Theor Chem Account (2007) 118:557–561
DOI 10.1007/s00214-007-0367-6
REGULAR ARTICLE
A re-statement of the Hohenberg–Kohn theorem and its extension
to finite subspaces
Ramiro Pino · Olivier Bokanowski ·
Eduardo V. Ludeña · Roberto López Boada
Received: 18 December 2006 / Accepted: 19 January 2007 / Published online: 27 June 2007
© Springer-Verlag 2007
Abstract Bearing in mind the insight into the Hohenberg–
Kohn theorem for Coulomb systems provided recently by
Kryachko (Int J Quantum Chem 103:818, 2005), we present a re-statement of this theorem through an elaboration on
Lieb’s proof as well as an extension of this theorem to finite
subspaces.
Keywords Hohenberg–Kohn theorem · DFT ·
Finite subspaces
1 Introduction
Density functional theory, DFT [1–12] has become a basic
tool in contemporary quantum chemistry [13–15] but, as
shown some decades ago by Lieb [16] and more recently
Contribution to the Serafin Fraga Memorial Issue.
R. Pino
North American Scientific, Nomos Radiation Oncology,
200 W Kensinger Dr., Cranberry Township, PA 16066, USA
e-mail: ramiropino@yahoo.com
O. Bokanowski
Lab. J.-L. Lions, Université Pierre et Marie Curie,
175 R. Chevralet, 75013 Paris, France
e-mail: boka@math.jussieu.fr
E. V. Ludeña (B)
Química, IVIC, Apdo. 21827, Caracas 1020-A, Venezuela
e-mail: eludena@ivic.ve
R. L. Boada
Miami-Dade College Wolfson Campus,
Department of Natural Sciences, Health and Wellness,
300 NE 2nd Ave., Miami, FL 33132, USA
e-mail: rlopezbo@mdc.edu
by several authors [17–23] due to its subtleties, this theory
cannot be considered as yet to be entirely elaborated.
The Hohenberg–Kohn theorem [24,25] has played a
fundamental role in the development of DFT. In a recent
work, however, Kryachko [26] has pointed out that the usual
reductio ad absurdum proof of this theorem is unsatisfactory
since the would-be-refuted assumptions on the one-electron
density and the assumption on the external potential evince
incompatibilities with the Kato cusp condition. Nevertheless, as shown by Kryachko [26], application of the Kato
cusp conditions actually leads to a satisfactory proof of this
theorem.
In the present work, within the context of Kryachko’s analysis, we advance an alternative proof of the Hohenberg–Kohn
theorem, which is based on the rigorous examination of the
original formulation of this theorem made by Lieb [16], a
number of years ago. In Lieb’s proof, it is required that the
N -particle wavefunction Ψ not vanish in a set of positive
measure. This condition, however, cannot be easily fulfilled.
In order to avoid this difficulty we present below an essentially algebraic proof of the Hohenberg–Kohn theorem which
dispenses with the latter condition.
In addition, we propose an extension of the present reformulation of the Hohenberg–Kohn theorem to the case of finite
subspaces. This finite subspace problem has been treated in
a restricted sense by Epstein and Rosenthal [27] and by Katriel et al. [28,29] and in a general sense by Harriman [30].
More recently, Görling and Ernzerhof have reexamined this
problem in relation to the linear response method to determine Kohn–Sham orbitals (and, purportedly, Kohn–Sham
wavefunctions; strictly speaking, it is not possible to attach a
rigorous meaning to Kohn–Sham wavefunctions as through
the application of the variational principle there only result
Kohn–Sham single-particle equations and their corresponding single-particle orbitals) from electron densities [31].
123
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Theor Chem Account (2007) 118:557–561
In order to set the proper background for our discussion,
we review in Sect. 1 both the original proof given by
Hohenberg and Kohn, in the context of Kryachko’s work,
as well as Lieb’s reformulation. In Sect. 2 we discuss the
modifications introduced in our present proof. In Sect. 3, we
consider the conditions that must be fulfilled in order that
this theorem be extended to finite subspaces.
