Chapter 33 non calculusQuantum Mechanics IIIThe Uncertainty Principle –Time and EnergyCHAPTER 33 QUANTUM MECHANICS IIITHE UNCERTAINTY PRINCIPLE – TIME AND ENERGY

In the last chapter we saw that if all particles had aparticle-wave nature obeying the de Broglie rela-tionship p = h/λ , then there would be a fundamentalrestriction to the accuracy with which one couldmake experimental measurements. This limit isknown as the uncertainty principle which comes intwo forms represented by the equations

ΔpΔx ≥ h

ΔEΔt ≥ h

The first form ΔpΔx ≥ h tells us that if we make aposition measurement to within an accuracy Δx , wehave made the momentum uncertain by an amount atleast as large as Δp . The second form ΔEΔt ≥ htells us that if we have only a limited time Δt inwhich to make an energy measurement, the energywill be uncertain at least as much as ΔE.

In the appendix to the last chapter, we derived thetime energy form of the uncertainty principle from

ΔpΔx ≥ h for a non relativistic particle whereE was the kinetic energy 1/2mv2. We asked you toapply this to an electron dropped from rest to seethat it would be impossible to accurately track themotion of an electron in the first 1/30 of a second.

In this chapter we apply Fourier analysis to a shortlaser pulse and end up with a fully relativisticderivation of the uncertainty principle for thephotons, or photon, in the pulse. In this derivationwe more clearly see how the uncertainty principle isa direct consequence of the particle-wave nature ofmatter.

Perhaps the most interesting and important part ofthe chapter comes when we reconcile the time-energy form of the uncertainty principle with the lawof conservation of energy. The result, which istotally outside the realm of classical physics, mayhave implications related to the origin of the uni-verse.

33-2 The Uncertainty Principle: Time and Energy

A FEMTOSECOND LASER PULSEIn the last chapter, we used an analysis of the two slitelectron diffraction experiment to introduce the po-sition-momentum form of the uncertainty principle.In this chapter we will use an experiment involvingthe analysis of a pulsed laser beam to introduce thetime-energy form.

Since the 1990s, it has been possible to build lasersthat send out very short pulses, pulses only a fewwavelengths long. Such pulses are useful for effec-tively taking “flash” pictures of the behavior of theelectrons in chemical reactions.

The experiment we will discuss involves the pulseshown in Figure (1a). This is a graph of the intensityof the electric field in the pulse. The wave is calleda femtosecond laser pulse because the main part ofthe pulse is about 20 femtoseconds (fs) long. (Onefemtosecond, fs, is 10– 15 seconds, or one millionthof a nanosecond. In Figure (1a), the main part of thepulse starts at –10 fs and goes up to +10 fs.)

The second graph, Figure (1b), shows the spectrumof radiation in the pulse. The center frequency hasa wavelength of 800 nanometers (nm) which islonger than the visible spectrum wavelengths thatrange from 300 nm for blue, up to 400 nm for red.Thus the pulse is in the longer wavelength infraredpart of the spectrum.

A wave with an 800 nm wavelength λ has a periodT = λ /c of 2.67 femtoseconds. Thus in the rangefrom –10 fs to + 10 fs, there should be about 7 cycles.In Figure (1a) you will count 14 maxima in this 20fs range because you are looking at the intensity ofthe electric field. The intensity is proportional to thesquare of the field, and when you square a sine wave,you get two maxima per cycle.

What may be surprising about the spectrum in Fig-ure (1b) is that the spectrum is not sharp. The rangefrom 750 nm to 850 nm is a third as wide as the rangefrom 400 nm to 700 nm for visible light. The laserswe used in Chapter 25 to study diffraction patternshad a distinct frequency. Send that laser light througha diffraction grating and you would get a sharp linerather than the spread out spectrum we see in Figure(1b).

One might be tempted to ask “What is wrong withthe laser used in Figure (1a)?” “Why doesn’t it givethe pure single wavelength waves we expect fromlasers?”

What is wrong is that the laser pulse is only a fewwavelengths long. In the section after next, we willuse Fourier analysis to demonstrate that such a shortpulse must contain a spectrum of wavelengths. Wewill see that the spread in wavelengths is needed tocancel out the waves outside the pulse. In the mean-time we will see that the spectrum in Figure (1b) isconsistent with the time-energy form of the uncer-tainty principle.

Figure 1aIntensity of the electric field in a 20 femtosecond (fs)laser pulse. (From F. Hajiesmaeilbaigi and A. AzimaCan. J. Phys. 76: p498 (1998).)

