1. Popularvetenskaplig sammanfattning

For att uppna en acceptabel modell for en fysiskalisk teori, ar tva saker viktiga. Dels attden kan bekraftas experimentiellt, dels att den underbyggande matematiken ar motsagelsefri.I denna uppsats undersoks den senare delen, dvs matematiska grunderna for kvantfysiken, dardet t.ex. visas att teorin genererar en unik losning for olika typer av fysikaliska situationer.For vateatomen visas t.ex. att unik losning existerar och att den ar stabil (vilket inte arfallet i den klassiska mekaniken!). Ett forsok gors ocksa att ge en fysikalisk tolkning av teorinpa ett intuitivt satt, med utgangspunkten att fotoner och materia beter sig pa samma satt.

2. Introduction

The fact that quantum theory stands on a firm mathematical ground is of fundamentalimportance because it is crucial that the theory is free from contradictions. Nonetheless thisis the case for quantum mechanics, and the purpose of this thesis will be to describe themathematical basis of it. The starting point will be a mathematically structure, that will beused to derive the phenomena observed in the nature of quantum mechanics. First we shallbriefly review the historical background of quantum mechanics. Then there is a descriptionof the mathematics that will be needed, and after that an overview of the subject. Forexample, we give a proper and complete description of the inner essence of the hydrogenatom, i.e. the existence of its wave function and its stability. Heisenberg’s inequality, whichis the main ingredient in the celebrated Heisenberg’s uncertainty principle is derived, andsome special cases such as the particle in the box and the tunneling effect, are also presented.Observe that most of the thesis is a litography, that is, material is taken from different kindsof texts. Though there are still some results that I have proven myself in the contents.

3. History

3.1. Black body radiation. At the beginning of the 20th century, there were still someimportant problems that could not be explained by classical physics. One of them was themystery of the black body radiation. Consider a container with a small window on its wall,which is put inside an oven with constant temperature. Suppose that no heat is leaving orentering the wall of the container, and that the only heat transfer is through the window.After a while, the temperature of the system will reach equilibrium, and hence it will beconstant everywhere. Now the radiation into and out of the container will (in absolutevalue) be the same. The problem here is to determine the frequencies of this radiation.We denote the total radiation per volume unit by U (the bigger the volume, the bigger theradiation, and hence radiation per volume makes more sense in this case). Furthermore, wedefine the radiation per volume with frequency u(f) in such a way that the energy radiated

by frequencies in an interval [f1, f2] will be given by∫ f2f1u(f)df . Hence

(3.1) U =

∫ ∞0

u(f)df.

In 1900, Lord Raylegh and James Jeans derived an expression for u(f) using classical electro-and thermodynamics. This is given by

uRJ(f) =8πf 2

c3kT,

1

where k is Boltzmann’s constant and c is the speed of light. Experimentally, this formulaseemed to work, when one spoke of low frequencies, but at higher frequencies this didn’tseem to fit with the experiments anymore (in fact, the integral in 3.1 would then diverge).Hence the scientists had to find some other explanation. In 1900 Max Planck was the onethat came up with the innovative solution that the radiation (and hence the energy) wasabsorbed and emitted with discrete values, and that this energy was proportional to thefrequency. That is, E = hf , for some constant h that were to be experimentally estimated.One can then derive the following expression for u:

u(f) =8πh

c3

f 3

ehf/kT − 1.

Furthermore, this expression can be used to evaluate the total energy emitted and absorbedfrom the container. Evaluating Equation 3.1 yields

(3.2) U =8πh

c3

∫ ∞0

f 3

ehf/kT − 1df =

[x =

h

kT

]=

8πh

c3

(kT

h

)4 ∫ ∞0

x3

ex − 1dx.

The integrand can be expressed as a Taylor expansion in the following way:

(3.3)x3

ex − 1= x3e−x

1

1− e−x= x3e−x

∞∑n=0

e−nx =∞∑n=0

x3e−(n+1)x,

which is valid for |e−x| < 1, or x > 0, which is valid in our case. Now since the integrandx3e−(n+1)x is positive in the interval [0,∞), the monotone convergence theorem (see e.g. [10])yields

∫ ∞0

∞∑n=0

x3e−(n+1)xdx =∞∑n=0

∫ ∞0

x3e−(n+1)xdx = [y = (n+ 1)x] =

∞∑n=0

1

(n+ 1)4

∫ ∞0

y3e−ydy =∞∑n=1

6

n4=π4

15

(3.4)

The result is:U = cT 4,

which is the Stefan-Boltzmann radiation Law. This expression has been experimentallyconfirmed as well

3.2. The photoelectric effect. Independent of Plancks discovery, it was discovered byHeinrich Hertz that a polished metal, when exposed to radiation, emits electrons. Moreoverit turned out that there was a threshold frequency that had to be attained in order to obtaina current out of the metal. This was a contradiction to the classical view of radiation, whichwas that it behaved as a continuous wave. First of all, the classical view could predictemission of electron, but then it had to be independent of the frequency. If an electron getshit by radiation, sooner or later it would have gathered enough energy to leave the surfaceof metal. Secondly, if an electron gets hit by radiation it would take some time before it getsenough energy. Such time delay could not be measured. Once again the solution, invokedby Albert Einstein, was that the radiation was divided in discrete energy packages. Theenergy of these packages were assumed to be proportional to the frequency of the light, thatis E = hf for some fundamental constant h. Later these energy packages were known as

2

photons, and the constant h was shown to be same as in the case of black body radiation.These could explain the phenomenon of the photoelectric effect in a correct way. Only aphoton of high enough energy could be able to hit an electron so that it would take off. Italso explained why there was no time delay between the radiation to hit the electron andthe electron to get out.

3.3. Compton scattering and the momentum of the photon. The next step was theprediction that the photons actually had momentum, even though it didn’t have any mass.This was done by Einstein in 1916, and would yield the fact that the photons were not onlypackages of energy, but also particles themselves. The experimental confirmation was doneby Arthur H. Compton in 1923. Compton sent radiation in the X-ray region through ametallic foil and examined the scattering. From classical point of view the rays would makethe electrons start oscillating and send out rays with the same frequency again with theintensity of 1+cos2(θ), where θ is the angle between the emitted ray and an unabsorbed ray.However the result of the experiment appeared to consist of two components: one with thesame frequency as the incoming radiation, the other with a frequency that varied with theangle θ. Compton explained this phenomenon with the assumption that the rays consistedof photons, which hit the electrons under conservation of both energy and momentum. Thatis, the photons are to be treated as particles. From the point of view of theory of relativity,it was already known that for a relativistic particle with restmass m, the relation betweenthe energy and momentum is given by E =

√(mc2)2 + (pc)2. However, the photon has no

restmass, so the corresponding relation for the photon (just set m = 0) would be E = pc,or assuming the energy to be proportional to the frequency, p = hf

c= h

λ. This value of the

momentum seemed to fit well with the Compton scattering experiment.

3.4. de Broglie’s assumption about the wave properties of matter. In 1923 Louisde Broglie came up with the next radical assumption: if radiation could behave both asparticles and waves, the corresponding property should should hold for matter as well. Therelation between the momentum and the wavelength of the particle was assumed to behavesimilar to those of the photon, that is λ = h

p. To prove the wave behavior, the main sugges-

tion was to somehow detect diffraction of electrons. This was done independently by JosephDavisson and George Thomson who achieved this by sending electrons towards a crystal.The result turned out to be diffraction of the reflected electrons. This was the fact whichled Schrodinger to his equation, which will be motivated later on.

The facts in this section was taken from [7].

4. classical mechanics

4.1. Hamiltons formulation. When we introduce the quantum mechanics our startingpoint will be Hamiltons principle of least action. First of all, consider the function φ(t), whichis called a generalised coordinate. This could for example be the position for a particle, oran angle that describes the displacement of the particle. According to Hamilton’s principle,φ(t) will be the function which minimizes the action S

(4.1) S(φ) =

∫ T

0

L(φ(t), φ(t))dt,

3

where L : X×V → R is a twice differentiable function, called the Lagrangian, andX, V ⊆ R3.The Lagrangian is given by the difference between the kinetic energy of the system and thepotential energy of the system. One can show (see [8]) that the least value of S would beachieved for those φ that satisfy the following differential equation:

(4.2)∂

∂t

(∇φL

)−∇φL = 0.

This is the famous Euler-Lagrange equation.

Example 4.1. Let x(t) = φ(t) be given by the position and v(t) = φ(t) be the velocity of a

particle as a function of time so that L = m|v|22− V (x). Then

(4.3) 0 =∂

∂t(∇vL)−∇xL =

∂

∂t(mv) +∇xV (x) = mx+∇xV (x)

That is mx = −∇xV (x), which is Newton’s second law.

4.2. The equivalence of Euler-Lagrange and Hamilton equations. Here we shallobtain a set of equations (due to William Hamilton) which are equivalent to the Euler-Lagrange equations. These will be used later on, when we examine the limit between classicaland quantum mechanics.

