Quantum Coulomb systems at the Brink Dirk Hundertmark (based on - - PowerPoint PPT Presentation

Quantum Coulomb systems at the Brink

Dirk Hundertmark (based on joint work with Markus Lange (UBC) and Michal Jex (KIT))

CIRM: The Analysis of Complex Quantum Systems: Large Coulomb Systems and Related Matters.

Karlsruhe Institute of Technology – Institute for Analysis // CRC 1173 Wave Phenomena: Analysis and Numerics

supported by the Deutsche Forschungsgemeinschaft (DFG) through CRC 1173.

Outline The approach

Outline

Some results. Decay of eigenstates at thresholds: Repulsion is your friend The physical heuristic. Making the heuristic into a proof: One body operators. Helium.

1 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

An atom

The operator for an atom with nuclear charge Z is H =

N

• j=1
• 1

2mP2

j − Ze

|xj|

• +
• 1≤j<k≤N

e2

|xj − xk|

Pj = −i∇j = −i∇xj, the kinetic energy of particle j, m electron mass, Z nuclear charge. Real atoms: Z = ze, z ∈ N, helium: Z = 2e, but in theory any Z > 0 allowed . Acts on L2(R3N) (either with or without symmetry constraints) No spin, but this can be easily handled.

2 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Scaling

Let (Uλψ)(x) ≔ λ3Nψ(λx), x = (x1, ·, xN) ∈ R3N. Then Uλ : L2(R3N) → L2(R3N) is unitary and H(Uλψ)(λx) = λ3N

• λ22

2m

N

• j=1
• (−∆jψ)(λx) − Zeλ

|λxj| ψ(λx)

• +
• 1≤j<k≤N

λe2 |λxj − λxk| ψ(λx)

• = λ3N λ22

2m

N

• j=1
• (−∆2

j ψ)(λx) −

2mZ

2λ|λxj| ψ(λx)

• +
• 1≤j<k≤N

2me2

2λ|λxj − λxk| ψ(λx)

• Choose λ = 2mZ/2, then

= λ3N 2mZ2 / 2 [HUψ] (λx) = Uλ

• 2mZ2

/ 2HUψ

• (x)

with HU =

N

• j=1
• P2

j − 1

|xj|

• +
• 1≤j<k≤N

U

|xj − xk|,

U = 1/Z, Pk = −i∇k.

3 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Letś concentrate on helium, N = 2. So we will consider the operator HU =

2

• j=1
• P2

j − 1

|xj|

• +

U

|x1 − x2|,

with U > 0 from now on.

4 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Letś concentrate on helium, N = 2. So we will consider the operator HU =

2

• j=1
• P2

j − 1

|xj|

• +

U

|x1 − x2|,

with U > 0 from now on. Known results: Essential spectrum: σess(HU) = σ(P2 −

1

|x |) = [− 1

4, ∞), HVZ Theorem (Hunziker, van Winter,

Zhislin).

4 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Letś concentrate on helium, N = 2. So we will consider the operator HU =

2

• j=1
• P2

j − 1

|xj|

• +

U

|x1 − x2|,

with U > 0 from now on. Known results: Essential spectrum: σess(HU) = σ(P2 −

1

|x |) = [− 1

4, ∞), HVZ Theorem (Hunziker, van Winter,

Zhislin). For Physicists: You reach the continuum spectrum once you kicked one electron to infinity!

4 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Letś concentrate on helium, N = 2. So we will consider the operator HU =

2

• j=1
• P2

j − 1

|xj|

• +

U

|x1 − x2|,

with U > 0 from now on. Known results: Essential spectrum: σess(HU) = σ(P2 −

1

|x |) = [− 1

4, ∞), HVZ Theorem (Hunziker, van Winter,

Zhislin). For Physicists: You reach the continuum spectrum once you kicked one electron to infinity! If U = 1 there are infinitely many bound states below −1/4. (Kato: More than 40.000, see e.g., Zhislin for full result ).

4 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

What about U > 1?

Let EU = ground state energy of HU. Easy to see: 0 < U → EU is strictly increasing. So by convexity, it must ‘really increase’.

5 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

What about U > 1?

