Table Of ContentSYMMETRY OF BIPOLARON BOUND STATES
FOR SMALL COULOMB REPULSION
RUPERT L. FRANK, ELLIOTT H. LIEB, AND ROBERT SEIRINGER
Abstract. WeconsiderthebipolaroninthePekar–Tomasevichapproximationand
address the question whether the ground state is spherically symmetric or not. Nu-
2
1 merical analysis has, so far, not completely settled the question. Our contribution
0 is to proverigorouslythat the groundstate remains sphericalfor small values ofthe
2
electron-electron Coulomb repulsion.
n
a
J
9
1 1. Introduction
] In this paper we shall be concerned with properties of the bound state of two
h
p polarons. WedothisinthecontextofthePekar–Tomasevich model[15,16]ofthelarge
-
h polaron, which, in turn, is based on Fr¨ohlich’s polaron model [5]. In the latter model
t
a two electrons interact with a quantized electric field generated by the displacement of
m
the nuclei in a polar lattice. There are two coupling constants in Fr¨ohlich’s model.
[
The coupling to the field, α, and the Coulomb repulsion among the electrons, denoted
1 by U. Pekar’s approximation is to assert that the wave function is a product of a
v
4 two-particle electron wave function ψ times a field function Φ. After eliminating the
5
field one is led to Pekar’s energy expression for ψ,
9
3 U
. [ψ] = ψ 2 + ψ 2 + ψ 2 dxdy 2αD[ρ ,ρ ]. (1.1)
1 U x y ψ ψ
E |∇ | |∇ | x y | | −
0 ZZR3×R3(cid:18) | − | (cid:19)
2
The electron coordinates are x and y and the electron spin does not appear explicitly,
1
: except that ψ is symmetric for the ground state, which is a singlet state. (The reason
v
i that it is a singlet is that the ground state is a positive function and must, therefore,
X
be symmetric – an observation that goes back to Wigner many years ago.)
r
a
In (1.1) ρ denotes the electron density, given by
ψ
ρ (x) = ψ(x,y) 2dy + ψ(y,x) 2dy,
ψ
| | | |
R3 R3
Z Z
and D[ρ,ρ] is the Coulomb energy of a charge distribution ρ,
1 ρ(x)ρ(x′)
D[ρ,ρ] = dxdx′.
2 x x′
ZZR3×R3 | − |
Note the minus sign in (1.1); the induced interaction is attractive.
c 2012 by the authors. This paper may be reproduced, in its entirety, for non-commercial
(cid:13)
purposes.
1
2 R. L. FRANK,E. H.LIEB, AND R.SEIRINGER
The Pekar energy is
e = inf [ψ] : ψ H1(R6), ψ = 1 , (1.2)
U U
{E ∈ k k }
with ψ denoting the L2 norm of ψ and H1 denoting the Sobolev space, i.e., square-
k k
integrable functions whose gradient is also square-integrable. This energy is more than
just an approximation, for it is asymptotically exact as α and U tend to infinity with
U/α fixed. This is stated in [14], following the technique of [13]; see also [2].
A benchmark for the bipolaron problem is the energy of a single polaron. It is
defined in the manner of equation (1.1)
[ψ] = ψ 2dx 2αD[ ψ 2, ψ 2],
E |∇ | − | | | |
R3
Z
and
e = inf [ψ] : ψ H1(R3), ψ = 1 . (1.3)
{E ∈ k k }
It is known that there is a minimizing ψ for this single polaron problem and that it is
unique, up to translations in R3 and multiplication by a constant phase [10]. Since it
is unique, it is a radial function.
By scaling we can always reduce to the case α = 1/2 and we shall do so henceforth.
There is a considerable literature on the subject of rotation invariance of the bipo-
laron energy minimizer, usually formulated in the language of ‘one-center bipolaron
versus two-center bipolaron’. The analyses are all based on variational calculations.
