University of Amsterdam - FNWI

Institute for Theoretical Physics Amsterdam

Bachelor Thesis in Physics

On the uncertainty principle:

The quantum debate betweenBohr and Einstein

August 19, 2009

Author:Luc Blom

Supervisor:Prof.dr. Erik Verlinde

Preformed between: May 1, 2009 and July 28, 2009Extent: 12 EC

2

Abstract

The forerunner of the EPR-paradox is the less known photon-box thought experiment of Bohrand Einstein. This thesis will start with a popular scientific introduction on a debate includingHeisenberg’s discovery of matrix mechanics. A new solution shows that the experiment is inagreement with the uncertainty principle. The time-energy and position-momentum uncertaintyrelations become equivalent also seen in Bohr’s solution. We obtain the equality of the Poissonbracket between H,t and x,p and show that that it is closely related to the quantum conditions.Further study might prove and understand why the uncertainty relations are equal.

Contents

1 Introduction 41.1 The start of the debate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Heisenberg’s revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Eintein’s quest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Gravitational time dilatation 72.1 The equivalence principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Clocks inside a gravitational field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Bohr’s triomph 103.1 The solution of the paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 The quantum theory 124.1 The quantization of physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Heisenberg’s revolution; Matrix mechanics . . . . . . . . . . . . . . . . . . . . . . . 134.3 Heisenberg’s γ-ray microscoop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.4 Wave mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.5 Position and momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.6 Energy and time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Analyses of the debate 195.1 A new solution of the paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.2 The Poisson Bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.3 The Poisson bracket of H,t and x,p . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.4 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

A Gravitational time dilatation from General Relativity 25A.1 The geodesic equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25A.2 The weak field limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

B The new quantization condition 28

C Nederlandse samenvatting 30

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Chapter 1

Introduction

1.1 The start of the debate

Albert Einstein boarded a train to Solvay (Belgium) on October 1927. His journey would take himto the greatest physical conference of the 19th century. All pioneer thinkers of the quantum theorywhere present at the international Solvay institute, where the conference was held (see figure 1.1).The theme was photons and electrons. Niels Bohr, also present, had the opportunity to discussthe latest developments in physics with Einstein. For Bohr it was clear that all problems in thedomain of the quanta would soon be solved. Einstein had a different view concerning this newtheory.

Figure 1.1: Official photograph of the sixth Solvay conference. Albert Einstein sitting in themiddle and Niels Bohr second row most left[1].

Abraham Pais has written the following on the matter[2]. The conference started with the officialsessions where presentations where given on the new quantum ideas. Discussions followed wereEinstein did not participate much. But the next day he was much livelier in the dining roomof the hotel. He started his breakfast with a conversation over the new quantum theory. It wasincorrect according to his conclusions. The quantum physicist Pauli and Dirac did not paid muchattention, ’ach was, das stimmt schon, das stimmt schon’. But Niels Bohr, reflected on it withcare. In the evening he cleaned up the matter and showed that the new theory was right all along.This was the starting point of a heated debate about the indeterminacy of energy and time inphysics.

4

1.2. HEISENBERG’S REVOLUTION 5

1.2 Heisenberg’s revolution

A revolution in kinematics started after the discovery of the dual wave/particle character of matter.The classical description of Newton, for point particles had to be replaced. Not long after thediscovery of the quantum theory the twenty-four year old Werner Heisenberg developed matrixmechanics. Later saying about this discovery[3]:

”It was about three o’ clock at night when the final result of the calculation lay before me. Atfirst I was deeply shaken. I was so excited that I could not think of sleep. So I left the house andawaited the sunrise on the top of a rock.”

Heisenberg saw that the orbits of an electron in an atom could never be measured. He foundthat day the correct mathematical rules for a quantum theory that relied only on what can beobserved. These are the frequencies and the intensities of the spectral lines. To difficult to solvethe problem for real atoms he solved the equations of motion of the anharmonic oscillator[4]. Butthere was a price to pay for this success.

Max Born, the mentor of Heisenberg saw that the discovery was governed by matrix multiplicationand found that the matrix product XP 6= PX[5]. It implied an uncertainty because the matricesdid not commute for the position X and momentum P . Experimentally it means that there isno observation that will ever measure the position and momentum of a particle simultaneously.The same holds for the energy and time. It’s because every measurement changes the particle, orbetter, it’s quantum mechanical state. The second observation does not correspond anymore toothe initial particle. All is summarized into the equations:

σEσt ≥~2

(1.1)

σxσp ≥~2

(1.2)

with σi the standard deviation of variable i.

Bohr discussed with Einstein the discovered principle of Heisenberg during the 1927 Solvay con-ference. Einstein was interest and impressed by this discovery, but not convinced. Calling thereplacement of Cartesian co-ordinates with calculus of infinite matrices ’a real witch. Most in-genious and protected by it’s complexity. The new theory was for him intrinsically statisticaland therefore meaningless. Any attempt to obtain a deterministic theory would be fruitless.[6]’Therefore he delivered simple thought experiments to obtain a paradox if the uncertainty principlewas taken into consideration. Bohr was able to point out the errors in his reasoning but unable toconvince him of his view. During this heated but friendly debate Einstein spoke his famous andwell known words ’God doesn’t play dice with the universe’[6].

6 CHAPTER 1. INTRODUCTION

1.3 Eintein’s quest

The debate continued in the hallway during the next sixth Solvay conference in 1930 and tookquite a dramatic turn according to Bohr.

Figure 1.2: Depiction of Einstein’s photonbox[1]

Einstein proposed a device where the exigencies of special relativity were taken into consideration[1].He considered a box with a small hole on one side, see figure 1.2. A clock inside the box con-trolled a shutter that opened and closed the hole. At start the box contained a certain amountof radiation. If the clock opened the shutter for a very short chosen time, a single photon wouldbe released through the hole. The change in energy is according to the mass energy equivalence(E = mc2) equal to the change in mass of the box. It would therefore be possible, by weighingthe box before end after the event, to determine the change in energy. This with any degree ofaccuracy if the errors of the instruments where neglected. The clock inside the box would alsomeasure (in principle) an exact time-interval. Therefore the thought experiment measures theenergy of the escaped photon precisely and also the time it takes. It would mean that Einstein’spropose contradicts the indeterminacy of time and energy according to Heisenberg’s uncertaintyprinciple.This was a bombshell. Bohr could not find any solution to this paradox. During the evening, goingfrom one to the other, knowing that it would mean the end of physics if Einstein were right[2].He tried to find out what happened to the clock during the energy loss. Let’s figure out whatrelativity tells us about a clock inside a gravitational field.

Chapter 2

Gravitational time dilatation

2.1 The equivalence principle

Einstein’s general theory of relativity is a very difficult subject but grew out of a very simple idea.This is in fact so easy that anyone could have figured it out. The story is as follows:

Albert Einstein was sitting at the Bern post office one day in 1907. During his work as a patentexaminer he daydreamed about an very unfortunate house painter that fell off the roof[7]. Heasked himself: How would the painter experience gravity during his downward fall? Later that dayhe found his answer calling it ’der gluckischste Gedanke meines Lebens’ or his happiest thought[2].He suspected that the painter would be weightless or experienced no gravitational field during hisfall towards the earth.