Let EU = ground state energy of HU. Easy to see: 0 < U → EU is strictly increasing. So by convexity, it must ‘really increase’. Bethe (1929): For some U > 1 one still has EU < −1/4. ⇒ there exists a critical Uc > 1 with EUc = −1/4. ⇒ for any 1 < U < Uc, the operator HU has a ground state ψU with ground states energy EU < −1/4.

5 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

What about U > 1?

Let EU = ground state energy of HU. Easy to see: 0 < U → EU is strictly increasing. So by convexity, it must ‘really increase’. Bethe (1929): For some U > 1 one still has EU < −1/4. ⇒ there exists a critical Uc > 1 with EUc = −1/4. ⇒ for any 1 < U < Uc, the operator HU has a ground state ψU with ground states energy EU < −1/4. What happens with these ground states as U ր Uc?

5 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

What about U > 1?

Let EU = ground state energy of HU. Easy to see: 0 < U → EU is strictly increasing. So by convexity, it must ‘really increase’. Bethe (1929): For some U > 1 one still has EU < −1/4. ⇒ there exists a critical Uc > 1 with EUc = −1/4. ⇒ for any 1 < U < Uc, the operator HU has a ground state ψU with ground states energy EU < −1/4. What happens with these ground states as U ր Uc? Theorem (Maria and Thomas Hoffmann-Ostenhof - Barry Simon (H2O2-S) 1984) There is binding for Helium at Uc, that is, HUc has a ground state eigenfunction ψUc ∈ L2(R6) and HUcψUc = − 1

4ψUc.

5 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

The result says nothing about how fast ΨUc decays.

6 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

The result says nothing about how fast ΨUc decays. Recall that with Agmon’s method one can show that as long as

−1

4 > EU = −εU − 1 4 then

|ψU(x1, x2)| exp (−(1 − ε)ρ(x))

for all 0 < ε < 1, ρ the Agmon distance

ρ(x) = √εUx∞ + 1

2 x0 with x∞ = max(x1, x2) (distance of the outer particle to the nucleus) and x0 = min(x1, x2) (distance of the inner particle to the nucleus). This says nothing when U = Uc, since then εUc = 0.

6 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Frank–Lieb–Seiringer found a different proof from the one by H2O2-S, which also yields a kind of localization result: Theorem (Frank-Lieb-Seiringer 2012) For any δ > 0, there is a constant Cδ > 0 such that for all 1 + δ ≤ U < Uc and for all normalized ground states ψU of HU one has

ψU, x∞−1ψU ≥ Cδ

7 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Frank–Lieb–Seiringer found a different proof from the one by H2O2-S, which also yields a kind of localization result: Theorem (Frank-Lieb-Seiringer 2012) For any δ > 0, there is a constant Cδ > 0 such that for all 1 + δ ≤ U < Uc and for all normalized ground states ψU of HU one has

ψU, x∞−1ψU ≥ Cδ

By a compactness argument, this shows that a ground state at critical repulsion U = Uc exists. They have a similar result for bi-polarons and multi-polarons (for near energy minimizers). Helium case got extended to atoms by Bellazzini–Frank–Lieb–Seiringer (2014). Polaron case: no existence of bound states (in zero total momentum channel) Atoms: They need infinite nucleus mass approximation.

7 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Our result for Helium

Theorem (51e, Markus Lange, Michal Jex 2019) There exist constants c1, c2 > 0 such that the ground state at critical coupling obeys

ψUc(x) exp

• −c1
• x∞ + c2 log x∞
• 8

September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Our result for Helium

Theorem (51e, Markus Lange, Michal Jex 2019) There exist constants c1, c2 > 0 such that the ground state at critical coupling obeys

ψUc(x) exp

• −c1
• x∞ + c2 log x∞
• These constants are quantitative (have a ‘physical meaning’....)

Similar result for atoms. We do not need the infinite nuclear mass aproximation! Similar result for multipolaron systems at critical coupling in the zero total momentum channel.

8 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Even more is true

Theorem (51e, Markus Lange, Michal Jex 2019: Exponential decay in almost all directions) Given x = (x1, x2) ∈ R6 with x1 0, x2 0, and x = 1, we have

lim

t→∞

1 t log ψUc(tx) = −1 2 = −

• |EUc|

9 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Even more is true

Theorem (51e, Markus Lange, Michal Jex 2019: Exponential decay in almost all directions) Given x = (x1, x2) ∈ R6 with x1 0, x2 0, and x = 1, we have

lim

t→∞

1 t log ψUc(tx) = −1 2 = −

• |EUc|

Using the partition of unity from Frank–Lieb–Seiringer, this cannot be proven. Our result says if one stays away from from cones of arbitrary small angle around R3 × {0} and

{0} × R3, then the wave function decays even exponentially!

