Basic concepts in quantum mechanics

Laszlo Erdos∗

Nov 18, 2008

The emergence of quantum physics in the mid 1920’s was a fundamental change; probablythe most important one in the long history of the physics. Moreover, it was so strikinglynew and different from anything known before, that several of the best scientific minds havedoubted its validity. There are many ways to explain why we should believe in it, and “foun-dations of quantum mechanics” has become a subject in itself for those who represent thefundamentalist’s point of view. The more pragmatically minded mainstream approach, how-ever, starts from a few basic axioms and focuses on results one can obtain from the theory.

The only difference between the axioms of quantum mechanics and the standard axiomsin mathematics (say, the axioms of elementary geometry [Euclid] or the axioms of set theory[Zermelo-Frenkel] or the axioms of integer arithmetics [Peano]) is that they are not at allobvious at first sight, and even not at second or third sights... So their justification is moreindirect, but a powerful one: they work. We will follow the most pragmatical point of view:whatever crazy its axioms sound, quantum mechanics, as a matter of fact, has correctlypredicted essentially any experiments in an enormous energy range. Quantum mechanicalprinciples seem to be valid from subatomic physics to astrophysics. They correctly accountfor many phenomena on large and small scales that no other theory could tackle. A prettyrewarding price for accepting some axioms that may sound unbelievable at the beginning....

In this course we will do quantum mechanics, which is the most basic part of quantumphysics. Quantum physics includes many other disciplines, such as quantum field theory, quan-tum statistical mechanics, quantum gravity etc., but they all originate in quantum mechanics,similarly as all classical physics (e.g. thermodynamics, fluid dynamics etc.) originates in clas-sical mechanics. The main goal of quantum mechanics is to describe the motion of quantumparticles.

∗Part of these notes were prepared by using the webnotes by Michael Loss: “Stability of Matter” and

the draft of a forthcoming book by Elliott Lieb and Robert Seiringer: “The Stability of Matter in Quantum

Mechanics”. I am grateful for the authors to make a draft version of the book available to me before publication.

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1 Classical mechanics

1.1 Phase space

The basic object in classical mechanics is a massive point particle. It has two intrinsic andpermanent characteristics: its mass m and its charge q. The mass is always positive. Thestate of the particle can be described by its position (location) in the d-dimensional Euclideanspace, x ∈ Rd, and by its momentum p ∈ Rd. Unless we say otherwise, we will always considerd = 3. The space of the possible x’s is called configuration space or position space, the spaceof possible momenta is momentum space. Depending on the physical situation, they may berestricted to a subset of Rd (e.g. if the particle is confined to a container, Ω ⊂ Rd, thenx ∈ Ω). The product space of the configuration space and momentum space, Rd ×Rd (or itsnatural subspace, e.g. Ω×Rd) is called the phase space, the pairs (x, p), describing a possibleposition and momentum of a particle, are called phase space points.

It is a deep fact of Nature, that the phase space point determines the state of the particle,i.e. knowing its position and momentum (two d-dimensional vectors) is sufficient to describeall its future; in other words, all what the particle “remembers” from its past, before a fixedtime t0, is given via the position and momentum at time t0. Actually there are two differentstatements combined in this sentence. One is that exactly two vectors (points) determineeverything, the other one is that it is sufficient to know these two vectors at a fixed time andwe can forget about the whole past. Both of these somewhat surprising facts are consequencesof Newton’s equation

md2x

dt2= F

where F = F (x, p) is the (instantenous) force acting on the particle; the force may dependon the phase space point. The fact that Newton’s equation is a differential equation is equiv-alent to the fact that the past influences the future through the state at present. The factthat Newton’s equation is of second order postulates that only two quantities are sufficient,there is no need for more, because a second order equation needs two initial data to have aunique solution. Traditional Newtonian kinematics considers position and velocity as thesetwo quantities (velocity being defined as the time derivative of position, x = dx

dt), it turns out

that momentum is a more canonical second quantity instead of velocity.More generally, we want to describe N massive point particles in Rd or in Ω ⊂ Rd. We

label the particles by 1, 2, . . . , N . Each particle has a position and a momentum, indexed bythe particle label. The location of the particles is given by x = (x1, x2, . . . , xN) ∈ RdN andthe momenta p = (p1, p2, . . . , pN) ∈ RdN .

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1.2 Hamiltonian

The state of the system may change with time, and at time t it is described by a time depen-dent phase space point (x(t), p(t)). The time is always a real variable. The dynamics (timeevolution) of the system is described by energy function or Hamilton function or Hamiltonianof the system. The Hamiltonian is a real function defined on the phase space

H : Rd × Rd → R (1.1)

Its value H(x, p) represents the energy of the physical system in state (x, p). A basic axiomof classical mechanics is that the Hamilton function determines the time evolution via theHamiltonian equations of motion

x = ∇pH(x, p), p = −∇xH(x, p) (1.2)

(dot denotes time derivative). Being a system of first order ordinary differential equations(ODE’s), the equations (1.2), under some mild regularity condition on H , fully determine thewhole future (and also past) trajectory (x(t), p(t)) once an initial data is given, i.e. once thestate of the system (x(t0), p(t0)) is known at some time t0. In other words, the Hamiltonianfunction comprises all the physical laws relevant for the system.

One important property of the Hamilton equations of motion is that the Hamiltonian(energy) is conserved with time

d

dtH(x(t), p(t)) = ∇xH · x+ ∇pH · p = ∇xH · ∇pH −∇pH · ∇xH = 0

When we build a physical model, we usually give its Hamiltonian. The standard Hamil-tonian of a single particle in classical mechanics (without magnetic fields) has the form

H =p2

2m+ U(x) (1.3)

where m > 0 is the mass of the particle and U(x) a real valued function, the potential. Thefirst term represents the kinetic energy of the particle, the potential describes the interactionwith an (unspecified) environment (e.g. container).

For N particles, we have

H =

N∑

j=1

p2j

2mj+ U(x) (1.4)

where mj is the mass of the j-th particle and U(x) is the potential. The term p2j/2mj represents

the kinetic energy of the j-th particle, the potential describes the interactions both among theparticles and with a possible environment.

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In many cases, the potential function simplifies into a one-particle and a two-particle part:

U(x) =∑

j

Vj(xj) +∑

j<k

Wjk(xj − xk). (1.5)

The first term is called background potential, the second one is the interaction potential. Notethat the interaction is assumed to be translationally invariant – not a necessity, but a conditionthat is satisfied in most cases.

The Hamiltonian (1.4), especially with the choice (1.5) may look a bit ad hoc, especiallyone may note the asymmetric role of the momenta and positions. It is, however, a fact of lifethat the two most “visible” interactions in real life, gravity and electrostatics, depend only onthe position and not on the momentum of the particles; thus the momenta are not directlycoupled.

If magnetic fields are present, then the kinetic energy of the jth particle is modified to

1

2mj

(pj −

qjcA(xj)

)2

(1.6)

where qj is the charge of the particle, c= 300,000 km/sec is the speed of light and A : Rd → Rd

is a vector field, representing the magnetic vector potential such a way that the magnetic fieldB is given by

B = ∇× A = curl A

(in dimension d = 3). We remark that the magnetic field (and quantities derived from it, likethe flux, which is the integral of A over closed loops) is the physically measurable quantity,the vector potential is not directly measurable. Maxwell’s equation dictates that any physicalmagnetic field is divergence free, ∇ · B = 0, thus it can be written as a curl. Notice that incase of the presence of a magnetic field, the velocity of the jth particle is

vj = xj = ∇pjH =

1

m

(pj −

qjcA(xj)

).

The formula (1.6) identifies the Lorenz force acting on the jth particle:

Fj =qjcvj ×B

Exercise 1.1 Check this formula for the Lorenz force from Newton’s law, from the Hamiltonequations and from the identity ∇A(v ·A)− (v ·∇)A = v× (∇×A) from vector calculus [wherethe first gradient acts only on A].