To make this more clear imagine that the painter let’s his paintbrush go during his fall. Bothobjects would fall with the same acceleration towards the earth. Therefore the painter will observehis paintbrush stationary, at rest, next to him. He will be in an free-float (inertial) frame. It meansthat the person feels the sensation of the absence of a normal force. He cannot feel gravity withoutany opposing forces. Therefore he would feel weightless during his free fall. This idea was essentialfor Einstein to come to the following conclusion[8]:

The equivalence principle (EP):

All physical experiments inside a small region of spacetime cannot make a distinction between agravitational field and a uniformly accelerated frame.

Correspondingly, all physical experiments inside a small region of spacetime cannot make a dis-tinction between a freely falling frame in a gravitational field and an inertial frame.

The modern version of this thought experiment is obviously an astronaut freely falling around theearth in a spaceshuttle1. Everything becomes weightless caused by the absence of a normal force.All objects fall with the same acceleration. Now to get more inside accelerate the spaceshuttleupward with the same strength as the gravitational field of the earth(g). The astronaut will bepushed against the floor and if he throws an object it will curve downward, making a parabolicpath. These observations are the same made by Galileo on the earth. The EP tells that objectswould move the same under the influence of acceleration or to the effects of gravity[8].

1In those day this was not so simple. For example the French science fantasy writer Jules Verne tells in his novel:De la Terre a la Lune about a passenger inside unpowered spaceship[9]. He found himself less pressed against thefloor on there trip towards the moon. Instead they should be weightless during there hole flight.

7

8 CHAPTER 2. GRAVITATIONAL TIME DILATATION

From the EP we can also define the inertial frame. It is a trajectory on which unacceleratedobjects travel or move in a state were the object is only subjected to gravity. This universalitymeant to give up gravity as a force propagating through spacetime. Einstein realized that hecould extend special relativity by including accelerating frames into his theory. Gravity will thenalso be included. The EP was the clue for Einstein to this discover this result. After seven yearshe finished his masterpiece called the General Theory of Relativity. The solution was given in amathematical framework, were gravity should be seen as a curved geometry of spacetime. Whatdoes this principle tells us about the clock?

2.2 Clocks inside a gravitational field

Now we will derive gravitational time dilatation equivalent to Carrol’s book[10]. Imagine we havetwo clocks in outerspace far from any gravitational object. The clocks are accelerating constantwere there distance remains the same, see figure 2.1. Call this distance z.

Figure 2.1:

The velocity of the clocks stay far below the speed of light. We therefore assume we have to workonly in first order approximations. At time t = 0 the clock on the left side will send a lightpulstowards the other clock. The pulse will travel a distance z according to

z = ct (2.1)

When the right clock receives the lightpuls both stop ticking. Now we measure the time interval.During the flight the receiver clock has gained velocity because of the acceleration.

v = at =az

c(2.2)

In our approximation we can make this substitution for z = ct. The acceleration will causea Dopplershift that will produce a change in wavelength. Therefore received lightpuls will beredshifted according to our well known Dopplershift equation.

λ

λ0=

√c+ v

c− v(2.3)

The additional velocity is small, so we can Taylor expand this formula about vc 1. Again

working only in first order we obtain

λ

λ0≈ 1 +

v

c= 1 +

az

c2(2.4)

The sending and receiving of the lightpuls inside the frames of both clocks takes some time whichis related by the wavelength. This is

λ =c

f= ct (2.5)

2.2. CLOCKS INSIDE A GRAVITATIONAL FIELD 9

were t is the time interval of both clocks. According to eq. 2.4 is the sending time different fromthe second clock that receives the lightpuls. Hence,

t

t0≈ 1 +

az

c2(2.6)

But according to the insight of Einstein the same should happen inside the gravitational field ofthe earth. The EP tells us we can replace the constant acceleration a by the gravitational fieldg. This by knowing that our measuring instruments (the clocks) are small and therefore cannotmeasure inhomogeneities inside the gravitational field. The equivalent thought experiment shouldtherefore be two persons carrying the same clocks. One is at ground floor, the other on the top ofsome building with height z. Again if the clock on the ground floor emitted the lightpuls the topfloor clock will measure a different receiving time given by

t

t0≈ 1 +

gz

c2(2.7)

This formula is called Gravitational time dilatation. Relativity therefore tells us that the clock inEinstein’s box will tick at a different rate after a position change1.

1General relativity gives the exact same answer if one insists that the equation of motion must reduce to Newton’sthird law for a gravitational field that is static and weak. This is given in appendix A but is quite cumbersome incomparison with the given thought experiment

Chapter 3

Bohr’s triomph

3.1 The solution of the paradox

Bohr found a solution the next day after Einstein proposed the thought experiment during the1930 Solvay conference. His reasoning was as follows:Suspended the box from a balance support with a mechanical spring[1] to make it more realistic.This because a weight measurement is just a height observation inside a known gravitational field.The hole set-up is placed inside the earth’s stationary field, where a pointer is added to read itsposition on a scale to the fixed frame support. After some time the box will be in balance. Theinitial weighing is performed by recording the position of the pointer, see figure 3.1.

Figure 3.1: Drawing of Einstein’s photonbox experiment[1]

After the photon are escaped, the loss in weight will be compensated by a load hung underneath thebox. The pointer returns to its initial position with an uncertainty in latitude of ∆x. Therefore theweighing process has an uncertainty in mass of ∆m. The added load transfers a momentum whichwe can measure with an accuracy of ∆p given by ∆x∆p ' h. Now Bohr makes the assumptionthat ∆p is smaller then the total momentum imposed by the gravitational field during the weighingprocess.

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3.1. THE SOLUTION OF THE PARADOX 11

This gives us an upper bound on the uncertainty in momentum on the box.

∆p < p = v(∆m) = gt(∆m) (3.1)

were t is the time taken to readjust the pointer and g the acceleration of the gravitational field.After using ∆x∆p ' h we obtain

∆m∆xgt > h (3.2)

Now Bohr used the red shift formula knowing that the clock has moved inside the gravitationalfield of the earth after the photon was released. The clock must, according to Einstein’s ownEquivalence principle, tick at different rates. This is caused by the photon loss that changes themass of the box.

∆t =gt

c2∆x (3.3)

Hence,∆E∆t = (∆m2)c2∆t > h (3.4)

Bohr therefore used Einstein’s own relativity theory to show the flaw in his thought experiment.After this refutation, he ceased his search for inconsistencies in Heisenberg’s uncertainty princi-ple according to Pais[11]. He no longer believed that the quantum theory was wrong, but wasincomplete. Some additional information should then give the total description of the quantumparticle. The indeterminacy is just caused by the experimenter that is unable to observe the exactquantities.

The view of the quantum community was the so-called Copenhagen interpretation[12]. Sayingthat the particle was produced by the measurement. It forced the particle to be defined at theobserved position. If one does not measure the particle it is nowhere.

Einstein and Bohr were both subscribed to there view for the rest of there lives and a settlementwas never reached. Even today this debate is alive in the physics community.