Figure 1bSpectrum of the radiation in the laser pulse of Figure(1a). Visible light has a range from 400 nm to 700 nm,thus this spectrum, centered on 800 nm, is in the infrared.

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The Uncertainty Principle: Time and Energy 33-3

We can say that the photon’s energy is uncertain byan amount

ΔE = E+ – E–

uncertainty of theenergy of anyphoton in the pulse

= hf+ – hf–

= h(.400 – .352) × 1015sec– 1

= h × (.048) × 1015sec– 1

(2)

Using the uncertainty principle in the form Δt = h/ΔE we get

Δt = hΔE

= hh×(.048)×1015sec– 1

The factors of h cancel and we are left with

Δt = 20.8 × 10–15sec ≈ 20 femtoseconds (3)

which is the length of the pulse seen in Figure (1a).Thus the uncertainty principle, applied to the spec-trum in Figure (1b) correctly predicts the length ofthe laser pulse.

The fact that Planck’s constant cancelled inEquation (3) suggests that the uncertainty principleis something more than just some weird quantumeffect. It suggests that it is a consequence of a moregeneral behavior of waves. We will demonstratethat this is true by showing that the uncertaintyprinciple is seen as a natural wave phenomena whenwe take the Fourier transform of a wave pulse.

USING THE UNCERTAINTY PRINCIPLEAs a laser pulse goes by us we have very little time,call it Δt , in which to measure the frequencies orenergies of the photons in the pulse. The time-energy form of Heisenberg’s uncertainty principlesays that if only a short time Δt available for makingan energy measurement, there will be an uncertainty

ΔE in the result, where ΔE must be as large as ΔE = h/Δt.

To apply the uncertainty principle to the laser pulsein Figure (1), we note that the pulse goes by in 20 fs.Thus we have only a time Δt = 20 fs in which tomeasure the spectrum of the laser light. But measur-ing the spectrum means measuring the frequenciesor energies of the photons in the pulse. Thus measur-ing the spectrum is an energy measurement whichmust be uncertain by an amount ΔE = h/Δt.

It is easier if we turn the argument around, firstsolving for the uncertainty ΔE in the energy of thephotons in the pulse. Then we will use the uncer-tainty principle written as Δt = h/ΔE to see if wepredict the correct length Δt of the pulse.

In Figure (1b) we see that the laser pulse consists ofwavelengths ranging from as short as 750 nm up to850 nm. The longer wavelength, lower frequencyphotons have a frequency f– given by

f– = c

λ–= 3 × 108m/sec

850 × 10– 9m= .352 × 1015sec– 1

(1a)(The dimensions of frequency, cycles/sec, is is actu-ally 1/sec or sec–1 because cycles are dimension-less.) The shorter wavelength photons have thehigher frequency f+ given by

f+ = c

λ+= 3 × 108m/sec

750 × 10– 9m= .400 × 1015sec– 1

(1b)From Einstein’s photoelectric effect formula

Ephoton = hf we see that photons in the pulse canhave an energy as low as E– = hf– up to an energy ashigh as E+ = hf+ . We do not know what the energyof any given photon in the pulse is. All we know isthat it is somewhere in the range between E– and E+ .

33-4 The Uncertainty Principle: Time and Energy

PULSE FOURIER TRANSFORMWe have recently modified the MacScope II pro-gram so that we can easily study the Fourier trans-form of short wave pulses like the laser pulse inFigure (1a). We will use this section to explain howthe pulse Fourier transform works. Then we willgo on to see how the time-energy form of theuncertainty principle emerges as a consequence.

In Figure (2) we whistled into a microphone at-tached to MacScope, and recorded the wave shown.A whistle produces a fairly good sine wave. In thatfigure we have also selected one cycle of the sine wave.When we press the button labeled Fourier, we get thestandard Fourier analysis window shown in Figure (3).

The way MacScope handles Fourier analysis is toassume that the selected section of the curve repeatsindefinitely. In Figure (3) we selected one completecycle of the sine wave. When that cycle is repeatedindefinitely, we end up with a complete pure sinewave containing only the one frequency componentwhich we see in the Fourier analysis window.

If, instead of pressing the Fourier button, we pressedthe Pulse button of Figure (4a), we first get thewindow shown in Figure (4b) asking whether wewant our pulse centered, left edge, or Gaussian. Forthis demonstration we chose Zero at Center and gotthe curve shown in Figure (5).

Now you can see what the Pulse Fourier Transformdoes. Instead of assuming that the selected sectionof the pulse repeats forever, the program zeros outall but the selected section of curve as shown inFigure (5). When we selected Zero at Center, theprogram drew the zero line at the center of ourselected section of the curve.