Definition 4.1. We define the Hamiltonian by the so called Legendre transform of theLagrangian

(4.4) h(x, p) = supv∈V

(p · v − L(x, v))

Remark 4.2. If Hv(L(x, v)) > 0,∀(x, v) ∈ X × V , that is, the Hessian with respect to v ispositive definite for all x and v, then supremum is attained in

(4.5) p = ∇vL(x, v).

Furthermore, for each p and x there is an unique v satisfying this equation.

Theorem 4.3. Let L(x, v) and h(x, p) be related as in 4.4, where Hv(L(x, v)) > 0,∀x, v.Then 4.2 is equivalent to Hamiltons equations, defined by

x = ∇ph

p = −∇xh(4.6)

Proof. First, assume that 4.2 holds. Then by 4.5

∇ph = v + (p−∇vL)∇p(v) = v = x

∇xh = −∇xL+ (p−∇vL)∇x(v) = −∇xL =∂

∂t(−∇vL) = −p

(4.7)

Conversely, assume that 4.6 holds. Then by 4.5 again

(4.8) ∇xL = −∇xh = p =∂

∂t(∇vL)

5. Mathematics

In this section some of the underlying mathematics is described. It is necessary when thevalidity of the theory of quantum mechanics is going to be confirmed.

4

5.1. Functional analysis. First are some general definitions and results of functional anal-ysis presented.

Definition 5.1. A nonempty set X is called a vector space over a scalar field K if one candefine the two operations of addition and multiplication by a scalar on it. The additionoperation must satisfy the following properties:

x+ y = y + x

x+ (y + z) = (x+ y) + z

(denoted x+ y + z)Furthermore there exists a vector 0 ∈ X called the zero vector and for every vector x ∈ X,there exists a vector −x ∈ X such that

x+ 0 = x

x+ (−x) = 0

For the multiplication with scalars, one requires the following properties:

α(βx) = (αβ)x

1x = x

α(x+ y) = αx+ αy

(α + β)x = αx+ βx

Remark 5.2. In the context of quantum mechanics the scalarfield K = C, where C is thecomplex plane.

Definition 5.3 (Normed space, Banach space). A normed space (X, ‖ · ‖) is a vector spaceX with a norm ‖ · ‖X defined on it. The norm is a real-valued function defined on X andhas the following properties (assume that x, y ∈ X and α ∈ K):

(5.1) ‖x‖ ≥ 0

(5.2) ‖x‖ = 0⇔ x = 0

(5.3) ‖αx‖ = |α|‖x‖

(5.4) ‖x+ y‖ ≤ ‖x‖+ ‖y‖A complete normed space (See Definition 5.5 below) is called a Banach space.

Definition 5.4. A sequence (xn)∞n=1 in a normed space X is said to be convergent to a pointx ∈ X if

limn→∞

‖xn − x‖ = 0

Definition 5.5. A sequence (xn)∞n=1 is called a Cauchy sequence if for every ε > 0 thereexists a number N , such that

m,n > N => ‖xm − xn‖ < ε

A space X is called complete if every Cauchy sequence converges to a point in X.5

Definition 5.6. An inner product space (X, 〈·, ·〉) is a vector space X with an inner product〈·, ·〉X defined on it. The inner product is a complex-valued function defined on X ×X andhas the following properties (assume that x, y, z ∈ X and α ∈ K):

(5.5) 〈x, y + z〉 = 〈x, y〉+ 〈x, z〉

(5.6) 〈x, αy〉 = α〈x, y〉

(5.7) 〈x, y〉 = 〈y, x〉

(5.8) 〈x, x〉 ≥ 0 and 〈x, x〉 = 0⇔ x = 0

A complete inner product space is called a Hilbert space.

Remark 5.7. With the inner product space one can always define a norm by ‖x‖ =√〈x, x〉.

One can check that this satisfies all the properties of a norm in Definition 5.3. Hence aHilbert space is a Banach space, see e.g. [6].

Lemma 5.8 (Cauchy-Schwarz’s inequality). If X is an inner-product space, then for everyx, y ∈ X(5.9) |〈x, y〉| ≤ ‖x‖‖y‖.

Proof. If x = 0 the proof is trivial. Suppose x 6= 0. Then for every α ∈ C0 ≤ ‖y − αx‖2 = 〈y − αx, y − αx〉 =

〈y, y〉 − 〈y, αx〉 − 〈αx, y〉+ 〈αx, αx〉 =

〈y, y〉 − α〈x, y〉 − α (〈y, x〉+ α〈x, x〉)(5.10)

Set α = 〈y, x〉/〈x, x〉. Then

(5.11) 0 ≤ 〈y, y〉 − 〈y, x〉〈x, x〉

〈x, y〉 = ‖y‖2 − |〈x, y〉|2

‖x‖2,

and from this the inequality follows.

Definition 5.9. A subset M ⊆ X is said to be dense in X if X = M ∪M ′, where M ′ =x ∈ X| every neighbourhood of x contains y ∈M distinct from x

In quantum mechanics, it turns out the notion of an operator is important.

Definition 5.10. Let X and Y be vector spaces over the same scalar field K and let D(A) ⊆X (called the domain of A). A mapping A : D(A) → Y is called a linear operator if for allx, y ∈ D(A) and α ∈ K(5.12) A(αx+ βy) = αA(x) + βA(y)

We define operations (addition and multiplication by a scalar) of operators by:

(αA1 + βA2)x = αA1x+ βA2x

and hence the set of all operators from D(A) to Y form a vector space. The set R(A) =A(D(A)) = y ∈ Y |y = Ax,∀x ∈ D(A) is called the range of A.

6

Remark 5.11. From now on, the notation Ax will be used instead of A(x). The former isthe standard notation in functional analysis.

Definition 5.12. If (x 6= y)⇒ (Ax 6= Ay) holds, then we say that A is invertible and definethe inverse A−1 : R(A)→ D(A) as y = A−1x if x = Ay

Definition 5.13. The commutator of two operators A and B is defined by

(5.13) [A,B] = AB −BA

Example 5.1. The identity operator 1 : X → X, defined by 1x = x for x ∈ X is linearsince 1(αx + βy) = αx + βy = α1x + β1y. From now on, we denote the identity operatorby 1.

Example 5.2. The partial differentiation operator ∂∂xj

on the space of all fuctions with

continuous derivatives, C1(Rn) is linear since ∂∂xj

(αf + βg)(x) = α ∂∂xjf(x) + β ∂

∂xjg(x)

Definition 5.14. Let X and Y be normed spaces and A : D(A) → Y a linear operator,where D(A) ⊆ X. The operator is said to be bounded with the norm ‖A‖if

(5.14) ‖A‖ = supx∈D(A)

‖Ax‖‖x‖

<∞

Note that if there exists some c such that ‖Ax‖ ≤ c‖x‖, then A is bounded by the supremumproperty of R.

Definition 5.15. Let A : D(A) → X be an operator, D(A) ⊆ X is dense. We define theoperator A∗ with the following properties:

D(A∗) = y ∈ X|∀x ∈ D(A), ∃y∗ ∈ X such that 〈Ax, y〉 = 〈x, y∗〉For each y ∈ D(A∗) let A∗ be defined by

y∗ = A∗y

The operator A∗ is called the Hilbert-adjoint (or simply the adjoint) of A.

Definition 5.16. Let X be a Hilbert space. An operator A : D(A)→ X, which satisfies

(5.15) 〈x,Ay〉 = 〈Ax, y〉for all x, y ∈ X, is called a symmetric operator. Moreover, if D(A) = D(A∗) the operator iscalled self-adjoint.

Lemma 5.17. Suppose A : D(A)→ D(A) is a bounded operator. Then for all n ∈ Z+

(5.16) ‖An‖ ≤ ‖A‖n

Proof. We prove this only for n = 2. The general case follows by induction. Denote y = Ax.By definition

‖A2‖ = supx∈D(A)

‖A2x‖‖x‖

= supx∈D(A)

‖A2x‖‖Ax‖

‖Ax‖‖x‖

=

supx∈D(A)

‖Ay‖‖y‖

‖Ax‖‖x‖

≤ supx∈D(A)

‖Ax‖2

‖x‖2= ‖A‖2

(5.17)

7

Theorem 5.18 (Neumann series). Suppose that X is a normed space and A : D(A)→ D(A)is an operator with ‖A‖ < 1, where D(A) ⊆ X. Then (A+ 1)−1 exists and satisfies

(5.18) (A+ 1)−1 =∞∑n=0

(−A)n

Proof. By Lemma 5.17, the series in 5.18 is absolutely convergent, and hence convergent:

(5.19)∞∑n=0

‖(−A)n‖ ≤∞∑n=0

‖(−A)‖n <∞.