This does not imply good localization!! The decay starts late, depending on how close one is to either R3 × {0} or {0} × R3

9 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

The heuristic

A picture of a tunneling problem.

10 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

The heuristic

A picture of a tunneling problem. Take home message: Repulsion makes the classically forbidden region stickier....

10 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

How to make the physics into a proof

Let H = P2 + V be a usual (one-particle) Schrödinger operator. IMS localization formula:

Re(ξ2ψ, Hψ) = ξψ, Hξψ − ψ, |∇ξ|2ψ

for any reasonable real-valued function ξ.

11 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

How to make the physics into a proof

Let H = P2 + V be a usual (one-particle) Schrödinger operator. IMS localization formula:

Re(ξ2ψ, Hψ) = ξψ, Hξψ − ψ, |∇ξ|2ψ

for any reasonable real-valued function ξ. Now assume the energy inequality

ϕ(H − E)ϕ ≥ ϕ, Wϕ

for all ϕ supported outside of a large enough ball BR(0). W > 0 (on BR(0)c), a local potential

11 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Let χ be a smoothed cut-off function for BR(0)c: χ(x) = χ(|x|/R) with χ ∈ C∞(0, ∞) with 0 ≤ χ ≤ 1 and χ(r) = 0 for 0 < r ≤ R, χ(r) = 1 for r ≥ 2R. With F a well-behaved function, set

ξ = χeF

If Hψ = Eψ, in a quadratic form sense,

ϕ, (H − E)ψ = 0,

then use ϕ = χ2ψ and note that, since E and ϕ, ψ = ξ2ψ, ψ = ξψ2 are real, we have Eξψ2 = E Reξ2ψ, Hψ = ξψ, Hξψ − ψ, |∇ξ|2ψ from the IMS localization formula. That is, 0 = ξψ, (H − E)ξψ − ψ, |∇ξ|2ψ

12 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Using also the energy inequality, we have 0 = ξψ, (H − E)ξψ − ψ, |∇ξ|2ψ ≥ ξψ, Wξψ − ψ, |∇ξ|2ψ Now ∇ξ = ∇(χeF) = ∇χeF + χeF∇F, so

|∇ξ|2 = χ2e2F |∇F|2 + 2e2F∇F · ∇χ + e2F |∇χ|2

Hence, putting everything together,

χeFψ, (W − |∇F|2)χeFψ ≤ ψ, e2F(∇F · ∇χ + |∇χ|2)ψ

13 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Using also the energy inequality, we have 0 = ξψ, (H − E)ξψ − ψ, |∇ξ|2ψ ≥ ξψ, Wξψ − ψ, |∇ξ|2ψ Now ∇ξ = ∇(χeF) = ∇χeF + χeF∇F, so

|∇ξ|2 = χ2e2F |∇F|2 + 2e2F∇F · ∇χ + e2F |∇χ|2

Hence, putting everything together,

χeFψ, (W − |∇F|2)χeFψ ≤ ψ, e2F(∇F · ∇χ + |∇χ|2)ψ

If W − |∇F|2 ≥ δ > 0, then

δ χeFψ2 ≤ CRψ2

since ∇χ is supported on an annulus and eF is bounded on such an annulus.

13 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Example: V goes to zero at infinity, so σess(H) = σ(P2) = [0, ∞) and Hψ = Eψ with E < 0: Then with W = V and P2 ≥ 0, we have

ϕ, (H − E)ϕ ≥ ϕ, (V − E)ϕ ≥ ϕ, (|E| − ε)ϕ

for any 0 < ε < |E| and all ϕ supported in the complement of BR(0), as long as R is large enough, depending on ε > 0. Thus

χeFψ, (|E| − ε − |∇F|2)χeFψ ≤ CRψ2

and choosing F(x) = µ|x|, we have |∇F| = µ, so

χeFψ, (|E| − ε − µ2)χeFψ ≤ CRψ2

hence x → eµ|x |ψ(x) ∈ L2 for any µ <

• |E|

(WKB asymptotic!)