Chapter 4

The quantum theory

4.1 The quantization of physics

The quantum theory started with the ultraviolet catastrophe. Max Planck showed that the energy-density of a black body can only be equated if the radiation is quantized in discrete ”packets”.This in the order of the well known Planck constant, according to

E = hf (4.1)

Einstein elaborated this idea and showed the photoelectric effect in his wonder year 1905. Thephenomenon that tells us that electron-emission inside an atom can only happen if the atomis radiated with the right frequency/energy. This idea was experimentally supported by theCompton effect. The discoveries showed that light can also be seen as particles as opposite to thewave phenomena seen by Maxwell’s equations. Particle-wave duality was extended by the inside ofde Broglie saying that quantum matter like an electron can also be seen as a wave. Therefore themomentum of the particle is related by it’s wavelength according to p = h

λ , with λ the wavelengthof the particle.

In 1913, Niels Bohr comes into play. He solved the riddle: How can electrons maintain it’s orbitaround the nucleus? This by knowing that accelerated charged particles must radiate. The classicaltheory predicted that an electron should continuously radiate electromagnetic waves. It will giverise to an energy loss and makes the orbit of the electron smaller and smaller, thereby eventuallyspiraling into the nucleus. The emitted radiation of the atom will then give a continuous spectra.It is in contradiction with the experimentally found emission lines of the hydrogen spectrum givingby the Rydberg formula.

1λ

= Rh(1n2− 1m2

) (4.2)

where Rh = 1.097 10−7m−1 and m,n ∈ N | n < mBohr assumed that electrons can travel on stable orbits around the nucleus, called stationarystates. The electron can make an transition from one orbit to the other but must radiate energy.This in agreement with the photoelectric effect of Einstein, according to

ω(n, n− α) =1~

[E(n)− E(n− α)] (4.3)

where ω(n, n − α) is the frequency of the emitted light from the atom. The labels n and α givethe transition of the electron from the nth orbit to the stable orbit n−α. Using these postulates,Bohr was able to theoretically calculate Rh and the fingerprint of the hydrogen atom. One canalso see the fundamental law of spectroscopy called Ritz rule from eq. 4.3.

ω(n, n− α) + ω(n− α, n− β) = ω(n, n− β) (4.4)

12

4.2. HEISENBERG’S REVOLUTION; MATRIX MECHANICS 13

The laws for an freely moving electron where already described by the classical mechanics andelectromagnetism. Therefore the quantum theory should describe the classical laws when theelectron becomes detached from the nucleus. This is clearly in the limit when the quantumnumber becomes infinite n → ∞. The quantized spectrum should then also become continues,which means that h→ 0. This is called the correspondence principle. If you now use the equationsof motion

x+ f(x) = 0 (4.5)

the quantization condition will follow from the classical limit1.∮pdq = J = nh (4.6)

When you solve this integral you find the equation that replaces the classical law by a newquantized law. There was only one problem. For each transition with a given frequency theremust also be a intensity xα(n). Einstein found in 1917 a different derivation of Planck’s energydensity by relating the probabilities of absorption, spontaneous and stimulated emission of anenergy level. It made transparently clear that the intensity of the radiation is related by thetransition probability X(n, n− α) of the atom energy levels[5].

4.2 Heisenberg’s revolution; Matrix mechanics

In 1925 Heisenberg wrote a paper to establish a new basis for the quantum theory[4]. It wasfounded only upon relations between observable quantities. He reasoned that the position andenergy of the electron around the nucleus can never be measured. Only the transitions can beobtained from the spectrum. Therefore the model, Bohr discovered, had to be replaced.

The energy of a stationary state E(n) related by the classical fundamental frequency ω(n)α is anunobservable quantity. We can only measure the transitions of these states, such as the spectrumof the hydrogen atom. Therefore Heisenberg argued that in his new quantum theory only theBohr-Einstein frequency condition (eq. 4.3) should be used:

ω(n)α→ ω(n, n− α) (4.7)

The motion of the electron around the nucleus is described in classical physics as a function oftime x(t). One can represent this function as an Fourier series.

x(n, t) =∑α

Xα(n)eiω(n)αt (4.8)

Here Xα(n) is the intensity of the nth stationary orbit. Only orbit transitions are measurable.Therefore must Xα(n) must be replaced in Heisenberg’s theory by

Xα(n)→ X(n, n− α) (4.9)

with α represents the transition to an orbit that has an quantum number α lower as n.The new representation made the Fourier decomposition

x(n, t) =∞∑

α=−∞X(n, n− α)eiω(n,n−α)t (4.10)

Now Heisenberg looked at the kinematics and asked himself: ’How is the quantity x(t)2 to berepresented in this new quantum theory?’ He first looked at the classical fourier series. Using eq.4.8 we obtain

x(n, t)2 =∑α

∑β

Xα(n)Xβ(n)eiω(n)(α+β)t (4.11)

1See syllabi: Van klassiek naar quantum (UvA - FNWI)

14 CHAPTER 4. THE QUANTUM THEORY

After relabeling α+ β = γ.x(n, t)2 =

∑γ

Yγeiω(n)γt (4.12)

whereYγ =

∑α

Xα(n)Xγ−α(n) (4.13)

Heisenberg argued that the ’most natural assumption’ for the new quantum theory must be

Y (n, n− γ)eiω(n,n−γt) =∑β

X(n, n− α)eiω(n,n−α)t X(n− α, n− γ)eiω(n−α,n−γ)t (4.14)

This because of the frequency summation rules. The exponentials cancel on both sides by usingRitz rule (eq.4.4).

Y (n, n− γ) =∑α

X(n, n− α)X(n− α, n− γ) (4.15)

This was Heisenberg’s multiplying law for transition amplitudes. Now he asked: If there are twoquantities x(t), z(t) what is there product x(t)z(t). It shows that there is a difference between theclassical and new representation because∑

α

X(n, n− α)Z(n− α, n− γ) 6=∑α

Z(n, n− α)X(n− α, n− γ) (4.16)

This ’significant difficulty’ were the non-commutative quantities in his new theory and was whatkept him awake2. He continued his paper by showing how to solve the harmonic and anharmonicoscillator for the new quantum replacements. Notice that this law is exactly matrix multiplication.Max Born also found this, a week after Heisenberg gave him the paper for publication. He appliedthe rules to the quantum condition3 and found for the diagonal elements∑

α

X(n, n− α)P (n− α, n)− P (n, n− α)X(n− α, n) = i~ (4.17)

This has become the new quantum condition, and provided the new quantization law for thetheory nowadays called quantum mechanics. I find it very remarkable how Heisenberg found thebasic laws of quantum mechanics especially because matrix manipulation was unknown to himwhen he wrote the paper. This derivation shows the necessity of non-commutative quantities in aquantum theory.

Heisenberg and Born presented these discoveries during the sixth Solvay conference[1]. They alsonoted that the time t and the energy/frequency (E = hω) should be conjugate variables. Theystated that these quantities could not commute in the new quantum theory.