Moreover

(A+ 1)∞∑n=0

(−A)nx =∞∑n=0

(A+ 1)(−A)nx =

∞∑n=0

((−A)n+1 + (−A)n

)x = x

(5.20)

since all terms but the first in the series cancel out. Moving (A+ 1) to the other side we getthe expression in 5.18.

Lemma 5.19. If A is invertible and B is bounded, and satisfies ‖BA−1‖ < 1. Then A+B,defined on D(A+B) = D(A) is invertible.

Proof. Rewrite (A+B) = (1+BA−1)A which is invertible, since A is invertible by assumptionand (1 +BA−1) is invertible by Theorem 5.18

Theorem 5.20. Let A : D(A)→ X be a self-adjoint operator, with D(A) ⊆ X, where X isa Hilbert space. If A is bounded, then the domain D(A) = X. If A is unbounded the domainD(A) is dense in X but not equal to X.

Proof. See [6]

Later we are going to have use of the following theorem, which makes it much easier toinvestigate whether an operator is self-adjoint or not:

Theorem 5.21. Let A be a linear, symmetric operator, where D(A) ⊆ X. X is a Hilbertspace. Then D(A) = D(A∗)⇔ R(A± i) = X.

Proof. See [9]

Lemma 5.22. Let A be a self-adjoint operator, where D(A) ⊆ X. X is a Hilbert space. IfR(A+ z) = X for some z ∈ C with Im z > 0, then it is true for every z with Im z > 0. Thesame statement holds for Im z < 0. Moreover, A+ z is invertible whenever Im z 6= 0 and

(5.21) ‖(A+ z)−1‖ ≤ 1

Im z.

Proof. We show it for Imz > 0. The other case is identical. Write z = α+βi, where α, β ∈ R,and let x ∈ D(A). Since A is symmetric, and since Re〈(A+ α)x, βx〉 = 0, we have

‖(A+ z)x‖2 = 〈(A+ α + iβ)x, (A+ α + iβ)x〉 =

〈(A+ α)x, (A+ α)x〉+ 2 Re 〈(A+ α)x, βx〉+ 〈iβx, iβ)x〉 =

‖(A+ α)x‖2 + ‖βx‖2 ≥ |β|2‖x‖2

(5.22)

8

This implies that (A + z)x = 0 ⇒ x = 0. This and R(A + z) = X yield that (A + z)−1 isdefined on all of X. Futher 5.22 yields

(5.23) ‖(A+ z)−1x‖ ≤ 1

|β|‖x‖

That is (A+ z)−1 is bounded, and 5.21 holds. Then, by Theorem 5.19

(5.24) A+ z′ = (A+ z) + (z′ − z)

is invertible fr |z − z′| < ‖(A+ z)−1‖−1= | Im z|. By this we can extend the invertibility of

A+ z′ to all Imz′ > 0

Remark 5.23. By Theorem 12.6 and Lemma 5.22, it will be enough to check that R(A±iλ) =X for some λ > 0 in order to prove the self-adjointness of the symmetric operator A.

Definition 5.24 (Lp-space). Consider the function f : D → C, D ⊆ Rn. We say thatf ∈ Lp(D) if f is measurable and

(5.25)

∫D

|f(x)|pdx <∞

The Lp-norm of f is defined by

(5.26) ‖f‖Lp(D) =

(∫D

|f(x)|pdx)1/p

Remark 5.25. If we define (αf + βg)(x) = αf(x) + βg(x), then Lp(D) becomes a vectorspace. One also has the triangle inequality

‖f + g‖Lp(D) ≤ ‖f‖Lp(D) + ‖g‖Lp(D),

which in this context is referred to as the Minkowski inequality.

Lemma 5.26. The space Lp(D) is a Banach space. Moreover, L2(D) is a Hilbert space withthe inner product

〈f, g〉L2(D) =

∫D

f(x)g(x)dx

Proof. See [6]

Definition 5.27. For a bounded operator A we define the exponential of A by

(5.27) eA =∞∑n=0

An

n!

There is no doubt about the convergence of this series, since it is clearly absolutely convergent(see Lemma 5.17).

(5.28)∞∑n=0

‖An‖n!≤

∞∑n=0

‖A‖n

n!<∞

Now we want to take care of the case with unbounded operators A. The exponentiationis not defined directly since the series 5.28 may diverge. However, it will still be possible todefine the operator eiA. To do this we use the following lemma:

9

Lemma 5.28. Let A be a self-adjoint operator, with D(A) ⊆ H dense in H, possibly un-bounded. Then for λ ∈ I = R+

(5.29) Aλ =1

2λ2((A+ iλ)−1 + (A− iλ)−1

)is self-adjoint and bounded. Moreover the sequence

eiAλ

λ∈I is a Cauchy sequence and

(5.30) limλ→∞

Aλx = Ax

Proof. The boundedness of Aλ follows immediately from Lemma 5.22. Since Aλ is symmetricand bounded it is self-adjoint by Theorem 5.20.

We prove that limλ→∞Aλx = Ax.

Aλ − A =λ2

2(A+ iλ)−1 − A

2+λ2

2(A− iλ)−1 − A

2=

(A+ iλ)−1

2

(λ2 − (A+ iλ)A

)+

(A− iλ)−1

2

(λ2 − (A− iλ)A

)=

(A+ iλ)−1A2

2+λ

2(A+ iλ)−1(λ− iA)+

+(A− iλ)−1A2

2+λ

2(A− iλ)−1(λ+ iA) =

(A+ iλ)−1A2

2+−iλ

2(A+ iλ)−1(A+ iλ)+

+(A− iλ)−1A2

2+iλ

2(A− iλ)−1(A− iλ) =

(A+ iλ)−1A2

2+

(A− iλ)−1A2

2=

1

2

((A+ iλ)−1 + (A− iλ)−1

)A2.

(5.31)

For all x ∈ D(A), by Lemma 5.22

‖(Aλ − A)x‖ =1

2

∥∥((A+ iλ)−1 + (A− iλ)−1)A2x

∥∥ ≤1

2

2

|λ|‖A2x‖ =

‖A2x‖|λ|

→ 0(5.32)

as λ → ∞. Now we prove thateiAλ

λ∈I is a Cauchy sequence. For all x ∈ D(A), rewrite

the expression ∥∥(eiAλ′ − eiAλ)x∥∥ =

∥∥∥∥∫ 1

0

∂

∂s

(eisAλ′ − eisAλ

)xds

∥∥∥∥ =∥∥∥∥∫ 1

0

eisAλ′ei(1−s)Aλi(Aλ′ − Aλ)xds∥∥∥∥ ≤∫ 1

0

∥∥eisAλ′ei(1−s)Aλi(Aλ′ − Aλ)x∥∥ ds ≤∫ 1

0

‖(Aλ′ − Aλ)x‖ ds→ 0

(5.33)

10

as λ, λ′ →∞, since the sequence Aλ is convergent and is therefore a Cauchy sequence.

Since D(A) is dense in H andeiAλ

λ∈I is a Cauchy sequence, it will converge to some

element in H. Thus we can state following definition.

Definition 5.29. For a self-adjoint operator A and for all x ∈ D(A) ⊆ H we define

(5.34) eiAx = limλ→∞

eiAλx

where Aλ is defined as in Equation 5.29.

Definition 5.30 (Spectrum of an operator). Let X 6= 0 be a complex normed space andA : D(A) → X be a linear operator, with D(A) ⊆ X. Then the resolvent set, ρ(A) is theset of all λ ∈ C such that the operator Rλ(A) = (A− 1λ)−1

(1) exists(2) is bounded(3) is defined on a set which is dense in X

The complement σ(A) = C \ ρ(A) is called the spectrum of A and a λ ∈ σ(A) is called aspectral value of A.

Theorem 5.31. Consider a self-adjoint operator A : D(A)→ X, where X is a Hilbert spaceand D(A) ⊆ X is dense. Then the spectrum σ(A) ∈ R

Proof. See [6]

5.2. Fourier analysis. Fourier analysis is a very important part of the wave theory becauseit provides tools to study the behaivor of the frequency of the waves. Since quantum me-chanics is built upon the fact that matter behaves like waves, Fourier analysis and especiallythe Fourier transform, will play a prominant role in quantum mechanics. Usually the Fouriertransform of a function is defined by the frequency of the angle velocity. However, since ithas been experimentally proven that the frequency is proportional to the momentum for aparticle, it is appropriate to define the Fourier transform in terms of the momentum.

Definition 5.32. The Fourier transform ψ of a function ψ ∈ L1(Rd) is defined by

(5.35) ψ(p, t) =1

(2π~)d/2

∫Rdψ(x, t)e−ip·x/~dx,

where ~ := h2π, and h is Planck’s constant. Below we give two important properties of the

Fourier transform:

Theorem 5.33 (Inverse of the Fourier transform). If ψ, ψ ∈ L1(Rd), then

(5.36) ψ(x, t) =1

(2π~)d/2

∫Rdψ(p, t)eip·x/~dp

Theorem 5.34 (Plancherel’s theorem). If ψ ∈ L1(Rd) ∩ L2(Rd). Then ψ ∈ L2(Rd) and

(5.37) ‖ψ‖2L2(Rd) = ‖ψ‖2

L2(Rd)

We will not prove Theorems 5.33 and 5.34 here. These are given in [2].11

Example 5.3.

