de Broglie standing waves did not radiate like other accelerated charged particles

http://hep.physics.indiana.edu/~hgevans/c105-iu/probs/ps04/solns.html

COLL C105: Homework 4 - Quantum Mechanics II

Note 1: it is difficult to enter pictures into OnCourse, so you are not required to post sketches in your OnCourse answers. However, you should make diagrams for yourself as part of the process of thinking about the problems. Note 2: several problems in this homework involve calculations. As for all such problems in this class, the math & numbers are not the important part. If you can do the math - great! - you'll get a better understanding of the concepts. But if you have problems with the technical details, don't worry, calculations will not appear on the exams. You should ask yourself the following questions (and include them in your answer): a) what concepts are involved in these calculations? b) how, generally, are they applied to get the answer?, c) why are these questions important?

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Problem 1

1. Which of the problems with the Bohr model of the atom did de Broglie address with his matter waves?
   De Broglie's matter waves were invented to motivate the requirement that the electrons in the Bohr atom inhabit only specific orbits. It did not directly address the question of why the electrons, which according to Bohr moved in circular orbits, did not radiate like other accelerated charged particles. However, by treating the electrons as waves the particle-like problem could be, to some extent, finessed.
2. Give a brief description of why treating the electon as a wave-like object resolved this problem.
   Waves involve a discrete length scale - the wavelength. In particular, standing waves can only exist over distances that are an integral multiple of their wavelength. Thus, by postulating that the electrons in Bohr's atoms were standing waves, de Broglie was able to give them a set of specific lengths, which he related to the circumferences of their allowed orbits. Electrons orbiting at different radii than these allowed states could not be standing waves, and were therefore forbidden in de Broglie's model.
3. How was de Broglie ble to set the wavelength of the electron within the planetary view of the atom?
   As mentioned above, de Broglie required that an integer number of wavelengths fit around the circumference of the electron's orbit. Using Einstein's relationship between the energy and frequency of the photon (which de Broglie assumed to hold for his wavelike electrons), he was then able to get Bohr's answer for angular momentum (and energy) quantization in the Hydrogen atom. For the mathematically inclined we have:
   n λ = 2π r
   De Broglie then used Einstein's relationship between the energy and frequency of a photon (E = hf) and the wave relation, c = λf to get a relationship between energy and wavelength:
   E = hc / λ
   Then using Maxwell's relation between energy and momentum of a wave (E = p/c) he got:
   λ = h/p
   Finally, putting it all together:
   n λ = 2π r ---> mvr = nh/2π
   which is just Bohr's quantization of angular momentum.

Problem 2

Electron microscopes take advantage of the small de Broglie wavelength of high energy electrons to allow very small objects to be resolved. This is done by directing a beam of these electrons toward the object to be probed. Depending on the type of electron microscope, one observes (using a detector) either electrons that are scattered back toward the source or those that pass through the object. The amount of scattering the electron wave experiences in its interaction with the object allows us to "see" the object, just as light scattered off of large objects allows us to see them. As a general guideline, waves tend to be scattered by objects with sizes greater than the wavelength of the wave. So, electron microscopes can resolve features down to about the wavelength of the electrons being used. You can get a feel for why this is the case by considering the effect of a small fixed object, like a reed, sitting in long wavelength water waves. The end result is that the reed does not perturb the pattern of crests and troughs in the water wave very much at all; it does not scatter the wave.

1. If I would like my electron microscope to be able to resolve features of 0.2 nm (2x10^-10 m) in size, what is the minimum kinetic energy of the electrons that I should use?
   The electron wavelength that we want here is the feature size - λ = 0.2 nm. We can relate this to the momentum using the de Broglie relationship:
   p = h/λ = (6.63x10^-34 J.s) / (2x10^-10 m) = 3.31x10^-24 kg.m/s
   The momentum can then be related to the kinetic energy (non-relativistically) using:
   K = 1/2 mv² = (1/2m) (mv)² = p²/2m
   = (3.31x10^-24 kg.m/s)² / 2(9.11x10^-31 kg ) = 6.0x10^-18 J = 38 eV
2. A wavelength of 0.2 nm for electromagnetic radiation corresponds to x-rays. Why don't we use 0.2 nm photons in the above microscope instead of 0.2 nm electrons?
   The problem with x-rays for this purpose is just their strength for medical uses - they penetrate normal matter. This penetrating property is why they can be used to see inside the body, however, it means that the x-rays don't interact very strongly with matter. If they did - they would be scattered by the skin. Electrons, on the other hand, are charged particles and therefore experience strong electro-magnetic interactions with matter. This would be a drawback if you wanted to use them to take a picture of a broken bone. But it is an advantage if you are try to probe a very small sample where you want the chances of the electron to interact with (scatter from) the sample to be high.