Niels Bohr also gave a presentation during the conference on the complimentarily of particleand wave physics. He stressed he same necessary of time-energy conjugation by considering thede Broglie-Einstein relations E = ~ω and p = h

λ [11]. A photon should be seen according toBroglie’s duality picture as a wave packet extended in space and time. In this interval lie a rangeof frequencies and wavelengthes given by σx and σ 1

λ . If f(x) is a function that represents theposition of the photon then the Fourier transformation represents the wave number F (k) = F ( 1

λ ).The widths of both graph’s can not both be made arbitrarily small. This is given by a theoremin Fourier analysis and classical optics. It states that

σxσ1/λ ≥12

2See quotation in section 1.23See appendix B for a derivation of the quantum condition

4.3. HEISENBERG’S γ-RAY MICROSCOOP 15

. The same holds for the time f(t) and the frequency F (ω).

σωσt ≥12

Now use the de Broglie-Einstein relations and you are there, finding the Heisenberg uncertaintyrelations.

σxσp ≈ h (4.18)σEσt ≈ h (4.19)

Einstein did not except the idea that positions in a spacetime could never be precisely known. Hetherefore rejected Heisenberg’s idea and started the debate between Bohr.

4.3 Heisenberg’s γ-ray microscoop

The question Heisenberg asked himself, after the position-momentum uncertainty was mathemat-ically proved, whether the relation is just a result of the used equations, or really build into everymeasurement[13]. He invented a thought experiment believing that every scientific concept mustbe based upon (actual or possible) experimental observations.

Imagine you have a microscope that can measure the position of an electron by using a high energyγ-ray beam. It sits directly beneath the center of the microscope’s focus point. (see figure 4.1). Acircular lens forms a cone of angle 2θ from the electron. The γ-ray light’s up the electron whichhas a short wavelength giving the measurement it’s highest possible resolution.

Figure 4.1: The imaginary microscope of Heisenberg.

The relation between the objects size and the wavelength is given by Bragg’s law using wave optics:

∆x =λ

2Sin(θ)(4.20)

There will be a transfer of momentum when the ray strikes the electron. Using the de Brogliewavelength we can calculate this momentum.

p =h

λ(4.21)

The γ-ray must be scattered within the angle of 2θ of the cone to observe the electron. There aretwo cases. The diffraction of the ray will be towards the right edge of the lens. The momentum inthe x-direction would be the sum of the electrons momentum in the x-direction and the momentumof the γ-ray in the x-direction:

16 CHAPTER 4. THE QUANTUM THEORY

p(1)totalx

= p(1)x +

h

λ(1)Sin(θ) (4.22)

The other case is diffraction of a γ-ray towards the left side of the lens. The momentum in thex-direction would be:

p(2)totalx

= p(2)x −

h

λ(2)Sin(θ) (4.23)

But the initial momentum must equal the final momentum in both cases because this quantity isconserved and therefore never lost. Hence,

p(1)x +

h

λ(1)Sin(θ) = p(2)

x −h

λ(2)Sin(θ) (4.24)

If the angle θ is small, then the wavelengths are approximately the same λ(1) ' λ(2) ' λ, giving:

p(2)x − p(1)

x = ∆px =2hSin(θ)

λ(4.25)

Together with the uncertainty in position this gives:

∆px∆x ' h (4.26)

This is the Heisenberg uncertainty relation in the x-direction. It’s shows that the concept ofuncertainty is an element of the thought experiment.

4.4 Wave mechanics

The alternative was given by the Austrain physicist Erwin Schrodinger. A year after Heisenberg’sdiscovery (1926) he found an easier route towards a quantum mechanics. This method appealedmany physicists because it led to the same discoveries as matrix mechanics but much more easilywithout giving up the picture of electron orbits within an atom. He used a set of axioms to provehis Schrodinger equation. There is no derivation, it can only be made plausible why this equationis right[12]. Start with the constant energy of classical mechanics.

E =p2

2m+ V (4.27)

In the wave theory all matrices become operators. These are

p→ p =~i

∂

∂xx→ x = x (4.28)

E → H = i~∂

∂tt→ t = t (4.29)

(4.30)

We assume that these operators act on the the wave function Ψ(x, t). The quantum conditionsmust obviously hold for the wave theory. Using the operators you can see that this is true for anyarbitrary wave function

[x, p]Ψ(x, t) = (xp− xp)Ψ(x, t) = i~Ψ(x, t) (4.31)

[E, t]Ψ(x, t) = i~Ψ(x, t) (4.32)

The energy and momentum of the classical equation must then also be replaced by operators inthe wave theory. This becomes the time-dependent Schrodinger equation.

(−~2

2m∇2 + V )Ψ(x, t) = i~

∂

∂tΨ(x, t) (4.33)

4.5. POSITION AND MOMENTUM 17

Max Born published a paper that same year concerning the scattering of electrons on an heavyatom. He solved the wave function Ψ(x, t) for this problem. This function must be seen, accordingto his thoughts, as the probability that the electron will scatter in an azimuthal angle between Ωand Ω + dΩ. In a footnote he made the following statement[11]. ’A more precise considerationshows that the probability is proportional to | (Ψ(x, t)) |2.’

Now we can see what the meaning of this wave function Ψ(x, t) is and especially why the followingcondition must hold. ∫ ∞

−∞| (Ψ(x, t)) |2 dx = 1 (4.34)

The wave function is normalized. It means that the totality of the particle is confined in a finitespace (in our case a line). We can also obtain the average value of an operator (which has beengiven the badly chosen name: the expectation value).

〈A〉 =∫ ∞−∞

Ψ(x, t)∗AΨ(x, t) dx ≡ 〈Ψ | A | Ψ〉 (4.35)

From this definition we can find the standard derivation

σ2A = 〈Ψ | (A− 〈A〉)2 | Ψ〉 (4.36)

4.5 Position and momentum

The commutation relation [x, p] will give rise to the uncertainty relation in the position andmomentum. This can be seen from the following derivation in Griffiths book[12].

First make the following substitution | f〉 = (A − 〈A〉) | Ψ〉. Then the product of the standardderivation becomes the innerproduct of f .

σ2A = 〈f | f〉 (4.37)

The same holds for any other observable B with | g〉 = (B − 〈B〉) | Ψ〉

σ2B = 〈g | g〉 (4.38)

Now multiply both and use the Schwarz inequality to obtain

σ2Aσ

2B = 〈f | f〉〈g | g〉 ≥| 〈f | g〉|2 (4.39)

The innerproduct z = 〈f | g〉 is a complex number, with

| z|2 = [Re(z)]2 + [Im(z)]2 ≥ [Im(z)]2 = [12i

(z − z∗)]2 (4.40)

This means thatσ2Aσ

2B ≥ (

12i

[〈f | g〉 − 〈g | f〉])2 (4.41)

We only have to understand what the meaning of the innerproducts of f and g is.