ψ(p) =1

(2π~)d/2

∫Rdψ(x)e−ip·x/~dx =

1

(2π~)d/2

∫Rdψ(x)eip·x/~dx = ψ(·)(−p)

(5.38)

The Fourier transform combined with Definition 5.29 enable us to define what is meantby f(A) for a real-valued function f .

Definition 5.35. Let A be a self-adjoint operator, f : R → R be a function with the

property that f ∈ L1(R). We define the operator f(A) as

(5.39) f(A) =1

(2π~)1/2

∫Rf(t)eitA/~dt

This operator will be well defined, bounded and self-adjoint if f is real.

5.3. Probability theory. In quantum mechanics it is never possible to exacly determinea quantity. There is only a probability that a quantity is measured to a specific value.Therefore it is appropriate to discuss some of the basic probabilistic features of the quantumtheory. We will se later that a quantity will be described with an operator, and the followingdefinition will be motivated later.

Definition 5.36. The expectation value of an operator A is given by

(5.40) 〈A〉ψ = 〈ψ,Aψ〉L2(R3) =

∫R3

ψ(x, t)Aψ(x, t)dx

We also want to know of how much a measure is expected to differ from the expectationvalue. Therefore we have following definition.

Definition 5.37. The standard deviation ∆A of an operator A is given by

(5.41) ∆A =√〈(A− 〈A〉)2〉

If we want to determine the probability to find some quantity in a specific region Ω, wecan use the characteristic function χΩ(λ) and the expectation value to define it.

Definition 5.38. The probability to measure an operator A in the region Ω is given by

(5.42) P (A ∈ Ω) = 〈χΩ(A)〉,where χΩ(A) is an operator defined in Definition 5.35.

Remark 5.39. In ordnary probability theory it is usual to first define a probability densityfunction for a random variable, and then define the expectation value starting from theprobability density. The problem with this is that here our random variables are operators,and hence there is no easy way to define a probability density function. Therefore it is betterto start from the expectation value definition.

Remark 5.40. It can be shown that the probability defined in Definition 5.38 is close con-nected to the spectrum of the operator defined in Definition 5.30. However, this subject istoo big to get included in this thesis. The interested reader should see [4].

12

6. Schrodinger representation

According to de Broglie, matter behaves like waves and we can only calculate the probabil-ity to find the position for it. We want to develop some mathematical theory for these waves,and since there is a wave equation for classical electromagnetical waves, we can assume thatthere will exist some sort of wave equation for the matter as well. Such an equation isimpossible to derive in a proper way without making assumptions, and therefore one canonly give a motivation of the equation (for example it is assumed that the potential energyis time independent even though the equation is valid when it is time dependent as well).First of all we state some (more or less intuitive) conditions which the wave equation mustobey:

(1) Causality: if we know the wave function at one moment we want to know it at anytime. That is because we want the probability density for the particle to be deter-ministic (see section 7).

(2) Superposition principle: since the optical waves are linearly added when they coin-cide, we want the waves that solve our equation to behave in the same way. That is,if ψ1 and ψ2 are solutions to the equation, then ψ1 + ψ2 is also a solution.

(3) Wave-particle duality: the photon behaves both as a wave and as a particles withthe relations E = hf and p = ~k, where k is the wave number of the wave. We wantthe matter waves to obey these equations as well.

(4) Correspondance principle: if we let Planck’s constant go to zero (or ~ → 0) inquantum mechanics, we should arrive at classical mechanics.

The first property says that there can only be one initial condition and therefore the equationmust be first order in time. The second says that the equation must be linear. Therefore wewill arrive at an equation on the form

(6.1)∂ψ

∂t= Aψ

where A is some linear operator.

Since the solution of our equation is supposed to behave like waves, we can assume thatit can be written as a linear combination of ei(k·x−ωt), in the sense of the Fourier transform,where k is the wave number and ω is the angle velocity. From the third property we wantto use the momentum p = ~k and the energy E = ~ω. Then Theorem 5.33 yields

(6.2) ψ(x, t) =1

(2π~)3/2

∫R3

ψ(p)ei(x·p−Et)/~dp.

Equation 6.1 now suggests us to differentiate with respect to t. We get

(6.3)∂ψ

∂t= (−i)E

~1

(2π~)3/2

∫R3

ψ(p)ei(x·p−Et)/~dp = −iE~ψ.

From this equation, and the fact that E = Ekin + V (x, t) = |p|22m

+ V (x, t), where V (x, t) isthe potential energy of the particle, we see that

13

(6.4) i~∂ψ

∂t=|p|2

2mψ + V (x, t)ψ

What is still left is to find an expression for the momentum. To do this we let the Laplacian∆ act on 6.2. We get

(6.5) ∆ψ = −|p|2

~2

1

(2π~)3/2

∫R3

ψ(p)ei(x·p−Et)/~dp = −|p|2

~2ψ,

Substituting this into 6.4 we get the famous Schrodinger equation:

(6.6) i~∂ψ

∂t(x, t) = − ~2

2m∆ψ(x, t) + V (x, t)ψ(x, t)

Remark 6.1. The equation is valid when the potential, V , is time dependent, that is V (x, t).We will, however, only investigate the cases when it is time independent, since the timedependent case is much more difficult.

We still have to check that the fourth property is fulfilled, that is when ~ → 0, wewill obtain the classical mechanical picture. To do this, write the solution as ψ(x, t) =a(x, t)eiS(x,t)/~, where a and S are real-valued functions and substitute into the Schrodingerequation. We will first calculate necessary derivatives:

(6.7)∂ψ

∂t= eiS/~

(∂a

∂t+ ia~−1∂S

∂t

)

∆ψ = ∆(aeiS/~) = ∇ · ∇(aeiS/~) =

∇ · (eiS/~(∇a+ i~−1a∇S)) =

eiS/~i~−1(∇S) · (∇a+ i~−1a∇S)+

+ eiS/~(∆a+ i~−1∇a · ∇S + i~−1a∆S) =

eiS/~(∆a+ 2i~−1∇a · ∇S − ~−2a|∇S|2 + i~−1a∆S)

(6.8)

Substitute these into Equation 6.6 to yield

i~eiS/~(∂a

∂t+ ia~−1∂S

∂t

)= − ~2

2meiS/~(∆a+ 2i~−1∇a · ∇S−

~−2a|∇S|2 + i~−1a∆S) + aeiS/~V = i~∂a

∂t− a∂S

∂t=

− 1

2m(~2a+ 2i~∇a · ∇S − a|∇S|2 + i~a∆S) + aV

(6.9)

Letting ~→ 0 and dividing by a we get

(6.10)∂S

∂t+|∇S|2

2m+ V = 0.

14

In the theory of partial differential equation one studies equations of the form

(6.11)∂S

∂t+ h(x,∇S) = 0,

for a suitable function h(x, p), which is known as the Hamilton-Jacobi equation. This equa-tion can be solved by

(6.12) S(a, b, t) =

∫ t

0

(dx

ds· p(s)− h(x(s), p(s))

)ds

where a = x(0) and b = x(t), where x(s) and p(s) satisfy Hamilton’s pair of equations

dx(s)

ds= ∇ph(x(s), p(s))

dp(s)

ds= −∇xh(x(s), p(s))

(6.13)

Observe that in our case h(x,∇S) = |∇S|22m

+ V . Applying this to 6.10 and using the firstpart of 6.13

S(a, b, t) =

∫ t

0

(p(s) · ∇ph(x(s), p(s))− h(x(s), p(s)) ds =∫ t

0

(p(s) · p(s)

m− |p(s)|

2

2m− V (x(s))

)ds =∫ t

0

(|p(s)|2

2m− V (x(s))

)ds =

∫ t

0

L(x(s), p(s))ds

(6.14)

This is equivalent to Hamiltons principle of least acton, see Equation 4.1. By this we haveshown that if we let ~→ 0 we will come back to classical mechanics again.

