6.1

Chapter 6: Wavelike Properties of Particles

The Bohr model is not very satisfying for several important reasons: (i) although it does explain and describe atomic spectra for certain atoms well, there are those stationary orbits – what the hell are they? (ii) Bohr model only is valid for hydrogen-like only; multi-electron atoms the model fails fast. We are in search of a new model that describes atomic structure well. We are going to go back to the photon, which may seem a step backwards in atomic structure, to explain atomic structure. The new idea, put forward by Louis de Broglie, was an unexpected and major step in the forward direction.

Review of Photon What is the photon? (i) The Photoelectric and Compton Effects clearly show that light is a particle-like thing – the photon. However, (ii) interference and diffraction (for the past 200 years) clearly show that light is a wave-like thing. Are we not contradicting ourselves when we make such statements? That is, how can something be spread out like a wave and localized like a particle at the same time? This creates a logical dilemma that is very unsettling. In classical physics (pre-1900’s), the two big distinctions of physics are particles and waves. It is clear that a microscopic view implies “waves” must sometimes obey the rules that where established for “particles.” Question: is there "symmetry" between particles and waves? Can particles also behave as waves?

Quantum weirdness beginnings: as we get more and more into this chapter, you will quickly be beginning to realize that there will appear to be no logical reasoning for what I will be saying. The ideas will appear illogical and inconsistent; however, they are the basis for the electronic revolution our modern society today. The only thing I have to say to you is TOO BAD! Of course, your intuition is going to be shocked, there is no way for you to experience any of these behaviors because your world does not exist at the microscopic level (h ≈ 1 J s).

Website Two-slit interference, single-slit diffraction

Particle or Wave? In 1923, the young French nobleman Prince Louis de Broglie added a new element to the quantum fray, one that would shortly help to usher in the mathematical framework of modern quantum mechanics and that earned him the 1929 Nobel Prize in physics. It sounds so simple, yet it struck to the heart of the matter. “If light waves also behave like particles, why shouldn’t electrons also behave like waves?” De Broglie suggested that the particle-wave duality applied not only to light but to matter as well. In his Ph.D. thesis, inspired by a chain of reasoning rooted from two equations that Einstein had derived for light quanta he reasoned if E = mc2 relates mass to energy and E = hf related energy to the frequency of waves, then by combining the two, mass should have a wave-like incarnation as well:

22

associated with mass associatedfrequency associated

wavelength

1E mc m c h f E hf

= → = ∝λ=

The de Broglie equation is

de BrogliePlanck 's constantwavelength

momentumh or p

= λ =

De Broglie suggested that just as light is a wave phenomenon, quantum theory shows to have an equally valid particle description; an electron—which we normally think of as being a particle—might have an equally valid description in terms of waves. De Broglie’s great achievement was to take the idea of particle/wave duality and to carry it through mathematically, describing how matter waves ought to behave and suggest ways in

6.2

which they might be observed. Nothing is a substitute for experimental proof and such proof was soon to come from the work of Davisson and Germer.

Davisson & Germer (1920’s) were studying how a beam of electrons bounces off of a chunk of nickel. Atomic crystals act like a diffraction grating (numerous slits).

When x-rays are shined at these crystals, light interferes constructively and destructively.

It is Bragg’s law (2dsinθ = nλ) that shows how interference occurs due to path-difference between the crystal layers. Davisson & Germer accelerated electrons using a voltage source, and shot them at a nickel crystal:

Applet http://phet.colorado.edu/en/simulation/davisson-germer

Using Bragg’s law and de Broglie’s relation (suggested by de Broglie), we write

de Broglierelationship

path difference interference accelertion throughde Broglie

voltage V

h h 2dsin n 2dsin n n p 2m e V

∆

∆

θ = λ → θ = λ = =

⋅

Davisson’s & Germer’s data results found interference very similar to x-rays reflecting from the nickel crystal: • Electrons produce an interference pattern similar

to light with an intense central maximum (many electrons arrive in this region) as well as secondary maxims.