〈f | g〉 = 〈 Ψ | (A− 〈A〉)(B − 〈B〉) | Ψ〉 (4.42)

The observable A is hermitian, therefore we can change the operator and the wave function

〈f | g〉 = 〈 Ψ | (AB − A〈B〉 − 〈A〉B + 〈A〉〈B〉) | Ψ〉 (4.43)= 〈AB〉 − 〈A〉〈B〉 (4.44)

The same holds for〈g | f〉 = 〈BA〉 − 〈A〉〈B〉 (4.45)

18 CHAPTER 4. THE QUANTUM THEORY

Therefore we obtain the following equation

σ2Aσ

2B ≥ (

12i〈AB〉 − 〈BA〉)2 = (

12i〈[A, B]〉)2 (4.46)

Now if x and p are substituted for A,B respectively, we obtain after using the quantum conditionof eq. 4.31

σx2σp

2 ≥ (12i〈[x, p]〉)2 = (

~2

)2 (4.47)

or since standard derivations are always positive

σxσp ≥~2

(4.48)

This is the position-momentum uncertainty relation.

4.6 Energy and time

Time is an outsider in wave quantum mechanics. The schodinger equation does not treat positionand time on equal footing. It’s has a first-order differential in time but a second-order differentialin position. Therefore it not easy to obtain the time-energy uncertainty relation. We will give thederivation of Mandelshtam and Tamm[12].

d

dt〈Q〉 =

d

dt〈Ψ | QΨ〉 = 〈∂Ψ

∂t| QΨ〉+ 〈Ψ | ∂Q

∂tΨ〉+ 〈Ψ | Q∂Ψ

∂t〉 (4.49)

Using the Schodinger equation i~∂Ψ∂t = HΨ we obtain

d

dt〈Q〉 =

i

~〈HΨ | QΨ〉 − i

~〈Ψ | QH Ψ〉+ 〈∂Q

∂t〉 (4.50)

But H is hermitian, so 〈HΨ | Q〉 = 〈Ψ | HQ Ψ〉. Then

d

dt〈Q〉 =

i

~〈[H, Q]〉+ 〈∂Q

∂t〉 (4.51)

Operators that depend on time are very rare, we can therefore assume that ∂Q∂t = 0 making

σ2Hσ

2Q ≥ (

12i〈[H, Q]〉)2 = (

12i

~i

d〈Q〉dt

)2 = (~2

)2(d〈Q〉dt

)2 (4.52)

Now define σH ≡ σE , andσt ≡

σQ| d〈Q〉/dt |

(4.53)

This factor has the same dimensions as time, but it means the time it takes for the system tochange substantially. Or in more abstract words: σt represents the time interval in which theoperator Q has changed one standard deviation. Notice that this time is not the same as thethe duration of an measurement. Now using this definition we obtain the time-energy uncertaintyrelation.

σEσt ≥~2

(4.54)

Chapter 5

Analyses of the debate

5.1 A new solution of the paradox

Looking back at Bohr’s derivation one can see that the change in height gives a change in theobservation of the time interval. I believe that that Einstein could not persist his reasoning afterhe heard that time dilatation was not included. But there is an subtlety in Bohr’s derivation.The upperbound in the momentum seems odd. Einstein wanted to make a very accurate positionmeasurement in order to obtain the right mass loss of the box. This should according to theposition-momentum uncertainty relation give a large uncertainty in the momentum. The positionobservation can make the momentum uncertainty therefore large and even bigger then the mo-mentum self during the time to readjust the pointer. It is therefore not right to give the upperbound ∆p < p in his reasoning ones Einstein makes a very accurate position measurement. Inorder to get rid of this subtlety we start our thought experiment over and look more closely to thesetup.

Consider again the box with a small opening on one side. Again the clock controls the shutterwhich covers the hole. The box is suspended from a stand with a mechanical spring hoveringinside the stationary gravitational field of the earth. We only consider quantum uncertainties andtherefore neglect all observational errors caused by the instruments. In order to get some moreinside on the position measurement we will not add a load to the box after the energy loss. Tostart the experiment Einstein adjust his clock for two time intervals. The first time interval isbetween the setup of the clock and the opening of the shutter and the second between the openingand closing time of the box.

After setup Einstein put’s his clock inside the box, he measures the position before the openingof the shutter. This must be very accurate in order to obtain the right initial mass and energy ofthe box. Using the classical force balance one sees that the spring force cancels the gravitationalforce.

mg = kx (5.1)

The initial position measurement causes a large uncertainty in the momentum given by Heisen-berg’s uncertainty relation.

∆p >h

∆x(5.2)

It produces a momentum after the measurement. Then the position of the box must also change,giving rise to a force.

F = p = kx−mg (5.3)

After a coordinate transformation this becomes

p = kz (5.4)

19

20 CHAPTER 5. ANALYSES OF THE DEBATE

where z is the change in height from it’s initial position. After some time the shutter will open.But then the height of the box is not longer the measured position. Therefore the clock will tickdifferent then it should on the measured position. This change in time, created by the uncertaintyin momentum, is equated by the gravitational time dilatation formula.

t0 = [1 +gz

c2]t (5.5)

But this should be changed into an integral because the position changes during the point of setupand the moment when the shutter opens. Therefore t is not the right time interval but the integralover time dt is. Now using the force on the box (eq. 5.4) we obtain∫ t0

0

dt′ −∫ t

0

dt′ =∫ t

0

gx

c2dt′ =

∫ t

0

gp

kc2dt′ (5.6)

Since g and kc2 are constants for our problem we can remove them out of the integral.

δt =g

kc2

∫ t

0

p =g

kc2p|p(t)p(0) =

g

kc2[p(t)− p(0)] =

g

kc2δp (5.7)

This all means that there is a time difference caused by the large uncertainty in the momentumduring the position measurement, according to

δt =g

kc2δp (5.8)

Now we look at the second time interval between the opening and the closing of the shutter. Theposition measurement moved the box before the shutter opened. Therefore the measured positiondoes not agree anymore with the real height of the box when the shutter opened. After thephotons moved out, the box will oscillate and after some time the friction will give the box a newequilibrium point where the spring force cancels the gravitational force. Now we again measurethe height of the box and find the difference in height caused by the escaped photons.

[m(t′)−m(t)]g = k[x(t′)− x(t)] (5.9)

But this length does not agree with the real height change of the box. This because the positionmeasurement caused the box to move before the shutter opened. Therefore the quantum error inthe height causes an error in the mass measurement, according to

(δm)g = k δx (5.10)

or

δE = (δm)c2 =kc2

gδx (5.11)

This means that there is a mass difference caused by the large uncertainty in momentum. Thechange in initial position causes an quantum error in the mass observation of the box.

Now both position and time have an certain error. Therefore if we remeasured these quantitieswe should obtain different values for the change in time and height of the photon-box. If the mea-surement was repeated many times we could find a average value. But we know how the quantumerrors of the position and time are related by the energy and momentum errors, respectively.Therefore the average value in the energy change of the box is

〈E〉 =kc2

g〈x〉 (5.12)

The same holds for the time measurement and the momentum average value.

〈t〉 =g

kc2〈p〉 (5.13)

5.2. THE POISSON BRACKET 21

The standard derivation of an measurement (A) is given by

σA2 = 〈A〉2 − 〈A2〉 (5.14)

So we only need the average value of the time and energy squared. This can be found by squaringthe time and energy values given by the position of every measurement. Compute again theaverage values to obtain

〈E2〉 = (kc2

g)2〈x2〉 (5.15)

and〈t2〉 = (

g

kc2)2〈p2〉 (5.16)

Now we have the standard deviation for the time and energy caused by the uncertainty in theinitial position measurement

σE2 = 〈E〉2 − 〈E2〉 = (

g

kc2)2(〈x〉2 − 〈x2〉) = (

g

kc2)2σx

2 (5.17)

and because this quantity and the constants are always positive we can lose the square withoutany loss of generality.