While there seems to be general agreement that the one-center bipolaron has the lower
energy, it is not completely clear that a more sophisticated variational treatment will
preserverotationalsymmetry, especiallynearthevalueofU wherethebipolaronceases
to be bound. We have shown rigorously that there is such a critical U [3]. Numerical
c
variational data seems to indicate that the critical U is rather close to 1, namely,
c
U 1.15. This tells us that binding is a delicate matter and, indeed, the existence
c
∼
of such a U was an open question for some time.
c
A minimizer also exists for a bipolaron provided the energy is below the energy of
twice the single polaron energy [8]. Interestingly, the bipolaron has a minimizer with
finite radius at the critical value U = U [4]. (The same holds for a helium atom for
c
the critical value of the nuclear charge.) The existence of a minimizer implies that the
translation invariance of the Pekar minimization problem is broken. It is, therefore,
not out of the question that the minimizer for two polarons might break rotational
symmetry as well in order to lessen the Coulomb repulsion.
The value of U determined by physical electrostatic considerations is always U 1.
≥
Nevertheless, one can consider the mathematical question for small, but positive U
and ask whether there is a possible lack of rotational invariance in that case. After
all, a rotating object like the earth becomes oblate even for the smallest amount of
rotation.
Inthispaper wewill provethatthereisnobreaking ofrotationalsymmetry forsmall
U. Our strategy for proving the lack of symmetry-breaking for small U is based on
SYMMETRY OF BIPOLARON BOUND STATES — January 18, 2012 3
the following consideration. For any small U the minimization problem, restricted to
rotation invariant functions, has an energy minimizer, as we shall prove. The question
is whether there is better minimizer without rotation invariance. Since U is small,
both minimizers would have to be very close to the unique (up to translations) U = 0
minimizer, and so we can discuss the existence of a symmetry breaking minimizer
by means of rigorously controlled perturbation theory. It is evident that the benefit
of symmetry breaking to the repulsive energy will be proportional to δ2, where δ
−
measures the non-sphericity, but this contribution is multiplied by U. On the other
hand, the increase in the rest of the energy is presumably also of the form Aδ2, where
A 0 is some U independent number. If A > 0 then δ wants to be zero for small U.
≥
The problem with this argument is that A could be zero, in which case the Uδ2
−
energy would always win, no matter how small U is. Most of what we do in the paper,
from the mathematical point of view, is to show rigorously that A is not zero, and
thus there is no distortion for small U.
Even utilizing the result that A > 0 for the single polaron [7], the proof given here
for the bipolaron will not be a short one. One of the complexities faced in this proof
is the fact that a simple translation is a distortion that costs no energy, i.e., there are
zero modes. While these are physically trivial distortions it is not a trivial matter to
separate their contribution, mathematically, from the relevant ones.
Theorem 1. There is a U > 0 such that for all U < U the minimizer of is
s s U
E
unique up to translations and multiplication by a constant phase. In particular, after
a translation it is rotation invariant, that is, ψ( x, y) = ψ(x,y) for any x,y R3
R R ∈
and any O(3).
R ∈
It remains an open problem to decide whether the ground state ceases to be rotation
invariant for U close to the critical value U .
c
The rest of this paper has two parts. In Part A, we reduce the proof of Theorem 1 to
a problem in second-order perturbation theory, namely the question of the distortion
coefficient A. In Part B, we complete the proof of Theorem 1 by showing that A is
positive and that zero modes play no important role.
Acknowledgments. We are grateful to Herbert Spohn for making us aware of this
problem. Partial financial support fromthe U.S. National Science Foundationthrough
grants PHY-1068285 (R.F.), PHY-0965859 (E.L.) and the NSERC (R.S.) is acknowl-
edged.
2. Proof of Theorem 1. Part A
2.1. Some preparations. Step 1. In searching for the minimum in (1.2) we can
confine our attention to non-negative, symmetric (i.e., ψ(x,y) = ψ(y,x)) functions.
The reason is that we can replace any ψ by
1
ψ˜(x,y) = ( ψ(x,y) 2+ ψ(y,x) 2). (2.1)
2 | | | |
r
4 R. L. FRANK,E. H.LIEB, AND R.SEIRINGER
The potential energy terms remain the same and the kinetic energy term does not
increase [11, Thm. 7.8].
