semiclassical theory [10pt] with self-generated magnetic field...

Semiclassical theory

with self-generated magnetic fieldConference Analyse microlocaleen memoire de Bernard Lascar,

Jussieu, Paris, December 3, 2013

Victor Ivrii

Department of Mathematics, University of Toronto

Victor Ivrii (Math., Toronto) Self-generated magnetic field December 3, 2013 1 / 32

Table of Contents

Table of Contents

1 ProblemHistoryAnswerIncluding magnetic fieldSelf-generated magnetic field

2 Microlocal analysisLocalizationSingularity

3 Combined magnetic fieldLocalizationSingularity


Problem History

Problem

Let us consider quantum Hamiltonian

HV = (−ih∇)2 − V (x) (1)

with Thomas-Fermi potential V = V (x).

20+ years ago I was involved inthe problem:

Problem 1 (old)

Calculate semiclassical asymptotics of Tr(H−V ) as h → +0 where H−

V is anegative part of HV so we are looking for a sum of negative eigenvalues.


Problem History

Problem

Let us consider quantum Hamiltonian

HV = (−ih∇)2 − V (x) (1)

with Thomas-Fermi potential V = V (x). 20+ years ago I was involved inthe problem:

Problem 1 (old)

Calculate semiclassical asymptotics of Tr(H−V ) as h → +0 where H−

V is anegative part of HV so we are looking for a sum of negative eigenvalues.


Problem History

This one-particle problem arises in the multi-particle problem:

H = HN :=∑

1≤j≤N

HV ,xj +∑

1≤j<k≤N

|xj − xk |−1 (2)

on

H =⋀

1≤n≤N

H , H = L 2(R3,Cq) (3)

describing N same type particles in the external field with the scalarpotential −V and repulsing one another according to the Coulomb law.


Problem History

Here xj ∈ R3, and (x1, . . . , xN) ∈ R3N , potential V (x) is assumed to bereal-valued. Except when specifically mentioned we assume that

V (x) =∑

1≤m≤M

Zm

|x − ym|(4)

where Zm > 0 and ym are charges and locations of nuclei.

Mass is equal to12 and the Plank constant and a charge are equal to 1 here. The crucialquestion is the quantum statistics.

Quantum statistics

We assume that the particles (electrons) are fermions. This means thatthe Hamiltonian should be considered on the Fock space H defined by (3)of the functions antisymmetric with respect to variables x1, . . . , xN .


Problem History


V (x) =∑

1≤m≤M

Zm

|x − ym|(4)

where Zm > 0 and ym are charges and locations of nuclei. Mass is equal to12 and the Plank constant and a charge are equal to 1 here.

The crucialquestion is the quantum statistics.

Quantum statistics



Problem History


V (x) =∑

1≤m≤M

Zm

|x − ym|(4)

where Zm > 0 and ym are charges and locations of nuclei. Mass is equal to12 and the Plank constant and a charge are equal to 1 here. The crucialquestion is the quantum statistics.

Quantum statistics



Problem History

Thomas-Fermi theory

If electrons were not interacting between themselves but the field potentialwas −W (x) then they would occupy lowest eigenvalues and ground statewave functions would be (anti-symmetrized) 𝜑1(x1)𝜑2(x2) . . . 𝜑N(xN)where 𝜑j and 𝜆j are eigenfunctions and eigenvalues of H = −Δ−W (x)with energy Tr(H−

W+𝜈)− 𝜈N.

Then the local electron density would be 𝜌Ψ =∑

1≤j≤N |𝜑j(x)|2 andaccording to the pointwise Weyl law

𝜌Ψ(x) ≈q

6𝜋2(W + 𝜈)

32+ (5)

where 𝜈 = 𝜆N .

This density would generate potential −|x |−1 * 𝜌Ψ and we would haveW ≈ V − |x |−1 * 𝜌Ψ.


Problem History

Thomas-Fermi theory


W+𝜈)− 𝜈N.


1≤j≤N |𝜑j(x)|2

andaccording to the pointwise Weyl law

𝜌Ψ(x) ≈q

6𝜋2(W + 𝜈)

32+ (5)




Problem History

Thomas-Fermi theory


W+𝜈)− 𝜈N.


1≤j≤N |𝜑j(x)|2 andaccording to the pointwise Weyl law

𝜌Ψ(x) ≈q

6𝜋2(W + 𝜈)

32+ (5)




Problem History

Replacing all approximate equalities by a strict ones we arrive toThomas-Fermi equations:

V −W TF = |x |−1 * 𝜌TF, (6)

𝜌TF = P ′(W TF + 𝜈) :=q

6𝜋2(W TF + 𝜈)

32+, (7)∫

𝜌TF dx = min(N,Z ), Z = Z1 + . . .+ ZM (8)

where 𝜈 ≤ 0 is called chemical potential and in fact approximates 𝜆N ;q = 1 so far.