• Angular width of central maximum is momentum dependent (sinθ ∝ 1/p) and

behaves as decreases the width gets narrow

sin increases the width gets wider

λ ⇒θ ∝ λ→λ ⇒

6.3

• Conclude the momentum of the electron is related to the wavelength of the interfering wave according to de Broglie postulate:

Broglie

Broglie

Larger momentum (smaller width) shorter h Smaller momentum (larger width) longer p

→ λ ⇒ λ =→ λ

Their experiment therefore showed that electrons exhibit interference phenomena, the telltale sign of waves. At dark spots, electrons were somehow "canceling each other out." Even if the beam of fired electrons was "thinned" so that, for instance, only one electron was emitted every ten seconds, the individual electrons still built up the bright and dark bands—one spot at a time. Somehow, as with photons, individual electrons "interfere" with themselves in the sense that individual electrons, over time, reconstruct the interference pattern associated with waves. We are inescapably forced to conclude that each electron embodies a wave-like character in conjunction with its more familiar depiction as a particle. Two-Slit Interference of de Broglie Waves This was harder accomplished and took longer to do (1976). The picture below is actual electron interference pattern filmed from a TV monitor as the electron beam densities increase. They start off by shooting a few electrons, and increasing the numbers until the interference pattern is observed. Once again, the electron’s must interference with itself in order to produce interference patterns – that is, the electron spreads out like “ocean waves” and does not have behave as a localized particle. Talk about Feynman's approach to the double slit.

Wave Diffraction To get an idea of when de Broglie wavelike behavior is observed, I first need to talk about wave diffraction. Suppose you walk out this lecture and walk around the corner. The fact that you cannot see me around a corner implies light does not go around corners very well. On the other hand, the fact that you can hear my sexy voice around the corners implies that sound does go around corners well. However, both light and sound have wave properties – so why the difference?

The wave properties of a wave are most noticeable when it interacts with objects that are themselves roughly comparable to the size of a wavelength. That is, • if waves interact with objects that are much larger than the λ, then you don’t notice

the wavelike aspects of it • if waves interact with objects comparable to the λ, then you do notice the wavelike

aspects of it

APPLET http://www.ngsir.netfirms.com/englishhtm/Diffraction.htm

6.4

small bending large bending large bending small bending (λ ≪ opening) (λ ≈ opening) (λ ≈ opening) (λ ≪ opening)

Applying this understanding to how light sound interact with the doorway, visible light

sound

doorway opening (does not bend well round corners)

(does not bend well round corners)=

λ≈ λ

visible light

sound

does not bend well round corners

and cannot see around the corner

does bend well round corners

and can hear around the corn

size of doorway

size of doorway

λ << →

λ ≈ →

visible light

sound

er

doorway opeening (does not bend well round corners)

(does not bend well round corners)=

λ≈ λ

Historical aside: why was there a debate between Huygens and Newton whether light was a wave or a particle? They could not see the wave aspect of light because everyday objects are much, much larger than the wavelength of visible light. Therefore, light did not diffract well and the wavelike aspects went unnoticed until Young’s Double-slit experiment some 200 years later. Physical Interpretation of de Broglie Waves Let’s go back to de Broglie equation. Note that the de Broglie wavelength depends on two things: (1) mass and (2) velocity.

Mass Affects

large masses small de Brloglie waves

small masses large de Brloglie waves

hh v

mv hmv

m⇒ λ

⇒

→→ →

=λ =

=λ

“Big objects” like a tennis ball or your brave instructor have masses considerably larger than the mass of an atom or electron (mtennis ball ≈ 1 kg >> melectron ≈ 1031 kg). One can use the de Broglie equation to estimate the matter wavelength of a tennis ball moving at a speed of 1 m/s:

343410 10 m

1 1h

mv

−−≈ ≈

⋅λ =

As you can see, this is a very tiny, tiny number and we will never notice the wavelike aspects of this tennis ball, though in principle, according to quantum physics they exist. Why? Planck’s constant is so small and ordinary masses are very large compared to atomic masses. On the other hand, if we want to see wavelike aspects of matter, two things must happen:

6.5

• we need very, very small masses in order to make the de Broglie wavelength as large as possible.

• if the de Broglie wavelength becomes comparable to the size of the object then the wavelike aspects will become very dramatic for those objects.

For example, for an electron in orbit in a typical atom, you’ll find the wavelength associated with that electron is roughly comparable to the size of the atom itself. So if de Broglie hypothesis is correct, we ought to notice wavelike aspects at the subatomic world but not in the macroscopic world. That is exactly what is observed in the Quantum Corral.