Step 2. An important step is to reformulate the energy minimization problem in
the following way: Define
U
[ψ,Φ] := ψ 2 + ψ 2 Φ(x) ψ 2 Φ(y) ψ 2 + ψ 2 dxdy
U x y
E |∇ | |∇ | − | | − | | x y | |
ZZR3×R3(cid:18) | − | (cid:19)
1
+ Φ 2dx ψ 2dxdy (2.2)
8π |∇ | | |
ZR3 ZZR3×R3
for ψ H1(R6) and Φ H˙ 1(R3). Then
∈ ∈
[ψ,Φ] ψ 2 [ ψ −1ψ] (2.3)
U U
E ≥ k k E k k
with equality if and only if Φ = x −1 ρ .
ψ/kψk
| | ∗
One advantages of this reformulation is that one can see immediately that there is a
unique (up to translations) minimizer for the U = 0 problem. For a given Φ we have,
in this case, a Schro¨dinger minimization problem for two independent particles, so the
best ψ is a product, ψ(x,y) = f(x)f(y), in which case ρ = 2 f 2 and the optimal Φ
ψ
| |
is 2 x −1 f 2. The problem therefore becomes
| | ∗| |
e = inf 2 f 2dx 4D[ f 2, f 2] : f = 1 ,
0
|∇ | − | | | | k k
R3
(cid:26) Z (cid:27)
which coincides, up to a rescaling, with problem (1.3). We conclude that e = 8e and
0
that the minimizing f is unique up to translations and multiplication by a constant
phase [10].
Another advantage of this reformulation is that one sees that for any U 0 an
≥
optimizer ψ, if it exists, is the ground state of a two-body Schro¨dinger operator.
Therefore, ψ is a multiple of a strictly positive function [11, Thm. 9.10]. This, in turn,
implies that ψ is symmetric (i.e., ψ(x,y) = ψ(y,x)), because otherwise replacing ψ by
(2.1) would strictly lower the kinetic energy [11, Thm. 7.8].
Step 3. In addition to the global minimization problem (1.2) one can define the
rotationally symmetric minimization problem, that is,
esymm = inf [ψ] : ψ H1(R6), ψ = 1, ψ rotation invariant with respect to 0 .
U EU ∈ k k
(2.4)
(cid:8) (cid:9)
N.B.: From now on ‘symmetry’ refers to ‘rotation symmetry’ and not to symmetry in
x and y. We recall that the rotation invariance of ψ means that ψ( x, y) = ψ(x,y)
R R
for any x,y R3 and any O(3). The density ρ of such ψ is, of course, radial
ψ
∈ R ∈
and, by Newton’s theorem, its potential ρ x −1 is a symmetric decreasing function
ψ
∗| |
bounded by 2 x −1.
| |
Of course, e esymm, and our goal in this paper is to investigate whether equality
U ≤ U
holds. We collect some properties of these energies. Both e and esymm are non-
U U
decreasing, concave functions of U (as infima of non-decreasing, linear functions).
Because of Step 2, e = esymm for U = 0. Moreover, simple trial function arguments
0 0
SYMMETRY OF BIPOLARON BOUND STATES — January 18, 2012 5
showthat e 2eandesymm efor allU. Lewin[8] hasshown that theinfimum e is
U ≤ U ≤ U
attained provided e < 2e. In the appendix of this paper we shall prove an analogous
U
result for the rotation invariant problem, with a different condition, however, namely,
esymm < e.
U
Proposition 2. If esymm < e, then the infimum in (2.4) is attained.
U
The reason for the discrepancy between 2e for e and e for esymm is that spherical
U U
symmetry prevents the formation of two more or less separate polarons. In other
words, the second polaron density has to be far away from the first, and in a radial
shell, which makes it impossible to retain an energy 2e with an essentially unbound
pair of polarons, both of which are spherically symmetric with respect to a common
center.