Assuming that Zj = zjZ with N ≍ Z ≫ 1 we discover that

W TF(x) = Z43 W TF(Z

13 x) (and 𝜈 = Z

43 𝜈) with W TF, 𝜈 calculated as if

Z = 1 and scaling x ↦→ Z13 x we arrive to e (1) with V := W TF + 𝜈 and

h = Z− 13 (and the result must be multiplied by Z

43 ).


Problem Answer

Answer

It was proven for Problem 1 that

Tr(H−V ) = κ0h

−3 + κ1h−2 + O(h−1), (9)

The second term Scott is due to Coulomb-like singularities in ym (this wasa challenging part); we need to assume that |ym − ym′ | & 1 ∀m = m′ (afterrescaling); later under assumption |ym − ym′ | ≫ 1 ∀m = m′ the next termSchwinger= κ2h

−1 was recovered and remainder was improved to o(h−1).


Problem Answer

Answer


Tr(H−V ) = κ0h

−3 + κ1h−2 + O(h−1), (9)

The second term Scott is due to Coulomb-like singularities in ym (this wasa challenging part); we need to assume that |ym − ym′ | & 1 ∀m = m′ (afterrescaling);

later under assumption |ym − ym′ | ≫ 1 ∀m = m′ the next termSchwinger= κ2h



Problem Answer

Answer


Tr(H−V ) = κ0h

−3 + κ1h−2 + O(h−1), (9)

The second term Scott is due to Coulomb-like singularities in ym (this wasa challenging part); we need to assume that |ym − ym′ | & 1 ∀m = m′ (afterrescaling); later under assumption |ym − ym′ | ≫ 1 ∀m = m′ the next termSchwinger= κ2h



Problem Answer

Remark

In asymptotics (9)

κ0 = −∫

P(V ) dx , P(V ) :=q

15𝜋2V

52+ , (10)

κ1 =∑

1≤m≤M

qz2mS (11)

but for original problem we need to take

κ0 = −∫

P(V ) dx − 1

2

x|x − y |−1𝜌TF(x)𝜌TF(y) dxdy (9)′

to avoid double counting of the energies of electron-electron interactionand add Dirac= κ′

2h−1 to avoid counting of electron self-interaction.


Problem Including magnetic field

Including magnetic field

To accommodate magnetic field we need to consider H = L 2(R3,Cq)with q = 2 and

HV ,A =((i∇− A) · σ

)2 − V (x) (12)

where σ = (σ1,σ2,σ3), σj are Pauli matrices.

Such problem with constant magnetic field (linear A(x)) was investigated20− y.a. and one should replace P(V ) by

PBh(V ) = (3𝜋2)−1q(12V

32+ +

∑j≥1

(V − 2jBh)32+

)Bh (13)

(which leads to magnetic Thomas-Fermi potential W TFBh and density 𝜌TFBh .

Here B = |∇ × A| is an intensity of magnetic field).Results for Bh . 1 and Bh & 1 are really different.





HV ,A =((i∇− A) · σ

)2 − V (x) (12)



PBh(V ) = (3𝜋2)−1q(12V

32+ +

∑j≥1

(V − 2jBh)32+

)Bh (13)


Here B = |∇ × A| is an intensity of magnetic field).

Results for Bh . 1 and Bh & 1 are really different.





HV ,A =((i∇− A) · σ

)2 − V (x) (12)



PBh(V ) = (3𝜋2)−1q(12V

32+ +

∑j≥1

(V − 2jBh)32+

)Bh (13)


Here B = |∇ × A| is an intensity of magnetic field).Results for Bh . 1 and Bh & 1 are really different.


Problem Self-generated magnetic field

Finally a case of study: self-generated magnetic field

A couple of y.a. it was proposed to consider arbitrary magnetic field but toinclude its energy to a final count which amounts to

Problem 2 (new)

Find E*𝜅 := infA∈H 1(R3) E𝜅(A) with

E𝜅(A) := Tr(H−A,V ) + 𝜅−1h−2

∫|𝜕A|2 dx . (14)

Remark

In our assumptions to V one can prove by functional analysis that suchminimizer exists as 𝜅 ≤ 𝜅* (small enough constant) but we have no idea ifit is unique! (life would me much easier then).