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Using the explicit form of H , we can write the equations of motion as

xj(t) =pj(t)

mj

, pj(t) = −∇xjU(x) = −∇Vj(xj) −

∑

k 6=j

∇W (xj − xk)

The first equation just says that the velocity (defined as the time derivative of x) is themomentum divided by the mass, the second equation is Newton’s equation if the negativegradient of the potential is interpreted as the force.

We remark that we presented the Hamilton formalism of classical mechanics. This formal-ism uses the assumption that there is an absolute concept of time. In relativistic systems suchan assumption cannot hold, and a more general formalism, the Lagrangian formalism, hasbeen developed. The two formalisms are equivalent if time is absolute. While the Lagrangianformalism is more general, its quantized version is much harder to define in a mathematicallyrigorous way, although it is necessary for doing e.g. relativistic quantum field theory. In thiscourse we will consider only non-relativistic quantum systems, so we will use the Hamiltonformalism and enjoy its advantages.

1.3 Coulomb systems

The most basic objects of study in quantum mechanics are massive, charged point particlesinteracting with electrostatic forces. This means that the Hamiltonian (1.4) holds (if nomagnetic fields are present) with a potential (1.5), where the interaction between the jth andkth particle is the Coulomb potential

Wjk(xj − xk) =qjqk

|xj − xk|

Note that the potential is negative for opposite charges and it is zero for infinitely distant par-ticles. If a potential goes to zero at infinity, then a negative potential is also called attractive,positive potential is repulsive. Note that the physics (equations of motion) is insensitive toadding an overall constant to the potential U . Using this freedom we will always (implicitly)assume that the potential goes to zero at infinity, whenever this is possible (i.e. wheneverlimx→∞ U(x) exists and is finite). We use the same implicit convention for all constituents ofthe potential, i.e. for Vj and Wjk as well.

The background potential originates from a fixed background charge distribution (x), i.e.it is also of Coulomb type:

Vj(x) = qj

∫

Rd

(y)

|x− y|dy = qj

(| · |−1 ⋆

)(x)

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Here the star denotes the convolution; in general

(f ⋆ g)(x) =

∫

Rd

f(x− y)g(y)dy =

∫

Rd

f(y)g(x− y)dy

In most cases is a sum of Dirac delta masses

(x) =K∑

k=1

Qk δ(x− Rk)

representing fixed point charges Qk sitting at the points Rk. In this case

Vj(x) =

K∑

k=1

qjQk

|x−Rk|

The simplest possible model is one single point charge moving in a zero potential field. Itis called the free particle and its Hamiltonian is

H =p2

2m.

With initial position xin and momentum pin at time t0, the equations of motion

x =p

m, p = 0

have the trivial solution

x(t) = xin +pin

m(t− t0), p(t) = pin

If several free particles are moving in a zero potential field, then

H =

N∑

j=1

p2j

2mj

and each particle follows its own trajectory

xj(t) = xinj +pinjmj

(t− t0), pj(t) = pinj

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without ever noticing each other.

The next simplest model is a single charged particle, with mass m and charge q, movingin the background of another particle with charge Q that is considered fixed at R ∈ Rd. TheHamiltonian is

H =p2

2m+

qQ

|x− R|Notice that R is considered as a parameter, i.e. it is not a dynamical variable. By shifting theorigin, we can assume R = 0. If the single charged particle is an electron with charge q = −e,and the fixed particle is a nucleus with proton number Z, i.e. with charge Q = Ze, then

H =p2

2m− Ze2

|x| (1.7)

This is the Hamiltonian of a hydrogenic atom; if Z = 1 then it is exactly the Hydrogen atom.

In full generality, we can consider K nuclei with charges Qk = Zke and masses Mk,k = 1, 2, . . . , K, located at positions R = (R1, R2, . . . RK), and N electrons, each with chargeq = −e and mass m at locations x1, . . . xN . Then the potential of the N electrons and Knuclei is

VC(x,R) = −N∑

j=1

K∑

k=1

Zke2

|xj − Rk|+

∑

k<ℓ

ZkZℓe2

|Rk −Rℓ|+

∑

j<ℓ

e2

|xj − xℓ|(1.8)

The first term represents the attraction between the electrons and the nuclei, the secondterm is the nuclei-nuclei repulsion while the last term is the electron-electron repulsion. Theattractive terms are negative, the repulsive ones are positive.

The full Hamiltonian of the N electrons is

H =N∑

j=1

p2j

2m+ VC(x,R) (1.9)

if the nuclei are considered fixed. In this case their positions are parameters, and H is definedon the phase space of the N electrons, i.e. on RdN ×RdN .

If we consider the nuclei dynamical as well, we need to introduce their momentum variables,call them (P1, P2, . . . , PK). The Hamiltonian of the N electrons and K nuclei thus is given by

H =N∑

j=1

p2j

2m+

K∑

k=1

P 2k

2Mk

+ VC(x,R) . (1.10)

This formula is the complete Hamiltonian of a molecule consisting of N electrons and Knuclei. Since in reality Mk ≫ m (the mass of the proton is about 1800 times of the mass of

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the electron), we often consider the simplified model (1.9) where the nuclei are considered soheavy (formally Mk = ∞) that they are treated as static particles.

One key feature of all these Coulombic Hamiltonians is that the range of the Hamiltonianfunction H is the whole R, in particular arbitrarily negative energies can be achieved. As-suming some radiation mechanism that is able to suck energy out of the system, the energy ofa Hydrogen atom (1.7) can be driven arbitrarily negative; just by placing the electron closerand closer to the attractive nuclei. Thus the Hydrogen atom would not be stable; the electronwould collapse into the nucleus. Moreover, it could release an infinite amount of energy – thisis clearly unphysical.

This problem has been noticed well before the discovery of quantum mechanics. Onepossible explanation is that the assumption about point particles is wrong; indeed the nucleihas a nonzero diameter of about 10−13 cm. However, the typical size of the Hydrogen atomis 10−8 cm, i.e several orders of magnitude bigger than the nucleus. Therefore the extendedshape of the nucleus cannot explain the non-collapse of the electron on a much bigger scale.

This question is known as the problem of the stability of Hydrogen, and similarly one canask whether a molecule of N electrons and K nuclei are stable in the sense that whetherinf H > −∞ or infH = −∞. As the formula (1.8) immediately shows, the Hamiltonians(1.9), (1.10) are unstable in this sense. Such a scenario would have dramatic consequenceson the world; it would indicate that after a long time the electrons of atoms and moleculeswould fall into their respective nuclei and matter would look rather like a dense soup insteadof consisting of fairly well separated particles and a huge energy would be released.

It was one of the great triumphs of early quantum mechanics that it could explain whyin the quantum model of the Hydrogen such a collapse does not occur. It took more than40 years after that before the similar but stronger stability statement (“stability of matter ofsecond kind” – see the definition later) was discovered and rigorously proven for molecules orfor any Coulomb system.

2 Quantum mechanics

2.1 States

The state of a single particle in quantum mechanics is given by a complex valued wave function

ψ : Rd → C

defined on the classical configuration space, Rd, or on a subset Ω ⊂ Rd. Unlike in classicalmechanics, where altogther 2d numbers were sufficient to specify the state (d position and d

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momentum coordinates), in quantum mechanics the state is given by a whole function, i.e.infinitely many numbers.

The non-negative function x → |ψ(x)|2 on Rd is interpreted as a probability density, i.e.for any subset Ω ⊂ Rd,

∫

Ω

|ψ(x)|2dx = Prob the particle is in Ω (2.1)

Since we wish to interpret |ψ(x)|2 as a probability density, we always assume the normalizationcondition ∫

Rd

|ψ(x)|2dx = 1 .

Therefore, the natural state space of a single quantum particle is the unit sphere in L2(Rd),the space of square integrable functions:

L2(Rd) =ψ : Rd → C :

∫

Rd

|ψ(x)|2dx <∞

(the integral here is understood in Lebesgue sense). We recall that the L2-space is equippedwith a natural scalar product

〈f, g〉 =

∫

Rd

f(x)g(x)dx

and with a norm‖f‖ = ‖f‖2 =

√〈f, f〉

and it is a Hilbert space, i.e. it is complete with respect to this norm. Since we will mostlyuse this L2-norm, we usually omit the subscript 2, i.e. ‖f‖ will always denote the L2-norm,by convention.