Proposition 2 shows the fact that for some values of U the rotation invariant mini-
mizer is not the true minimizer. A rotationinvariant minimizer is necessarily a critical
point of the Pekar–Tomasevich functional. If it is a ‘false’ minimum (i.e., its energy
is bigger than the true ground state energy), its existence can possibly lead to com-
putational difficulties for the true minimizer.
In the appendix we show two things about the rotation invariant minimization
problem: (1) There is a critical constant Usymm with 1 < Usymm 4 such that
c c ≤
esymm < e for U < Usymm and esymm = e for U Usymm; see Proposition 8.
U c U ≥ c
(2) The rotation invariant minimizer, if it exists, is a function of x , y and t =
| | | |
x y/ x y . We will show that for fixed x and y the minimizer is non-increasing as
· | || | | | | |
a function of t. That is, the two particles try to avoid each other; see Proposition 9.
2.2. Beginning of the proof of Theorem 1. After these preparations we are now
ready to give the first part of the proof of Theorem 1. Our strategy is as follows:
Let U > 0 be a sequence such that U 0. For every sufficiently large n there is a
n n
→
global minimizer ψ and a rotation invariant minimizer ϕ corresponding to (1.2) and
n n
(2.4) with U = U . We shall prove that for all large n, ϕ and ψ coincide up to a
n n n
translation and a constant phase.
This clearly implies the theorem. Indeed, if the theorem were not true we could find
˜
a sequence U tending to zero, and associated minimizers ψ and ψ which are not
n n n
translates or multiples of each other. By what we are going to show, however, they
are both translates and multiples of a rotation invariant minimizer ϕ if n is large,
n
which is a contradiction.
Thus, from now on we fix a sequence U > 0 tending to zero. We will only consider
n
n sufficiently large such that there are a global minimizer ψ and a rotation invariant
n
minimizer ϕ corresponding to (1.2) and (2.4) with U = U . By Steps 1 and 2
n n
above we may assume that ψ and ϕ are positive and permutation symmetric (i.e.,
n n
ψ (x,y) = ψ (y,x) and similarly for ϕ ). Since the global minimization problem is
n n n
6 R. L. FRANK,E. H.LIEB, AND R.SEIRINGER
translation invariant, we may translate ψ in such a way that
n
α := ψ (x,y)ϕ (x,y)dxdy = max ψ (x a,y a)ϕ (x,y)dxdy.
n n n n n
ZZR3×R3 a∈R3 ZZR3×R3 − −
(2.5)
With this normalization, our goal is to show that ψ = ϕ for all large n.
n n
We decompose
ψ = α ϕ +j ,
n n n n
where, according to the definition of α and to the maximizing property in (2.5),
n
j ϕ dxdy = 0 and j ~e ( + )ϕ dxdy = 0 for any ~e S2.
n n n x y n
· ∇ ∇ ∈
ZZR3×R3 ZZR3×R3
(2.6)
The derivative term comes from differentiating the last double integral in (2.5) with
respect to a.
We next claim that
jn H1 0 as n . (2.7)
k k → → ∞
Indeed, both ψ and ϕ are minimizing sequences for the e problem. Since the
n n 0
minimizer for this problem is unique up to translations (see Step 2 above), the results
of [8] imply that
ψ ( a , a ) f f and ϕ ( b , b ) f f in H1(R6)
n n n n n n
·− ·− → ⊗ ·− ·− → ⊗
for some sequences a and b in R3. Here f f is the e minimizer with f chosen
n n 0
⊗
spherically symmetric about the same origin that we fixed to formulate the problem
(2.4). Since the density of ϕ is spherically symmetric, one easily concludes that
n
b 0 and, therefore, ϕ f f in H1 as n . This, together with the
n n
→ → ⊗ → ∞
maximizing property of α in (2.5), implies also that a 0 and ψ f f in H1
n n n
→ → ⊗
as n . Thus α 1 and j = ψ α ϕ 0 in H1, as claimed in (2.7).