7. The basics of quantum mechanics

7.1. Probabilistic interpretation. How should we interpret the wave function of a parti-cle? We will find the answer by looking at the photon, considering how the wave functionis interpreted there. For radiation it is well known that the intensity is proportional to thesquare of the wave function. That means, for example, that if radiation is sent with only onedirection of the amplitude (say along z-axis) and if it hits a polarizing filter with an angleθ, then the intensity will be proportional to cos2 θ. But what happens if we interpet theradiation as a beam of photons? A photon cannot divide into two parts at the filter. Whatwill instead be the case is that there is a probability for the photon to either pass throughthe filter or get interrupted by it. This probabilty must of course be the ratio of the intensityfor passing through and the total intensity. In the same way we can interpret the square ofthe wave function as probability density for the particle. That is

(7.1) P (x ∈ Ω) =

∫Ω

|ψ(x, t)|2dx15

We want the probability to find the particle somewhere at all to be 1, and therefore we claimthat ψ ∈ L2(R3), and

(7.2) ‖ψ‖2L2(R3) =

∫R3

|ψ(x, t)|2dx = 1

If this holds for all t > 0 we say that probability is conserved. In Theorem 7.2 we will seewhen this is the case.

7.2. Operators. Next question is how to calculate physical quantities (such as momentum)if the wave function is known. Since ψ ∈ L2(R3) by assumption, the natural way wouldbe to use operators defined on dense subsets of L2(R3), and define the expectation valueof an operator A as 〈A〉ψ = 〈ψ,Aψ〉. If we want to interpret this from a physical point ofview it is important that the expectation value is real. If A is a self-adjoint operator, then〈ψ,Aψ〉 = 〈Aψ,ψ〉 = 〈ψ,Aψ〉, and this claim is true. Therefore the study of self-adjointoperators with domain dense in L2(R3) is central in the mathematical theory of quantummechanics. Such operators are called observables. Below we give some examples of operatorsand prove their self-adjointness.

7.2.1. Position operator. The probabilty density of finding a particle is given by |ψ(x, t)|2.From probability theory we know that the expectation value is given by

(7.3) 〈x〉 =

∫R3

x|ψ(x, t)|2dx =

∫R3

ψ(x, t)xψ(x, t)dx = 〈ψ, xψ〉.

From here we see that the position operator is just multiplication with the real-valued func-tion x. It is obvious that it is symmetric. To see that this is self-adjoint, for every componentxj in x we solve the equation (xj ± i)ψ(xj, t) = f(xj, t), and see that there exists a solutionψ for every f ∈ L2(R3). Hence R(x± i) = L2(R3).

7.2.2. Momentum operator. By Plancherel’s theorem (Theorem 5.34), we know that

(7.4)

∫R3

ψ(p, t)ψ(p, t)dp =

∫R3

ψ(x, t)ψ(x, t)dx = 1

This leads us to interpret |ψ(p, t)|2 as the probability density of the momentum, and hencewe can define the expectation value as

(7.5) 〈p〉 =

∫R3

ψ(p, t)pψ(p, t)dp

Though we want to find the momentum operator acting on ψ and not ψ.

Theorem 7.1. Suppose ψ, pψ ∈ L2(R3). Then

(7.6)

∫R3

ψ(p, t)pψ(p, t)dp =

∫R3

ψ(x, t) (−i~∇x)ψ(x, t)dx

Proof. First recall that ψ(p, t) = ψ(·, t)(−p, t) (see Example 5.3), and then we find an ex-

pression for pψ(p, t):16

pψ(p, t) =1

(2π~)3/2

∫R3

pψ(p, t)(x, t)eip·x/~dx =

1

(2π~)3/2

∫R3

(1

(2π~)3/2

∫R3

ξψ(ξ, t)e−iξ·x/~dξ

)eip·x/~dx =

1

(2π~)3/2

∫R3

−i~∇x

(1

(2π~)3/2

∫R3

ψ(ξ, t)e−iξ·x/~dξ

)eip·x/~dx =

1

(2π~)3/2

∫R3

−i~∇x (ψ(−x, t)) eip·x/~dx

(7.7)

Here we substitute our expressions into 7.5, change the order of integration and do a variablesubstitution: ∫

R3

ψ(p, t)pψ(p, t)dp =

∫R3

ψ(·, t)(−p, t)(1

(2π~)3/2

∫R3

−i~∇x (ψ(−x, t)) eip·x/~dx)dp =∫

R3

−(∫

R3

ψ(·, t)(−p, t)eip·x/~dp)i~∇x (ψ(−x, t)) dx =∫

R3

−ψ(−x, t)i~∇x (ψ(−x, t)) dx =∫R3

ψ(x, t) (−i~∇x)ψ(x, t)

(7.8)

This shows that the momentum operator is the differentiation operator: p = −i~∇x. Onecan use partial integration to check that it is symmetric. To show that it is self-adjoint,

consider (−i~∇x ± i)ψ(x, t) = f(x, t). The Fourier transform yields (p± i)ψ(p, t) = f(p, t),which yields the same situation as for the position operator, and thereby the self-adjointnessfollows.

7.2.3. The Hamiltonian. Due to the particular form of the Schrodinger equation, it is ap-propriate to define the Hamiltonian H as

(7.9) H = − ~2

2m∆ + V (x)

and hence the Schrodinger equation will be given by i~∂ψ∂t

= Hψ. As we will see later,an important part of the mathematical theory of quantum mechanics is to investigate howsolutions to the Schrodinger equation depend on the Hamiltonian H. An important questionis of course whether H is self-adjoint or not. However, the operator H0 = − ~2

2m∆ is self-

adjoint. The fact that it is symmetric is derived by partial integration, and the domainquestion is answered exactly in the same way as for the momentum operator above. Since〈H〉 = 〈− ~2

2m∆〉 + 〈V (x)〉 the expectation value of the Hamiltonian must measure the total

energy of the system. Now we state an important theorem for the Hamiltonian17

Theorem 7.2. Consider the Schrodinger equation

(7.10) i~∂ψ

∂t= Hψ,

where H is given by Equation 7.9. Solutions ψ(x, t) conserves probability if and only if H issymmetric.

Proof.

d

dt〈ψ, ψ〉 =

⟨dψ

dt, ψ

⟩+

⟨ψ,dψ

dt

⟩=

⟨1

i~Hψ,ψ

⟩+⟨

ψ,1

i~Hψ

⟩=

1

i~(〈Hψ,ψ〉 − 〈ψ,Hψ〉) = 0,

(7.11)

if H is symmetric. Conversely if probability is conserved, then

(7.12) 0 =d

dt〈ψ, ψ〉 =

1

i~(〈Hψ,ψ〉 − 〈ψ,Hψ〉)

and hence 〈Hψ,ψ〉 = 〈ψ,Hψ〉.

7.3. Time derivative of the expectation value. Now we will see what happens whenwe try to differentiate the expectation value of an observable A (i.e. a self -adjoint operatorA) with respect to the time. We use the Schrodinger equation as well as the conjugate of it

in order to find ∂ψ∂t

as well as ∂ψ∂t

∂

∂t〈A〉ψ =

∂

∂t

∫R3

ψ(x, t)Aψ(x, t)dx =∫R3

∂

∂t

(ψ(x, t)Aψ(x, t)

)dx =∫

R3

(∂ψ

∂t(x, t)Aψ(x, t) + ψ(x, t)A

∂ψ

∂t(x, t)

)dx =

i

~

∫R3

(Hψ(x, t)Aψ(x, t)− ψ(x, t)AHψ(x, t)

)dx =

i

~

∫R3

(ψ(x, t)HAψ(x, t)− ψ(x, t)AHψ(x, t)

)dx =

i

~

∫R3

ψ(x, t)[H,A]ψ(x, t)dx =

⟨i

~[H,A]

⟩ψ

(7.13)

We will now use 7.13 to find two important relations with expectation values. First, by easycalculations, we claim that [H, x] = −~2

m∇. From this, we see that

(7.14)∂

∂t〈x〉ψ =

⟨i

~

(−~2

m∇)⟩

ψ

=1

m〈−i~∇〉ψ .

For the other expression we notice that [∆,∇] = 0 and hence

(7.15)∂

∂t〈p〉ψ =

⟨i

~[H, p]

⟩ψ

=

⟨i

~[V, p]

⟩ψ

= 〈−∇V 〉ψ .

18

One can immediately see that this is the quantum analogue of the classical Hamilton equa-tions:

m ∂∂t〈x〉ψ = 〈p〉ψ mx(t) = p(t)

m ∂∂t〈p〉ψ = 〈−∇V 〉ψ p(t) = −∇V (x(t))

8. Existence of dynamics

The Schrodinger equation is a partial differential equation, which is supposed to be ful-filled for nonrelativistic particles. Hence it is important for the solution of the Schrodingerequation, given initial conditions, to be unique. That is in fact that there exists an uniquesolution to

i~∂ψ

∂t= Hψ

ψ(x, 0) = ψ0(x)(8.1)

and

(8.2)

∫R3

|ψ(x, t)|2dx = 1, ∀t ≥ 0

Therefore following definition is important for physical interpretation:

Definition 8.1 (Dynamics). We say that the dynamics if for all ψ0 ∈ L2(R3) if 8.1 has andunique solution, and 8.2 holds.

The remaining problem is now to find out for which H the dynamics exist. We will seethat this is the case if and only if H is self-adjoint. But first we state an another importanttheorem.