Figure 2Selecting one cycle of a sine wave.

Figure 5Selecting the Pulse Fourier Transform, whichcreates a pulse by zeroing all but the selectedsection of the curve.

Figure 3Fourier analysis of a sine wave. The assumptionis that this cycle repeats indefinitely.

Figure 4aSelecting the Pulse Fourier Transform.

Figure 4bChoosing the Zero line.

The Uncertainty Principle: Time and Energy 33-5

Fourier Analysis of a PulseIn Figure (6), we show the complete MacScopeCurve A window. At the top we see our one cyclepulse with the rest of the curve zeroed out. Below inthe Fourier analysis window we see a whole bunchof harmonics. Let us see where they came from.

In Figure (7a) we have selected the center harmonic,and in the upper window see that MacScope hasdrawn a pathetic little sine wave that has the samewavelength as our selected cycle. This is telling usthat we are not going to reconstruct our pulse froma single harmonic.

In Figure (7b) we selected the five biggest harmon-ics. In the upper window we see the result of addingthese five sine waves together. They are beginningto build a pulse centered on our selected pulse.

In Figures (7c) and (7d) we see that as we select moreharmonics, the closer we come to reconstructing ourpulse. The lesson here is that to build a sharp pulseout of harmonic sine waves, takes a lot of sine waves.All these waves must add up at the pulse, andcompletely cancel each other elsewhere.

Figures 7Fourier analysis of apulse. Here we seehow a short pulse isconstructed from longsinusoidal waves.

In (a) we selected thelargest harmonic andall it represents is asmall sine wave.

When we add togetherthe five biggestharmonics in (b), apulse begins to form.

When we add up the 32biggest harmonics, weget a closerepresentation of thepulse in (d). We need alot of harmonics tocancel the waveoutside the pulse.

Figure 6Harmonics in a short pulse.

(a)

(d)

(c)

(b)

33-6 The Uncertainty Principle: Time and Energy

Intensity vs AmplitudeIn our use of Fourier analysis so far, we have studiedthe amplitudes of the harmonics contained in asignal. But when we study the spectrum of light ina light wave, we measure the intensity of the electricfield, which is the square of the field. If we want tocompare the spectrum in the Fourier analysis of ashort pulse, with the experimental spectrum of thelaser pulse in Figure (1b), we should have MacScopeplot the intensities rather than the amplitudes of theharmonics. This is easily done by looking at thebuttons below the Fourier Analysis Plot, and select-ing the one labeled Intensity, rather than the onelabeled Amplitude.

Spectrum of a PulseNow we are ready to compare Figure (7d) where wehave selected most of the harmonics, with Figure (1b)showing the spectrum of wavelengths in our sixcycle laser pulse. In both cases we see a similarspectrum of harmonics. You should now see that thereason why the short laser pulse contained a spec-trum of wavelengths ranging from 750 nm up to 850nm. It takes all these different wavelengths to add upto a short pulse.

CHANGING ΔtAt the beginning of the chapter, we used the uncer-tainty principle in the form Δt = h/ΔE to calculatethe length Δt of the laser pulse. We found thatwith ΔE = h(f+ – f–) we got a time of Δt equal to20 femtoseconds which accurately describes thelength of the laser pulse.

What we want to do now is write the uncertaintyprinciple equation in the form

ΔE = hΔt (4)

and look at the consequences of changing the pulselength Δt.

The clear prediction of Equation (4) is that everytime we double the length Δt of the pulse, we willcut the energy uncertainty ΔE in half. With MacScope,we can show that this is precisely what happens.

We can change the length of Δt by going back to ourrecording of a whistle, selecting more than one cycleof the sine wave, and doing a pulse Fourier trans-form. In Figure (8) we selected 2 cycles and see thatwe get a narrower spectrum.

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Figures 1 & 7dComparison of the spectra of short pulses.

Figure 8With 2 cycles the spectrum is narrower.

The Uncertainty Principle: Time and Energy 33-7

To get an overview of how the width of the spectrumof harmonics changes as we change Δt , in Figure (9)we kept doubling Δt from 1 cycle to 2 cycles to 4cycles to 8 cycles. We see that the frequency spreadin the main harmonics drops from 32 harmonics, to16 harmonics, to 8 harmonics, to 4 harmonics.Clearly doubling the length of the pulse cuts thespread in harmonics in half.

The spread in harmonics tells us the range of fre-quencies (f+ – f–) that are important in the pulse.Multiply this range of frequencies by Planck’s con-stant h, and you get the formula for the range ΔE ofphoton energies in the pulse.