σE =g

kc2σx (5.18)

The same holds for the time and the momentum

σt =kc2

gσp (5.19)

Now multiply both to obtain the time-energy uncertainty relation.

σEσt = σxσp ≥~2

(5.20)

5.2 The Poisson Bracket

Consider a one dimensional dynamical system made up of N particles and specified in the termsof the Hamiltonian dynamics. The equations of motion are given by

∂H

∂pi= qi (5.21)

and∂H

∂qi= −pi (5.22)

where the Hamiltonian is H(q, p; t). This function gives the energy at time t expressed in thecoordinate qi and momentum pi canonical variables. The lower index i runs from 1 till N for aone dimensional problem.

Dirac found in a classical analogy of the quantum conditions which we now shall find[14]. Classicalmechanics provides a valid description under certain conditions. This is when the particles andbody of the system are massive of have a high energy. The classical mechanics must be thecorrespondence limit of quantum mechanics. Thus when n→∞ and ~→ 0. We therefore expectthat a concept of classical mechanics must be equivalent to the commutation relations of quantummechanics.

22 CHAPTER 5. ANALYSES OF THE DEBATE

An important operator in the classical Hamiltonian dynamics is the Poisson bracket, which isclosely related to the time evolution of an dynamical variable. First we start with the definition.

Definition 1 Let f(qi, pi) and g(qi, pi) be two functions on phase-space. Then the Poisson bracketis

f, g =∂f

∂qi

∂g

∂pi− ∂f

∂pi

∂g

∂qi(5.23)

The Poisson bracket has the following properties. If f(qi, pi), g(qi, pi) and h(qi, pi) are functionson phase-space and ∀ α, β ∈ R is:

f, g = −g, f (5.24)Linearity : αf + βg, h = αf, h+ βg, h (5.25)Leibniz rules : fg, h = fg, h+ f, hg (5.26)

: f, gh = f, gh+ gf, h (5.27)Jacobi identity : f, g, h+ g, h, f+ g, f, g = 0 (5.28)

The first two are obvious. The third and forth properties follows from the chain rule, but theidentity is not so easy to prove1. You can write out all 24 terms (that you do not want to do) ormake us of fact that the Poisson bracket is a bilinear homogenous function. The most importantproperty is the following. Note that

df

dt=∂f

∂qiqi +

∂f

∂pipi +

∂f

∂t(5.29)

Now use the Hamiltonian equations, which are eq. 5.21 and eq. 5.22.

df

dt=∂f

∂qi

∂H

∂pi− ∂f

∂pi

∂H

∂qi+∂f

∂t= f,H+

∂f

∂t(5.30)

Now you see why the Poisson bracket is closely related to the time evolution of the variable f . Ifa function does not explicitly depends on time (g(q, p)) and satisfies

f,H = 0 (5.31)

we can find a constraint of motions. We say that f and H Poisson commute. It means that thefunction f is constant throughout the motion of the system.

We will now use the Poisson properties to evaluate the following bracket [u1u2, v1v2] by using thetwo different Leibniz rules. Using the first rule

u1u2, v1v2 = u1, v1v2u2 + u1u2, v1v2= (v1u1, v2+ u1, v1v2)u2 + u1(v1u2, v2+ u2, v1v2)= v1u1, v2u2 + u1, v1v2u2 + u1v1u2, v2+ u1[u2, v1v2)

and using the second rule

u1u2, v1v2 = u1u2, v1v2 + v1u1u2, v2= u1, v1u2v2 + u1u2, v1v2 + v1u1u2, v2+ v1u1, v2u2

Now subtract both to obtain

[u1, v1](u2v2 − v2u2) = (u1v1 − v1u1)[u2, v2] (5.32)

1For a prove see Lev Landau’s mechanics book[15]

5.3. THE POISSON BRACKET OF H,T AND X,P 23

The following condition must then hold if u1, v1 are independent of u2, v2

u1v1 − v1u1 = i~[u1, v1] (5.33)u2v2 − v2u2 = i~[u2, v2] (5.34)

where ~ is a real constant which according to experiments should be equal to the Planck constant.Now notice that the operators given in section 4.4 have the same properties as the Poisson bracketfor the simplest quantum condition.

qi, qj = 0 [qi, qj ] = 0 (5.35)pi, pj = 0 [pi, pj ] = 0 (5.36)pi, qj = δij [pi, qj ] = i~δij (5.37)

We therefore could make the assumption that the quantum commutator have the same value asthe classical Poisson bracket.

[x, p] = limn→∞

i~ x, p (5.38)

Paul Dirac had also read the paper of Heisenberg and saw that it was governed by matrix algebra.He looked at the diagonal terms of the matrix [XP − PX](n, n) seen in his paper shortly writtenHeisenberg’s and found the same algebraic structure[4] as Poisson brackets. Therefore he make theconclusion: The classical Poisson bracket can be regarded as the same as the quantum mechanicscommutator for the most simple operators x and p that are considered as classical objects.

5.3 The Poisson bracket of H,t and x,p

After seeing the equivalence of the Poisson bracket with the quantum commutator and Bohr’sderivation of the time-energy uncertainty relation, one mind think that in the classical correspon-dence limit it should be possible to obtain the relation

H, t = x, p (5.39)

First notice that time can be represented as a function of x and p. You can find an inverse t(x, p),if one has obtained the functions p(t) and x(t). Let the position of the particle start at the originon time t = 0, then

t =∫ t

0

dt =∫ x(t)

x(0)

dt

dx′dx′ =

∫ x

0

1∂H∂p (p′(x, p;x′), x′)

dx′ (5.40)

Now p′(x, p;x′) is the momentum at position x’ which get replaced after integration by the trueposition at time t. We assume that the equations of motion hold for the total path of the system,see figure 5.1.Now consider the poisson bracket of H, t were the time (t) is the integral of eq. 5.40.

H, t(x,p) =∂H

∂p

∂t

∂x− ∂H

∂x

∂t

∂p=∂H

∂p[∂t

∂x′∂x′

∂x+

∂t

∂p′∂p′

∂x]− ∂H

∂x

∂t

∂p(5.41)

If x′ becomes equal to x(t) then ∂x′

∂x = 1 and ∂H∂p

∂t∂p′

∂p′

∂x = ∂H∂x

∂t∂p , so

H, t(x,p) =∂H

∂p

∂t

∂x′= 1 (5.42)

Hence, there is a canonical transformation between H, t and x, p that leaves the Poisson bracketsinvariant for a one dimensional Hamiltonian.