Problem 3

If Planck's constant were significantly larger than what it actually is, the world would look very different than it does.

1. What is a sensible criterion for an object to be inherently "quantum" in nature? There is, of course, no single correct answer here, however, your criterion should address the dual particle-wave nature of matter.
   If an object's de Broglie wavelength (defined using its mass and velocity) is similar in magnitude to its spatial extent then the object clearly cannot be treated as a classical particle. Note that the spatial extent of such an object is not terribly well defined, but can be understood to be the area over which its probability density is large if all else fails. Of course, other definitions would work as well. Perhaps a more general way of stating the above definition would be to say that a problem is "quantum" when the de Broglie wavelength of the participating particles is approximately the same order as the length scale involved in the problem. In the following we will use λ = size as the "quantum" test.
2. What value would Planck's constant have to have for a 3 g pellet, with a diameter of 1 mm (in our world), travelling at 500 m/s to become "quantum"?
   If we want λ = D then we must have:
   λ = h_new / p
   ⇒ h_new = Dp = (0.001 m) [(0.003 kg)(500 m/s)] = 1.5x10^-3 J.s = 9.4x10¹⁵ eV.s
   This is more than 31 orders of magnitude larger than the real value of Planck's constant.
3. For the value of Planck's you found above, what would be the time over which virtual electron-positron pairs could be produced from the vacuum in accordance with Heisenberg's uncertainty principle?
   Heisenberg's uncertainty relation for energy-time is:
   ΔE Δt >= h/2π
   For this problem ΔE corresponds to the energy required to produce and electron-positron pair at rest (2 m_e c² - that is twice the rest-energy of the electron - 511 keV). Heisenberg tell us that if the time over which this production happens is smaller than the limit given by his uncertainty relation, then we fundamentally can't observe that energy imbalance. Energy conservation has been cheated (energy spontaneously appeared in the form of the electron and positron) but no one can notice - so it doesn't matter. The maximum time that this swindle can go on (in the new universe) is:
   Δt_max = [h_new / 2π] / ΔE = [(9.4x10¹⁵ eV.s) / 2pi] / [2 x (511x10³ eV)] = 1.5x10⁹ s = 46 years

Problem 4

1. What was Heisenberg's view of the reality of Bohr's electron orbits? How did his formulation of quantum mechanics address this view?
   Heisenberg considered Bohr's orbits to be unmeasurable and therefore not a valid way of describing nature. He concentrated, in his theory, on predicting things that could be measured - like frequency of atomic emission spectra. Making assumptions about the "true" nature of the atom, as Bohr had by modeling them after the solar system, seemed to him to be not justified.
2. Why did physicists prefer Schrödinger's formulation of quantum mechanics to Heisenberg's even though they contained fundamental points that were not understood.
   Schrödinger's wave mechanics was more widely accepted than Heisenberg's matrix mechanics because it was far easier to do calculations using the wave equation than it was to do them using infinite dimensional matrices. This, by the way, is far from being a point of philosophy or sociology. The point of physics is to predict things about the physical world. If your theory doesn't allow you to make predictions - then it's not a very satisfying theory.
3. What was a weakness shared by Heisenberg's and Shrödinger's quantum theories? This was the point that Dirac was trying to address.
   Neither Heisenberg's matrix mechanics nor Schrödinger's wave mechanics obeyed the principle of relativistic invariance. They weren't invariant under Lorentz transformations. Neither of the formulations could be used for problems involving high speeds.

