±1. In case it is -1, we can change the sign of γ1 without changing the commutation relations and arrive at the same relation with +1. Hence we may adopt the convention that γ1⁢ϕ=ϕ.

The point about introducing this formalism is the following result.

Lemma A.6.

Introduce a (2⁢ν+1)×(2⁢ν+1) matrix ω with entries

ωα,β={0𝑖𝑓⁢α=1⁢𝑜𝑟⁢β=1⁢𝑜𝑟⁢α=β,〈0|i⁢γα⁢γβ|0〉𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒.

Then

ω=diag⁢(0,-i⁢σ2,…,-i⁢σ2),

where the zero is a number and there are ν 2×2-blocks -i⁢σ2.

Proof.

Since ω is skew and vanishes on the diagonal, it suffices to compute ωα,β when α<β. Moreover, since γ1⁢|0〉=|0〉 and γ1 is selfadjoint, we have ω1,β=0 for all β>1. For the remaining entries, we need to distinguish whether α and β are even or odd. When both are even, we have for 1≤j<k≤ν,

ω2⁢j,2⁢k=〈0|i⁢(cj+cj*)⁢(ck+ck*)|0〉=i⁢〈0|cj⁢ck*|0〉=0.

Similarly, when both are odd, we have for 1≤j<k≤ν,

ω2⁢j+1,2⁢k+1=-〈0|i⁢(cj-cj*)⁢(ck-ck*)|0〉=-i⁢〈0|cj⁢ck*|0〉=0.

Next, we consider α is even and β is odd. If α=β-1, we get

ω2⁢j,2⁢j+1=〈0|(cj+cj*)⁢(cj-cj*)|0〉=-1.

Otherwise, for 1≤j<k≤ν,

ω2⁢j,2⁢k+1=〈0|(cj+cj*)⁢(ck-ck*)|0〉=0.

Finally, we have the case where α is odd and β is even. For 1≤j<k≤ν, we get

ω2⁢j+1,2⁢k=〈0|(cj-cj*)⁢(ck+ck*)|0〉=0.

This proves the claimed formula for the entries of the matrix ω. ∎

After these preliminaries we discuss now an alternative approach to the Dunne–Min generalization [11] of [30]. The following example is relevant. It is the higher-dimensional analog of choice for the vector potential in [30]. Consider

(A.1)((1+i⁢γ→⋅x)⁢η,γ→⁢(1+i⁢γ→⋅x)⁢η),

where η∈ℂ2ν is normalized. Recall that for d=3 the γ matrices are the Pauli matrices and there is the well-know identity

|(η,σ→⁢η)|2=|η|4.

This leads to the identity

σ→⋅(η,σ→⁢η)⁢η=η,

which is very useful for constructing zero modes. It turns out that this identity also holds for d=5, but not in higher dimensions. In particular, it does not hold for (A.1) for general η.

Things simplify considerably if we choose the constant spinor η to be the vacuum ϕ. We compute

(1-i⁢γ→⋅x)⁢γj⁢(1+i⁢γ→⋅x)=γj-i⁢γ→⋅x⁢γj+i⁢γj⁢γ→⋅x+γ→⋅x⁢γj⁢γ→⋅x

=γj-2⁢i⁢x⋅γ→⁢γj+i⁢x⋅(γj⁢γ→+γ→⁢γj)-(x⋅γ→)2⁢γj

+x⋅γ→⁢x⋅(γj⁢γ→+γ→⁢γj)

=γj-2⁢i⁢x⋅γ→⁢γj+2⁢i⁢xj-|x|2⁢γj+2⁢x⋅γ→⁢xj

=(1-|x|2)⁢γj+2⁢x⋅γ→⁢xj-2⁢i⁢x⋅γ→⁢γj+2⁢i⁢xj.

Taking expectation we get

((1+i⁢γ→⋅x)⁢ϕ,γj⁢(1+i⁢γ→⋅x)⁢ϕ)=(1-|x|2)⁢(ϕ,γj⁢ϕ)+2⁢x⋅(ϕ,γ→⁢ϕ)⁢xj-2⁢∑k≠j,1xk⁢(ϕ,i⁢γk⁢γj⁢ϕ).