The definition (2.1) leaves a lot of room for discussion, especially what do we mean by theprobability here. As we said, we are not going into fundamentalist issues; we just mentionthat quantum mechanics does not allow to determine the precise position of the particle inany measurement (uncertainty principle). Moreover, we point out the experimental fact thatthe outcome of a quantum experiment is not a deterministic quantity, but rather a randomnumber: if the same experiment is repeated several times, the measuring apparatus may showdifferent numbers: it is only their statistics that is meaningful, i.e. we can ask what theprobability that the gauge in the apparatus shows number 1 is, or what the expectation valueof the shown number is if many identical experiments are performed.

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2.2 Observables

It is a fact that not everything can be measured in quantum mechanics. The wave functionin principle contains all information about the state, nevertheless not every property of ψ isaccessible by measurements. By definition, the measurable quantities are those that can berepresented by self-adjoint (linear) operators O acting on L2(Rd), i.e. O : L2(Rd) → L2(Rd);these are called observables. The result of the measurement on the state ψ is given by

〈ψ,Oψ〉 = Expected value of the measurement O in state ψ . (2.2)

and it is always a real number. Without the normalization condition ‖ψ‖ = 1, the expectedvalue of the measurement is given by

〈ψ,Oψ〉〈ψ, ψ〉 .

Recall that apart from (non-trivial and non-negligible domain questions) self-adjointnessmeans that O is symmetric, i.e.

〈ψ,Oχ〉 = 〈Oψ, χ〉, ψ, χ ∈ D(O) ⊂ L2(Rd)

and it is defined on the same domain as its adjoint, D(O) = D(O∗). To facilitate the introduc-tion, we do not worry about domain questions for the moment. For those who feel cheated,just think about bounded symmetric operators for the moment; it is a fact that any boundedoperator can be extended uniquely to the whole Hilbert space, even if it were originally definedonly on a dense subset (see, Theorem I.7. of Reed-Simon Vol I.) thus any symmetric boundedoperator is self-adjoint.

Note that any measurable quantity (2.2) is quadratic in ψ, e.g. it does not make sense toask, for example, for the integral of

∫ψ(x)dx. Moreover, an overall phase factor is invisible

for experiments, i.e. if we multiply the wave function by a phase factor eiα, α ∈ R, thenclearly

〈eiαψ,O(eiαψ)〉 =

∫

Rd

eiαψ(x)O(eiαψ(x))dx = e−iαeiα∫

Rd

ψ(x)Oψ(x)dx = 〈ψ,Oψ〉

where the linearity of O has been used. This means that no measurement can distinguishbetween the state ψ and the state eiαψ, so one may even identify these states; in mathematicallanguage take the factor space with the equivalence relation ψ ∼ χ iff there is α ∈ R suchthat χ = eiαψ.

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2.3 Position and momentum

The observable measuring the position is the multiplication operator by the variable x, i.e.

〈ψ, xψ〉 =

∫

Rd

ψ(x) xψ(x)dx =

∫

Rd

x |ψ(x)|2dx

which is clearly the first moment of the probability distribution |ψ(x)|2. [Remark: Since x isa d-vector, this is actually a vector-valued observable, so each coordinate of x = (x(1), . . . x(d))is a real valued observable and 〈ψ, xψ〉 is interpreted as a d-vector whose components the realvalued observables 〈ψ, x(j)ψ〉, j = 1, 2 . . . d.]

The observable measuring the momentum is −i times the derivative operator, p = −i∇ =−i∇x, i.e.

〈ψ, (−i∇x)ψ〉 = −i∫

Rd

ψ(x)∇xψ(x)dx (2.3)

Remark 2.1 Later we will insert a constant – the Planck’s constant ~ (3.2) – into the defini-tion, i.e. p = −i~∇. This is clearly necessary even for dimensional reasons: the momentumhas a dimension (mass) · (length) · (time)−1, while the derivative has dimension (length)−1,thus we need a constant with dimension (mass) · (length)2 · (time)−1 to compensate. Its exactvalue, determines the relation between classical world (derivative, length) and quantum world(quantum momentum). However, even later (see Section 6) we will choose units where ~ = 1,and in most of the course we will not see ~ at all. So in this section, for simplicity, we drop~.

Simple integration by parts in (2.3) shows that −i∇ is (formally) self-adjoint, i.e.

〈χ, (−i∇x)ψ〉 = 〈(−i∇x)χ, ψ〉

This relation certainly holds for sufficiently smooth (e.g. once continuously differentiable)functions that sufficiently decay at infinity (for example, compactly supported). Later we willsee how to determine the correct domain of self-adjointness of −i∇. The meaning of −i∇x

in position space is more obscure than that of the position operator x, but if we rewrite it inFourier space, we see a complete duality between position and momentum.

We recall that the Fourier transform of ψ is defined (formally) as

ψ(k) =

∫

Rd

ψ(x)e−2πix·kdx

This definition is meaningful if ψ is an integrable function, i.e. ψ ∈ L1(Rd), but it can beextended to L2(Rd), moreover, it turns out to be an isomorphy in L2(Rd) (i.e. the map

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“taking the Fourier transform” is a bijection from L2(Rd) onto L2(Rd) and it preserves thescalar product). In particular, we have Plancherel’s formula (theorem)

∫

Rd

ψ(k)χ(k)dk =

∫

Rd

ψ(x)χ(x)dx

in particular‖ψ‖ = ‖ψ‖ .

The inverse Fourier transform is given by

f(x) =

∫

Rd

f(k)e2πix·kdk

(note the positive sign in the exponent!) and it can be shown that indeed

ψ = (ψ) = (ψ) (2.4)

Formally the first relation can be seen from

(ψ) (x) =

∫

Rd

e2πix·k(∫

Rd

ψ(y)e−2πiy·kdy)dk

=

∫

Rd

ψ(y)(∫

Rd

e2πi(x−y)·kdk)dy

=

∫

Rd

ψ(y) δ(x− y)dy

=ψ(x)

(2.5)

however, here the application of the Fubini theorem and also the usage of the delta functionis not fully rigorous, nevertheless, (2.4) can be established rigorously as an identity betweenL2 functions (in particular, one does not expect it to hold for every x, only for almost everyx).

Moreover, it follows directly from the definition of ψ that

〈ψ, (−i∇x)ψ〉 = −i∫

Rd

ψ(x)∇xψ(x)dx = 2π

∫

Rd

k|ψ(k)|2dk

and similarly

〈ψ,−∆ψ〉 = 〈ψ, (−i∇) · (−i∇)ψ〉 =

∫

Rd

|∇ψ(x)|2dx = (2π)2

∫

Rd

k2|ψ(k)|2dk (2.6)

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In summary, the action of the momentum operator −i∇x on ψ(x) is just multiplication by

2πk in the Fourier representation ψ(k). This correspondance works in the other direction aswell:

〈ψ, xψ〉 = 〈ψ, (−i∇k)ψ〉i.e. there is a complete duality between position and momentum and between position spacerepresentation of the state, i.e. ψ(x), and its momentum space representation ψ(k). Derivativein one representation corresponds to multiplication by the variable (times 2π) in the otherrepresentation.

This correspondance holds even pointwise via the following formulas:

[−i∇ψ(x)](k) = (2π)kψ(k), [(2π)xψ(x)](k) = −i∇kψ(k) (2.7)

Remark 2.2 (VERY IMPORTANT) It is a useful rule of thumb to think of x carryingthe dimension of a length, while the Fourier variable, k, carries the dimension of (length)−1.In general, large scale properties of a function ψ(x) (i.e. behaviour for |x| ≫ 1) is reflected

in the short scale properties of ψ(k), i.e. |k| ≪ 1; e.g. for a function ψ that decays slowly in

x-space, we will have a singularity at k = 0 in its Fourier transform ψ(k). Vice versa: short

scale properties of ψ are reflected in large scale properties of ψ. In physics, the first regime iscalled infrared (IR) (short wavelength = large distance) regime, the second one is ultraviolet(UV) regime (large wavelength = short distance).