2 The original Hohenberg–Kohn proof and Lieb’s
reformulation
Let us consider a system formed by N -electrons interacting
with a positive background through an “external" potential
V (r1 , . . . , r N ) =
N
(1)
v(ri ).
i=1
The many-electron Hamiltonian for such a system is
v = H
o + V
H
(2)
o is defined by
where H
o = − 1
H
2
∇r2i +
N
−1
N
i=1 j=i+1
1
.
|ri − r j |
(3)
It is assumed that the selected class {v(r)} of single-particle
external potentials is such that it possesses a ground-state
wavefunction {Ψov }. The one-electron density ρov (r) associated with Ψov is defined by
v
3
ρo (r1 ) = N d r2 · · · d 3 r N |Ψov (r1 , . . . , r N )|2 .
(4)
For such systems, the Hohenberg–Kohn theorem states
that there exists a one to one correspondence between an
external potential v(r) and the exact ground-state density
ρov (r). The original proof of this theorem [24] is carried out
by reductio ad absurdum. Consider two potentials v(r) and
v ′ (r) differing by more than a constant. The exact groundv
state wavefunctions for the corresponding Hamiltonians H
v ′ ) are assumed to be different (actually, these assump(or H
tions immediately evoke the Kato theorem and show the
way to a proof that dispenses with the reductio ad absurdum
argument) and for this reason the following strict variational
inequalities hold:
′
′
v |Ψov > Ψov | H
v |Ψov ≡ E ov
Ψov | H
(5)
v ′ |Ψov ′ ≡ E ov ′ .
v ′ |Ψov > Ψov ′ | H
Ψov | H
(6)
and
Adding these inequalities and carrying out the integration
over all coordinates but one, one obtains
′
(7)
d 3 r(v ′ (r) − v(r))(ρ0v (r) − ρ0v (r)) < 0 .
123
Because Eq. (7) is a strict inequality, a contradiction ensues
(0 < 0) when it is assumed that different potentials yield the
same one-particle density. Thus, it follows that there is a one
to one correspondence between the exact ground-state oneparticle densities and their corresponding external potentials.
In the present notation, Lieb’s statement of this theorem
(Theorem 3.2 of Ref. [16]) is the following: suppose Ψov
′
(respectively, Ψov ) is a ground state for v (respectively, v ′ )
′
and v = v ′ + constant. Then ρ0v (r) = ρ0v (r). Lieb’s proof
′
starts from the suppositions that ρ0v (r) = ρ0v (r) = ρ0 and
′
Ψov = Ψov because they satisfy different Schrödinger equations, and proceeds as in the original proof showing that
this leads to a contradiction. As it was mentioned above,
the argument for writing the strict inequalities [Eqs. (5) and
(6)] in Hohenberg–Kohn’s paper [24] is based on the assump′
tion that Ψov and Ψov satisfy different Schrödinger equations,
′
namely, that Ψov = Ψov .
The fact that the space of single particle potentials is not
specified in the original Hohenberg–Kohn proof was remedied in Lieb’s proof [16] by selecting this space as Y =
L 3/2 (R3 ) + L ∞ (R3 ) (where f (x) ∈ L m if d x | f (x)|m <
m if f ∈ L m and it is integrable in any bounded
∞. f ∈ L loc
set; f ∈ H 1 if f, ∇ f ∈ L 2 ) and by demanding that v(r) ∈
Y . This choice—which follows from the requirement
that
ρ 1/2 ∈ H 1 (R3 )—guarantees that the integral d 3 rρ(r)v(r)
(in fact, the essentially self-adjoint character of the
Hamiltonian [32]) is well defined.