n n n n n
→ ∞ → − →
We now expand the energy of ψn to second order in jn H1. A simple but tedious
k k
computation shows that
[ψ ] = [ϕ ]+(j ,L j )+ ( j 3 ) (2.8)
EUn n EUn n n n n O k nkH1
with the linear operator
U 1 1
n
L = ∆ ∆ + ρ ρ µ 4X .
n − x − y x y − ϕn ∗ x − ϕn ∗ y − n − ϕn
(cid:18) | − | | | | | (cid:19)
Here ρ x −1 is an abbreviation for (ρ −1)(x), we introduced µ = [ϕ ]
ϕn∗| | ϕn ∗|·| n EUn n −
D[ρ ,ρ ] and X is the integral operator on L2(R6) with the integral kernel
ϕn ϕn ϕn
ϕ (x,y)ϕ (x′,y′) ϕ (x,y)ϕ (x′,y′)
X (x,y,x′,y′) = n n + n n .
ϕn x x′ y y′
| − | | − |
SYMMETRY OF BIPOLARON BOUND STATES — January 18, 2012 7
In order to see that there is no linear term in j in the computation, we used the
n
Euler–Lagrange equation for ϕ , that is,
n
U 1 1
n
∆ ∆ + ρ ρ ϕ = µ ϕ . (2.9)
− x − y x y − ϕn ∗ x − ϕn ∗ y n n n
(cid:18) | − | | | | |(cid:19)
(From the minimizing property of ϕ one obtains this equation only when integrated
n
against rotation invariant functions but, since both sides are rotation invariant func-
tions, it is true even when integrated against any function.) We also used the fact
that
1
α = 1 j 2 1/2 = 1 j 2 + ( j 4).
n n n n
−k k − 2k k O k k
In deriving (2.8) we app(cid:0)lied standa(cid:1)rd estimates to bound all terms of higher than
quadratic order by a constant times j 3 .
k nkH1
The way forward is now clear: The quantity A referred to in the introduction can
be identified as (j,L j)/ j 2 for j’s satisfying (2.6), and we need to prove that this
n
k k
is non-zero. We will prove that there are constants N 1 and c > 0 such that
≥
(j,L j) c j 2 for all n N and all j satisfying the orthogonality conditions (2.6).
n
≥ k k ≥
Then (2.8) will imply that
[ψ ] [ϕ ]+c j 2 + ( j 3 ).
EUn n ≥ EUn n k nk O k nkH1
Since ψ is, by assumption, an energy minimizer, we have necessarily j = 0 for all
n n
large n, which means ψ = ϕ , as we intended to prove.
n n
Thusweareleftwithprovingalowerboundon(j,L j). Weshowthisperturbatively
n
by analyzing the U = 0 case.
2.3. The Hessian. Instead of working directly with the operator L , it is more con-
n
venient to work with a closely related operator, namely the Hessian H of at ϕ .
n EUn n
That is, for any normalized, real-valued j H1(R6) we define
∈
d2 ϕ +εj
n
= (j,H j). (2.10)
dε2(cid:12)ε=0EUn" 1+2ε(ϕn,j)+ε2# n
(cid:12)
A similar computation (cid:12)as beforepshows that
(cid:12)
H = L + k ϕ + ϕ k +β ϕ ϕ ,
n n n n n n n n n
| ih | | ih | | ih |
where
U
n
β = 4 ϕ , ∆ ∆ + ϕ 3D[ρ ,ρ ]
n n − x − y x y n − ϕn ϕn
(cid:18)(cid:18) (cid:18) | − |(cid:19) (cid:19) (cid:19)
and
U
k = 2 ∆ ∆ + n ϕ 2ρ x −1ϕ 2ρ y −1ϕ .
n − − x − y x y n − ϕn ∗| | n − ϕn ∗| | n
(cid:18)(cid:18) | − |(cid:19) (cid:19)
The expressions for β and k can be somewhat simplified using equation (2.9) for ϕ ,
n n n
but we will not need this. The only thing that is relevant for us is that (j,H j) =
n
(j,L j) if j satisfies the first orthogonality condition (2.6).