ΔE = h(f+ – f–)

But this range of photon energies is the uncertaintyin our knowledge of the energy of any one photon inthe pulse. Thus we see that the uncertainty in thephoton energy is cut in half every time we double thelength of the pulse.

(a)

(d)

(c)

(b)

Figure 9When we doublethe length Δt ofthe pulse, we cutthe spread ΔE ofthe harmonics inhalf. The productΔEΔt remainsconstant.

33-8 The Uncertainty Principle: Time and Energy

PROBABILITY INTERPRETATIONBy now we should be quite comfortable with seeingthe frequency or wavelengthspectrum of the laser pulse inFigure (1b). The graph tells usthat most of the photons have awavelength around λ = 800 nm,while some photons have a wave-length as long as λ+ = 850 nm,or as short as λ– = 750 nm. If we change the hori-zontal scale from photon wavelengths to photonenergies hc/λ , then we get a plot of the distributionof the energies of the photons in the pulse.

We can note that when we did a Fourier analysis ofa pulse, we got a spectrum of the harmonicfrequencies fn . To get the spectrum of photon ener-gies h fn , you just multiply by Planck’s constant h.

Now we are going to ask a question that, at first, mayseem controversial. Suppose the laser pulse ofFigure (1a) contained only one photon. Then howshould we interpret the spectrum in Figure (1b)?What does it mean to have a spectrum of energies fora single photon?

The answer to this question lies in the probabilityinterpretation of the photon wave. If there were onlyone photon in the Figure (1) laser pulse, then thespectrum tells us the distribution of probabilitiesthat the photon has the corresponding frequencyor energy. What was an energy spectrum when wehad many photons, becomes a probability spectrumwhen we have one or a few photons.

Even if there are many photons in the pulse, we canstill interpret the spectrum as a probability spectrumfor each photon. Then, for example, if each photonin the laser pulse is most likely to have a wavelengthof around 800 nm, then when we measure photonwavelengths, we should find the most photons in the800 nm wavelength range.

With the probability interpretation, any given pho-ton has some probability of being at the high energypart of the spectrum, and some probability of beingat the low energy end. Thus the width ΔE of thespectrum is truly the uncertainty in the photon’senergy.

Measuring Short TimesWe have said that pulsed lasers can produce pulsesas short as 2 femtoseconds. How do we know that?Suppose we gave you the job of measuring thelength of the laser pulse, and the best oscilloscopeyou had could measure times no shorter than ananosecond. This is a million times too slow to seea femtosecond pulse. What do you do?

We answered that question at the beginning of thischapter. We used the uncertainty principle

Δt = h/ΔE to correctly calculate the 20 femtosecondlength of the Figure (1) pulse. Essentially we areusing a diffraction grating as a clock!

Exercise 1

An electron is in an excited state of the hydrogenatom, either the second energy level at –3.40 eV, orthe third energy level at –1.51 eV. You want to do anexperiment to decide which of these two states theelectron is in. What is the least amount of time youmust take to make this measurement?

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The Uncertainty Principle: Time and Energy 33-9

Short Lived Elementary ParticlesTo drive home the fundamental nature of the rela-tionship ΔE = h/Δt , we will discuss an unstableelementary particle that lives for such a short time

Δt that its rest mass energy m0c2 is demonstrablyuncertain. We will then use the equation Δt = h/ΔEto measure the very short lifetime of the particle.

We usually think of the rest energy of a particle ashaving a definite value. For example the rest energyof a proton is 938.2723 × 106eV . The proton itselfis a composite particle made of 3 quarks, and thenumber 938.2723 MeV represents the total energyof the quarks in the allowed wave pattern thatrepresents a proton. This rest energy has a verydefinite value because the proton is a stable particlewith plenty of time to settle into a precise wavepattern.

A rather different particle is the so-called“Lambda (1520)” or “ Λ (1520)”, which is anothercombination of 3 quarks, but very short lived. Thename comes partly from the fact that the particle’srest mass energy is about 1520 million electron volts(MeV). As indicated in Figure (10), a Λ (1520) canbe created as a result of the collision between a K–

meson and a proton.

We are viewing the collision in a special coordinatesystem, where the total momentum of the incomingparticles is zero. In this coordinate system, theresulting Λ (1520) will be at rest. By conservation ofenergy, the total energy of the incoming particlesshould equal the rest mass energy of the Λ (1520).Thus if we collide K– particles with protons, weexpect to create a Λ (1520) particle only if theincoming particles have the right total energy.