Theorem 8.2. Suppose H is self-adjoint. Then there exist family of operators U(t) =e−itH/~, which obey following properties (s, t ∈ R):

(8.3) i~∂

∂tU(t) = HU(t) = U(t)H

(8.4) U(0) = 1

(8.5) U(t)U(s) = U(t+ s)

(8.6) ‖U(t)ψ‖ = ‖ψ‖

Proof. If H is bounded, since the series in Equation 5.28 is uniformly convergent we can useit as if it where a polynomial (differentiate term by term etc.) and hence 8.3 - 8.6 is easilyproven. For unbounded operators, just use the Definition 5.29.

Theorem 8.3. The dynamics exist if and only if H is self-adjoint.

Proof. If H is self-adjoint, the existence of a solution follows imedietly from Theorem 8.2.To show the uniqueness, suppose ψ1(x, t) and ψ2(x, t) are both solutions to the Schrodinger

19

equation with the initial data ψ0. Then ψ1(x, t) − ψ2(x, t) is a solution as well and byTheorem 7.2 ∫

R3

|ψ1(x, t)− ψ2(x, t)|2dx =

∫R3

|ψ1(x, 0)− ψ2(x, 0)|2dx =∫R3

|ψ0(x)− ψ0(x)|2dx = 0.

(8.7)

Since ψ1(x, t)−ψ2(x, t) ∈ L2(R3), it follows that ψ1(x, t) = ψ2(x, t). For the other direction,see [5].

9. Heisenberg representation

In the Schodinger representation we assumed that the operators were time-independentand the wave-function were time-dependent. We could instead treat the operator as time-depentent and the wave-function as time-independent. We want to define this operator A(t)such that 〈A(t)〉ψ0 = 〈A〉ψ(t) holds. We use Equation 7.13.

⟨∂A(t)

∂t

⟩ψ0

=∂

∂t〈A(t)〉ψ0

=∂

∂t〈A〉ψ(t) =

⟨i

~[H,A]

⟩ψ(t)

=⟨i

~[H,A](t)

⟩ψ0

=

⟨i

~[H,A(t)]

⟩ψ0

(9.1)

Since 9.1 must hold for all ψ0 that satifies the time-independent Schrodinger equation wehave:

(9.2)∂A(t)

∂t=i

~[H,A(t)] =

i

~HA(t)− i

~A(t)H

(9.3) 〈A(t)〉ψ0 = 〈A〉ψ(t)

To solve this equation we can interpret it as a differential equation of first order (the differenceis that this is an operator equation and the operators will not necessarily commute andtherefore the usual ansatz for homogenuous, first order equations with constant coefficientswill not do here). Hence we make the ansatz more general. Let A(t) = eiC1tCeiC2t, definedas in Definition 5.29, where C, C1 and C2 are self-adjoint. Then

∂A(t)

∂t= iC1e

iC1tCeiC2t + eiC1tCiC2eiC2t =

iC1eiC1tCeiC2t + eiC1tCieiC2tC2 = iC1A(t) + A(t)iC2

(9.4)

Inserted in Equation 9.2 we have iC1A(t)+A(t)iC2 = i~HA(t)− i

~A(t)H. Since we want this

to hold for all A(t), we can compare 9.4 to 9.2. This yields C1 = H~ and C2 = −H

~ , whichare self-adjoint by the assumption that H is self-adjoint. Finally, we have to determine thevalue of C. To do this, we see that eitH/~ is an unitary operator and use Equation 9.3:

20

〈A(t)〉ψ0 =

∫R3

ψ0(x)A(t)ψ0(x)dx =∫R3

ψ0(x)eitH/~CeitH/~ψ0(x)dx =∫R3

e−itH/~ψ0(x)CeitH/~ψ0(x)dx =∫R3

ψ(x, t)Cψ(x, t)dx = 〈C〉ψ(t)

(9.5)

This yields C = A, and gives us an expression for the time evolving operator A(t):

Definition 9.1. Let A : D(A) → X and H : D(H) → X be self-adjoint operators, withdomain dense in the Hilbert space X. Then we define the time-dependent operator A(t) as

(9.6) A(t) = eitH/~Ae−itH/~

10. Heisenbergs uncertainty principle

One of the most famous relations in quantum mechanics is the uncertainty principle. Itis stated by Werner Heisenberg and predicts that position and momentum cannot both bemeasured with arbitrary accuracy at the same time.

Theorem 10.1 (Heisenbergs uncertainty principle). Let A and B be self adjoint operatorswhere the domains and ranges are dense subsets of L2(R3). Then

(10.1) ∆A∆B ≥ 1

2|〈[A,B]〉ψ|

Proof. Let S = A− 〈A〉 and T = B − 〈B〉. Then 〈S〉 = 〈T 〉 = 0 and [A,B] = [S, T ].

|〈[A,B]〉ψ| = |〈[S, T ]〉ψ| = |〈ψ, STψ〉 − 〈ψ, TSψ〉| =|〈Sψ, Tψ〉 − 〈Tψ, Sψ〉| ≤ |〈Sψ, Tψ〉|+ |〈Tψ, Sψ〉| =2|〈Sψ, Tψ〉| ≤ 2‖Sψ‖L2(R3)‖Tψ‖L2(R3) =

2√〈(A− 〈A〉)2〉

√〈(B − 〈B〉)2〉 = 2∆A∆B

(10.2)

This relation has very important physical applications.

Example 10.1. The commutator of the operators x and p is given by

(10.3) [x, p] = i~and hence

(10.4) ∆x∆p ≥ 1

2|i~| = ~

2

Physically this means that the position and momentum never can be determined exactly atthe same measure. The more precisely one of them is known the more uncertain the otheris.

21

11. Special solutions of the Schrodinger equation

There are very few cases where there is an analytic solution to the Schrodinger equation.Often one has to prove the self-adjointness of the Hamiltonian and then solve the Schrodingerequation numerically. Though there are some special cases where there still is a possibilityto find an analytic solution. In the examples below, will only provide explicit solutions it inone dimension. For several dimensions it is easy to use the method of separation of variablesto break down the problem into one dimension. First we will state a general theorem thatstates that the Schrodinger equation has a unique solution if the time independent potentialV (x) is bounded.

Theorem 11.1. Assume that the function V (x) is real and bounded. Then the operator

H = − ~22m

∆ + V (x) with D(H) = D(∆) is self-adjoint on L2(R3)

Proof. It is easy to show that H is symmetric. We show that R(H + iλ) = L2(R3). That isthe same as showing that the equation

(11.1) (H ± iλ)ψ = f

has an unique solution for every f ∈ L2(R3). We show it only for (H + iλ)ψ = f . The

second statement is similar. We know that H0 = − ~22m

∆ is self-adjoint and hence (H0 + iλ) :D(H0)→ L2(R3) is bijective. Therefore, we can apply (H0 + iλ)−1 to equation 11.1 and getψ+ (H0 + iλ)−1V ψ = (H0 + iλ)−1f . To simplify the expression, write K(λ) = (H0 + iλ)−1Vand g = (H0 + iλ)−1f and we get ψ + K(λ)ψ = g (Note that for every f , there alwaysexists an unique g). By Lemma 5.22, if ‖ · ‖ denotes the operator norm in L2(R3), then‖K(λ)‖ ≤ 1

λ‖V ‖L2(R3). Now choose |λ| > ‖V ‖L2(R3). Then ‖K(λ)‖ < 1, and by Theorem

5.18, (1+K(λ))−1 exists and is bounded. Therefore ψ = (1+K(λ))−1g = (1+K(λ))−1(H0 +iλ)−1f is the unique solution to 11.1 and we are done.

We see that whenever the potential is real and bounded, the Hamiltonian will be self-adjoint, and hence the Schrodinger equation has an unique solution. In order to continuewith the special cases, we have to define the time independent Schrodinger equation, whichis obtained from separation of variables when the potential V (x) is time independent.

Definition 11.2. The time independent Schrodinger equation is given by

(11.2) − ~2

2m∆ψ(x) + V (x)ψ(x) = Eψ(x),

where E is a constant to be determined (physically interpreted as the total energy of thesystem).