n
8 R. L. FRANK,E. H.LIEB, AND R.SEIRINGER
We collect two facts about the operators H . First, H (which commutes with
n n
angular momentum since ϕ is rotation invariant) is non-negative on the subspace of
n
angular momentum zero. This follows from the minimizing property of ϕ . Second,
n
thefunctionsϕ and~e ( + )ϕ ,e S2, areinthekernelofH . Forϕ , thisfollows
n x y n n n
· ∇ ∇ ∈
immediately from the Euler–Lagrange equation (2.9) for ϕ , and for ~e ( + )ϕ ,
n x y n
· ∇ ∇
this follows by differentiating the equation (2.9) for ϕ (x + t~e,y + t~e) (which is the
n
same as that for ϕ ) with respect to t at t = 0.
n
The following proposition says that, for small U , H is strictly positive away from
n n
the zero modes found above.
Proposition 3. There is a number c > 0 and an N 1 such that for all n N
≥ ≥
and for all j which are orthogonal to ϕ and to ~e ( + )ϕ , ~e S2, the following
n x y n
· ∇ ∇ ∈
estimate holds:
(j,H j) c j 2.
n
≥ k k
Equivalently, if P denotes the projection onto the four-dimensional space spanned by
n
ϕ and ~e ( + )ϕ , e S2, then for n N
n x y n
· ∇ ∇ ∈ ≥
P⊥H P cP⊥.
n n n⊥ ≥ n
Note that, as remarked above, (j,H j) = (j,L j) if (ϕ ,j) = 0 and, therefore,
n n n
(cid:3)
Proposition 3 concludes the proof of Theorem 1.
Since ϕ f f in H1(R6) as n , the operators H tend to the corresponding
n n
→ ⊗ → ∞
operator at U = 0 in norm resolvent sense. In particular, the eigenvalues converge
and, therefore, it suffices to prove Proposition 3 in the case U = 0. This is the topic
of the next section.
The reason that we have chosen to work with the operator H instead of L is that
n n
the operator L has a negative eigenvalue. (This follows from the variational principle
n
since (ϕ ,L ϕ ) = 4(ϕ X ,ϕ ) < 0 by (2.9).) The positivity of (j,L j) asserted
n n n − n ϕn n n
in Proposition 3 therefore crucially relies on the orthogonality condition (2.6). This
condition is not easy to use, however, since ϕ is not an eigenfunction of L . In
n n
contrast, it is an eigenfunction of H . We also note that H is the operator that
n n
automatically takes care of the normalization condition, without any reference to
orthogonality, and, therefore, ismost directly connected tothe coefficient Amentioned
in the introduction.
3. Proof of Theorem 1. Part B
As explained in Step 2 of the previous section, at U = 0 we have ϕ(x,y) = f(x)f(y),
where f is a radial decreasing function on R3 with f2dx = 1 and
R 1 µ
hf = 0, h = ∆ 2f2 . (3.1)
− − ∗ x − 2
| |
SYMMETRY OF BIPOLARON BOUND STATES — January 18, 2012 9
Here, µ = e 4D[f2,f2] = 8e 4D[f2,f2]. The Hessian of at f f is the operator
0 0
− − E ⊗
on L2(R6) given by
H = L+ r f +f r f f + f f r f +f r +β f f f f ,
| ⊗ ⊗ ih ⊗ | | ⊗ ih ⊗ ⊗ | | ⊗ ih ⊗ |
where
L = h +h 4X
x y f⊗f
−
and where β is a constant and r is a function in L2(R3). More precisely,
β = 8 (f, ∆f) 6D[f2,f2]
− −
and (cid:0) (cid:1)
r = 2 ∆ 4f2 x −1 f .
− − − ∗| |
The main result of this section is the following U = 0 analogue of Proposition 3. We
(cid:0) (cid:1)
note that the three-dimensional space spanned by ~e ( + )ϕ, ~e S2, coincides
x y
· ∇ ∇ ∈
with the space spanned by f f′Y + f′Y f, m = 1,0,1, where Y are
1,m 1,m 1,m
⊗ ⊗ −
spherical harmonics of degree one.