Figure (11) shows the results of some collisionexperiments, where a K– meson and a proton col-lided to produce a Λ and two π mesons. The prob-ability of such a result peaked when the energy of theincoming particles was 1,520 MeV. This peak oc-curred because the incoming K– meson and protoncreated a Λ (1520) particle, which then decayed intoan ordinary Λ and two π mesons, as shown inFigure (10). The Λ (1520) was not observed di-rectly, because its lifetime is too short.

Figure (11) shows that the energy of the incomingparticles does not have to be exactly 1520 MeV inorder to create a Λ (1520). The peak is in the rangefrom about 1510 to 1530 Mev, which implies thatthe rest mass energy of the Λ (1520) is 1520 MeVplus or minus about 10 MeV. From one experiment toanother, the rest mass energy can vary by about 20MeV. (The experimentalists quoted a variation of16 MeV.)

K-

π–

π+

pp +p

1520Λ

Λ Figure 11

The probability that a K – meson and a proton collideto produce aΛΛ particle and two ππ mesons, peak at anenergy of 1520 MeV. The peak results from the factthat a ΛΛ (1520) particle was created and quicklydecayed into the ΛΛ and two ππ mesons. The probabil-ity peaks at 1520 MeV, but can be seen to spread outover a range of about 16MeV. The small circles areexperimental values, the vertical lines represent thepossible error in the value. (Data from M.B. Watson etal., Phys. Rev. 131(1963).)

a) A K – meson and a proton are about to collide. We arelooking at the collision in a coordinate system where thetotal momentum is zero (the so-called “center of mass”system).

b) In the collision a ΛΛ (1520) particle is created.It is at rest in this center of mass system.

c) The ΛΛ (1520) then quicklydecays into a lower energy ΛΛ particle and two π mesons.

Figure 10A ΛΛ (1520) particle can be created if the total energy(in the center of mass system) of the incoming particlesequals the rest mass energy of the ΛΛ (1520).

1460center of mass energy, MeV

1480 1500 1520 1540 1560

K + p Λ(1520) Λ+π +π++− −

33-10 The Uncertainty Principle: Time and Energy

Figure 12Consider a process where a 2 eV photon suddenlycreates a positron-electron pair. A short time later thepair annihilates, leaving a 2 eV photon. In the longterm, energy is conserved.

Why isn’t the peak sharp? Why does the rest massenergy of the Λ (1520) particle vary by as much as16 to 20 MeV from one experiment to another? Theanswer lies in the fact that the lifetime of the ΛΛ (1520)is so short, that the particle does not have enoughtime to establish a definite rest mass energy. The 16MeV variation is the uncertainty ΔE in the particle’srest mass energy that results from the fact that theparticle’s lifetime is limited.

The uncertainty principle relates the uncertainty inenergy ΔE to the time Δt available to establish thatenergy. To establish the rest mass energy, the time

Δt available is the particle’s lifetime. Thus we canuse the uncertainty principle to estimate the lifetimeof the Λ (1520) particle. With Δt = h/ΔE we get

Δt = h

ΔE=

6.63 × 10–34joule second

16×10 6eV × 1.6×10–19 jouleeV

Δt = 2.6 × 10– 22 seconds (5)

The lifetime of the Λ (1520) particle is of the orderof 10– 22 seconds! This is only about 10 timeslonger than it takes light to cross a proton! Only byusing the uncertainty principle could we possiblymeasure such short times.

THE UNCERTAINTY PRINCIPLEAND ENERGY CONSERVATIONThe fact that for short times the energy of a particleis uncertain, raises an interesting question aboutbasic physical laws like the law of conservation ofenergy. If a particle’s energy is uncertain, how dowe know that energy is conserved in some processinvolving that particle? The answer is — we don’t.

One way to explain the situation is to say that naturewill cheat if it can get away with it. Energy does nothave to be conserved if we cannot do an experimentto demonstrate a lack of conservation of energy.

Consider the process shown in Figure (12). It showsa red, 2 eV photon traveling along in space. Sud-denly the photon creates a positron-electron pair.The rest mass energy of both the positron and theelectron are .51 MeV. Thus we have a 2 eV photoncreating a pair of particles whose total energy is

1.02 × 106eV , a huge violation of the law of conser-vation of energy. A short time later the electron andpositron come back together, annihilate, leavingbehind a 2 eV photon. This is an equally hugeviolation of the conservation of energy.