An important difference arises, however, from the fact that
Lieb notes that in order to prove the statement that Ψov and
′
Ψov satisfy different Schrödinger equations it is necessary to
show that the equivalence
V (r1 , . . . , r N )Ψ (r1 , . . . , r N )
= V ′ (r1 , . . . , r N )Ψ (r1 , . . . , r N )
implies that v(r) = v ′ (r). Fulfillment of this condition
requires that the Ψov corresponding to the external potential v ∈ Y not vanish on a set of positive measure. As has
been indicated by Lieb [16] (p. 255), the unique continuation
theorem may be invoked to guarantee that Ψov does not vanish
in an open set. However, this theorem strictly holds only for
v ∈ L 3loc although it is believed to hold also for v ∈ Y . But
let us mention that there are subtle problems related to the
space to which a single particle potential belongs and to its
relation to the wavefunction. Thus, for example, as shown by
Englisch and Englisch [33], for a one particle case there exists
a non-vanishing density ρ (or equivalently, a non-vanishing
wavefunction given as Ψ = ρ 1/2 ) which does not arise from
any v, in the sense that for a v = ρ −1/2 ∇ 2 ρ 1/2 , −∇ 2 + v
cannot be defined as a semibounded operator. Precisely in
order to avoid these difficulties, we advance an algebraic
proof of the Hohenberg–Kohn theorem where these issues
are avoided.
Theor Chem Account (2007) 118:557–561
559
3 A re-statement of the Hohenberg–Kohn theorem
The present proof is essentially based on Lieb’s version of
the HK theorem (Theorem 3.2 and Remark (ii) in p. 255 of
Ref. [16]). But as mentioned above, in order to avoid some
mathematical complications, we have, however, removed the
′
assumption that Ψov = Ψov , i.e., we consider the case where
′
v = v ′ + constant but Ψov = Ψov (Case I of Kryachko [26])
and have added the condition on the ground state wavefunco be
tion that it vanishes at most on a zero-measure set. Let H
the Hamiltonian of an electronic Coulomb system without
o is not
external potential [cf. Eq. (3)]. In fact, the form of H
very important, as the proof is essentially algebraic. Let us
v given by Eq. (2).
consider the many-electron Hamiltonian H
We denote Y as in the above Section. We assume that ρov is
v if there exists a ground-state
the ground-state density of H
v
wavefunction Ψo of Hv . We denote by E ov the corresponding
eigenvalue.
Theorem 1 (Hohenberg–Kohn) Let v, v ′ be in Y . Let ρov
v and ρov ′ a ground state denbe a ground state density of H
v ′ . We assume that the ground state wavefunction
sity of H
v
Ψo of Hv vanishes at most on a Lebesgue’s zero-measure set
′
of R3N . Suppose that ρov = ρov . Then almost everywhere in
the Lebesgue’s measure sense (a.e.)
′
v(r) − v ′ (r) = (E ov − E ov )/N .
(8)
Proof We essentially make explicit what was implicit in
′
Lieb’s proof [16]. Let us introduce the notation ∆E = E ov −
N
v
′
E o , ∆v = v − v and ∆V = i=1 ∆v(ri ). We have then
v = H
v ′ − ∆V and
H
v |Ψov ≤ Ψov ′ | H
v |Ψov ′ = E ov ′ − ρov ′ ∆v. (9)
E ov = Ψov | H
where the equal sign must be included as we are not assum′
ing that for v = v ′ + constant the condition Ψov = Ψov
holds.
So we get a ≥ 0 where a = ∆E − ρo ∆v, and ρo = ρov =
′
ρov . Reversing v and v ′ we get similarly a ≤ 0. So a = 0
and this implies also that all the preceding inequalities are
′
v |Ψov ′
in fact equalities. In particular, we have E ov = Ψov | H
′
′
v : H
v Ψov = E ov Ψov ′ . In
so Ψov is also a ground state of H
v Ψov = E o Ψov
v ′ Ψov = E ov ′ Ψov . Using also H
the same way: H
v = ∆V , by subtraction we obtain
v ′ − H
and H
∆V Ψov = ∆E Ψov .