Recall that the Fourier transform expresses oscillations in a function. The oscillation hasa natural lengthscale (wavelength), and its inverse (frequency) is the corresponding Fourier

variable k (often called also mode). Thus ψ(k) tells us how much oscillation with wavelength

k−1 occurs in ψ. Oscillation is closely related to derivative: higher frequency content (big ψ(k)for some large k) implies higher derivative of ψ, this is clear from (2.7).

You can read more about the Fourier transform in Lieb-Loss: Analysis, Chapter 5 or in ahandout to be published later. A final remark is that Fourier transform always carries 2π andthere are different conventions where one tucks 2π in. We used the convention of the bookLieb-Loss: Analysis, while Reed-Simon defines the Fourier transform and its inverse as

ψ(k) =1

(2π)d/2

∫

Rd

ψ(x)e−ix·kdx, f(x) =1

(2π)d/2

∫

Rd

f(k)eix·kdk

The discrepancy is irrelevant as long as one is aware of it and checks at the beginning of eachbook which convention is used.

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3 Hamiltonian: the generator of the time evolution

The energy is the most important measurable quantity, the corresponding quantum observableis the Hamilton operator H . It is a self-adjoint operator defined on L2(Rd), thus its expectedvalue is always real (analogously, the Hamilton function (1.1) is real valued). The Hamiltoniangenerates the time evolution of the state of the system, i.e. the time evolution of the timedependent wave function ψ(t) via the Schrodinger equation

i~∂tψ = Hψ (3.1)

where~ = 1.05 × 10−34 Joule×sec = 1.05 × 10−27g cm2 sec−1 (3.2)

is a universal physical constant (Planck’s constant divided by 2π) – we will discuss the unitslater.

Similarly to classical mechanics, the Hamiltonian (which is now an operator and not afunction) contains all physical information about the system, so any modelling in physicsstarts with determining H . We remark, that similarly to classical mechanics, this formalismapplies only to non-fully-relativistic situations, i.e. where there is an absolute time. Otherwise,the quantized version of the Lagrangian formalism is needed.

The Schrodinger equation is a first order evolution equation. With a given initial data,ψ(t0) = ψ0, at a fixed time t0, it has a unique solution

ψ(t) = e−i(t−t0)~−1Hψ0 (3.3)

Formal substitution shows that (3.3) indeed solves (3.1). The main question, however: whatexactly the exponential on the right hand side of (3.3) is and whether the formal rules ofdifferentiations really apply.

Even before we try make sense of the formula for the solution, the first question is whetherthe solution exists at all, and if yes, whether it is unique. Being a simple evolution equa-tion, from standard ODE theory we know that existence and uniqueness (at least locally) isguaranteed by Lipschitz continuity, i.e. if there is a constant K such that

‖Hψ −Hψ‖ ≤ K‖ψ − ψ‖which, by the linearity of H is equivalent that H is bounded.

If H were a bounded operator (in particular a matrix acting on the finite dimensionalHilbert space CN) then one could define eitH for any constant t ∈ R by a power series:

eitH =

∞∑

n=0

(it)nHn

n!(3.4)

14

Exercise 3.1 Check that this series converges in the operator norm and that the usual rulesof differentiations apply, in particular d

dteitH = iHeitH , actually the same holds for any power

ecH with c ∈ C. Along the way, you will have to check directly from (3.4) that

eitHeisH = ei(t+s)H

As we will see in a moment, the most important Hamilton operators are unbounded, sincethey contain derivatives, and derivative operators are never bounded in L2, inequality of theform ‖∇ψ‖ ≤ K‖ψ‖ could NEVER hold (WHY?). In particular, typical Hamiltonians arenot defined on the whole L2(Rd) [Recall Hellinger-Toeplitz theorem, Corollary to TheoremIII.12 in Reed-Simon Vol. I]. Therefore not only the series (3.4) does not converge, but it iseven questionnable whether there is any element of the Hilbert space for which the right handside even term by term could be applied (in principle it could be that the intersection of alldomains D(Hn), n = 1, 2, 3, . . . is trivial).

It turns out that the symmetry of H is not sufficient to define eitH , and to define thedynamics, we will need self-adjointness. The definition will go through the spectral theorem,which is a generalization of the diagonalization of hermitian matrices to unbounded operators.Recall that if H is a finite hermitian matrix on CN , H = H∗, then, alternatively to (3.4), onecould define eitH as

eitH = UeitDU∗ (3.5)

where H = UDU∗ is the diagonalization of H , i.e. U is a unitary matrix (U−1 = U∗)containing the orthonormalized eigenbasis of H and D = diag(λ1, λ2, . . . , λN) is a diagonalmatrix containing the eigenvalues (with multiplicity). The exponential of the diagonal matrixeitD is defined as the diagonal matrix with entries eitλj

Exercise 3.2 Check that the two definitions (3.4) and (3.5) coincide for hermitian matrices

The precise formulation of the spectral theorem for unbounded operators is fairly long;we will do it only later, when we will really need it. You can read the statement in SectionVIII.3 of Reed Simon, but it may seem scary for the moment. The essence is that with itshelp one can define functions of self-adjoint operators, i.e. not only polynomials of H makesense (like H2, H3, . . .), but essentially for any function f(λ) with real argument λ ∈ R onecan define an operator f(H) such that for polynomials it coincides with the usual definitionand all standard “calculus” rules apply. E.g. with the function ft(λ) = eitλ one can define theoperators eitH for any t in such a way that e.g. eitHeisH = ei(t+s)H holds. It is quite remarkablethat such a powerful calculus exists with operators that are more complicated objects thanfunctions. The spectral theorem says that all these are possible, if H is self-adjoint.

15

We remark, that for certain special Hamiltonians, eitH can be easily computed withoutreference to spectral theorem. For example if H = V (x), i.e. no kinetic energy is present,then clearly

ψt(x) = e−itV (x)/~ψ0(x)

solves the Schrodinger equation (3.1). Similarly, if H = −~2∆, then eit~∆ is a multiplicationin Fourier space, so

ψt(k) = eit~∆ψ(k) = eit~(2πk)2ψ(k)

is the Fourier transform of the solution to (3.1). [See Homework problem]. Unfortuntely,such explicite formulas are not available for the general case, H = −∆ + V , and e−itH cannotbe “put together” (at least not easily) from e−itV (x) and eit∆, because V (x) and ∆ do notcommute, thus

e−it(−∆+V ) 6= eit∆eitV

4 Hamiltonian: the energy

Similarly to the classical case, the energy Eψ of a single quantum particle in state ψ is thesum of two parts: a kinetic energy and potential energy

Eψ = Tψ + Vψ

where the kinetic energy (without magnetic fields) is

Tψ =~2

2m

∫

Rd

|∇ψ(x)|2dx

and the potential energy is

Vψ =

∫

Rd

V (x)|ψ(x)|dx

where V (x) is a real valued function (potential). Written with the observable notation

Tψ =1

2m

∫

Rd

|(pψ)(x)|2dx =⟨ψ,

p2

2mψ

⟩=

1

2m〈−i~∇ψ,−i~∇ψ〉 =

~2

2m〈ψ,−∆ψ〉

where we used the notationp = −i~∇x

to replace the classical momentum p with an operator (the notation is a bit sloppy, one reallyshould distinguish the operator p from the classical momentum, e.g. by putting a hat on

16

it, p = −i~∇x, but we will not use the classical momentum later). We see that with thisreplacement, the quantum kinetic energy formally is the same as the classical kinetic energyin (1.3).

The potential energy is even easier, in observable form it is

Vψ = 〈ψ, V ψ〉

Thus the total energy is represented by the expectation value

Eψ =⟨ψ,

[ p2

2m+ V

]ψ

⟩=

~2

2m

∫

Rd

|∇ψ(x)|2dx+

∫

Rd

V (x)|ψ(x)|2 (4.1)

of the Hamilton operator

H =p2

2m+ V = − ~2

2m∆ + V

which acts on any function ψ as follows

(Hψ)(x) = − ~2

2m(∆ψ)(x) + V (x)ψ(x)

i.e. the potential acts as a multiplication operator. Notice that the second identity in (4.1)requires an integration by parts, so one should worry about the domain problem, i.e. preciselyfor which ψ’s are both expressions well defined.