But have we really violated the conservation ofenergy? During its lifetime, the positron-electronpair is a composite object whose total energy isuncertain. If the pair lived a long time, its totalenergy would be close to the expected energy of

1.02 × 106eV . But suppose the pair were in exist-ence only for a very short time Δt,a time so short thatthe uncertainty in the energy could be as large as

1.02 × 106eV . Then there is some probability thatthe energy of the pair might be only 2 eV and theprocess shown in Figure (12) could happen.

electron

positron2 eV photon

2 eV photon

The Uncertainty Principle: Time and Energy 33-11

Figure 13Quantum fluctuation. The uncertainty principle allowssuch an object to suddenly appear, and then disappear.

The length of time Δt that the pair could exist andhave an energy uncertain by 1.02 MeV is

Δt = h

ΔE=

6.63 × 10–34joule sec

1.02×10 6eV×1.6×10–19 jouleeV

Δt = 4 × 10– 21sec (6)

Another way to view the situation is as follows.Suppose the pair in Figure (12) lasted only 4 × 10– 21

seconds or less. Even if the pair had an energy of 1.02 × 106eV , the lifetime is so short that any mea-

surement of the energy of the pair would be uncer-tain by at least 1.02 × 106eV , and the experimentcould not detect the violation of the law of conserva-tion of energy. In this point of view, if we cannotperform an experiment to detect a violation of theconservation law, then the process should havesome probability of occurring.

Does a process like that shown in Figure (12) actu-ally occur? If so, is there any way that we can knowthat it does? The answer is yes, to both questions. Itis possible to make extremely accurate studies of theenergy levels of the electron in hydrogen, and tomake equally accurate predictions of the energyusing the theory of quantum electrodynamics.

We can view the binding of the electron to the protonin hydrogen as resulting from the continual ex-change of photons between the electron and proton.During this continual exchange, there is some prob-ability that the photon creates a positron-electronpair that quickly annihilates as shown in Figure (12).In order to predict the correct values of the hydrogenenergy levels, the process shown in Figure (12) hasto be included. Thus we have direct experimentalevidence that for a short time the particle-antiparticlepair existed.

QUANTUM FLUCTUATIONSAND EMPTY SPACEWe began the text with a discussion of the principleof relativity—that you could not detect your ownmotion relative to empty space. The concept ofempty space seemed rather obvious—space withnothing in it. But the idea of empty space is not soobvious after all.

With the discovery of the cosmic background radia-tion, we find that all the space in this universe isfilled with a sea of photons left over from the bigbang. We can accurately measure our motion rela-tive to this sea of photons. The earth is movingrelative to this sea at a velocity of 600 kilometers persecond toward the Virgo cluster of galaxies. Whilethis measurement does not violate the principle ofrelativity, it is in some sense a measurement of ourmotion relative to the universe as a whole.

Empty space itself may not be empty. Consider aprocess like that shown in Figure (13) where aphoton, an electron, and a positron are all created atsome point in space. A short while later the threeparticles come back together with the positron andelectron annihilating and the photon being absorbed.

One’s first reaction might be that such a process isridiculous. How could these three particles justappear and then disappear? To do this we wouldhave to violate both the laws of conservation ofenergy and momentum.

But, of course, the uncertainty principle allows us todo that. We can, in fact, use the uncertainty principleto estimate how long such an object could last. Thearguments would be similar to the ones we used inthe analysis of the process shown in Figure (12).

electron

positron

photon

33-12 The Uncertainty Principle: Time and Energy

In the theory of quantum electrodynamics, a com-pletely isolated process like that shown in Figure(13) does not affect the energy levels of the hydro-gen atom and should be undetectable in electricalmeasurements. But such a process might affectgravity. A gravitational wave or a graviton mightinteract with the energy of such an object. Somecalculations have suggested that such interactionscould show up in Einstein’s theory of gravity (as acontribution to his famous cosmological constant).

An object like that shown in Figure (13) is anexample of what one calls a quantum fluctuation.Here we have something that appears and disap-pears in so-called empty space. If such objects cankeep appearing and disappearing, then we have torevise our understanding of what we mean by empty.

The uncertainty principle allows us to tell the differ-ence between a quantum fluctuation and a realparticle. A quantum fluctuation like that in Figure(13) violates conservation of energy, and thereforecannot last very long. A real particle can last a longtime because energy conservation is not violated.

However, there is not necessarily that much differ-ence between a real object and a quantum fluctua-tion. To see why, let us take a closer look at the πmeson. The π+ is a particle with a rest mass energyof 140 MeV, that consists of a quark-antiquark pair.The quark in that pair is the so-called up quark thathas a rest mass of roughly 400 MeV. The other is theantidown quark that has a rest mass of about 700MeV. (Since we can’t get at isolated quarks, thequark rest masses are estimates, but should not betoo far off). Thus the two quarks making up the πmeson have a total rest mass of about 1100 MeV.How could they combine to produce a particlewhose rest mass is only 140 MeV?