Problem 2 (new)


E𝜅(A) := Tr(H−A,V ) + 𝜅−1h−2

∫|𝜕A|2 dx . (14)

Remark

In our assumptions to V one can prove by functional analysis that suchminimizer exists as 𝜅 ≤ 𝜅* (small enough constant)

but we have no idea ifit is unique! (life would me much easier then).





Problem 2 (new)


E𝜅(A) := Tr(H−A,V ) + 𝜅−1h−2

∫|𝜕A|2 dx . (14)

Remark

In our assumptions to V one can prove by functional analysis that suchminimizer exists as 𝜅 ≤ 𝜅* (small enough constant) but we have no idea ifit is unique! (life would me much easier then).


Microlocal analysis Localization

Microlocal analysis

First consider a simplified problem: V ∈ C 2+ and HA,V replaced by𝜓HA,V𝜓 with 𝜓 ∈ C 2

0 (B(0, 1)).

Then a minimizer must satisfy

1

𝜅h2ΔAj(x) = Φj(x) :=

− Re tr(σj

((hD − A)x · σ

)(𝜓(x)e(x , y , 0)𝜓(y)

))y=x

(15)

where e(x , y , 𝜏) is the Schwartz kernel of the spectral projectorθ(𝜏 − 𝜓HA,V𝜓).



Microlocal analysis

First consider a simplified problem: V ∈ C 2+ and HA,V replaced by𝜓HA,V𝜓 with 𝜓 ∈ C 2

0 (B(0, 1)). Then a minimizer must satisfy

1

𝜅h2ΔAj(x) = Φj(x) :=

− Re tr(σj

((hD − A)x · σ

)(𝜓(x)e(x , y , 0)𝜓(y)

))y=x

(15)

where e(x , y , 𝜏) is the Schwartz kernel of the spectral projectorθ(𝜏 − 𝜓HA,V𝜓).



Estimate to a minimizer: smooth case

On the right we have a spectral expression with corresponding Weylexpression 0.

So in fact we have a remainder.

Trouble: we do not know how regular A is.

Good news: I was able to adopt microlocal analysis to deal with thisnon-smoothness (combining rough microlocal analysis of earlierdevelopment with successive approximations). Observe that |Φj | = O(h−3)(it cannot be worse than this–one can prove it). But then |ΔAj | = O(h−1)and we have an estimate to a minimizer |𝜕2Aj | = O(h−𝜃| log h|) with𝜃 = 1 (and factor 𝜅 definitely does not hurt!).

This is enough to improve estimate of |Φj | which is enough to push anestimate to a minimizer down: |𝜕2Aj | = O(h𝛿−𝜃| log h|) which is enoughto push it further down, . . . , until non-smoothness is not a problem and|Φj | = O(h−2) and

|𝜕2Aj | = O(| log h|). (16)




On the right we have a spectral expression with corresponding Weylexpression 0. So in fact we have a remainder.




|𝜕2Aj | = O(| log h|). (16)







This is enough to improve estimate of |Φj | which is enough to push anestimate to a minimizer down: |𝜕2Aj | = O(h𝛿−𝜃| log h|)

which is enoughto push it further down, . . . , until non-smoothness is not a problem and|Φj | = O(h−2) and

|𝜕2Aj | = O(| log h|). (16)







This is enough to improve estimate of |Φj | which is enough to push anestimate to a minimizer down: |𝜕2Aj | = O(h𝛿−𝜃| log h|) which is enoughto push it further down, . . . ,

until non-smoothness is not a problem and|Φj | = O(h−2) and

|𝜕2Aj | = O(| log h|). (16)








|𝜕2Aj | = O(| log h|). (16)



Trace estimate: smooth case

Then I can prove by the same arguments

Tr((𝜓HA,V𝜓)−) + h−3

∫P(V ) dx = O(h−1). (17)

In the process every time I have estimate O(h−1−𝜃) we conclude thatE𝜅(A) = κ0h

−3 + O(h−1−𝜃) and thenE𝜅(A) = κ0h

−3 + 𝜅−1h−2‖𝜕A‖2 + O(h−1−𝜃) butE𝜅(0) = κ0h

−3 + O(h−1−𝜃) and since A is a minimizer E𝜅(A) ≤ E𝜅(0)and ‖𝜕A‖2 = O(h1−𝜃), and ‖𝜕A‖L ∞ = O(h(1−𝜃)/5).

Boring!

Therefore we arrive to a solid but not very exciting result: in these settings

E*𝜅 = κ0h

−3 + O(h−1), ‖𝜕A‖ = O(h12 ) and ‖𝜕A‖L ∞ = O(h

15 ).