Velocity Affects

large velocities small de Brloglie waves

small velocities large de Brloglie waves

hh m

mv hmv

v⇒ λ

⇒

→→ →

=λ =

=λ

Question: Why have we not noticed the wavelike behavior in baseballs or cars in everyday situations? It depends on the wavelength and the size of the object we are trying to observe.

Example 6.1

The best way to answer this is by calculating some of the wavelengths of moving objects. a. 500-g billiard ball moving at 10 m/s (23 mph)

3434

billiard ballh 6.63 10 1.33 10 mp (0.5 kg)(10 m/s)

−−×

λ = = = ×

This is such a tiny wavelength! When comparing to the size of the billiard ball the de Broglie wavelength it is so small that we do not have the eyes to see the difference. It looks and acts like a particle.

b. Electron accelerated through a 10 volt potential

10e 2 6

p 2mK K eV

h hc (1240 nm eV) 3.9 10 mp 2mc e(10 V) 2(0.511 10 eV) (10 eV)

−

= =

⋅λ = = = = ×

× ⋅

This wavelength is small on the everyday scale of things but it is not small on the atomic scale. It turns out that the typical diameter of an atom is about 10-10 m. So when studying physics at the atomic scale, we must consider the wavelike behavior of the electrons. That is, the de Broglie wavelength is comparable in size to the atom and therefore, one can tell the difference. If you treat the electron as a particle, you will get into trouble because at this level it displays wavelike characteristics.

c. 10 MeV electron 13

eh 1.2 10 mp

−λ = = ×

6.6

The typical nuclear size is 1510 m−≈ . (somewhere between the atomic and nuclear size.) The higher the momentum, the smaller the wavelength becomes. This is why particle accelerators are so long. By increasing the momentum of the electron, the wavelength of the electron gets smaller and smaller until it gets down to the size of a quark. Therefore, it makes it an excellent probe.

So what does this all mean? When considering objects and whether wavelike behavior is a significant effect, it is Planck’s constant that determines this. That is, Planck’s constant sets the scale at which wave behavior occurs ( h ≈ 10−34). The Orbits for the Bohr Atom What do we gain from de Broglie’s wave hypothesis? We immediately gain an understanding of the Bohr atom. In the Bohr atom electrons where stuck in certain allowable quantized orbits, however, the model gave no explanation of why. Now with the de Broglie hypothesis, there is a very good explanation of why these allowed orbits occur.

Analogy: standing waves on a string In order to form standing waves, two things must happen:

• a reflected wave must be inverted upon the fixed right hand end • an integer number of half-wavelengths must “fit” in between the posts such that

31 2 42 2 2 2length of string whole number of half waves , , , , ...= = λ λ λ λ

As the wave is sent down to the fixed right hand end, it reflects back inverted and these incoming and reflected wave interference to produce a series of standing waves. That is, the wave on a string is reinforced by its successive reflections. One cannot create a standing wave pattern such that the wave does not reflect inverted upon itself. It is just not possible. The only phenomena involving integers in Physics were those of interference and of normal modes of vibration.

Applying this to the Bohr orbitals, the electron wave would travel around the orbit, reinforcing itself constructively at each turn, just as the wave on a string is reinforced by its successive reflections. The de Broglie waves must fit evenly into the circumference of the orbits. The explanation of the allowed quantized electron orbits using de Broglie matter waves is exactly analogous except these orbits are circular orbits around the nucleus and therefore, we are talking about how many waves can fit in this circular type of orbit. • Atomic orbits have an integer number of these half-de Broglie matter wavelengths

that fit around a particular orbit. That would be an allowed orbit. circumference whole number of

, 2 , 3 , ...of allowed orbit de Broglie wavelengths

= = λ λ λ

In other words, the electron wave would travel around the orbit, reinforcing itself constructively at each turn, just as the wave on a string is reinforced by its successive reflections.

6.7

• If there are not an integer number of whole waves, that is, the wave does not come back on itself, then that orbit is not a possible situation and destructive interference eliminates that orbit.

De Broglie matter waves explain Bohr model of the atom in terms of which allowed orbits can exist versus those that cannot exist. The energy is related to the frequency of vibration and these frequencies are quantized according to the standing waves formed by de Broglie waves.