As an example, the Hamiltonian of a hydrogenic atom, assuming that the nucleus withcharge Ze is fixed at the origin and we describe only the single electron with charge −e, isgiven by

HHydr = − ~2

2m∆x −

Ze2

|x| (4.2)

The Z = 1 case is the usual Hydrogen atom.In almost all cases in quantum mechanics, the Hamilton operator can be decomposed into

a kinetic energy operator H0 and a potential energy operator:

H = H0 + V

In most cases H0 is a (pseudo)differential operator, while V is a multiplication operator. Onemay think that the potential is the “easy” part and the kinetic energy is the complicatedone, eventually multiplication by a function seems easier than differentiation. However, if onewrites the action of H in Fourier space then, by duality, differentiation and multiplicationget interchanged. In particular, as (2.7) shows, derivative in position space corresponds tomultiplication in Fourier space and vice versa.

17

Independenty of the representations, the key point is that typically H0 and V do notcommute, since p and x do not commute. This is the basic source in quantum mechanics whythe theory cannot be described by classical objects like functions that commute; so one needsmatrices, or “generalized matrices”, i.e. operators. One of the fundamental observation of the“founding fathers” of quantum mechanics was that non-commutativity is needed.

We mention that in the presence of a magnetic field B = ∇× A in d = 3 dimensions thekinetic energy of a particle with mass m and charge q is modified

TA,ψ =1

2m

∫

R3

∣∣∣(− i~∇− q

cA(x)

)ψ(x)

∣∣∣2

dx = 〈ψ,HAψ〉 (4.3)

with

HA =1

2m

(p− q

cA(x)

)2

=1

2m

(− i~∇− q

cA(x)

)2

beind the kinetic energy operator. Notice that the formula is gauge invariant, i.e. for anyreal function χ : R3 → R the kinetic energy operator with A and with A + ∇χ are unitarilyequivalent:

HA+∇χ = U(χ)HAU∗(χ) (4.4)

where U(χ) is the multiplication operator by the complex phase exp(−i(q/~c)χ(x)).

Exercise 4.1 Check the formula (4.4)!

We add two remarks that will be relevant later on. First, the magnetic field itself has anenergy (price to be paid to the power company to generate this field [E.Lieb]). In standardCGS units (and in d = 3) it is

1

8π

∫|B(x)|2dx (4.5)

wher the magnetic field has a dimension (mass)1/2 · (length)−1/2 · (time)−1, i.e. measured ing1/2cm−1/2sec−1. If we allow the system to adjust its own magnetic field, the field energymust also be included in the total energy of the system, although this is not an operator butjust a number. Second, we remark that the magnetic kinetic energy (4.3) applies only tospinless particles, in particular, strictly speaking, not to electrons, that have spin 1

2. We will

later introduce the corresponding kinetic energy (Pauli operator) that takes also spins intoaccount.

18

5 Ground state energy

In the previous section we explained that in order to define the time evolution, one first needsto establish the self-adjointness of H (on a suitable domain) and second one needs the spectraltheorem. Both steps require some non-trivial preparation from functional analysis and theyare not very intuitive. So we will postpone their discussion, and we first focus on an aspectof quantum mechanics that can be presented with much less technicalities.

We mentioned that due to radiations, systems tend to favor their low energy states, espe-cially their ground states. Finding out the lowest energy state of a system is also importantbecause it tells us at most how much energy it can release. If we had a system whose energycould be unbounded from below, then driving this system into lower and lower energy states,we could gain infinite amount energy out of it – clearly a cheap solution to all our energyproblems.

So one of the very first questions about a quantum Hamiltonian is whether it is boundedfrom below, more precisely if its ground state energy, defined as

E0 = infEψ = 〈ψ,Hψ〉 : ‖ψ‖ = 1

is finite or minus infinity. The infimum may not be achieved even if E0 > ∞, neverthelesswe still call it ground state energy. If there is a minimizer, ψ0 such that E0 = Eψ0

, then ψ0

is called the ground state. If E0 > −∞, then we say that the system satisfies the stability offirst kind.

Note that to pose the question of stability, one can avoid defining the domain ofH precisely,we can simply use the following definition of the energy:

Eψ = Tψ + Vψ =~2

2m

∫

Rd

|∇ψ(x)|2dx+

∫

Rd

V (x)|ψ(x)|2dx

The advantage of this definition (called the quadratic form of H), is that the kinetic energyterm is the integral of a non-negative quantity, and if ψ is non-differentiable, more precisely ifits derivative does not coincide almost everywhere with an L2-function, then we simply defineTψ = ∞. The potential term may have both positive and negative part; we decompose it

V (x) = [V (x)]+ − [V (x)]−

where for any real number a,

[a]+ = maxa, 0, [a]− = −min0, a

denote the positive and the negative parts, respectively.

19

As long as the negative part of the potential is controlled by the kinetic energy, in thesense that ∫

Rd

[V (x)]−|ψ(x)|2dx ≤ Tψ +K‖ψ‖2 (5.1)

for some finite constant K, independently of ψ, then Eψ is bounded from below:

Eψ = Tψ +

∫

Rd

[V (x)]+|ψ(x)|2dx−∫

Rd

[V (x)]−|ψ(x)|2dx ≥ −K‖ψ‖2 = −K

taking into account the normalization ‖ψ‖ = 1. We thus managed to reduce the problemof stability to the proof of inequality (5.1) and notice that this inequality requires no worryabout domains. If Tψ = ∞, then the inequality always holds, so we can restrict our attentionto those ψ’s such that Tψ < ∞ (this space will be called the H1-Sobolev space). If the lefthand side of (5.1) is infinite, then the inequality does not hold, otherwise we have to comparetwo finite numbers.

6 Units

Before we go on with more complicated formulas and concepts, we will fix which physical unitswe use. The idea is that by choosing the units properly, all physical constants like electroncharge, mass, Planck constant etc. can be set equal 1 and in this way we can focus on thestructure of the formulas undisturbed by irrelevant constants.

In CGS units we have

• m= mass of the electron = 9.11 × 10−28 g

• e = (-1)× charge of the electron = 4.8 × 10−10 g1/2cm3/2sec−1

• c= speed of light = 3 × 1010 cm sec−1

• ~ = Planck’s constant divided by 2π = 1.055 × 10−27 g cm2sec−1

The first three constants are known from classical physics. Planck’s constant is the fun-damental constant of quantum mechanics, essentially by connecting the momentum with thephysical space scale through p = −i~∇x.

Out of these four constants, there is a unique way to form a single dimensionless constant

α =e2

~c=

1

137.04

20

which is called the fine structure constant. Since the electrostatic interaction potential isquadratic in the charge, α can be thought as measuring the strength of this interaction.

The natural lengthscale is the Compton wavelength of the electron, defined as

ℓC =~

mc= 2.43 × 10−10cm

and the natural energy scale is the rest mass energy of the electron

Er = mc2 = 8.2 × 10−7ergs = 8.2 × 10−7g cm2sec−2

These are the natural units in relativistic quantum mechanics. In most of this course wewill deal with non-relativistic quantum mechanics, so the appearance of the speed of light isunnatural.

Our basic unit of length thus will be

ℓ =ℓC2α

=~2

2me2= 1.06 × 10−8cm (6.1)

which is twice the Bohr radius, the typical size of the Hydrogen atom. Our energy unit willbe

2mc2α2 =2me4

~2= 4 Ry = 8.73 × 10−11ergs (6.2)

where 1 Rydberg (Ry) is the ground state energy of the Hydrogen atom.Changing the unit of length requires rescaling the wave function ψ; note that the normal-

ization condition must be respected. Therefore, if ψ(x) is the wavefunction in a certain unit,and we rescale the space by a factor λ, i.e. x → X = x/λ, then the wave function must berescaled as

ψ(x) → λ−3/2ψ(xλ−1) = λ−3/2ψ(X) =: ψ(X)

in the new unit. We can define a unitary transformation

(Uλψ)(x) = λ−3/2ψ(xλ−1)

representing the change of unit of length. It is easy to see that

U∗λ∂xUλ = λ−1∂x

andU∗λ | · |α Uλ = λα| · |α

where | · |α is a multiplication operator with |x|α. Comparing these two formulas, we see thatthe derivative operator scales as (length)−1, i.e. exactly as the Coulomb potential.