(10)
or, equivalently,
(∆V − ∆E)Ψov = 0.
(11)
Since we have by assumption that Ψ vanishes at most on
a set of zero measure (we take it to be a nodeless ground
state wavefunction) it follows from Eq. (11) that ∆V = ∆E
almost everywhere for (r1 , . . . , r N ) ∈ R3N , except for a set
of zero measure. Then setting r1 = · · · = r N = r we obtain
N ∆v(r) = ∆E (see also Harriman’s comments in p. 641
and in the Appendix of Ref. [30]). The present argument is
rigorous provided v is continuous; otherwise, the proof can
be completed using Lemma 1 proved in the Appendix.
⊔
⊓
4 The Hohenberg–Kohn theorem in finite subspaces
We first state a Hohenberg–Kohn theorem that holds in subspaces which are not necessarily finite-dimensional.
Theorem 2 (Infinite-dimensional subspaces) Let v, v ′ be
in Y . Let F be some subspace of the antisymmetric N -particle
v ′ ) such that F be
v and H
Hilbert space (in the domains of H
v F ⊂ F and
′
stable under the action of Hv and Hv , i.e., ( H
v
Hv ′ F ⊂ F). Take ρo a ground state density of the restriction
v | F and ρov ′ a ground state density of H
v ′ | F . Again, assume
H
that the ground state wavefunction vanishes at most on a set
′
of zero measure. Suppose that ρov = ρov . Then
′
v(r) − v ′ (r) = (E 0v − E 0v )/N .
(12)
Proof It is carried out along the same steps as in Theorem 1,
′
except for the fact that Ψov and Ψov must be in F in order to
apply the variational principle and obtain a = 0, and, hence,
′
v |Ψov ′ implying that Ψov ′ is a ground state of
E ov = Ψov | H
v | F .
⊔
⊓
H
We see, therefore, that it is possible to extend the HK formulation of Density Functional Theory to a subspace F as
long as the conditions of stability of Theorem 2 are satisfied.
However, as shown in Theorem 3 below, it is not possible,
in general, to satisfy the assumptions of Theorem 2. First
v ′ (F) ⊂ F then by taking the
v (F) ⊂ F and H
note that if H
difference we obtain ∆V (F) ⊂ F. We recall also that the
associated to a scalar potential V is defined by
operator V
(V (Ψ ))(x) := V (x)Ψ (x).
Theorem 3 (Finite-dimensional subspaces) Let F be a
finite-dimensional subspace of L 2 (Rn ) (n ≥ 1). We suppose that F = V ect(u 1 (x), .. . , u M (x)) where the (u i (x))
is an orthonormal set (i.e., u i u ∗j = δi j ) and such that
M
2
n
i=1 |u i (x)| > 0 a.e. for x ∈ R . Let V (x) be real-valued
potential, and continuous. Then
(F) ⊂ F) ⇒ (V (x) = const on Rn ).
(V
We note that this theorem also holds with weaker assumptions, such as, for instance, F ⊂ L 1loc (Rn ) (the space of
1 (Rn ) [i.e.
locally integrable functions on Rn ), and V ∈ Hloc
2
n
V, ∇V ∈ L loc (R )].
Proof We first remark that V behaves on F as an M × M
matrix since it is a linear operator. So there exists M = (m i j )
such that
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Theor Chem Account (2007) 118:557–561
V (x)u i (x) =
m i j u j (x).
(13)
j=1,...,M
Since u j is orthonormal, we have
dxu i (x)∗ V (x)u j (x)
mi j =
6 Appendix
Rn
using (13). Since V is real we obtain m i j = m ∗ji and thus
M is an hermitian matrix. So, we can diagonalize M in an
orthonormal basis: there exists a unitary matrix P (P † P =
P P † = Id ) and a diagonal matrix D = diag(λ1 , . . . , λ M )
such that M = P † D P.