H, t(x,p) = 1 (5.43)

24 CHAPTER 5. ANALYSES OF THE DEBATE

Figure 5.1: Depiction of the object moving over the time path

H,H(x,p) = t, t(x,p) = 0 (5.44)

The Poisson bracket H, t must remain constant for all times t, so ddt (H, t) = d

dt (x′, p′) must

be equal to zero.First notice that

H, p′(x,p) =∂H

∂p

∂p′

∂x− ∂H

∂x

∂p′

∂p=∂H

∂p[∂p′

∂x+∂p′

∂p

∂p

∂x]− ∂H

∂x[∂p′

∂x

∂x

∂p+∂p′

∂p] (5.45)

=∂H

∂p

∂p′

∂x+∂H

∂x

∂p′

∂p− ∂H

∂p

∂p′

∂x− ∂H

∂x

∂p′

∂p= 0

and

H,x′(x,p) =∂H

∂p

∂x′

∂x− ∂H

∂x

∂x′

∂p=∂H

∂p[∂x′

∂x+∂x′

∂p

∂p

∂x]− ∂H

∂x[∂x′

∂p+∂x′

∂x

∂x

∂p] (5.46)

=∂H

∂p

∂x′

∂x+∂H

∂x

∂p′

∂p− ∂H

∂p

∂x′

∂x− ∂H

∂x

∂p′

∂p= 0

Now use the Jacobi equation

0 = H, x′, p′+ x′, H, p′+ p′, x′, H = H, x′, p′ (5.47)

=d

dt(x′, p′)− ∂

∂t[x′(x, p), p′(x, p)] =

d

dt(x′, p′)

The partial time derivative is zero since x′ and p′ are implicit function of time. Therefore thePoisson bracket between x and p are equivalent to those of between the energy H and t. The alsoremain constant throughout the motion for the classical object.

5.4 Conclusions

We have seen that there is a different method to obtain the time-energy uncertainty relation forthe escaped photon in section 5.1. It shows that the laws of quantum mechanics are in agreementwith the setup of Einstein’s thought experiment. The uncertainty in position and momentumbecome there equal to the uncertainty in energy and time. However it is not clear why these twoshould be equivalent in the setup of the thought experiment. We think that since the Poissonbracket of x, p and H, t are equal it might be possible to obtain the equivalence of the uncertaintyrelations for classical object. This by seeing that the Poisson brackets are closely related to thecommutation relations and the uncertainty principle (see eq. 4.46). Further study is howeverneeded to prove this.

Appendix A

Gravitational time dilatation fromGeneral Relativity

A.1 The geodesic equation

The action is for any dynamical system equal to the integral of the lagrangian over time. TheLagrangian has units of energy. Then the action has the unit of energy times time. Note that ithas the same unit as ~ and as the time-energy uncertainty relation. Now we will find the equationfor gravitational time dilatation from Carrol’s book[10]. First we start to construct a relativisticaction.

The action is a functional. It takes a set of functions that describe a world-line and gives as outputa number. If the particle starts at the origin and ends at some spacetime point there are manypossible trajectory’s. We need an action were all observers agree for any world-line. Well, allobservers agree on the time elapsed on the clock carried by the moving particle. Naturally thiswould mean that the action calculates the proper-time.

dτ2 = −gµνdxµdxν (A.1)

In order to get units of action we need an additional Lorentz invariant factor with unit energy.This is of course mc2.

S = mc2τ =∫mc2dτ =

∫mc2

√−gµν

dxµ

dτ

dxν

dτdτ (A.2)

The minus sign inside the square root does not look very nice. We can simplify the expressionwith a trick. The variation of the action is

δτ =∫mc2δfdτ = −mc

2

2

∫1√−f

δf dτ (A.3)

were f = gµνdxµdxν = −1. Look at equation A.2. The third and forth line represent the same

integral with an√f difference. Then f = −1.

δτ = −mc2

2

∫δfdτ (A.4)

The extremization condition for variation of the path with fixed endpoints can be translated, bycalculus of variation into a partial differential equation.

δτ = −mc2

2

∫[∂σ(gµν)

dxµ

dτ

dxν

dτδxσ + gµν

d(δxµ)dτ

dxν

dτ+dxµ

dτ

d(δxν)dτ

] dτ (A.5)

25

26 APPENDIX A. GRAVITATIONAL TIME DILATATION FROM GENERAL RELATIVITY

The terms inside the bracket can be integrated by parts; for example,

δτ = −mc2

2

∫gµν

dxµ

dτ

d(δxν)dτ

dτ = +mc2

2

∫[gµν

d2xµ

dτ2+d(gµν)

dτ

dxµ

dτ]δxν dτ (A.6)

=mc2

2

∫[gµν

d2xµ

dτ2+ ∂σ(gµν)

dxσ

dτ

dxµ

dτ] δxν dτ

The boundary terms vanish because the variation δxν = 0 at the endpoints. After rearrangingsome dummy indices the extremization becomes

δτ = mc2∫

[gµσd2xµ

dτ2+

12

(∂µgνσ + ∂νgµσ − ∂σgµν)dxµ

dτ

dxν

dτ]δxσ dτ (A.7)

Now δτ must vanish for every variation δx, the expression inside the brackets must vanish. Aftermultiplying by the inverse metric gρσ we obtain the geodesic equation.

0 =d2xρ

dτ2+gρσ

2(∂µgµσ + ∂νgµσ − ∂σgµν)

dxµ

dτ

dxν

dτ≡ d2xρ

dτ2+ Γρµν

dxµ

dτ

dxν

dτ(A.8)

This is the equation of motion in a gravitation field. The symbol Γρµν is called the Christoffelconnection. It describes the curvature of space-time. Note that when Γρµν = 0 it becomes anEuclidean space where d2xρ

dτ2 = 0.

A.2 The weak field limit

There is a more universally derivation of the gravitational time dilatation[10]. This by forcingthat the equation of motion of general relativity must reduce to Newton’s third law when the fieldis static and weak. The object must also move with non-relativistic velocities (v c).First we look at geodesic equation:

d2xρ

dτ2= Γρµν

dxµ

dτ

dxν

dτ(A.9)

For small velocities much smaller then c is dxi

dτ dtdτ . Meaning that the object moves mostly in

the time-part of spacetime. Only the dominant term in the double sum remains in the geodesicequation

d2xρ

dτ2= Γρ00

dt

dτ

dt

dτ(A.10)

The gravitational field of the earth does’t change in time and makes it static. Therefore all timederivatives must become zero ∂gµν

∂x0 = 0. This simplifies the Christoffel symbol:

gµνΓρ00 = −12∂g00

∂xµ(A.11)

We also assumed the weakness of the field. The metric is not too different from the flat spacetimemetric. Therefore there is a small perturbation or change (hµν) from the flat spacetime metic(nµν = diag(−1 1 1 1)).

gµν = nµν + hµν with |hµν |2 1 (A.12)

Here hµν(x) is a small correction field that depends on the height from the surface of the earth.This will simplify the Christoffel symbol even more:

nρλΓρ00 = −12∂h00

∂xλ(A.13)

A.2. THE WEAK FIELD LIMIT 27

On the left hand side is the term with hµν negligible because it is small. The spacelike componentsof the flat spacetime metric is just a identity matrix. We now get

d2xµ

dτ2− 1

2∂h00

∂xµdt

dτ

dt

dτ= 0 (A.14)

The zero component or time component shows

dx0

dτ= constant (A.15)

and the spacelike component with i.