11.1. The infinite potential well. Assume that the potential V is on the following form:

(11.3) V (x) =

∞ x < 00 0 ≤ x ≤ a∞ a < x

,

where ∞ is being interpreted as that the solution ψ must be constant 0 there. In the region0 ≤ x ≤ a the time independent Schrodinger equation yields

22

− ~2

2m

∂2ψ

∂x2= Eψ

ψ(0) = ψ(a) = 0.(11.4)

Here the boundary conditions come from the continuity requirement of ψ. If we let E ≤ 0one can (by the boundary conditions) easy show that ψ ≡ 0. If E > 0, we see immediatelythat the solution must be on the form

(11.5) ψ = A cos

(√2mE

~x

)+B sin

(√2mE

~x

)The requirement ψ(0) = 0 yields A = 0 and hence

(11.6) ψ = B sin

(√2mE

~x

)The reuirement ψ(a) = 0 yields

(11.7) 0 = B sin

(√2mE

~a

)

The case B = 0 is uninteresting and hence sin(√

2mE~ a

)= 0. This is the same as stating

(11.8)

√2mE

~a = πn n ∈ Z

We want our solution ψ to be normalized, that is (after some calculations) ‖ψ‖2 = |A|2 a2

= 1

and hence A =√

2a. From here we see that the energy E can only take certain discrete

values with their corresponding ψn, namely

En =~2π2n2

2ma2

ψn(x) =

√2

asin(nπax)

n ∈ Z(11.9)

11.2. The potential barrier and the tunneling effect. Now concider the case

(11.10) V (x) =

0 x < 0V0 0 ≤ x ≤ a0 a < x

,

that is when there is a barrier in the middle, reducing or blocking the incoming particles’momentum. In the classical case, if the total energy E of the particle would have been lowerthan V0, it would have bounced on the wall and hence reflected back. We will see that thisis not always the case in quantum mechanics. Our mission is here to derive reflexion andtransmission coefficients (denoted R and T ).

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In order to find a solution we divide the problem into three cases, and get solutions ofthe following form:

(11.11)ψ = A1e

ip1x + A2e−ip1x x < 0

ψ = B1eip2x +B2e

−ip2x 0 ≤ x ≤ aψ = C1e

ip1x + C2e−ip1x a < x,

where

p1 =

√2mE

~

p2 =

√2m(E − V0)

~

(11.12)

Now, in order to be a solution to the Schrodinger equation on the entire real axis, ψ and itsderivative must be continuous everywhere, even in those points where V is not continuous,i.e. in x = 0 and x = a. Therefore we get

A1 + A2 = B1 +B2

ip1(A1 − A2) = ip2(B1 −B2)

B1eip2a +B2e

−ip2a = C1eip1a + C2e

−ip1a

ip2(B1eip2a −B2e

−ip2a) = ip1(C1eip1a − C2e

−ip1a)

(11.13)

We will now make some assumptions. First of all we assume that no particles are arrivingfrom the left, so we set C2 = 0. Futhermore, to detect the tunneling effect we are onlyinterested in what is happening before and after the ”wall”. It doesn’t matter what ishappening inside. Therefore we substitute B1 and B2 in the equations and after somecalculations we arrive at

A2 = A1(p2

1 − p22) sin(ap2)

2iP1p2 cos(ap2) + (p21 + p2

2) sin(ap2)

C1 = A14p1p2e

−ia(p1−p2)

(p1 + p2)2 − e2iap2(p1 − p2)2

(11.14)

The coefficients we are looking for must be given by the relative intensity between theoutgoing particle and the incoming particle. Since the intensity is given by the square of theabsolute value of the amplitude, the coefficients will be given by

R =|A2|2

|A2|2

T =|C1|2

|A2|2

(11.15)

Now there is only one problem to solve. The solution ψ that is derived is in L2(R3)-function,and hence it cannot be used as a physical interpretation! The problem is based on the fact

24

that we assumed the momentum of the particle to be p1, that is an exact value! Hence∆p = 0. But then, according to Heisenberg’s uncertainty principle, Theorem 10.1, ∆x =∞.The probability to find such a particle will not vanish in ±∞ and hence this is not realistic.To avoid this we introduce an uncertainty to the momentum p, with the probability density

|ψ(p)|2. Then we can write the solution as ψ(x) = 1(2π~)3/2

∫R ψ(p)eipx/~dp. By Plancherel’s

theorem, 5.34, we have∫R |ψ(x)|2dx =

∫R |ψ(p)|2dp, and hence ψ ∈ L2(R3) if we choose

suitable ψ, that is ψ ∈ L2(R3).

12. The Hydrogen atom

Since all matter is built of atoms and molecules, it is essential to show in which cases wecan apply the theory of quantum mechanics to the molecules. Here we will only treat thesimplest of them all, that is the Hydrogen atom where the potential is the Coulomb potentialV (x) = − e2

|x| . This can of course the extended to more complicated atoms and molecules.

The following lemma is crucial in showing the existence of the wave function for the hydrogenatom.

Lemma 12.1. The potential on the form V (x) = α|x| fulfills the inequality

(12.1) ‖V ψ‖ ≤ a‖H0ψ‖+ b‖ψ‖

for some a, b > 0 with a < 1

Proof. We rewrite V (x) as V (x) = V1(x) + V2(x), where

V1(x) =

V (x) if |x| ≤ 1

0 if |x| > 1V2(x) =

0 if |x| ≤ 1

V (x) if |x| > 1

and hence divide the proof into two cases. In the first case, consider V1(x). Using the Fouriertransform, for any c > 0, we get an estimate for ψ(x, t):

|ψ(x, t)| =∣∣∣∣∫

R3

ψ(p, t)eix·p/~dp

∣∣∣∣ =∣∣∣∣∫R3

(c+ |p|2)−1(c+ |p|2)ψ(p, t)eix·p/~dp

∣∣∣∣ =∣∣∣∣∫R3

(c+ |p|2)−1 (cψ −∆ψ)(p, t)eix·p/~dp

∣∣∣∣ =⟨(c+ | · |2)−1, (cψ −∆ψ)eix·p/~

⟩≤

‖(c+ | · |2)−1‖L2(R3)‖ (cψ −∆ψ)‖L2(R3) =

‖(c+ | · |2)−1‖L2(R3)‖cψ −∆ψ‖L2(R3) ≤‖(c+ | · |2)−1‖L2(R3)(‖cψ‖L2(R3) + ‖∆ψ‖L2(R3))

(12.2)

25

Here we have used Cauchy-Schwarz inequality and the triangle inequality. From this, it iseasily checked that

‖V1ψ‖L2(R3) ≤ sup |ψ|‖V1‖L2(R3) ≤‖V1‖L2(R3)‖(c+ | · |2)−1‖L2(R3)(‖cψ‖L2(R3) + ‖∆ψ‖L2(R3))

(12.3)

which proves the first part of the statement. Note that there is no problem with the conver-gence of ‖V1‖L2(R3) since it is an integral of three dimensions.

Next, consider V2(x). Since 1|x| < 1 on its domain

‖V2ψ‖2L2(R3) =

∫|x|>1

|ψ(x)|2 α2

|x|2dx ≤ α2

∫|x|>1

|ψ(x)|2dx ≤

α2

∫R3

|ψ(x)|2dx = α2‖ψ‖2L2(R3)

(12.4)

which proves the second part. All in all we have

‖V ψ‖L2(R3) = ‖(V1 + V2)ψ‖L2(R3) ≤ ‖V1ψ‖L2(R3)+

‖V2ψ‖L2(R3) ≤ a‖∆ψ‖L2(R3) + b‖ψ‖L2(R3)

(12.5)

where we choose c such that a < 1, and this proves the lemma.

Theorem 12.2. Assume that H0 is a self-adjoint operator with D(H0) = L2(R3) and V isa symmetric operator satisfying Equation 12.1 with a < 1. Then the operator H = H0 + Vwith D(H) = D(H0) is self-adjoint.

Proof. It is easy to show that H is symmetric. What remains is to prove that R(H ± iλ) =L2(R3) for some λ > 0. That is,

(12.6) (H ± iλ)ψ = (H0 + V (x)± iλ)ψ = f

has a solution ψ for every f ∈ L2(R3). We will show it only for the case when (H+iλ)ψ = f .The other statement is shown in a similar way. Since H0 is self-adjoint, R(H0 +iλ) = L2(R3),and hence H0 + iλ : D(H)→ L2(R3) is bijective. Therefore it is possible to formally rewrite12.6 in the following way:

(1 + V (H0 + iλ)−1)(H0 + iλ)ψ = f

⇔(H0 + iλ)ψ = (1 + V (H0 + iλ)−1)−1f(12.7)

Therefore, since (H0 + iλ) is invertible (with a bounded inverse), to solve the equation abovefor ψ, it remains to show that (1 + V (H0 + iλ)−1) is also invertible for λ big enough. ByTheorem 5.18 this is the case if the operator norm ‖V (H0 + iλ)−1‖ < 1. Since by theself-adjointness of H0 one has Re〈H0θ, iλθ〉 = 0 for all θ ∈ L2(R3), it holds that

‖H0θ‖2L2(R3) ≤ ‖H0θ‖2

L2(R3) + |λ|2‖θ‖2L2(R3) =

‖H0θ‖2L2(R3) + 2 Re〈H0θ, iλθ〉L2(R3)+

|λ|2‖θ‖2L2(R3) = ‖(H0 + iλ)θ‖2

L2(R3).