The Quantum Wave Function As is often the case, the dilemma between the wave and particle nature of light is self-imposed. The terms particle-like and wave-like are ideal representations that we have contrived. LIGHT is the basic physical reality; it is NOT a particle or a wave. It is simply NOT correct to state that “light is a stream of particles” or “light is a wave.” Maybe a more limited statement approach is more correct: “light behaves in some ways like a particle” or “light behaves some ways like a wave.” (Principle of Complementary) Such a statement doesn’t get us into trouble with the idealizations that we have placed upon light. Take a pragmatic viewpoint. As you will find out, QM does this anyways. De Broglie had no clear idea what his matter waves really were; in other words, he did not know the nature of their wave function. If electrons are waves, what’s waving? To answer this question, let look at what a classical wave function is.

For a complete description of any wave, we must discuss its WAVE FUNCTION. This is the mathematical function that specifies the wave disturbance at each point of space and time. Some examples are

• waves on a string have wave function y(x,t) = Asin(kx-ωt) • sound waves have wave function ∆p(x,t) = ∆pmax sin(kx-ωt) • light waves have wave function (r, t) = max sin(kr-ωt)]

The interpretation of the matter wave function that is generally accepted today was proposed by the German physicist Max Born in 1926. Born's ideas were taken up and extended by Bohr and his associates in Copenhagen and, for this reason, are often called the Copenhagen interpretation of quantum mechanics. The distinctive feature of Born's proposal is that the matter wave function specifies only probabilities - rather than specific values - of a particle's properties, as we now describe. To understand Born's proposal, it is helpful to consider the connection between the EM wave function Emax(r,t) and the photon. We know that the energy of an EM wave is carried by discrete photons. The number of photons in a small volume dV is proportional to the intensity |Emax(r,t)|2 of the light: intensity = (number of photons in dV at r) ∝ |Emax(r,t)|2 dV

WEB Use the 2-slit simulation for light

The interference patterns produced by a double-slit occur when two light waves constructively/destructively interference at the screen. As a result, interference maximas and minimas are seen. An alternative interpretation of the interference intensity is where a

6.8

• maximum intensity occurs means you are most likely to find a photon • minimum intensity means that you are less likely to find a photon.

What I am really saying doesn’t seem unrealistic and makes good sense because greatest light intensity really means more photons landing at a particular location. Keep this in the back of you might.

This result cannot be exactly true as written for the following reason: If we were to choose a small enough volume dV, then it would predict a fractional number of photons, and this is impossible. A correct statement is that it gives the probable number of photons in the volume dV

( ) 2probable number of photons in dV at r | E(r, t) | dV∝

To illustrate what this means, suppose that we shine a steady beam of light across a room. We select some definite small volume dV and imagine somehow counting the number of photons in dV (at any instant t). Further suppose that the number predicted in a certain small volume dV is 1.5. Of course, we cannot find 1.5 photons in dV in any one observation. On the contrary, each observation will yield some whole number, and we might get 1 photon, then 2, then 0, and so on. After many repeated observations, our average result will be 1.5.

Born proposed that a relation similar for photons should apply to electron or matter waves; that is, there must be an electron wave function, usually denoted by the Greek capital letter Ψ(r, t), whose square gives the probable number of electrons in a small volume dV:

2

2

P(finding particle in dV at r ) (r, t) dV

(r, t) probability density for finding particle at r

= Ψ

Ψ =

6.9

See Side notes on Double-slit experiment

Important: Interpretation of the Quantum Wave Function Suppose that we can somehow arrange an experiment with a succession of quantum particles, all with the same wave function Ψ. If we measure one particle's position (by letting it run into a photographic film, for example), we will find some definite result (in the form of a single dark spot on the film). However, after measuring several particles, all with exactly the same wave function, we will generally get several different answers. What the quantum wave function tells us is the frequency of occurrence of the various possible positions. The particles will be found most often at those points where |Ψ|2 is largest and least often where |Ψ|2 is smallest.

Even when we know a particle's wave function Ψ exactly, we cannot specify a unique position of the particle; rather, we can give only the respective probabilities that the particle will be found at the various possible positions. For this reason, modern quantum mechanics is often described as a probabilistic theory. This is perhaps the most profound difference between classical and quantum mechanics. In classical mechanics, every experiment has an outcome that can, in principle, be predicted unambiguously; in quantum mechanics the same experiment repeated under the same conditions can produce different outcomes. Nonetheless, the probabilities of the various results can be predicted, and it is these probabilities that are measured in an experiment such as the two-slit experiment.

Question: What is the wave function Ψ of a quantum particle?

Answer: Ψ is a function whose intensity |Ψ|2 gives the probability of finding the particle at any particular position.