The answer lies in the potential energy of thegluon force that holds the quarks together. As wehave seen many times, the potential energy of anattractive force is negative. In this case the potentialenergy of the gluon force is almost as big in magni-tude as the rest mass of the quarks, reducing the totalenergy from 1100 MeV to 140 MeV.

Suppose we had an object whose negative potentialenergy was as large as the positive rest mass energy.Imagine, for example, that the object consisted of acollection of point sized elementary particles soclose together that their negative gravitational po-tential energy was the same magnitude as the posi-tive rest mass and kinetic energy. Suppose such acollection of particles were created in a quantumfluctuation. How long could the fluctuation last?

Since such an object has no total energy, the viola-tion ΔE of energy conservation is zero, and there-fore the lifetime Δt = h/ΔE could be forever.

Suppose the laws of physics required that such afluctuation rapidly expand, greatly increasing boththe positive rest mass and kinetic energy, whilemaintaining the corresponding amount of negativegravitational potential energy. As long as ΔE re-mained zero, the expanding fluctuation could keepon going. Perhaps such a fluctuation occurred 13.7billion years ago and we live in it now.

Figure 13 (repeated)Quantum fluctuation. The uncertainty principle allowssuch an object to suddenly appear, and then disappear.

electron

positron

photon

The Uncertainty Principle: Time and Energy 33-13

CHAPTER 33 REVIEWThis is one of the most important chapters in the text,because it gives an entirely new perspective on thebasic nature of physics. In a sense, many of theprevious chapters have been included in order toprovide the background needed for this chapter.

We began the chapter with the experiment of Figure(1) which showed that a short laser pulse, about 12cycles long, contained a broad spectrum of wave-lengths. The pulse was in the infrared, and thespread of wavelengths, from 750 to 850 nanometerswas a third as wide as the spread of visible wave-lengths which range from 400 to 700 nanometers.

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Wavelength (nm)

One usually thinks of a laser as having a pure color,like the red lasers that are so common. But thespread in wavelengths of the laser pulse was due notto poor equipment but to a fundamental property ofwaves. When we used MacScope to study the Fou-rier transform of a short pulse, we saw that any shortpulse had to be made up from a spread of wave-lengths or frequencies. The reason for this is that asine wave is an infinitely long wave. If you are goingto construct a short pulse out of sine waves, you needa distribution of wavelengths to get complete can-cellation beyond the ends of the pulse.

Figure (9) shows the main feature of the spread infrequencies required to create a short pulse. As wego down from (9a) to (9d), we keep doubling thelength of the pulse and see that the spread in fre-quencies is cut in half. In other words, if we call Δtthe length of the pulse, and Δ f the width of thespread in frequencies, the product ΔtΔ f does notchange as we keep doubling Δt .

(d)(c)

(b)

Fig

ure

9

Let us look at the product ΔtΔ f for the laser pulsein Figure (1). Since the pulse is 20 femtosecondslong, we have for the length of the pulse

Δt = 20 × 10– 15 seconds

The wavelengths range from λ–=750×10– 9metersto λ+=850×10– 9meters . The related frequenciesare

f+ = cλ–

= 3×108

750×10– 9 = 4.0×1014sec–1

f– = cλ+

= 3×108

850×10– 9 = 3.5×1014sec–1

and the spread in frequencies is

Δ f = f + – f– = 0.5×1014sec–1

Thus the product ΔtΔ f has the surprisingly simpleresult

ΔtΔ f = 20×10–15sec × 0.5×1014 1sec = 1.0

As a final step, note that the laser pulse is made upof photons that obey Einstein’s formula E = hf. Thus

hΔ f is the energy spread ΔE of the photons in thelaser pulse. With ΔE = hΔ f or Δ f = ΔE/h , we get

Δ fΔt = ΔEhΔt = 1

ΔEΔt = h

for the photons in the pulse.

Figure 1

(a)

33-14 The Uncertainty Principle: Time and Energy

Nature’s answer to this question appears to be—energy only has to be conserved when you canmeasure it.

The most convincing evidence for this point of viewcomes from a detailed calculation of the energy ofthe electron in the hydrogen atom. In the theory ofQuantum Electrodynamics, the electric force hold-ing the electron in the atom is caused by the ex-change of photons between the electron and theproton. Since the electron binding energy is only13.6 eV, the photons on the average must have asmall energy, not much larger than 13 eV.