Problem 5

1. What is complementarity and how does it relate to wave-particle duality and Heisenberg's uncertainty principle?
   A pair of observable quantities are complementarity if there are limits on our ability to simultaneously determine them. Wave-particle duality is an example of complementarity in that wave-like and particle-like characteristics of an object cannot both be measured at the same time. This duality lies at the heart of all other examples of complementary variables. The Heisenberg uncertainty relations are quantitative estimates of the effect of complementarity on specific pairs of observables: momentum-position, energy-time, etc.
2. Give a qualitative example of how Heisenberg's uncertainty principle for position and momentum follows naturally from wave-particle duality.
   The most straightforward seems, to me, to take the wave-like nature of all matter as its starting point. Consider an object whose position we know quite well. This means that the wavefunction of the object is non-zero only over a fairly small area of space. According to Schrodinger, such a wavefunction can be constructed from the sum of many waves, all with different wavelengths. Since de Broglie showed us that an object's wavelength is related to its momentum, if we don't know exactly what the wavelength is - then we don't know very well what the momentum is. Conversly, if we know an object's momentum with extremely high precision, then we must know its wavelength quite well. This means that our wavefunction will be a "normal" wave that extends over a large area of space with a fixed wavelength. Since the wavefuction is non-zero over a large region, we can't localize where the object is with an accuracy.
3. I have mentioned several times that the absolute zero of temperature is the point at which all molecular motion ceases. Why is this an incorrect statement given what you now know about the uncertainty principle?
   For any object that we can see, we can define a region in which the molecules making up the object must exist. At the coarsest level this is just the physical size of the object that we observe. (Of course, we could probably do better than this if we tried by thinking a bit more about the structure of the object - but the coarse level is enough for the argument here.) Heisenberg says that the uncertainty in the momenta of the molecules in the object is then related to our uncertainty in their positions by Δp ≥ h / [2π Δx]. So, since Δx is finite, so will be Δp. This is in contradiction with the statement that the molecules in an object at absolute zero are at rest. Since, in that case, we would know the momenta of those molecules with zero uncertainty. Of course, the deviation of the molecules' speeds from zero due to the uncertainty relation is tiny and can safely be neglected in almost all cases.
4. The spacing between copper atoms in solid copper is approximately 2.6x10^-10 m. If we consider the position of the atoms to be unknown to within that spacing, how well can we know the speed of the atoms. Copper atoms have a mass of approximately 1.06x10^-25 kg.
   Δx Δp >= h/2π ⇒ Δp >= h / [2π Δx] = (6.63x10^-34 J.s)) / [(2π) (2.6x10^-10 m)] = 4.06x10^-25 kg.m/s Δp = m Δv ⇒ Δv = Δp / m = (4.06x10^-25 kg.m/s) / (1.06x10^-25 kg) = 3.8 m/s

Problem 6

1. What did the Stern-Gerlach experiment measure and what caused this effect?
   The Stern-Gerlach experiment (see the diagram to the right for an illustration of the setup) was designed to measure the deflection of a beam of silver atoms by a non-uniform magnetic field. This deflection was measured by observing where the silver atoms landed on a screen a small distance away from the region of magnetic field. The deflection was caused by the field exerting a force on the atoms whose magnitude depended on the direction of the angular momentum of the electrons in the atom with respect to the direction of the magnetic field. If the atoms' angular momentum could have any direction (as predicted by classical physics), then all deflections would be possible. If, on the other hand, quantum mechanics was correct, then only certain orientations of the angular momentum would be possible meaning that only certain values of the deflection would be observed.
2. How did the results of the experiment differ from the classical picture of angular momentum?
   The silver beam was observed to produce two distinct bands on the screen. This meant that only two orientations of the silver angular momentum were present in contrast to the predictions of classical physics described above.
3. What were the results of the experiment interpreted as evidence for?
   These results were interpreted as evidence for the concept of quantization of the direction of angular momentum. However, they applied only by extension to the standard quantum notion of orbital angular momentum because the quantization of this variable leads to predictions of an odd number of possible orientations in contrast to the even number (two) observed. The m quantum number is only allowed to take on 2l+1 possible values, where l is the integer orbital angular momentum quantum number. After Goudsmit and Uhlenbeck's proposal of spin, physicists realized that they were actually seeing the interaction of the spin angular momentum of the outermost electron in the silver atom with the magnetic field. Therefore, the experiment really gave evidence for the spin quantum number.

Problem 7

1. What are the four quantum numbers that completely describe an electron's orbit in an atom and what can be their possible values?
   A table summarizing these is given below.
   

   Quant No.	Related to	Possible	No. of Values
   n	Energy E = −(13.6eV)/n2 (for Hydrogen)	1,2,3,...	infinite
   l	Orbital Angular Mometum L2 = l(l+1) h/2π	0,1,2,,...,n-1	n
   m	Component of Ang. Mom. Lz = m h/2π	-l,...,0,...,+l	2l+1
   ms	Component of Spin Sz = ms h/2π	-1/2, +1/2	2

2. What is the maximum number of electrons that can be present simultaneously in the n=2 energy level?
   The n=2 energy level allows l = 0,1 as possible angular momentum quantum numbers. Counting of the m,m_s quantum states is given in the table below.
   

   l	m	ms	No. States
   0	0	−1/2, +1/2	2
   1	−1	−1/2, +1/2	2
   1	0	−1/2, +1/2	2
   1	+1	−1/2, +1/2	2
   Total			8

3. Discuss why there is a limit to the number of electrons that can be present in a given atomic energy level. What is the principle at work here?
   The number of electrons that can be present in a given atomic energy level is limited because of the Pauli Exclusion Principle. This principle states that, for spin 1/2 particles like electrons, no two such particles can inhabit the same quantum state in the same atom. Thus, each electron in a single atom must have a unique combination of values for the four quantum numbers: n, l, m, m_s. Note that the Pauli Exclusion Principle only applies to particles that have half-integer spin (1/2, 3/2, 5/2,....). Particles with integer values of spin (0,1,2,...), like the photon, are not governed by it.