Wave Packets and Fourier Analysis In order to arrive at some of the properties of the quantum wave function, we review some of the classical properties of waves and look at their properties from the view point of Fourier analysis. The de Broglie relations imply that if a particle has definite values for its energy and momentum, its wave function has corresponding definite values for its frequency and wavelength:

hE hf and p= =λ

A wave with definite frequency and wavelength is called a sinusoidal or harmonic wave. We review briefly the properties of these waves. The simplest of classical waves are waves on a string. Let us consider first a sinusoidal wave traveling to the right on a taut string. The wave function for such a wave has the form

( )y(x,t) A sin kx t= − ω

The various parameters characterizing a wave can be summarized as follows:

6.10

• Spatial parameters: wavelength

wave number

2 k πλ ←→ =

λ

• Time parameters: period frequency

angular frequency

2 T 1/ f 2 fTπ

= = ←→ ω = π =

• Wave speed: wave speed v f v

T kλ ω

= λ = ←→ =

• Energy and momentum: hE hf and p k= = ω = =λ

(In quantum mechanics the parameters k and ω are used frequently)

A sinusoidal wave, with definite frequency and wavelength, is a mathematical idealization that never occurs in practice. For example, we speak of a pure musical tone as a harmonic wave with one precise frequency, but a careful spectral analysis shows that any real musical note is a mixture of many different frequencies. It is easy to see why an exact sine wave cannot occur. A pure sine wave is perfectly periodic, repeating itself endlessly in time and in space. A real wave may repeat itself for a long time and over a large distance, but certainly not forever. This is especially clear in the case of matter waves. A matter wave function that was a pure sine wave would extend indefinitely and would represent a particle that was equally likely to be found anywhere. This is obviously absurd. Consider the following example to make this point clear.

TV Example Imagine an electron in a TV tube. At some time t0 it is ejected from the cathode, and at some later time t1 it hits the screen, creating a small flash of light. At any time between t0 and t1, the electron – being a wave – cannot have a precisely defined position. Nevertheless, the electron is definitely within some region that is tiny compared to the size of the TV tube. Thus, at any instant during its flight, the electron’s wave function might look something like

±∆x

The important points about this wave function are that 1. There is a region, labeled ±∆x, where the wave differs from zero and the electron

may be found. 2. Outside this region the wave is zero and the electron will not be found.

We refer to this kind of wave function that is localized within some region as a WAVE PACKET. My immediate goals are to describe the mathematical properties of these kinds of localized wave packets, which require a crash course in Fourier analysis. The premise of Fourier analysis is that any function can be written as a sum of sines and cosines. There are two broad categories: (1) periodic functions and (2) nonperiodic, nonrepeating functions such as wave packets.

Fourier series

6.11

An example of a periodic function is square wave of width a with a repeat distance of .

It turns out that any such periodic function can be expressed as a Fourier series

( )n n n 1 1 2 2n 0 n 0

2 nf(x) A cos x A cos k x 1 A cosk x A cosk x∞ ∞

= =

π = = = + + + λ ∑ ∑

where An are constants, called the Fourier coefficients for the function f(x). Notice that the different terms in the series correspond to different wavelengths. The first term (n = 0) is a constant term, and the succeeding terms have wavelengths λ, λ/2, λ/3… The longest wavelength term (n = 1) is called the fundamental and the other terms are called higher harmonics.

Now lets us consider the case of nonperiodic functions such as the wave packet

A nonperiodic function can be regarded as a periodic function in the limit that the repeat distance goes to infinity as shown below.

In other words, for larger and larger wavelengths, the wave number spacing gets closer together:

large small k2k small large k

∆λ ⇒ ∆π∆ = → ∆λ ⇒ ∆∆λ

In other words, while the periodic function is made up of a discrete sum of k’s, the nonperiodic function is made up of a continuous distribution of k’s. The discrete Fourier sum becomes a continuous Fourier Integral

distribution of wave numbers kf(x) A(k) coskx dk

that make up the wave packet

= ⋅ ⋅ ≡

∫

To view this nonperiodic wave packet in a picture, it means that the way a wave packet is formed is by adding up cosines with a distribution of wavelengths where the cosines on the ends cancel out due to destructive interference between those waves:

6.12

Classical Wave Uncertainty Relation: ΔxΔk / 2≈ π Without proof, I will state that a wave packet of size ∆x is made up of a range of wave numbers ∆k such that ∆x∆k ≈ π/2. This relationship is illustrated below:

• A narrow wave packet (small ∆x) corresponds to a large spread of wavelengths (large ∆k)

• A wide wave packet (large ∆x) corresponds to a small spread of wavelengths (small ∆k)

1 12 2

x k x k −

∆ ∆ ≈ π→ ∆ ∆ ≥wave packetuncertainty relation

Remarks 1. We can now understand this relation intuitively: The larger the spread in

wavelengths, the more rapidly the waves dephase as one moves away from the center of the wave packet and the more rapidly the packet decays to zero: Bigger ∆k means smaller ∆x and vice versa.

2. For uniformity, we will always use this form. This relation is sometimes called the wave-packet uncertainty relation since it relates the spreads, or uncertainties, in x and k. Many authors write 1x k∆ ∆ ≈ ; in practice, this inequality is used mainly to give rough estimates of various quantities, and in such cases, the odd factor of 2 is of little consequence.

3. Applying this inequality to a perfectly sinusoidal wave like Asin(kx−ωt). This wave has an exactly defined wave number k and hence has ∆k = 0.

k 0 k 0

1lim x limk∆ → ∆ →

∆ ≈ →∞ ∆

If ∆k approaches zero, ∆x must approach ∞, and this is exactly what we knew already: a perfectly sinusoidal wave must extend periodically, through all of space and therefore has ∆x = ∞ .

There is a second uncertainty relation that relates time and angular frequency. The analysis of such a function of time g = g(t) is precisely as discussed above, but with replacements

( ) ( )space to time x tf x g t

k 2 / 2 / T→

→ → = π λ → ω = π

The synthesis of a pulse localized in time requires a continuous spread ∆ω of angular frequencies. The smaller the interval ∆t in which the wave is localized, the larger must

6.13

be the spread ∆ω of angular frequencies. Specifically, the spreads ∆t and ∆ω satisfy an inequality exactly analogous to the one with position and wavelength:

12t∆ ⋅ ∆ω ≥

We will now that these two inequalities (∆x∆k ≥ ½ and ∆t∆ω ≥ ½ ) are especially important in the case of matter waves, for which they imply the celebrated Heisenberg uncertainty principle.

The Uncertainty Relation of Position and Momentum Historical Point on Heisenberg After earning his PhD at Munich, Heisenberg worked with Born and then Bohr. His many contributions to modern physics include an early formulation of quantum mechanics in terms of matrices, several ideas in nuclear physics, and the famous uncertainty principle, for which he won the 1932 Nobel Prize in physics. He remained in Germany during World War II and worked on nuclear reactor design. The possibility that he might be working on an atomic bomb for the Nazis so frightened the Allies that a plan was devised to have him as-sassinated. An American agent, named Moe Berg, posed as a physicist and met with Heisenberg when he was visiting neutral Switzerland in late 1944. After talking with Heisenberg, agent Berg decided that the Germans had made little progress toward a bomb and chose not to kill him.

We have seen that the wave function of a single particle is spread out over some interval. This means that a measurement of the particle's position x may yield any value within this interval. The standard interpretation of quantum mechanics is that this uncertainty is not just a reflection of our ignorance of the particle's position. Rather, the particle does not have a definite position. The uncertainty exists in nature, not just in the mind of the physicist. The uncertainty ∆x can be smaller in some state than in others; but for any given state, specified by a wave function Ψ(x, t) there is some nonzero interval within which the particle may be found, and the particle's position is simply not defined any more precisely than that.

In our Fourier analysis of waves, we saw that the wave function that describes a particle can be built up from sinusoidal waves that require a spread of different wave numbers k (or wavelengths). Applying this to matter or de Broglie waves, the de Broglie relation implies that a spread of wave numbers implies a spread of momenta:

hp k p k= = → ∆ = ∆λ

That is, the particle’s momentum p is uncertain. Plugging this into the wave packet uncertainty relation, we get the one (of two) inequalities called the Heisenberg uncertainty principle:

1 12 2x k x p∆ ∆ ≥ → ∆ ∆ ≥

Interpretation It implies that both the position and momentum of a particle have uncertainties in the sense just described. • One can find states for which ∆x is small, ∆p will be large; • One can also find states for which ∆p is small, but ∆x will be, large. • In all cases their product, ∆x∆p, will never be less than / 2 .