21

As an exercise, we compute the Hamiltonian of a hydrogenic atom (4.2) (written in CGSunits) in our new units. We write the space coordinate x in our unit of length (6.1) as

x = ℓX =~2

2me2X

then~2

2m∆x =

~2

2m

(~2

2me2

)−2

∆X =2me4

~2∆X

andZe2

|x| =Ze2

|X|(

~2

2me2

)−1

=2me4

~2

Z

|X|Thus, in the new units

H = − ~2

2m∆x −

Ze2

|x| =2me4

~2

(− ∆X − Z

|X|)

so the Hamiltonian of the hydrogenic atom is

H = −∆X − Z

|X| (6.3)

measured in our energy unit (6.2). Comparing with (4.2) we see that in our new units

2m = ~ = e = 1

Important rule of thumb: It is always good to check how various quantities scale withlength. The derivative scales as (length)−1, thus the Laplacian scales as (length)−2 and theCoulomb potential scales as (length)−1. Moreover, any integral

∫R3(. . .)dx scales as (length)3

and any Fourier variable (frequency) scales as (length)−1. This indicates how the appropriateterm behaves under scaling x→ λx. The scaling property of a term should not change alongany mathematical manipulations, so this gives rise to a quick check: after a long calculationone can simply check whether the initial formula and the final formula scales in the same waywith the length. If not, there is an error!

In particular, the two terms in the hydrogenic Hamiltonian scale differently with length,thus it gives rise to an intrinsic lengthscale, namely a lengthscale of order Z−1 (measured inthe units we use, in this case in ℓ). For example, we will see that the eigenfunctions of H liveon a lengthscale Z−1; this is the “built-in” lengthscale in H : if Z were (length)−1, then thetwo terms in H would both scale in the same way.

22

Exercise 6.1 (i) Prove that the Hamiltonian of a hydrogenic atom in the relativistic unitsℓC and Er is given by

H = −1

2∆x −

Zα

|x|

(ii) Show that a vector potential A has dimension g1/2cm1/2sec−1 (i.e. charge/length) andthe magnetic field B has dimension g1/2cm−1/2sec−1 (i.e. charge/(length)2). Chooseℓ−1C

√~c as the unit for A, and ℓ−2

C

√~c as the unit for the magnetic field and prove that

the Hamiltonian of a hydrogenic atom in a magnetic field is given by

H =1

2(p+

√αA)2 − Zα

|x|

in the relativistic units ℓC and Er.

(iii) Show that in units indicated in part (ii), the field energy (4.5) remains unchanged, i.e.it is still given by

1

8π

∫

R3

|B(x)|2dx

This exercise shows that in the relativistic units

m = ~ = c = 1

and√α is the elementary charge e.

Convention: To economize the formulas, we will often omit dx and the domain from theintegration, i.e. we write ∫

f for

∫

R3

f(x)dx

7 Stability of the hydrogenic atom

We now show that the hydrogenic atom, given by the Hamiltonian (6.3), is stable:

Theorem 7.1 Consider the set

M :=ψ : R3 → C :

∫

R3

|∇ψ(x)|2dx <∞,

∫

R3

|ψ(x)|2|x| dx <∞,

23

of (unnormalized) wave functions whose kinetic energy and Coulomb energy are finite. Then,for any Z > 0, the ground state energy

E0 = inf∫

R3

|∇ψ(x)|2dx−∫

R3

Z

|x| |ψ(x)|2dx : ψ ∈ M, ‖ψ‖ = 1

(7.1)

is finite and it is given by

E0 = −Z2

4

and the function

ψ0(x) =Z3/2

√8πe−Z|x|/2 (7.2)

is the unique minimizer.

The key ingredient of the proof is a lemma that establishes the bound (5.1) for our potential(after a Schwarz inequality ab ≤ 1

2(a2 + b2) for positive numbers a = ‖∇ψ‖, b = ‖ψ‖):

Lemma 7.2 Let ψ ∈ M, then

∫ |ψ(x)|2|x| dx ≤ ‖∇ψ‖ ‖ψ‖ (7.3)

and equality holds if and only if ψ0(x) = const. e−c|x| for some positive constant c > 0.

Remark. You may wonder how to figure out such an inequality. Recall that in (5.1) wewanted to control the negative part of the potential (in this case the complete potential) bythe kinetic energy (plus the L2-norm). Apart from the trivial constant Z, the left hand sideof (7.3) is the potential energy. We want to bound it in terms of ‖∇ψ‖ and ‖ψ‖ – what kindof inequality has any chance to be correct at all?

Here is a “back-of-the-envelope” test that checks whether an inequality can be correct. Itdoes not prove the inequality, but if an inequality does not pass this test, it is surely wrong.The idea is to test how the inequality scales with two different rescaling.

First, notice that if ψ is replaced by λψ with some λ > 0, then both sides of (7.3) changeswith a factor λ2. The two sides must scale in the same way with λ, otherwise the inequalitycannot be correct.

Second, replace ψ(x) with ψ(x/lambda). After a change of variables, the left hand sidescales as ∫ |ψ(x/λ)|2

|x| dx = λ−1λ3

∫ |ψ(y)|2|y| dy

24

Here λ3 comes from the change of variables, y = x/λ, the extra λ−1 comes from the denomi-nator. On the right hand side, we find

(∫|∇[ψ(x/λ)]|2dx

)1/2(∫|ψ(x/λ)|2dx

)1/2

= λ−1λ3(∫

|∇[ψ(y)]|2dy)1/2(∫

|ψ(y)|2dy)1/2

,

where, again, λ3 comes from the trivial volume factors and λ−1 comes from the derivative.Again, we see that the two sides scale in the same way, which is a necessary condition for theinequality to hold.

Actually, it is easy to see that no other combination of the form ‖∇ψ‖α‖ψ‖β would passboth tests apart from (7.3) with α = β = 1.

Proof of Lemma 7.2. By a standard density argument, it is sufficient to prove the inequalityfor all ψ ∈ C∞

0 , i.e. for all smooth, compactly supported functions. – See a remark at the endof the proof.

Using integration by parts, compute

2〈ψ, 1

|x|ψ〉 =

3∑

j=1

〈ψ,[∂xj

,xj|x|

]ψ〉

= −∑

j

(〈∂xj

ψ,xj|x|ψ〉 + 〈 xj|x|ψ, ∂xj

ψ〉)

≤ 2∑

j

|〈∂xjψ,

xj|x|ψ〉|

(7.4)

where [A,B] = AB −BA is the commutator. By Schwarz inequality

|〈∂xjψ,

xj|x|ψ〉| ≤ ‖∂xj

ψ‖ ‖ xj|x|ψ‖

After summing up, and using another Schwarz inequality for the sum, we have

∑

j

|〈∂xjψ,

xj|x|ψ〉| ≤ ‖∇ψ‖ ‖ψ‖

Thus the proof of (7.3) is complete.

Exercise 7.3 Prove the case of equality stated in the theorem.

25

Proof of Theorem 7.1. Using (7.3) and ‖ψ‖ = 1, we have

∫|∇ψ|2 − Z

∫ |ψ|2|x| ≥ ‖∇ψ‖2 − Z‖∇ψ‖ =

(‖∇ψ‖ − Z

2

)2

− Z2

4

after completing the square. Taking the infimum for all ψ ∈ M, we have

E0 ≥ −Z2

4

and it is easy to check that among the functions ψ0(x) = const. e−c|x| the equality is achievedonly if c = Z/2. The prefactor in (7.2) comes from the normalization. (CHECK!)