d2xi

dτ2− 1

2∂h00

∂xidt

dτ

dt

dτ= 0 (A.16)

The laws of Newton tells us that d2xi

dt2 = −g, so after rearranging the dt and dτ terms (multiplyby (dτdt )2)

g = −12∂h00

∂xi(A.17)

Thenh00 = −2

gz

c2(A.18)

making the gravitational metric, by using eq. A.12

g00 = −(1 + 2gx

c2) (A.19)

Notice that only the time component remains. Now use the proper time function given by eq. A.1end Taylor approximate about gx

c2 ≈ 0.

dτ =√−g00dt =

√1 + 2

gx

c2≈ 1 +

gx

c2(A.20)

This will give us the formula for the gravitational time dilatation

δt = dτ − dt =gx

c2dt (A.21)

Appendix B

The new quantization condition

We will derive the new quantum condition by using matrix mechanics given in the paper ofAitchison et al.[16]. First, see that the time derivative of x(n, t) is

x(n, t) =∑α

i X(n, n− α)ω(n, n− α)eiω(n,n−α)t (B.1)

Use the old quantum condition with p ≡ mx and the classical fourier decomposition

J =∮p dq =

∫ ∞−∞

mq2 dt =∑α

∑β

∫ ∞−∞−m αβXα(n)αXβ(n)ω2(n)

∫ ∞−∞

eiω(n)[α+β]t dt (B.2)

Make the following substitution t→ ω(n)t and equate the integral

J = nh =∑α

∑β

m αβ Xα(n)Xβ(n) ω(n)2π δ[α+ β] =∑α

2πα2 ω(n) Xα(n)X−α(n) (B.3)

The motion of the electron is a real quantity. Then the transition amplitude Xα(n) must beHermitian and Xα(n) = X∗−α(n). Hence,

nh

2π= n~ =

∑α

m α2 ω(n) | Xα(n) |2 (B.4)

Notice that the sum does not depend on the sign of α. Therefore the summand can be taken onlyover the positive α after multiplication by a factor two. Take the derivative of n to obtain

~ = 2∑α>0

m αd

dn(α ω(n) | Xα(n) |2) (B.5)

Now Heisenberg replaced the fourier decomposition of the old quantum condition by a differencein term of matrix elements

~ = 2m∑α>0

[ω(n+ α, n) | X(n+ α, n) |2 − ω(n, n− α) | X(n, n− α) |2] (B.6)

To make this more clear notice that the energy states and the frequency condition give

limh→0

ω(n, n− α) = limh→0

αE(nh)− E([n− α]h)

αh= α

dE

dn= αω(n) (B.7)

Born recognized that the matrix ω(n, n−α) is obtained from the classical frequency in the corre-spondence limit. He generalized this rule by postulating that

α∂φ(n)∂n

↔ φn− φ(n− α) (B.8)

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or it iteration

α∂φ(n, α)∂n

↔ φ(n+ α, n)− φ(n, n− α) (B.9)

This will give the difference in eq. B.6. Max Born looked at the diagonal matrix elements of[xˆx− ˆxx](n, n). The first term gives us using the multiplication rule of Heisenberg

xˆx =∞∑

α=−∞X(n, n− α) iω(n− α, n)X(n− α, n) (B.10)

Notice that the summand for α = 0 is zero because there is no transition between states. Thereforewe can separate the sum in a positive and negative α term.

=∑α<0

X(n, n− α) iω(n− α, n)X(n− α, n) +∑α>0

X(n, n− α) iω(n− α, n)X(n− α, n) (B.11)

Now make use of the fact that the matrix elements of X(n, n − α) must be real because it is anobservable quantity, so X(n, n−α) = X ∗ (n−α, n). Also notice that ω(n, n−α) = −ω(n−α, n).As last we change α→ −α for the first term without loss of generality. Hence,∑

α>0

iω(n+ α, n) | X(n+ α, n) |2 −∑α>0

iω(n, n− α, n) | X(n, n− α) |2 (B.12)

The same can be done with the term ˆxx since the elements of a matrix do commute. Thereforewe end up with

[ˆxx− ˆxx](n, n) = 2i∑α>0

[ω(n+α, n) | X(n+α, n) |2 −∑α>0

ω(n, n−α, n) | X(n, n−α) |2] (B.13)

Now use the expression for ~ found by Heisenberg and assume that p = mˆx to obtain the newquantum condition.

[xp− xp](n, n) = i~ (B.14)

It was shown by Born that the off diagonal elements are zero.

Appendix C

Nederlandse samenvatting

Een bekent citaat van Einstein is: God dobbelt niet met het universum. Minder bekent is hetverhaal achter deze woorden.

In oktober 1927, vond de vijfde Solvay conferentie plaats in Brussel. Al onze kwantumheldenwaren uitgenodigd. Ontdekkingen zoals: het onzekerheidsprincipe van Heisenberg, Schodingersgolfmechanica en complementariteit werden daar besproken. Albert Einstein, ook aanwezig, washet niet eens met deze revolutie in de fysica. Hij was ervan overtuigd dat elementaire objecten eenwel gedefinieerde positie en impuls dan wel energie en levensduur bezaten. Dit in tegenspraak metde onzekerheidsrelaties van Heisenberg. Daarom legde hij om zijn gelijk te bewijzen met simpelegedachten experimenten de kwantummechanica op de pijnbank. Na drie jaar leek het dat hij eenjuist tegenvoorbeeld had gevonden.

Einstein beschouwde een doos met daarin fotonen. Een klok in deze doos opent een sluiter opeen van te voren ingesteld tijdsinterval. Indien het interval minimaliseert zal er maar een fotonkunnen ontsnappen. Door de massa te meten voor en na het openen van de sluiter kan (metde relatie E = mc2) de energie van het foton met elke willekeurige precisie worden bepaald. Deontsnappingstijd wordt gegeven door de klok. Hierdoor lijkt de tijd-energie onzekerheidsrelatievan Heisenberg te zijn geschonden. Niels Bohr, de grondlegger van de kwantumtheorie, was hettelkens gelukt om de fout in Einsteins redenering te ontdekken. Deze keer was er niet zomaar eenoplossing voor de paradox.

Mijn scriptie vertelt over de geschiedenis van dit debat. Hoe Heisenberg matrix mechanica ontdekteen hoe dit in golfvergelijkingen kan worden omgeschreven. Een afleiding van de onzekerheidrelatieswordt gegeven. Het debat wordt opnieuw tot stand gebracht en Bohrs beredenering zorgvuldigbekeken, die hij vond na een nacht piekeren. Het bleek dat het equivalentie principe een onzek-erheid gaf in de tijdmeting. Dit principe vormt het beginsel van eigen Einsteins meesterwerk: dealgemene relativiteit! Door gravitationele tijdilitatie in rekening te brengen zal het experimentwel voldoen aan de tijd-energie onzekerheidsrelatie. Een andere afleiding wordt gegeven om telaten zien dat Bohr het echt juist had. Tot slot wordt bekeken waarom de Poisson haakjes in deklassieke limiet equivalent zijn, wat mogelijk aangeeft dat de onzekerheidsrelaties ook aan elkaargelijk zijn.

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