Example 6.2 (a) Macroscopic Example – The position x of a 0.01-g pellet has been carefully

measured and is known within ±0.5 µm. According to the uncertainty principle what are the minimum uncertainties in its momentum and velocity, consistent with our knowledge of x?

6.14

(b) Microscope Example – An electron is known to be somewhere in an interval of total width L≈ 0.1 nm (the size of a small atom). What is the minimum uncertainty in its velocity, consistent with this knowledge?

Solution a. If x is known within ±0.5 µm, the spread ±∆x in the position is certainly no larger than

0.5 µm: x 0.5 μm∆ ≤

According to the uncertainty relation, this implies that the momentum is uncertain by an amount

3428

610 J sp 10 kg m/s p

2 x 10 m

−−

−

⋅∆ ≥ ≥ = ⋅ ≤ ∆

∆

Therefore, the velocity v = p/m is uncertain by 28

235

p 10 kg m/sv 10 m/s vm 10 kg

−−

−

∆ ⋅∆ = ≥ = ≤ ∆

Interpretation Clearly, the inevitable uncertainties in p and v required by the uncertainty principle are of no practical important in this case. (To appreciate how small 10-23 m/s is, notice that at this speed our pellet would take about a million years to cross an atomic diameter.) Although the uncertainty principle is seldom important on the macroscopic level, it is frequently very important on the microscopic level, as the next example illustrates.

b. If we know the electron is certainly inside an interval of total width L, ∆x ≤ ½L. (Remember that ∆x is the spread from the central value out to either side.) According to the uncertainty relation, this implies that

p 2 x L

∆ ≥ ≥∆

This implies that

( ) ( )2

62 6

p c 200 eV nm cv c 10 m/s vm mL 250mc L 0.1 nm 0.5 10 eV∆ ⋅

∆ = ≥ = = ⋅ = ≈ ≤ ∆⋅ ×

Consequences of the Uncertainty Principle I will look at several examples where the Uncertainty Principle give some interesting insight into the quantum world.

• Zero-point energy • Width of Spectra Lines • Quantum Jungle • Heisenberg’s Microscope

Zero-Point Energy Perhaps the most dramatic consequence of the uncertainty principle is that a particle confined in a small region cannot be exactly at rest, since if it were, its momentum would be precisely zero, which would violate

p 2 x L

∆ ≥ ≥∆

Since its momentum cannot be precisely zero, the same is true of its kinetic energy. Therefore, the particle has a minimum KE, which we can estimate as follows: since the momentum is spread out by an amount by ∆p ≥ /L, the magnitude of p must be, on the average, at least of this same order. Thus the KE, whether it has a definite value or not, must on average have magnitude

6.15

( )22 2

2

ppK K2m 2m 2mL

∆= ≥ → ≥

The energy is called the zero-point energy. It is the minimum possible KE for a quantum particle confined inside a region of width L. The KE can, of course, be larger than this, but it cannot be any smaller. Quarks inside a nucleon As you may already know, nucleons are made of quarks; protons (uud) and neutrons (udd) were the masses are 938.3 MeV and 939.6 MeV, respectively. The key point is that the majority of the nucleon mass comes from quark interactions. To see this, consider the following points:

1. if you were to mass a “free” up or down quark, they would have a mass of only a few MeV.

2. if you restricted these quarks to within the nucleus (about 10^(-15) m), they would have masses on the order of several hundred MeV. Memory tells me somewhere on the order of a few hundred MeV (300-400 MeV). This is derived from the zero point energy for 10^(-15) m well.

3. So to account for the missing 600-700 MeV, one has to look at the quark interactions within the nucleons. And the current ability to calculate this (via QCD) is currently beyond our theoretical abilities.

So to say that the nucleons have different masses because of the differences in the masses of the neutrinos, is not correct. The majority of the nucleon masses comes from quark interactions! If we take a native model to try to explain the differences in mass, the down quark is a few MeV more massive than the up quark (no one knows why). Because of the zero point energy of being trapped within a distance of 10^(-15) m, each nucleon should each have a mass of around 1 GeV. So the only difference between the proton and neutron will be on the order of MeV, and this difference can be thought of as a result of • the differences between the proton (udu) and neutron (udd) is that the has neutron’s second down quark is heavier than the proton’s second up quark. So the greater mass of this down quark gives the neutron a greater mass than the proton. • the electrostatic forces among two ups and a down (proton) will differ from those between those two downs and an up (neutron). So the instability of the neutron (beta decay) is due to it having a slightly greater mass than the proton. And as the saying goes, nature always seeks the lowest state of energy (mass).