∫P(V ) dx = O(h−1). (17)



−3 + 𝜅−1h−2‖𝜕A‖2 + O(h−1−𝜃)

butE𝜅(0) = κ0h


Boring!


E*𝜅 = κ0h

−3 + O(h−1), ‖𝜕A‖ = O(h12 ) and ‖𝜕A‖L ∞ = O(h

15 ).






∫P(V ) dx = O(h−1). (17)



−3 + 𝜅−1h−2‖𝜕A‖2 + O(h−1−𝜃) butE𝜅(0) = κ0h


Boring!


E*𝜅 = κ0h

−3 + O(h−1), ‖𝜕A‖ = O(h12 ) and ‖𝜕A‖L ∞ = O(h

15 ).


Microlocal analysis Singularity

But we have a singularity!

Now consider Coulomb-like singularity at 0. Consider ℓ-admissible partitionof unity with ℓ(x) = 1

2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥

∑k Tr((𝜓HA,V ′𝜓)−) with V ′ = V + ch2|x |−2.

Consider a ball B(z , r) with r = 12 |z |. Scaling x ↦→ (x − z)r−1,

h ↦→ ~ = hr−12 and 𝜏 ↦→ 𝜏 r we find ourselves in the framework of smooth

theory and then contribution of B(z , r) with r ≥ h2 to the remainder in

the trace asymptotics is O(r−1~−1) = O(r−12 h−1) and summation by

r ≥ h−2 results in O(h−2). One can prove that contribution of B(0, h2) isO(h−2) as well.

So instead of O(h−1) we get O(h−2). Exactly this happened 20+ yearsago–but then there was no magnetic field. And this was not a failure–thiswas manifestation of Scott correction term!





2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥




theory

and then contribution of B(z , r) with r ≥ h2 to the remainder in








2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥





the trace asymptotics is O(r−1~−1) = O(r−12 h−1)

and summation byr ≥ h−2 results in O(h−2). One can prove that contribution of B(0, h2) isO(h−2) as well.






2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥






r ≥ h−2 results in O(h−2).

One can prove that contribution of B(0, h2) isO(h−2) as well.






2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥












2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥







So instead of O(h−1) we get O(h−2).

Exactly this happened 20+ yearsago–but then there was no magnetic field. And this was not a failure–thiswas manifestation of Scott correction term!





2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥







So instead of O(h−1) we get O(h−2). Exactly this happened 20+ yearsago–but then there was no magnetic field.

And this was not a failure–thiswas manifestation of Scott correction term!





2 |x |: 1 =∑𝜓2k ; then

Tr(H−A,V ) ≥










Estimate to a minimizer: Coulomb case

First we conclude that ‖𝜕A‖ = O(1).

If we knew that the minimizer isunique then in spherically symmetric case it will be spherically symmetricvector field with 0 divergence and belonging to H 1 and therefore 0. Butwe do not know if a minimizer is unique!Using estimate to minimizer (15) but now with 𝜓 = 1 and estimate‖𝜕A‖ = O(1) I was able to derive

‖𝜕A‖ ≤ C𝜅, |𝜕A| ≤ C𝜅ℓ−32 , |𝜕A| ≤ C𝜅ℓ−

52 | log(ℓh−2)|. (18)




First we conclude that ‖𝜕A‖ = O(1). If we knew that the minimizer isunique then in spherically symmetric case it will be spherically symmetricvector field with 0 divergence and belonging to H 1 and therefore 0. Butwe do not know if a minimizer is unique!

Using estimate to minimizer (15) but now with 𝜓 = 1 and estimate‖𝜕A‖ = O(1) I was able to derive

‖𝜕A‖ ≤ C𝜅, |𝜕A| ≤ C𝜅ℓ−32 , |𝜕A| ≤ C𝜅ℓ−

52 | log(ℓh−2)|. (18)




First we conclude that ‖𝜕A‖ = O(1). If we knew that the minimizer isunique then in spherically symmetric case it will be spherically symmetricvector field with 0 divergence and belonging to H 1 and therefore 0. Butwe do not know if a minimizer is unique!Using estimate to minimizer (15) but now with 𝜓 = 1 and estimate‖𝜕A‖ = O(1) I was able to derive

‖𝜕A‖ ≤ C𝜅, |𝜕A| ≤ C𝜅ℓ−32 , |𝜕A| ≤ C𝜅ℓ−

52 | log(ℓh−2)|. (18)



Now we deal with the singularity exactly as I did 20+ y.a. Observe that

the contribution of zone {ℓ(x) ≥ r} to the remainder is O(r−12 h−1).