You may find the proof of Lemma 7.2 very special. The commutator trick works onlyfor the Coulomb potential. What if we consider a different potential? This will lead us toSobolev inequalities, which we will discuss later. One of them is the following lower boundon the L2-norm of the gradient in d = 3 dimensions: there exists a universal constant C suchthat

‖∇ψ‖ ≥ C‖ψ‖6

This inequality can be used to prove the bound (5.1), since, by Holder’s inequality (CHECKthe exponents!) ∫

[V ]−|ψ|2 ≤ ‖ψ‖26 ‖[V ]−‖3/2

and thus combining this with the Sobolev inequality, we have

∫[V ]−|ψ|2 ≤ ‖∇ψ‖2 = Tψ

as long as ‖[V ]−‖3/2 ≤ C2. If this latter condition is not satisfied (e.g. for the Coulombpotential V (x) = |x|−1 6∈ L3/2) one can still use this idea with a bit twist. Suppose that onecan write

[V ]− = V1 + V2

where V2 ∈ L∞ and V1 ∈ L3/2 with ‖V1‖3/2 ≤ C−2. Then

∫[V ]−|ψ|2 =

∫V1|ψ|2 +

∫V2|ψ|2 ≤ ‖∇ψ‖2 + ‖V2‖∞‖ψ‖2 = Tψ + ‖V2‖∞‖ψ‖2

i.e. (5.1) is still satisfied with K = ‖V2‖∞.

26

VERY IMPORTANT REMARK: We will often use the idea that to prove an inequal-ity like (7.3) for all functions for which the two sides are finite, it is sufficient to prove theinequality for “nice” functions, most often for C∞

0 functions. In mathematics literature, thisfact is referred to as “standard density argument”, and one usually does not waste more timeon it. We will do the same, but once for all, I want to show on this example how it goes.

We have to know that C∞0 is dense in H1-norm, i.e. in the L2 norm and in the norm of

the L2 of the gradient:‖ψ‖2

H1 := ‖ψ‖2 + ‖∇ψ‖2

This statement can be found in Theorem 7.6 of Lieb-Loss (the density in L2 is in Theorem2.16) and later we will define H1 precisely, for the moment: it is the space of all functionswhose gradient is in L2.

Armed with this information, the argument goes as follows: suppose that (7.3) is provenfor all C∞

0 functions, and let ψ ∈ M an arbitrary function. From the density of C∞0 in H1

can find an approximating sequence ψn ∈ C∞0 ∩M such that

‖ψ − ψn‖ → 0, ‖∇ψ −∇ψn‖ → 0 (7.5)

(in particular, ‖∇ψn‖ → ‖∇ψ‖ and ‖ψn‖ → ‖ψ‖) and we have

∫ |ψn(x)|2|x| dx ≤ ‖∇ψn‖ ‖ψn‖ (7.6)

Moreover, ψn is a Cauchy sequence in the weighted L2 space with measure dµ(x) = |x|−1dx,since

∫|ψn(x) − ψm(x)|2dµ(x) =

∫ |ψn(x) − ψm(x)|2|x| dx ≤ ‖∇(ψn − ψm)‖ ‖ψn − ψm‖ → 0

as n,m → ∞, where we used (7.3) for C∞0 functions. Using the Riesz-Fischer theorem

(Completeness of Lp, Theorem 2.7 of Reed-Simon) and passing to a subsequence (which wecontinue to denote by ψn) we can assume that ψn converges in L2(R3, dµ) as well. The limitmust be ψ (WHY? – because a subsequence converges almost everywhere pointwise, bothin L2 and in L2(R3, dµ), this fact is also part of Riesz-Fischer theorem, and a sequence offunctions has only one pointwise limit)

Taking now the n → ∞ limit on both sides of the inequality (7.6), we obtain that (7.3)holds for ψ as well.

27

8 Stability of atoms and molecules

8.1 Stability of first kind

In the previous section we proved that the very special hydrogenic atom is stable. But actuallythe same proof will immediately show that any atom or molecule or even any extended matterwith Coulomb interaction is stable.

The state of N particles is described by a wave function of N variables, i.e. by elements of

L2(R3N ) =ψ(x1, x2, . . . xN ) : R3N → C :

∫|ψ(x1, . . . xN)|2dx1 . . .dxN <∞

Later we will see that not every function is allowed, only those with a certain symmetry type(depending on whether the particles are bosons or fermions) but for the moment we considerall functions.

The Hamilton operator (formally) is given by

H =

N∑

j=1

(−∆xj) + VC(x,R)

where x = (x1, x2, . . . , xN ) and VC was given in (1.8) (setting e = 1). This is the naturalquantum analogue of (1.9) (in our chosen units), and we consider here the case of staticnuclei. The Laplacian ∆xj

acts only on the j-th variable, but it is still considered as anoperator acting on functions ψ with N variables. Often we will write ∆j for ∆xj

and similarlyfor the gradient ∇j = ∇xj

.Thus the energy of this molecule in state ψ is

Eψ =

N∑

j=1

∫|∇jψ(x)|2dx +

∫VC(x,R)|ψ(x)|2dx (8.1)

where dx = dx1dx2 . . .dxN .

Theorem 8.1 (Stability of first kind for atoms and molecules) For a fixed location ofnuclei at R = (R1, . . . RK), consider the set

MR :=ψ : R3N → C :

N∑

j=1

∫|∇jψ(x)|2dx <∞,

K∑

k=1

N∑

j=1

∫ |ψ(x)|2|xj − Rk|

<∞

28

Then the ground state energy of the molecule with N electron and K fixed nuclei at locationsR is given by

E0(R) = infEψ : ψ ∈ MR, ‖ψ‖ = 1

and the total ground state energy is given by

E0 = infR

E0(R)

i.e. the ground state energies minimized for all nuclear positions. Then E0 > −∞.

Proof. Since we are looking for a lower bound, we can drop the positive repulsion termsfrom VC , so it is sufficient to show that there exists a finite constant C, independent of R,such that

N∑

j=1

∫|∇jψ(x)|2dx −

K∑

k=1

N∑

j=1

∫Zk|ψ(x)|2|xj −Rk|

≥ −C‖ψ‖2 (8.2)

Let Z = maxk Zk. It is clearly sufficient to prove that there is a finite C ′ such that for eachfixed j and fixed k

1

K

∫|∇jψ(x)|2dx −

∫Z|ψ(x)|2|xj −Rk|

≥ −C ′‖ψ‖2 (8.3)

and then we just add up these inequalities (for all j = 1, 2, . . .N , k = 1, 2, . . .K) to get theresult with C = C ′NK. We can write

1

K

∫|∇jψ(x)|2dx −

∫Z|ψ(x)|2|xj − Rk|

=1

K

∫dx1 . . . dxj . . .dxNH(x1, . . . xj , . . . xN ) (8.4)

with

H(x1, . . . xj , . . . xN ) =[ ∫

dxj |∇ψ(x1, . . . xj , . . . xN)|2 − KZ

|xj − Rk||ψ(x1, . . . xj , . . . xN )|2

]

and with the convention that hat means omission (i.e. dxj means that the dxj integration ismissing). Here we view all but the xj variable fixed (as a parameter) and consider the function

g(xj) = gx1,...bxj ,...xN(xj) = ψ(x1, . . . xj , . . . xN )

as a function of one variable:

g : xj → ψ(x1, . . . xj , . . . xN )

parametrized by the others.

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This is an L2 function for almost all choices of (x1, . . . xj , . . . xN ) since

∫‖gx1,...bxj ,...xN

‖2dx1 . . . dxj . . . dxN =

∫ [∫dxj |ψ(x1, . . . xj , . . . xN)|2

]dx1 . . . dxj . . .dxN

= ‖ψ‖22 = 1

(8.5)

Thus

H(x1, . . . xj , . . . xN ) =

∫

R3

[|∇g(xj)|2 −

KZk|xj − Rk|

|g(xj)|2]dxj

where g = gx1,...bxj ,...xN. From the stability of the hydrogenic atom, we know that

H(x1, . . . xj , . . . xN) ≥ −1

4(KZ)2‖gx1,...bxj ,...xN

‖2

(the fact that the nucleus is at Rk and not at the origin does not change anything in theestimate, since the kinetic energy and the L2 norm are both translation invariant). Pluggingthis estimate into (8.4), we obtain (8.3) with C ′ = 1

4Z2K, so we have the lower bound (8.2)

with C = 14Z2NK2.