Problem 8

Consider a hypothetical version of Bell's experiment where an eta particle (with spin zero) decays to two muons (each having spin 1/2). Two detectors (A and B) that are sensitive to the direction of the muon's spin are set up far from each other, with the eta particle half-way between them.

1. What does angular momentum conservation tell us about the directions of the spins of the two muons?
   The initial angular momentum of the system (before the eta particle decays) is zero, since the eta has spin zero. Angular momentum conservation tells us that this value must be maintained in the final state where there are two muons. This means that the spins of the two muons must add up (as vectors) to zero - that is, they must point in opposite directions.
2. If detector A measures the spin of its muon to be "up" then we know that detector B will have measured the spin of its muon to be "down" according to angular momentum conservation. How did Einstein and Bohr differ in their interpetation of what this statement implied?
   Note: in the following I gloss over some of the subtleties of this problem. However, the main arguments are correct. If you're interested in the details, don't hesitate to come and talk with me during office hours. Einstein - Hidden Variables: Einstein clung to the belief in a universe that is both deterministic and causal. He claimed that the spins of each of the muons have objective reality independent of each other. They are linked by the fact that angular momentum is conserved when they are produced - but the detection of one has no effect on the physical existance of the other. This is a sensible interpretation. However, when Einstein continued using this line of reasoning to try to interpret other observations, like the correlation between spins measured in different directions, he found that his causal and deterministic views gave answers that disagreed with experimental facts. To preserve causality and determinism, he came up with the idea of Hidden Variables theories, which claim that a true theory, which is both deterministic, causal and successful at predicting all experimental results, exists, but we haven't found it yet. The hidden variables in the theories act to preserve causality in the same way that Einstein's general relativity, where the force of gravity is transmitted by gravitons at the speed of light, solved the problem of instantaneous action-at-a-distance that was present in Newton's theory of gravity. Bohr - Quantum Mechanics Bohr believed that the quantum universe is not deterministic in that the observation of one part of the wavefunction of a system changes the state of all parts of the wavefunction. In his view, the state of the spin of muon-2 is undetermined until muon-1's spin is observed. His answer to the question of causality was basically that the state of muon-2 has no objective reality until we actually measure it. Performing that measurement is clearly a causal act.
3. Bell proposed to check which of the two interpretations was correct by examining cross-correlations between the spins measured in detectors A and B. Briefly explain what Bell's inequality is and how it distinguished between Einstein's and Bohr's interpretation of the experiment.
   Stating Bell's inequality generally is quite difficult, which is why I asked you to think about it. Basically, though, it says that the correlation between two sets of related observations, such as the spins of the muons or polarizations of photons, will be different in hidden variables theories (Einstein) and in quantum mechanical theories (Bohr). Hidden variable theories attempt to preserve the objective reality of each measurement, independent of the other and always satisfy Bell's inequality. Quantum theories say that the entire sytem must be treated as a single wavefuction - that is measurements on one part influence all the rest, and often don't satisfy Bell's inequality. By making measurements of the correlation one can then experimentally check whether Bell's inequality is satisfied or not and distinguish between local hidden variable theories and quantum mechanics. The actual mathematical formulation of Bell's inequality depends on the exact nature of the system. I do not expect you to know that formulation or the details of Bell's inequality experiments. I only expect you to understand the significance of Bell's inequality to the debate between objective, local reality and quantum mechanics.

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Extra Credit Problems

The following problems will be covered on Midterm 2 - so you should understand them. You are not required to do them to get full credit for this homework though. If you do attempt them, one point for each will be added to your score for this homework, up to the maximum of 8 points total for homework 4.