Now assume temporarily that M = 1 and y1 = 0 and consider ourpotential as a perturbation of purely Coulomb potential V 0 = z |x |−1; thedifference near singularity is O(1). Therefore considering the differenceand writing Weyl expression for a perturbation we see that thecontribution of B(x , r) to an error is O(~−2) = O(rh−2) and summationby zone {ℓ(x) ≤ r} results in O(r h−2).

So the total error is O(r−12 h−1 + r h−2) and optimizing by r we get

O(h−43 ). So, if we consider the errors when we replace traces by their

Weyl expressions and consider the difference between these errors for HA,V

and HA,V 0 we get O(h−43 ).





Now assume temporarily that M = 1 and y1 = 0 and consider ourpotential as a perturbation of purely Coulomb potential V 0 = z |x |−1; thedifference near singularity is O(1).

Therefore considering the differenceand writing Weyl expression for a perturbation we see that thecontribution of B(x , r) to an error is O(~−2) = O(rh−2) and summationby zone {ℓ(x) ≤ r} results in O(r h−2).











O(h−43 ).

So, if we consider the errors when we replace traces by theirWeyl expressions and consider the difference between these errors for HA,V




Except for Coulomb potential trace is infinite and Weyl expression divergesat infinity, so it must be regularized, f.e. as

Tr(HA,V 0−𝜂) + h−3

∫P(V 0 − 𝜂) dx (19)

with 𝜂 > 0.

Then

E𝜅(V ,A) + h−3

∫P(V ) dx

= E𝜅(V0 − 𝜂,A) + h−3

∫P(V 0(x)− 𝜂) dx + O(h−

43 ) (20)

(uniformly by 𝜂). Then the left-hand expression is estimated from below by

lim𝜂→+0

(E*𝜅(V

0 − 𝜂) + h−3

∫P(V 0 − 𝜂) dx

)+O(h−

43 ).




Tr(HA,V 0−𝜂) + h−3

∫P(V 0 − 𝜂) dx (19)

with 𝜂 > 0. Then

E𝜅(V ,A) + h−3

∫P(V ) dx

= E𝜅(V0 − 𝜂,A) + h−3

∫P(V 0(x)− 𝜂) dx + O(h−

43 ) (20)

(uniformly by 𝜂).

Then the left-hand expression is estimated from below by

lim𝜂→+0

(E*𝜅(V

0 − 𝜂) + h−3

∫P(V 0 − 𝜂) dx

)+O(h−

43 ).




Tr(HA,V 0−𝜂) + h−3

∫P(V 0 − 𝜂) dx (19)

with 𝜂 > 0. Then

E𝜅(V ,A) + h−3

∫P(V ) dx

= E𝜅(V0 − 𝜂,A) + h−3

∫P(V 0(x)− 𝜂) dx + O(h−

43 ) (20)

(uniformly by 𝜂). Then the left-hand expression is estimated from below by

lim𝜂→+0

(E*𝜅(V

0 − 𝜂) + h−3

∫P(V 0 − 𝜂) dx

)+O(h−

43 ).



Scott correction term

In the selected expression we have three parameters–h, 𝜅 and z but due tohomogeneity of Coulomb potential we can exclude two and rewrite it as2h−2z2S(𝜅z) with unknown function S(𝜅z).

Then

E *𝜅(V ) + h−3

∫P(V ) dx ≥ 2h−2z2S(𝜅z) + O(h−

43 ). (21)

This is estimate from below. To derive estimate from above we plug testfunction A′

𝜂 which is minimizer for E𝜅(V0 − 𝜂,A) arriving to

E*𝜅(V )+h−3

∫P(V ) dx ≤ E𝜅(V

0−𝜂,A)+h−3

∫P(V 0−𝜂) dx+O(h−

43 )

= E*𝜅(V

0 − 𝜂) + h−3

∫P(V 0 − 𝜂) dx + O(h−

43 )

also uniformly by 𝜂 > 0 and then

E*𝜅(V ) + h−3

∫P(V ) dx ≤ 2h−2z2S(𝜅z) + O(h−

43 ). (22)




In the selected expression we have three parameters–h, 𝜅 and z but due tohomogeneity of Coulomb potential we can exclude two and rewrite it as2h−2z2S(𝜅z) with unknown function S(𝜅z). Then

E *𝜅(V ) + h−3

∫P(V ) dx ≥ 2h−2z2S(𝜅z) + O(h−

43 ). (21)

This is estimate from below.