Let us write u(x) = (u 1 (x), . . . , u M (x)). Then (13) reads
V (x)u = Mu. So, it follows that
V (x)Pu = P V (x)u = PMu = P P † D Pu = D Pu.
Hence if we define ψ(x) = Pu and denote its components
as (ψ0 (x), . . . , ψ M (x)), we obtain:
V (x)ψi (x) = λi ψi (x), i = 1, . . . , M.
(14)
We have simply diagonalized V (x) in an orthonormal basis
M
2
2
set. Then let us notice that
i=1 |ψi (x)| = ||ψ|| =
M
2
2
||u|| =
i=1 |u i (x)| since P is unitary. Obviously this
quantity is non-negative and thus we have a.e. x ∈ R n the
existence of an i ∈ {1, . . . , M} such that ψi (x) = 0. From
(14) we obtain V (x) = λi for this x. This implies finally that
the range of V is included in the finite set {λ1 , . . . , λ M }. For
1 this means that V
a regular V (x) such as continuous or Hloc
⊔
is a constant, which concludes the proof of Theorem 3. ⊓
A consequence of Theorem 3 is that, in general, it is not
possible to fulfill the stability conditions of Theorem 2 when
F is finite dimensional, except if we suppose that V (x) and
V ′ (x) are constants as then they would trivially satisfy the
main conclusion of Theorem 2, namely, ∆V = const. Let us
mention that this result is in agreement with the conclusion of
Görling and Ernzerhof for local potentials in finite subspaces
[see Eq. (A9) and the discussion below in Ref. [31]].
But in the infinite dimensional case Theorem 3 does not
hold and thus Theorem 2 becomes interesting. As an example, let F = L 2 (R3 ) and v(x) = 1/(1 + |x|2 ). Then we
have obviously v(F) ⊂ F (since v(x) ≤ 1) but v(x) is not
constant.
5 Conclusions
The main contribution of this article is to provide an algebraic
proof for the Hohenberg–Kohn theorem that allows us to discuss in a very simple way the extensions of this theorem to
both infinite-dimensional and finite-dimensional subspaces.
In the former case, such an extension is possible as long as the
v . In the latter case,
subspace is stable under the action of H
′
when the external potentials V and V , or their one-particle
components v and v ′ are constants.
123
Acknowledgments E.V.L. would like to express his gratitude to
FONACIT of Venezuela, for its support of the present work through
Project G-97000741.
Lemma 1 Suppose that
v(r1 ) + · · · + v(r N ) = 0, a.e. x = (r1 , . . . , r N ) ∈ R3N .
(15)
Then v(r1 ) = 0 a.e. r1 ∈ R3 .
Proof First note that we cannot (a priori) take r1 = r2 =
· · · = r N because {(r1 , . . . , r N ) ∈ R3N , r1 = r2 = · · · =
r N } is a set of zero measure in R3N . To bypass this difficulty,
we consider a real-valued
continuous function ρ(r ) > 0,
3 , such that ρ(r )dr = 1, and denote ρ ǫ (x) =
defined
on
R
1
ρ xǫ . We multiply Eq. (15) by ρ ǫ (x1 −r1 ) · · · ρ ǫ (x N −r N )
ǫ3
and then integrate over (r1 , . . . , r N ) ∈ R3N . We obtain
(16)
v ǫ (r1 ) + · · · + v ǫ (r N ) = 0, a.e.,
where v ǫ (x) = R3 v(y)ρ ǫ (x − y)dy (convolution product).
Then it is well known [34] that v ǫ is a continuous function and
thus Eq. (16) holds everywhere and not only almost everywhere. Then we can take r1 = · · · = r N = r and conclude
that v ǫ (r ) = 0 for all r . On the other hand it is also well
known that, as ǫ → 0+ v ǫ (r ) → v(r ) for a.e. r ∈ R3 (eventually for some subsequence v ǫn extracted from v ǫ , [34]).
⊔
⊓
Hence we conclude that v(r ) = 0 a.e. r ∈ R3 .
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