Problem 9

1. How was Dirac's approach to quantum mechanics different from Heisenberg's or Schrödinger's?
   There were several conceptual differences between Dirac's approach to quantum mechanics and the early work of Heisenberg and Schrödinger.
   • Dirac required that his equations be relativistically correct (Lorentz invariant) as an existance condition. Other people had abandoned this requirement when it proved to be difficult to simultaneously fulfill it and still have a theory that made correct predictions.
   • Dirac also explicitly considered the quantum nature of the electromagnetic field in its interaction with matter - hence the term quantum field theory. He wrote the Hamiltonian describing a given process as:
     H(total) = H(matter) + H(field) + H(interaction)
     All of these terms were considered quantum mechanically. Previous work had tended to only include the quantum nature of matter, which then led to quantized radiation. This incomplete approach to writing down the basic equations was the reason that it was impossible to make previous theories relativistically invariant.
2. What new phenomenon did Dirac's quantum electrodynamics predict?
   Dirac's theory reproduced all previous predictions of quantum mechanical effects (atomic spectra, etc.). It also explained why the phenomenon of spin was necessary. Dirac's equations did not work without explicitly including spin in the description of the electron, while previous theories worked fine without that concept and only included it because it was an observed property of the electron. The new thing that was predicted by Dirac's theory was anti-matter - the positron.
3. Where did Dirac's theory run into problems? Briefly explain the source of those problems.
   Dirac's theory ran into difficulties when it was applied rigorously because it predicted infinity for observables like the energy of a free electron and its mass. Generally speaking, the source of these infinities was in the interaction of the electron with its own field. (If that was ignored, the theory made quite reasonable predictions.) These self-interaction infinities fall into three broad classes, which can mix among themselves.
   • Self Energy involving the emission and reabsorption of virtual photons (i.e. photons that exist for such a short period of time that Heisenberg's uncertainty principle says they are unobservable). These types of corrections gave the electron infinite energy and mass.
   • Vertex Corrections which are similar to self-energy except that between the emission and reabsorption of the virtual photon, the electron interacts with an external photon. These corrections made the electron's interaction with an electro-magnetic field infinite.
   • Vacuum Polarization where photons (virtual or real) near the electron split into virtual electron-positron pairs, again governed by Heisenberg's uncertainty principle. The virtual positrons are attracted by the electron during their brief life, therefore forming a cloud of positive charge around the electron. This effect gave the electron an infinite charge.
   Of course, all of these effects are present at the same time for any given electron and their are an infinite number of ways in which they can happen - so sorting this all out mathematically was a huge task, which was only completed after WWII.

Problem 10

As we discussed in class, the positron was discovered in cosmic rays using cloud chamber tracks in a magnetic field by Anderson in 1932. His clever experimental technique was to insert a thin iron plate in his cloud chamber and consider tracks that passed through it such as the one shown to the right.

1. What is the direction in which the particle that produced the track shown above was moving? How do we know? Knowing the direction of the magnetic field in the cloud chamber - what is the particle's sign?
   The particle shown above must be travelling upward because the radius of curvature of its track is less (and hence its momentum is less) above the plate than below. Since particles lose energy going through matter, the original, high-momentum particle must have entered from below, lost energy and momentum while traveling through the plate and then exited at the top. Since F = qvxB, the particle's velocity is pointing upward and the particle is curving toward the right with B pointing out of the page, the particle's sign must be positive to make F point to the right.
2. How could we determine that the particle is similar to an electron (and is therefore not a proton)?
   There are two things that would allow us to determine whether the particle is a positron or a proton.
   • The amount of energy it loses in the iron plate. A positron will lose far less than a proton of the same energy.
   • The thickness of the particles track in the cloud chamber. A positron does not produce as much ionization as a proton and will therefore have a much fainter (thinner) track because it provides less ions for water vapor to condense around.
3. Using the radii of curvature of the tracks given in the diagram above, calculate how much energy the positron lost in its passage through the iron plate. Remember that the positron is probably relativistic so its energy is given by E² = (pc)² + m²c⁴.
   First, let's calculate how much momentum the particle has above and below the plate from the curvature of the tracks. Of course, I forgot to give you the magnitude of the magnetic field in this problem, which will make this calculation very difficult. Let's assume that it is 1.0 T.
   p(below) = qBR(below) = (1.6x10⁻¹⁹ C) (1.0 T) (0.1 m) = 1.6x10⁻²⁰ kg.m/s
   p(above) = qBR(above) = (1.6x10⁻¹⁹ C) (1.0 T) (0.057 m) = 9.1x10⁻²¹ kg.m/s
   The energies of the particles are then given by: E² = (pc)² + (mc²)², where m = 9.1x10⁻³¹ kg is the electron mass.
   E(below) = 4.8x10⁻¹² J = 3.0x10⁷ eV = 30 MeV.
   E(below) = 2.7x10⁻¹² J = 1.7x10⁷ eV = 17 MeV.
   ⇒ E(loss) = E(below) − E(above) = 13 MeV.

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Last Modification: 20-Mar-2012