To derive estimate from above we plug testfunction A′


E*𝜅(V )+h−3


0−𝜂,A)+h−3


43 )

= E*𝜅(V

0 − 𝜂) + h−3

∫P(V 0 − 𝜂) dx + O(h−

43 )


E*𝜅(V ) + h−3

∫P(V ) dx ≤ 2h−2z2S(𝜅z) + O(h−

43 ). (22)




In the selected expression we have three parameters–h, 𝜅 and z but due tohomogeneity of Coulomb potential we can exclude two and rewrite it as2h−2z2S(𝜅z) with unknown function S(𝜅z). Then

E *𝜅(V ) + h−3

∫P(V ) dx ≥ 2h−2z2S(𝜅z) + O(h−

43 ). (21)

This is estimate from below. To derive estimate from above we plug testfunction A′


E*𝜅(V )+h−3


0−𝜂,A)+h−3


43 )

= E*𝜅(V

0 − 𝜂) + h−3

∫P(V 0 − 𝜂) dx + O(h−

43 )


E*𝜅(V ) + h−3

∫P(V ) dx ≤ 2h−2z2S(𝜅z) + O(h−

43 ). (22)



M ≥ 2 and decoupling of singularities

If we have more than one singularity the same arguments work but in theestimate from below we have a term

−C𝜅−1h−2‖𝜕A′‖2𝒳 (23)

with minimizer A and 𝒳 = {x : 13a ≤ ℓ(x) ≤ a} where ℓ(x) is the distance

to the closest nuclei and a is the minimal distance between nuclei.

However for potential which decays at infinity sufficiently fast (W TF + 𝜈)+decays as ℓ−4 one I proved that |𝜕A| = O(𝜅ℓ−3) and then (23) does notexceed C𝜅a−3h−2 and we arrive to estimate

E*𝜅 = κ0h

−3 + κ1h−2 + O(h−

43 + 𝜅a−3h−2) (24)

with

κ1 =∑

1≤m≤M

2z2mS(𝜅zm). (25)



M ≥ 2 and decoupling of singularities

If we have more than one singularity the same arguments work but in theestimate from below we have a term

−C𝜅−1h−2‖𝜕A′‖2𝒳 (23)

with minimizer A and 𝒳 = {x : 13a ≤ ℓ(x) ≤ a} where ℓ(x) is the distance

to the closest nuclei and a is the minimal distance between nuclei.However for potential which decays at infinity sufficiently fast (W TF + 𝜈)+decays as ℓ−4 one I proved that |𝜕A| = O(𝜅ℓ−3) and then (23) does notexceed C𝜅a−3h−2 and we arrive to estimate

E*𝜅 = κ0h

−3 + κ1h−2 + O(h−

43 + 𝜅a−3h−2) (24)

with

κ1 =∑

1≤m≤M

2z2mS(𝜅zm). (25)



More delicate arguments based on refined analysis of propagation ofsingularities I developed 20− y.a. and recently adopted to current problemallow to improve this to

Theorem 1

For Thomas-Fermi potential rescaled with a ≥ 1

E*𝜅 = κ0h

−3 + κ1h−2 + κ2h

−1

+ O(h−1+𝛿 + h−1a−𝛿 + 𝜅| log 𝜅|13 h−

43 + 𝜅a−3h−2) (26)

with κ0, κ2 exactly as without self-generating magnetic field and with 𝜅1given by (25).



Remark

1 Term 𝜅a−3h−2 shows one extra difficulty as M ≥ 2: the loss oflocality due to self-generated magnetic field. Fortunately in free nucleimodel the last term in the remainder estimate could be dropped.

2 Apart from S(𝜅) being monotone decaying and S(𝜅) ≥ S(0)− c𝜅(and value S(0)) we don’t know a damn thing about it. It mayhappen that S(𝜅) = S(0) as 𝜅 < 𝜅* or that S(𝜅) = −∞ as 𝜅 is largeenough!


Combined magnetic field Localization

Combined magnetic field: work in progress

Currently I am working on the combined magnetic field when A = A0 + A′

with constant external magnetic field of intensity 𝛽 A0 (soA(x) = 1

2(−𝛽x2,+𝛽x1, 0) and unknown self-generated magnetic field A′

and only energy of A′ is counted:

E𝜅(A′) = Tr(H−

A,V ) + 𝜅−1h−2‖𝜕A′‖2. (27)

Let us start from local smooth theory as we did as A0 = 0. Let us explorefirst semiclassical approximation without justification:

ℰ𝜅(A) := −h−3

∫PBh(V )𝜓2 dx + 𝜅−1h−2‖𝜕A′‖2 (28)

where B = |∇ × A| is an intensity of the combined field.