Since γ1⁢ϕ=ϕ, we find that (ϕ,γj⁢ϕ)=0, j≠1. Hence we have that for this particular state

((1+i⁢γ→⋅x)⁢ϕ,γ1⁢(1+i⁢γ→⋅x)⁢ϕ)=(1-|x|2+2⁢x12)⁢|ϕ|2.

For the component j≠1 we find

((1+i⁢γ→⋅x)⁢ϕ,γj⁢(1+i⁢γ→⋅x)⁢ϕ)=(2⁢x1⁢xj+2⁢[ω⁢x]j)⁢|ϕ|2.

Here ω is the (2⁢ν+1)×(2⁢ν+1) skew matrix introduced above. We introduce the field

Uj⁢(x):=11+|x|2⁢((1+i⁢γ→⋅x)⁢ϕ,γj⁢(1+i⁢γ→⋅x)⁢ϕ)={1-|x|2+2⁢x121+|x|2if ⁢j=1,2⁢x1⁢xj+2⁢[ω⁢x]j1+|x|2if ⁢j≠1.

This can be written more concisely as

U⁢(x→)=11+|x|2⁢((1-|x→|2)⁢e→1+2⁢(e→1⋅x→)⁢x→+2⁢ω⁢x→),

where [ω⁢x→]k=∑j=1dωk⁢j⁢xj. A straightforward computation shows that

|U⁢(x→)|2=1(1+|x|2)2⁢((1-|x→|2)⁢e→1+2⁢(e→1⋅x→)⁢x→+2⁢ω⁢x→)2

=1(1+|x|2)2⁢((1-|x→|2)2+4⁢(e→1⋅x→)2⁢|x|2+4⁢|ω⁢x→|2+4⁢(1-|x|2)⁢(e→1⋅x→)2).

Since |ω⁢x→|2=(ω⁢x→,ω⁢x→)=(x→,ωT⁢ω⁢x→)=∑j=2dxj2, we get

|U⁢(x→)|2=1.

In other words, the vector

U→⁢(x→)=(1+i⁢x→⋅γ→(1+|x|2)1/2⁢ϕ,γ→⁢1+i⁢x→⋅γ→(1+|x|2)1/2⁢ϕ)

is a unit vector. Now consider the self adjoint matrix

M:=U→⁢(x→)⋅γ→,

whose square is |U→⁢(x→)|2=1. Hence the eigenvalues of M are ±1. Moreover,

(1+i⁢x→⋅γ→(1+|x|2)1/2⁢ϕ,M⁢1+i⁢x→⋅γ→(1+|x|2)1/2⁢ϕ)=|U→⁢(x→)|2=1

and hence we have that

M⁢1+i⁢x→⋅γ→(1+|x|2)1/2⁢ϕ=1+i⁢x→⋅γ→(1+|x|2)1/2⁢ϕ.

If we set

ψ:=1+i⁢x→⋅γ→(1+|x|2)d2⁢ϕ,

then a simple computation yields

-i⁢γ→⋅∇⁡ψ=d1+|x|2⁢ψ

and if we define

A⁢(x):=d1+|x|2⁢U⁢(x),

then

-i⁢γ→⋅∇⁡ψ=γ→⋅A⁢ψ

and we have constructed our zero modes.

B Generalization of the spin-field interaction term to arbitrary dimensions

Squaring the Dirac equation yields

[γ⋅(-i⁢∇-A)]2=∑j⁢kγj⁢γk⁢(-i⁢∂j-Aj)⁢(-i⁢∂k-Ak)

=∑jγj⁢γj⁢(-i⁢∂j-Aj)⁢(-i⁢∂j-Aj)

+∑j≠kγj⁢γk⁢(-i⁢∂j-Aj)⁢(-i⁢∂k-Ak).