But, the calculation shows that there is some finiteprobability for one of the photons to create anelectron-positron pair during an exchange. Sincethe rest mass energy of a positron-electron pair isone million electron volts, we have a situation wherea photon whose energy should be only a few electronvolts creates a million electron volt pair.

How can this happen? The answer is that the pairlasts for such a short time Δt that the violation ofenergy conservation ΔE cannot be directlydetected due to the uncertainty principle.

The uncertainty principle can also be used as a clockto measure very short times. We found that the restmass of the so called (Λ1520) particle varied by asmuch as 16 million electron volts when the particledecayed. The reason for this variation is that theparticle did not live long enough to accuratelydetermine its own rest mass energy. Calling thisvariation the uncertainty ΔE of the particle’s restmass energy, and Δt its lifetime, we got

Δt = h

ΔE=

6.63×1034 joule sec106eV × 1.6×10– 19 joule/eV

= 4×10– 21 seconds

There is no other way to measure such short times.

Probability InterpretationLet us take another look atthe short pulse in Figure(9a), which is made up of afairly wide distribution offrequencies. If this repre-sented a real laser pulsewith many photons, thebottom graph would rep-resent the distribution offrequencies or energies of the photons in the pulse.

Now comes the interesting question. Suppose thereis only one photon in the pulse. What is the meaningof the distribution of frequencies or energies meanfor a single photon? The answer is that it is thedistribution of probabilities that the photon hasthe corresponding frequency or energy. There issome probability that the photon has a frequency orenergy as low as the first few harmonics shown. Andsome probability that its energy is as high as thatrepresented by the 32nd harmonic. The photon’senergy can be anywhere in this range of frequencyor energy. Thus our calculation of ΔE representsthe uncertainty in the energy of the photon.

If you read Figure (9) from (9d) to (9a), you see thatas you make the laser pulse shorter, the uncertainty

ΔE of the energy of the photon in the pulse becomesgreater. The shorter the time Δt we have availableto observe the photon, the more uncertain our mea-surement of the photon’s energy becomes.

Energy ConservationWe introduced the concept of energy by saying thatits main property was that it was conserved. Energyconservation became one of the most important lawsin the text. Now we see that if you try to measure theenergy of an object, be it the non relativistic particlediscussed at the end of the last chapter, or a fullyrelativistic particle like a photon in a laser pulse,any measurement of the particle’s energy will leadto uncertain results. How do we know that energy isconserved if we cannot measure it precisely?

(a)

The Uncertainty Principle: Time and Energy 33-15

Quantum FluctuationsThe uncertainty principle affects our view of whatwe used to call empty space. To say that I have aregion of empty space, is to say that that region haszero energy exactly. But if I do an experiment tomeasure the energy in that region of space, myanswer must be uncertain by an amount ΔE = h/Δt .Thus for short times, something is probably there.

A candidate is the example of a quantum fluctuationshown in Figure (13). To have an electron, apositron, and a photon, suddenly appear in space,requires a violation ΔE of the conservation ofenergy of at least one million eV. But if theseparticles all annihilate each other in a time

Δt = h/ΔE , the violation of conservation of energycannot be observed and there should be some prob-ability of such a process occuring.

The Early UniverseRecent studies of the early universe suggest thatquantum flucuations are responsible for the diver-sity of structure we now see in the universe.

In Chapter 26 on photons, we discussed the threedegree radiation that was released when the uni-verse became transparent at the young age of 1/3 ofa million years. (The universe is now 13.7 billionyears old.) The main feature of this radiation is thatit is very uniform. It took specially designed satel-lites to detect any variation from point to point in thesky.

The latest satellite (NASA’s Wilkinson MicrowaveAnisotropy Probe [WMAP]) detected a pattern ofsmall variations that appear to have resulted fromquantum fluctuations in the very, very early uni-verse. The current theory is that once the universebecame transparent, these small variations allowedthe gravitational collapse of clouds of gas to formthe stars, galaxies, and the clusters of galaxies wesee today. If the universe had been perfectly uni-form, with no lumps created by quantum fluctua-tions, the universe would not have the structure weobserve.

Figure 13 (repeated again)Quantum fluctuation. The uncertainty principle allowssuch an object to suddenly appear, and then disappear.

electron

positron

photon

CHAPTER EXERCISEExercise 1 0n page 8An electron is in an excited state of the hydrogenatom, either the second energy level at –3.40 eV, orthe third energy level at –1.51 eV. You want to do anexperiment to decide which of these two states theelectron is in. What is the least amount of time youmust take to make this measurement?

33-16 The Uncertainty Principle: Time and Energy