Assuming that 𝜇 = |𝜕A′| ≪ 𝛽 (which could be ) observe thatB = 𝛽 + (𝜕1A

′2 − 𝜕2A

′1) + O(𝜇2𝛽−1) and then

PBh(V ) ≈ P𝛽h(V ) + 𝜕𝛽P𝛽h(V )(𝜕1A′2 − 𝜕2A

′1) and we arrive to

− h−3

∫P𝛽h(V )𝜓2 dx

−h−3

∫𝜕𝛽P𝛽h(V )(𝜕1A

′2 − 𝜕2A

′1)𝜓

2 dx + 𝜅−1h−2‖𝜕A′‖2 . (29)

Optimizing we get |𝜕A′| ≍ 𝜅𝛽h and selected expression ≍ −𝜅𝛽2 as 𝛽h ≤ 1and |𝜕A′| ≍ 𝜅 and selected expression ≍ −𝜅h−2 as 𝛽h ≥ 1.

So self-generated magnetic field appears on the semiclassical level and Ican actually prove that if 𝛽h . 1 but 𝜅𝛽2h ≫ 1 then‖𝜕(A′ − A′′)‖ ≪ ‖𝜕A′‖ where A′, A′′ are minimizers for E𝜅(A) and ℰ𝜅(A)respectively. The same is true in some instances even as 𝛽h & 1.




′2 − 𝜕2A




− h−3


−h−3


′2 − 𝜕2A

′1)𝜓

2 dx + 𝜅−1h−2‖𝜕A′‖2 . (29)

Optimizing we get |𝜕A′| ≍ 𝜅𝛽h and selected expression ≍ −𝜅𝛽2 as 𝛽h ≤ 1and

|𝜕A′| ≍ 𝜅 and selected expression ≍ −𝜅h−2 as 𝛽h ≥ 1.





′2 − 𝜕2A




− h−3


−h−3


′2 − 𝜕2A

′1)𝜓

2 dx + 𝜅−1h−2‖𝜕A′‖2 . (29)






′2 − 𝜕2A




− h−3


−h−3


′2 − 𝜕2A

′1)𝜓

2 dx + 𝜅−1h−2‖𝜕A′‖2 . (29)


So self-generated magnetic field appears on the semiclassical level

and Ican actually prove that if 𝛽h . 1 but 𝜅𝛽2h ≫ 1 then‖𝜕(A′ − A′′)‖ ≪ ‖𝜕A′‖ where A′, A′′ are minimizers for E𝜅(A) and ℰ𝜅(A)respectively. The same is true in some instances even as 𝛽h & 1.




′2 − 𝜕2A




− h−3


−h−3


′2 − 𝜕2A

′1)𝜓

2 dx + 𝜅−1h−2‖𝜕A′‖2 . (29)


So self-generated magnetic field appears on the semiclassical level and Ican actually prove that if 𝛽h . 1 but 𝜅𝛽2h ≫ 1 then‖𝜕(A′ − A′′)‖ ≪ ‖𝜕A′‖ where A′, A′′ are minimizers for E𝜅(A) and ℰ𝜅(A)respectively.

The same is true in some instances even as 𝛽h & 1.




′2 − 𝜕2A




− h−3


−h−3


′2 − 𝜕2A

′1)𝜓

2 dx + 𝜅−1h−2‖𝜕A′‖2 . (29)





Estimates to minimizer

The first step is to estimate minimizer starting from equation (15) whichstill holds:

1

𝜅h2ΔA′

j(x) = Φj(x) :=

− Re tr(σj

((hD − A)x · σ

)(𝜓(x)e(x , y , 0)𝜓(y)

))y=x

. (15)

Recall that the right-hand expression there is a pointwise spectralexpression and they are not easy under magnetic field: the obstacle are notonly periodic trajectories but also loops (especially short loops) and theyare plentiful in this case.

Luckily I already investigated similar asymptotics in Chapter 16 of[V. Ivrii, Future Book].



Trace estimates

Theorem 3

1 In the smooth local theory let |𝜕2A′| ≤ 𝜈 ≤ 𝜖𝛽, |𝜕A′| ≤ 𝜈 ≤ 𝜖𝛽. Then

|Tr((𝜓HA,V𝜓)−) + h−3

∫PBh(V )𝜓2|

≤ C

{h−1 + h−

13 𝜈

43 𝛽h ≤ 1,

𝛽 + 𝛽h23 𝜈

43 𝛽h ≥ 1

(31)

under modest non-degeneracy assumption

minj≥0

|V − 2j𝛽h|+ |∇V |+ |∇2V | ≍ 1; (32)

2 In the general case one needs to add C𝛽h−12 to the right-hand

expression.