(12.8)

26

Now, if we let θ = (H0 + iλ)−1φ, using 12.8, Lemma 5.22 and 12.1 we get

‖V (H0 + iλ)−1φ‖L2(R3) ≤ a‖H0(H0 + iλ)−1φ‖L2(R3)+

b‖(H0 + iλ)−1φ‖L2(R3) = a‖H0θ‖L2(R3)+

b‖(H0 + iλ)−1φ‖L2(R3) ≤ a‖(H0 + iλ)θ‖L2(R3)+

bλ−1‖φ‖L2(R3) = a‖φ‖L2(R3) + bλ−1‖φ‖L2(R3)

(12.9)

Since a < 1 by assumption, choosing λ such that a+ bλ−1 < 1 we get

‖V (H0 + iλ)−1‖ = sup‖V (H0 + iλ)−1φ‖L2(R3)

‖φ‖L2(R3)

≤

‖(a+ bλ−1)φ‖L2(R3)

‖φ‖L2(R3)

= a+ bλ−1 < 1.

(12.10)

Using this it follows that the solution ψ = (H0 + iλ)−1(1 +V (H0 + iλ)−1)f to Equation 12.6is well-defined and the theorem is thereby proven.

This and Theorem 8.3 shows that there exists an unique solution to the Schrodingerequation for the Hydrogen atom.

Theorem 12.3. On L2(R3)

(12.11) −∆ ≥ 1

4|x|2

Remark 12.4. The interpretation of A ≥ 0 is that 〈ψ,Aψ〉 ≥ 0 for every ψ ∈ D(A). Therefore

Theorem 12.3 should be interpreted as⟨ψ,(−∆− 1

4|x|2

)ψ⟩≥ 0.

Proof. First we show some useful relations. For self-adjoint operators A and B we have

〈ψ, i[A,B]ψ〉 = i〈ψ, (AB −BA)ψ〉 = i〈ψ,ABψ〉−i〈ψ,BAψ〉 = i〈Aψ,Bψ〉 − i〈Bψ,Aψ〉 = i〈Aψ,Bψ〉−

i〈Aψ,Bψ〉 = −2 Im〈Aψ,Bψ〉(12.12)

27

Using the product rule we have

3∑j=1

i[|x|−1pj|x|−1, xj

]f =

3∑j=1

i

[|x|−1(−i~)

∂

∂xj|x|−1, xj

]f =

3∑j=1

~(|x|−1 ∂

∂xj|x|−1xjf − xj|x|−1 ∂

∂xj|x|−1f

)=

3∑j=1

~(|x|−1 ∂

∂xjxj(|x|−1f)− |x|−1xj

∂

∂xj|x|−1f

)=

3∑j=1

~(|x|−1xj

∂

∂xj(|x|−1f) + |x|−1(|x|−1f)

∂

∂xj(xj)−

−|x|−1xj∂

∂xj|x|−1f

)=

3∑j=1

~(|x|−2f

∂

∂xj(xj)

)= 3~|x|−2f

(12.13)

Further

(12.14) pj|x|−1 = |x|−1pj + i~|x|−3xj

From these relations and the fact that |x|−1 is self adjoint, we get

3~∥∥|x|−1ψ

∥∥2

L2(R3)= 〈ψ, 3~|x|−2ψ〉 =

〈ψ,3∑j=1

i[|x|−1pj|x|−1, xj

]ψ〉 =

− 23∑j=1

Im⟨|x|−1pj|x|−1ψ, xjψ

⟩=

− 23∑j=1

Im⟨|x|−2pjψ + i~xj|x|−4ψ, xj

⟩=

− 23∑j=1

(Im⟨pjψ, xj|x|−2

⟩− ~ Im

(i⟨x2j |x|−4ψ, ψ

⟩))

(12.15)

28

The second term in the expression can be evaluated:

23∑j=1

~ Im(i⟨x2j |x|−4ψ, ψ

⟩)= 2~ Im

(i⟨|x|2|x|−4ψ, ψ

⟩)=

2~ Im(i⟨|x|−1ψ, |x|−1ψ

⟩)= 2~

∥∥|x|−1ψ∥∥2

L2(R3)

(12.16)

This term cancels out the term on the left hand side in 12.15. Now, using Cauchy-Schwarzinequality twice (once to the inner product containing the imaginary part, and once to thesummation) in the remaining expression we arrive at

~∥∥|x|−1ψ

∥∥2

L2(R3)= −2

3∑j=1

Im⟨pjψ, xj|x|−2

⟩≤2

∣∣∣∣∣3∑j=1

Im⟨pjψ, xj|x|−2

⟩∣∣∣∣∣ ≤ 23∑j=1

‖pjψ‖L2(R3)‖xj|x|−2ψ‖L2(R3)

≤2(3∑j=1

‖pjψ‖2L2(R3))

1/2‖|x|−1ψ‖L2(R3)

≤2~√〈ψ,∆ψ〉 ‖x|−1ψ‖L2(R3),

(12.17)

where we have also used partial integration in deducing the last line of the estimate above.Finally, this leads to

(12.18)∥∥|x|−1ψ

∥∥L2(R3)

≤ 2√〈ψ,∆ψ〉 ⇔

⟨ψ, |x|−1ψ

⟩≤ 4 〈ψ,∆ψ〉

which was the statement of the theorem.

A very important application of the lower bound estimate in Theorem 12.3 is the proofof the stability of the hydrogen atom. Recall that the total energy of a system is given bythe expectation value of the Hamiltonian. We have to prove that the energy can never reach−∞, which is the case when the electron reaches the nucleus. By Theorem 12.3 we havethat the Hamiltonian for the hydrogen atom satisfies

(12.19) H = − ~2

2m∆− α

|x|≥ ~2

8m|x|2− α

|x|,

where α > 0 is a constant. We want to find the minimum of the right hand expression. Todo this, define f(x) = ~2

8mx2− α

x. Observe that |x| ≥ 0, so we just need to look at the case

when x > 0. To find the minimum value, we differentiate f and find its zeroes.

f ′(x) = − ~2

4mx3+α

x2= 0

− ~2

4m+ αx = 0

x =~2

4mα

(12.20)

29

To prove that this is a minimum we differentiate f one more time and check its sign in thecrititcal point.

f ′′(

~2

4mα

)=

3~2

4mx4− 2α

x3

∣∣∣∣x= ~2

4mα

=192α4m3

~6− 128α4m3

~6> 0.

This means that f reaches its minimum at x = ~24mα

, with value

f

(~2

4mα

)= −2mα2

~2

This yields 〈ψ,Hψ〉 ≥ −2mα2

~2 . That is, the energy reaches it’s minimum at a value > −∞and therefore the electron cannot collapse into the nucleus. Hence we have proven that thehydrogen atom is stable! It can be mentioned that if one tries to apply Maxwell’s equationsto the hydrogen atom, since the electron is constantly accelerating, it will emit radiation andwill eventually that collapse into the nucleus.

13. Termination

Even though quantum mechanics is a good model when describing physics where classicalphysics is insufficient, there are still several situations when the quantum mechanical rea-soning is not quite enough. First of all, classical quantum mechanics is not relativistic. Thisproblem is not easily overcome by just replacing classical particle mechanics by relativisticmechanics. Secondly, quantum mechanics only takes care of a fixed number of particles.Therefore, it cannot describe the process of creation and annihilation of particles, which areexperimentally confirmed. Thirdly, in electromagnetic theory, the objects are described asfields, not particles, and therefore it is not possible to treat this with standard quantummechanical methods.

To solve these problems, the next step in the order is to develop the quantum field theory.As a starting point Heisenberg’s representation is used, rather than Schrodingers represen-tation. Just like the the quantum mechanics, quantum field theory also contains a lot ofmathematical machinery that must be verified in order to show that the theory is suitableas a realistic model. But it turns out that the quest for putting quantum field theory on acontradiction-free and rigorous mathematical ground is a very challenging and difficult taskand many problems in construction of a fully rigourous mathematical theory remain open.

References

[1] Alm, S.E. and Britton, T. Stokastik, Liber AB[2] Folland, G. B. Real Analysis, Wiley, 1984.[3] Fulling, S. A. Aspects of Quantum Field Theory in Curved Space-Time, Cambridge University Press

1989[4] Gustafson, S.J and Sigal, I. M. Mathematical Concepts of Quantum Mechanics, Springer-Verlag 2011[5] Hislop, P. and Sigal, I.M. Introduction to Spectral Theory. With Applications to Schrodinger Operators,

Springer-Verlag, 1996.[6] Kreyszig, E. Introductory Functional Analysis with Applications, University of Windsor[7] Liboff, R. L. Introductory Quantum Mechanics, Cornell University.[8] Marion, J. B. and Thornton, S. T. Classical Dynamics of Particles and Systems, 5th Edition, Thomson

Brooks/cole30

[9] Reed, M. and Simon, B. Methods of Modern Mathematical Physics, Vol I. Functional analysis, AcademicPress, 1972.

[10] Rudin, W. Principles of mathematical analysis. Second edition McGraw-Hill Book Co., New York 1964

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