14 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

No safety distance

What if we do not have a safety distance to the essential spectrum, that is, E = 0? Assume that V = −V1 +

U

|x |+1 with V1 ≥ 0, having compact support, such that for small enough

U > 0 the operator HU = P2 − V1 + U

|x| + 1

has a strictly negative ground state energy EU < 0. Then EU is strictly increasing in U There is a critical Uc > 0 with EUc = 0. Obviously σ(HUc) = [0, ∞).

15 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Claim: HUc has a non-trivial zero energy eigenfunction ψ ∈ L2. Moreover x → exp

• c1
• |x| + 1 − c2 log(|x| + 1)
• ψ(x) ∈ L2

Proof: Step1: Assume that HUcψ = 0. Take R > so large that supp (V1) ∈ BR(0). Then for all ϕ with support outside BR(0) we have 0 = ξψ, (HUcξψ) − ψ, |∇ξ|2ψ = ξψ,

• P2 +

Uc

|x| + 1

• ξψ − ψ, |∇ξ|2ψ

≥ ξψ,

Uc

|x| + 1ξψ − ψ, |∇ξ|2ψ

when ξ = χeF and χ = 0 on BR(0).

16 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

As before we get

χeFψ,

• Uc

|x| + 1 − |∇F|2

• χeFψ ≤ CRψ2

But now there is no chance that

Uc

|x |+1 − |∇F|2 ≥ δ > 0 for all |x| large enough.

17 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

As before we get

χeFψ,

• Uc

|x| + 1 − |∇F|2

• χeFψ ≤ CRψ2

But now there is no chance that

Uc

|x |+1 − |∇F|2 ≥ δ > 0 for all |x| large enough.

Instead choose F(x) = U1/2

c

(|x| + 1)1/2

Then |∇F| =

U1/2

c

2(|x |+1)1/2 , so

Uc

|x| + 1 − |∇F|2 =

3Uc 4(|x| + 1).

17 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

This gives

χeFψ,

3Uc 4(|x| + 1) χeFψ ψ2

• r, in other words,

x → exp

• U1/2

c

(|x| + 1)1/2 − log(|x| + 1)

• ψ(x) ∈ L2

Even stretched exponential growth wins over polynomial decay

18 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Step 2: HUc has a non-trivial zero energy eigenstate. Since the ground state energy EU is strictly increasing, there are normalized L2 ground states ψU, for any 0 < U < Uc, corresponding to ground state energies EU < 0. Rerunning the above argument for ψU and dropping some positive terms, yields

∫

• exp
• c(|x| + 1)1/2 − log(|x| + 1)
• ψ(x)
• 2

dx ≤ K

(∗)

for some constants c, K, as long as Uc − δ ≤ U < Uc, for some small enough δ > 0.

19 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Step 2: HUc has a non-trivial zero energy eigenstate. Since the ground state energy EU is strictly increasing, there are normalized L2 ground states ψU, for any 0 < U < Uc, corresponding to ground state energies EU < 0. Rerunning the above argument for ψU and dropping some positive terms, yields

∫

• exp
• c(|x| + 1)1/2 − log(|x| + 1)
• ψ(x)
• 2

dx ≤ K

(∗)

for some constants c, K, as long as Uc − δ ≤ U < Uc, for some small enough δ > 0. Now take Un ր Uc and set ψn = ψUn. The bound (∗) yields tightness in position space,

lim

R→∞ sup n

∫

|x |>R

|ψn(x)|2 dx = 0

19 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173

Outline The approach

Since the potential is form small with respect to the kinetic energy, one also has

ψn, P2ψn = ∫ |η|2| ψn(η)|2 dη ≤

K for some constant K. This yields tightness in momentum space,

lim

L→∞

∫

|η|>L

| ψn(]η)|2 dη = 0

and since ψn = 1, one can pick a weakly convergent subsequence, which, by tightness, the converges strongly to some normalized ψ, which is the zero energy ground state of HUc. For a simple proof of weak convergence plus tightness = strong convergence in L2, see 51e–Young-Ran Lee: On non-local variational problems with lack of compactness related to non-linear optics. Journal of Nonlinear Science 22 (2012).

20 September 12, 2019 Dirk Hundertmark – The Brink KIT – Institute for Analysis // CRC 1173