Trace estimates

Theorem 3

1 In the smooth local theory let |𝜕2A′| ≤ 𝜈 ≤ 𝜖𝛽, |𝜕A′| ≤ 𝜈 ≤ 𝜖𝛽. Then

|Tr((𝜓HA,V𝜓)−) + h−3

∫PBh(V )𝜓2|

≤ C

{h−1 + h−

13 𝜈

43 𝛽h ≤ 1,

𝛽 + 𝛽h23 𝜈

43 𝛽h ≥ 1

(31)

under modest non-degeneracy assumption

minj≥0

|V − 2j𝛽h|+ |∇V |+ |∇2V | ≍ 1; (32)

2 In the general case one needs to add C𝛽h−12 to the right-hand

expression.Victor Ivrii (Math., Toronto) Self-generated magnetic field December 3, 2013 27 / 32


Remark

1 In Theorem 3 threshold happens as it should at 𝛽 ≍ h−1 (but thereare hidden threshold for estimate of 𝜈);

2 Theorem 3 is proven by the same advanced non-smooth microlocalanalysis;

3 Combining Theorems 2, 3 we get remainder estimates; plugging into

(31) we can skip 𝜅𝛽12 in (30);

4 Then ‖𝜕(A′ − A′′)‖ ≤ C (𝜅h2Q)12 where Q is the remainder estimate

in Theorem 3 and ‖𝜕(A′ − A′′)‖L ∞ ≤ C‖𝜕2A′‖35L ∞‖𝜕(A′ − A′′)‖

25 .


Combined magnetic field Singularity

Singularity

Simple scaling shows that “near singularity” zone adds Scott correctionterm 2S(𝜅z)z2h−2 to the main part of asymptotics and

O(𝛽13 h−1 + 𝜅| log 𝜅|

13𝛽

29 h−

43 ) to the remainder as 1 ≤ 𝛽 ≤ h−2:

For 𝛽 ≤ 1 results are like there was no external magnetic field;

For 𝛽 ≥ h−2 results are as if there are no singularities at all.

However as 1 ≤ 𝛽 ≤ h−2 non-locality of self-generated magnetic fieldentangles singularity with the regular zone and other singularities. Still,adding O(𝜅h−2) to the remainder estimate–which is not necessary a lossat all–allows us to detach singularity.



Case 𝛽h ≤ 1

Theorem 4

1 Let M = 1 (single singularity), 1 ≤ 𝛽 ≤ h−1 and non-degeneracyassumption (32) be fulfilled. Then

E*𝜅 = κ0h

−3 + κ1h−2 + O(𝛽

13 h−1 + 𝜅| log 𝜅|

13𝛽

29 h−

43 + 𝜅𝛽h−1) (33)

with κ0 = −∫P𝛽h(V ) dx and κ1 = κ1(𝜅) is the same Scott

correction term as without external magnetic field;

2 Without non-degeneracy assumption one should add O(𝛽h−12 ) to the

remainder estimate;

3 If M = 2 and V decays fast enough from singularities (V = O(ℓ−4)would be enough) then one should add O(𝜅a−3h−2) to the remainderestimate.



Case 𝛽h ≥ 1

Theorem 5

1 Let h−1 ≤ 𝛽 ≤ h−2, 𝜅𝛽h2 ≤ | log h|−K and non-degeneracyassumption (32) be fulfilled. Then

E*𝜅 = κ0h

−3 + κ1h−2 + O(𝛽 + 𝛽

13 h−1 + 𝜅h−2 + 𝛽h

23 𝜈

43 ) (34)

with κ0 = −∫P𝛽h(V ) dx and κ1 = κ1(0) is the same Scott

correction term as without any magnetic field at all; here 𝜈 is theright-hand expression in (30);

2 Without non-degeneracy assumption one should add O(𝛽h−12 ) to the

remainder estimate.



Reference

Microlocal Analysis, Sharp Spectral, Asymptotics and Applications (inprogress)http://weyl.math.toronto.edu/victor2/futurebook/futurebook.pdf

Chapter 26. Asymptotics of the ground state energy of heavymolecules in self-generated magnetic field, pp 2350–2424;Chapter 27. Asymptotics of the ground state energy of heavymolecules in combined magnetic field, (in progress);

Large atoms and molecules with magnetic field, includingself-generated magnetic field (Results: old, new, in progress and inperspective)http://weyl.math.toronto.edu/victor2/preprints/Talk 10.pdf


http://weyl.math.toronto.edu/victor2/futurebook/futurebook.pdf

http://weyl.math.toronto.edu/victor2/preprints/Talk_10.pdf

semiclassical theory [10pt] with self-generated magnetic field...

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