This proof was presented for static nuclei. If they were also dynamical, the proof is eveneasier, since then we have additional positive kinetic energy terms in the Hamiltonian.

8.2 Stability of second kind, preliminary estimates

We have followed the dependence of the constant C = 14Z2NK2 on the parameters, especially

on the total particle number N +K. In a typical molecule or extended matter N is close to Kto ensure electrostatic (almost) neutrality. Thus our lower bound is cubic in the total numberof the particles. This gives stability (of first kind), but is not completely satisfactory, sincewe would like to show that the ground state energy is at most proportional to the number ofparticles, i.e. we would like to have a bound that is linear in N +K.

Definition 8.1 We say that a Coulombic system consisting of N electrons and K nucleisatisfies the stability of second kind, if there is a constant C = CZ that may depend on themaximal charge of the nuclei, Z = maxk Zk but is independent of the total number of particles,such that

Eψ ≥ −CZ(N +K)

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We will see that the extended matter with Coulomb interaction actually satisfies thisstronger criteria for stability, but we will have to restrict the set of admissible wave functionsin M to antisymmetric ones. Physically this reflects the fact that the electrons are fermionsand it is also called the Pauli principle. The nuclei can have arbitrary particle types.

There will be three ingredients for such a proof:

(i) Coulomb singularities can be controlled by the kinetic energy

(ii) Electrostatic screening

(iii) Pauli principle

So far we have seen (i), this is essentially the key inequality in Lemma 7.2. The electrostaticscreening is a special property of the Coulomb systems, very roughly saying, it expresses thefact that if we have a collection of particles (with both negative and positive charges) in abounded domain, then the electrostatic potential generated by these particles far away looksas if all charges were concentrated at one point. In particular, there is a strong cancellation.Finally, Pauli principle will strengthen the one particle Lemma 7.2 and will lead to the Lieb-Thirring inequalities, that replace Sobolev inequalities in the fermionic setup.

Why is Pauli principle important? The following explanation should give a first (non-rigorous) insight. Suppose we have N electron and one single nucleus, say, at the origin, withnuclear charge Z. Assume that we neglect all interaction among the electrons (which arepositive, so for a lower bound we are allowed to neglect them). The Hamiltonian is

H =N∑

i=1

(− ∆j −

Z

|xj |)

(8.6)

or, with the quadratic form

Eψ =

N∑

i=1

∫ [|∇jψ(x)|2 − Z|ψ(x)|2

|xj |

]dx (8.7)

What is the minimal energy? For one electron, the energy would be −Z2/4 and the minimizingfunction ψ0(x) = (const)e−Z|x|/2 (see Theorem 7.1). If the electrons do not interact, then thesecond electron can occupy the same state as well, etc, so

ψ(x) = ψ(x1, x2, . . . , xN ) = ψ0(x1)ψ0(x2) . . . ψ0(xN)

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would be a natural guess (note that the single particle functions ψ0(xj) should be multipliedand not added). It is easy to compute (EXERCISE), that the energy of this state is

Eψ = −NZ2

4

i.e. the energy is additive.

Exercise 8.2 Extend this argument to show rigorously that the atom with one nucleus ofcharge Z always satisfies the stability of second kind.

The problem is that a typical electrostatic matter has several nuclei. What kind of nuclearconfiguration yields the lowest energy? If we neglect the nucleus-nucleus repulsion as well(very crude assumption), then nothing forbids the K nuclei with charges Z1, Z2, . . . , ZK topile up atop of each other and forming a nucleus with total charge Z1 + ... + ZK . It canbe proven, that indeed the total pile-up is the lowest energy configuration and the minimalenergy is

E0 = −(Z1 + . . .+ ZK)2N

4≥ −1

4Z2NK2

where Z = maxk Zk, this bound is exactly the same as we got in Theorem 8.1. To convinceyourself, just check that the energy of a total pile up is lower than if the nuclei are concentrated,say, at two different centers, that are very far away. Let A be the index set of the nuclei aroundone center, then the energy is [WHY??]

E(A) = min0≤N1≤N

[−N1

(∑j∈A Zj

)2

4−

(N −N1)(∑

j 6∈A Zj

)2

4

]

Check [!!] that

E(A) ≥ −(Z1 + . . .+ ZK)2N

4

and equality is if and only if A = ∅ or A = 1, 2, . . . , N, i.e. there is only one center.

One may argue that we neglected the electrostatic repulsion among the nuclei (also alsoamong the electrons), and this forbids putting all nuclei atop of each other. This is true,but the correct electrostatic alone does not solve the problem, we will prove later that takingall electrostatics into account the ground state energy is of order −CZ(N + K)5/3. It is animprovement compared with the cubic behavior, but it is still not linear.

To achieve a lower bound that is linear in N+K, one additionally needs the Pauli principle.Very roughly, the Pauli principle forbids the two electrons occupy the same state; if the

32

first electron is in state ψ0(x1), then the second one cannot be in ψ0(x2), i.e. the functionψ0(x1)ψ0(x2) as a two-electron wavefunction is forbidden.

More precisely, Pauli principle will imply that if one electron is in state ψ0, then the secondone must be in a state that is orthogonal to ψ0, e.g. its wavefunction is ψ0(x1)ψ1(x2) whereψ1 ⊥ ψ0. Actually, the precise definition will require that the two particle wave functionψ(x1, x2) be antisymmetric, i.e. ψ(x1, x2) = −ψ(x2, x1), so the product ψ0(x1)ψ1(x2) is stillnot correct, but the antisymmetrized product

(ψ0 ∧ ψ1)(x1, x2) =1√2

[ψ0(x1)ψ1(x2) − ψ0(x2)ψ1(x1)

]

will do the job (the 1/√

2 is the correct normalization to make ψ0∧ψ1 have norm one, CHECK!)If we choose ψ1 to be the eigenfunction to the second lowest eigenvalue of the Hamiltonian

−∆ − Z/|x|, then ψ1 ⊥ ψ0 (think of the orthogonality of two eigenvectors with differenteigenvalues of a hermitian matrix). The energy (8.7) of the function ψ = ψ0 ∧ ψ1 will be thesum of the two lowest eigenvalues.

Exercise 8.3 Check this statement on the formal level, i.e. assuming that

(− ∆ − Z

|x|)ψj = Ejψj, j = 0, 1

and E0 6= E1, then〈ψ,Hψ〉 = E0 + E1

where ψ = ψ0 ∧ ψ1 and H is given in (8.6). You can use the formal self-adjointness of H.

For N electrons, the Pauli principle will imply that the lowest energy state of (8.6) is theantisymmetric product ψ = ψ0 ∧ ψ1 ∧ . . . ∧ ψN−1 of the N eigenfunctions, corresponding tothe N lowest eigenvalues (with multiplicity), i.e. to E0 ≤ E1 ≤ E2 . . .. The energy of ψ isE0 + E1 + . . .+ EN−1.

Exercise 8.4 Reviewing the degeneracy structure of the Hydrogen eigenvalues (with nuclearcharge Z) compute how the sum E0 + E1 + . . . + EN−1 behaves in N for large N . (Answer−CZ2N1/3)

Imagine again that all nuclei are concentrated at the origin (i.e. neglect nucleus-nucleusinteraction) and also neglect electron-electron interaction, but take into account the Pauliprinciple, we see that the ground state energy is of order −CZ2K2N1/3.

33

Summarizing: If we neglect repulsion (i.e. electrostatic is wrong), and neglect Pauli prin-ciple, then ground state energy is −CZ(N +K)3. If we neglect Pauli principle, but take intoaccount proper electrostatic, then the ground state energy is smaller than −CZ(N + K)5/3,i.e. electrostatic improves the power by 4/3. Finally, if we take into account Pauli principle,but neglect electrostatics, the ground state energy is −CZ(N + K)7/3 (always expressed interms of the total number of particles, N + K), i.e. Pauli principle improves the power by2/3. The goal will be that if we take both effects into account, then the bound is linear inN +K, i.e. the original cubic power is improved by 4/3 + 2/3 = 2.

34