Proposition 4. There is a number C > 0 such that for all j H1(R6) with j(x,y) =
∈
j(y,x) that satisfy
(f f,j) = (f f′Y +f′Y f,j) = 0 for all m = 1,0,1,
1,m 1,m
⊗ ⊗ ⊗ −
the following estimate holds:
(j,Hj) C j 2.
≥ k k
Equivalently, if P denotes the projection onto the four-dimensional space spanned by
f f and f f′Y +f′Y f, m = 1,0,1, then
1,m 1,m
⊗ ⊗ ⊗ −
P⊥HP⊥ CP⊥.
≥
We shall deduce this from a result about the Hessian of the one-particle functional,
which we discuss in the following subsection.
3.1. The one-particle Hessian. We recall that the function f minimizes the one-
polaron functional
ψ 2dx 2D[ ψ 2, ψ 2] (3.2)
|∇ | − | | | |
R3
Z
and that the corresponding Euler–Lagrange equation reads hf = 0 with h from (3.1).
Moreover, the Hessian of the above one-polaron functional at f reads
H(1) = L(1) + r f + f r +γ f f
| ih | | ih | | ih |
with
L(1) = h 4x .
f
−
Here x is the operator on L2(R3) with the integral kernel f(x) x x′ −1f(x′) and r is
f
| − |
the same function as in the expression of the two-particle operator H. The following
theorem is the one-particle analogue of Proposition 4. Its proof relies heavily on
previous work of Lenzmann [7].
10 R. L. FRANK,E. H.LIEB, AND R.SEIRINGER
Proposition 5. There is a number C′ > 0 such that for all j H1(R3) that satisfy
∈
(f,j) = (f′Y ,j) = 0 for all m = 1,0,1,
1,m
−
the following estimate holds:
(j,H(1)j) C′ j 2.
≥ k k
Similarly as in the discussion before Proposition 3, we note that f and f′Y are
1,m
zero modes of H(1). Thus the proposition says that there are no other zero modes.
Proof. WefirstnotethatH(1) isnon-negative, sincef isaminimizer oftheone-polaron
problem. We now argue that L(1) has exactly one negative eigenvalue. Indeed, since
(j,L(1)j) = (j,H(1)j) if (f,j) = 0 and since H(1) 0, the variational principle implies
≥
that L(1) has at most one negative eigenvalue. On the other hand, the Euler–Lagrange
equationforf impliesthat(f,L(1)f) = 4(f,x f) < 0, which, againbythevariational
f
−
principle, means that L(1) has at least one negative eigenvalue. This proves the claim.
We next recall Lenzmann’s result [7], which states that
kerL(1) = span f′Y : m = 1,0,1 . (3.3)
1,m
{ − }
We also note that L(1) commutes with angular momentum. Moreover, its essential
spectrum starts at µ/2 > 0 and, therefore, (3.3) implies that there is a constant
−
c′ > 0 such that
(j,H(1)j) = (j,L(1)j) c′ j 2
≥ k k
for all j with angular momentum l 2 and for all j with angular momentum l = 1
≥
satisfying the additional constraint that (f′Y ,j) = 0 for all m = 1,0,1. (Actually,
1,m
−
the argument of [7] shows that the best constant c′ is achieved either for l = 1 or for
l = 2.)
Since the function r appearing in the definition of H(1) is radial, the operator H(1)
commutes with angular momentum. Therefore, the previous discussion reduces the
proof of Proposition 5 to finding a lower bound on (j,H(1)j) for radial j satisfying
(f,j) = 0. In other words, we have to exclude the possibility that 0 is a degenerate
eigenvalue of H(1) restricted to l = 0.
As an aside, before completing the proof, we show that there is a radial function R
such that
L(1)R = µf
and
1
(R,f) = .
2
This is also contained in [7], but we include the short proof for the convenience of the
reader. We define f (x) = β2f(βx) and note that the Euler–Lagrange equation for f
β
implies the following equation for f ,
β
1 µ
∆ 2f2 f = β2 f .
− − β ∗ x β 2 β
(cid:18) | |(cid:19)