We have

∑jγj⁢γj⁢(-i⁢∂j-Aj)⁢(-i⁢∂j-Aj)=(-i⁢∇-A)2

and

∑j≠kγj⁢γk⁢(-i⁢∂j-Aj)⁢(-i⁢∂k-Ak)

=12⁢∑j≠kγj⁢γk⁢(-i⁢∂j-Aj)⁢(-i⁢∂k-Ak)+12⁢∑j≠kγk⁢γj⁢(-i⁢∂k-Ak)⁢(-i⁢∂j-Aj)

=12⁢∑j≠kγj⁢γk⁢[(-i⁢∂j-Aj)⁢(-i⁢∂k-Ak)-(-i⁢∂k-Ak)⁢(-i⁢∂j-Aj)]

=i2⁢∑j≠kγj⁢γk⁢[∂j⁡Ak-∂k⁡Aj].

For each fixed x∈ℝd, the matrix Bj⁢k:=∂j⁡Ak-∂k⁡Aj is an antisymmetric matrix and there is an orthogonal matrix R (depending on x) such that

RT⁢B⁢R=D,

where

D={diag⁢(D1⁢i⁢σ2,…,Dν⁢i⁢σ2,0)if⁢d=2⁢ν+1⁢is odd,diag⁢(D1⁢i⁢σ2,…,Dν⁢i⁢σ2)if⁢d=2⁢ν⁢is even.

Here there are ν 2×2 blocks i⁢σ2 and, if d is odd, an additional 1×1 “block” consisting of the number 0. For instance, in five dimensions

[0D1000-D10000000D2000-D20000000].

Since the trace of B is zero, we have

∑jγj⁢γj⁢Bj⁢j=0.

Hence

i2⁢∑j≠kγj⁢γk⁢Bj⁢k=i2⁢∑j⁢kγj⁢γk⁢Bj⁢k

=i2⁢∑α⁢β∑j⁢kγj⁢γk⁢Rj⁢α⁢Dα⁢β⁢Rk⁢β

=i2⁢∑α⁢β(∑jγj⁢Rj⁢α)⁢(∑kγk⁢Rk⁢β)⁢Dα⁢β.

If we set

Γβ:=∑jγj⁢Rj⁢β,

then, according to Corollary A.2, there exists a unitary matrix U such that

Γα=U*⁢γα⁢U,α=1,…,d,

and we can write

i2⁢∑j≠kγj⁢γk⁢Bj⁢k=U*⁢i2⁢∑α⁢βγα⁢γβ⁢Dα⁢β⁢U=U*⁢i⁢[γ1⁢γ2⁢D1+γ3⁢γ4⁢D2+…+γ2⁢ν-1⁢γ2⁢ν⁢Dν]⁢U.

The matrices γ1⁢γ2 and γ3⁢γ4, etc., are skew symmetric, commute with each other and we can simultaneously diagonalize them by a unitary matrix V, that is,

γ2⁢k-1⁢γ2⁢k=-i⁢V*⁢Σ2⁢k-1,2⁢k⁢V,k=1,…,ν,

with diagonal matrices Σ2⁢k-12⁢k. As (γi⁢γj)2=-1, the eigenvalues of Σ2⁢k-1,2⁢k must be ±1. Thus, all things considered, we get

i2⁢∑j≠kγj⁢γk⁢Bj⁢k=(V⁢U)*⁢[Σ12⁢D1+Σ34⁢D3+⋯⁢Σ2⁢ν-1,2⁢ν⁢Dν]⁢(V⁢U),

where the matrices Σi,i+1 are diagonal and have ±1 in the diagonal. Thus, if ψ is a spinor, we have that

|〈ψ,i2⁢∑j≠kγj⁢γk⁢Bj⁢k⁢ψ〉|≤|ψ|2⁢∑k=1ν|Dk|.

This fits with the three-dimensional case where ν=1 and |D1|=|B|.

Moreover, we have

∑k=1ν|Dk|≤ν12⁢(∑k=1ν|Dk|2)12=ν1/2⁢(∑j<k|Bj⁢k|2)12.

The last identity comes from the fact that conjugation by an orthogonal matrix R does not change the Hilbert–Schmidt norm of the matrix D=RT⁢B⁢R.

Acknowledgements

The authors would like to thank H. Kovarik and M. Lewin for helpful remarks.

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Received: 2021-01-29

Revised: 2021-11-29

Published Online: 2022-04-28

Published in Print: 2022-07-01

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https://doi.org/10.1515/crelle-2022-0015