The Uncertainty Relation for Time and Energy Just as the classical inequality ∆x∆k ≥ ½ implies the position-momentum uncertainty relation, ∆x∆p ≥ ½ , so does the inequality ∆t∆ω ≥ ½ implies a corresponding relation for time and energy. This is called the time-energy uncertainty relation

1 12 2t t E∆ ∆ω ≥ ⇒ ∆ ∆ ≥

Interpretation of ∆E and ∆t • ∆E is the uncertainty in the particle’s energy: a quantum particle generally does not

have a definite energy, and measurement of its energy can yield any answer within a range ±∆E.

• ∆t characterizes the time for which a particle is likely to be found at the position x. • One can find states for which ∆t is small, ∆E will be large; one can also find states for

which ∆t is large, but ∆E will be small.

6.16

If a particle has definite energy, then ∆E = 0 tells us that ∆t must be infinite. That is, a quantum particle with definite energy stays localized in the same region and in the same state for all time. State with this property are the quantum analog of Bohr’s stationary orbits and are called stationary states. If a particle does not remain in the same state forever, ∆t is finite and that means ∆E cannot be zero; that is, the energy must be uncertain.

Example: any unstable state of atom or nucleus lives for a certain finite time ∆t, after which it decays by emitting a particle. This means that the energy of any unstable atom or nucleus has a minimum uncertainty ∆E ≥ /2∆t. Since the energy of the original unstable sate is uncertain, the same is true of the ejected particle. In some cases, one can measure both the spread of energies of the ejected particles and the lifetime ∆t. In many applications one measures one of the quantities, then use the uncertainty principle to estimate the other. The shape of the Alpha Decay potential (http://physics.stackexchange.com/questions/10863/tunneling-of-alpha-particles?rq=1)

Consider this explanation of the alpha decay: It says

The Coulomb barrier faced by an alpha particle with this energy is about 26 MeV, so by classical physics it cannot escape at all. Quantum mechanical tunneling gives a small probability that the alpha can penetrate the barrier.

The explaining sentence and the attached picture suggest that the alpha particle has to overcome the coulomb barrier in order to escape - and as is hasn't enough energy to do that, it must tunnel through it.

But the Coulomb force is a repulsive force in this case as both the nucleus and the alpha particle are positively charged. So why is there any Coulomb barrier at all, shouldn't the Coulomb force "help" the alpha particle to escape? (The other way round, it's obvious - the particle wants to enter the nucleus, but it is hindered by the repulsive force. But when it wants to leave the nucleus, it is only retained by the nuclear force, as far as I can see.)

Why does the alpha particle have to go trough the coulomb barrier - or am I just misinterpreting the explanation and the tunneling occurs somewhere else?

6.17

This model is something that I've seen a lot of students (including myself) struggle with, and a big part of it is that calling it the "Coulomb barrier" is quite over-simplified.

Going back to fundamental physics principles, the Coulomb force is repulsive as you say, so I'll write this as F∝1/r2, where F is the force and r is the distance from the center of the nucleus and the positive sign indicates that the force is in the positive r direction, away from the nucleus. The potential, which is what's shown on the graph, is the negative of the integral of the field (which is proportional to the force), this results in E ∝1/r. For an analog, imagine lifting a weight where "up" is the positive z-axis. The path involves an increase in z-coordinate and the force from the field (gravitational) is negative. The negative integral of the field over distance is the potential.

The missing part is why they "chop off" this 1/r-function at some radius, presumably the radius of the nucleus. Well, this allows us to infer some things about the other force present, which is the nuclear force. It would appear that the force acts weakly beyond the radius of the nucleus, since the 1/r form is unaltered beyond that point. It also absolutely must have a potential that increases faster than 1/r as r→0. Both of these would be loosely satisfied if the nuclear potential was −1/r n (note the negative) where n > 1. The alpha particle, while in the nucleus, by the way, has kinetic energy which is the reason its energy is higher than the hypothetically lowest energy level possible for it. This has to do with the fact that quantum mechanics only allows certain (quantized) energy levels.

Just For an example of how this is possible, I'll say Enuclear=−1/r4 (I am not saying this is how the force acts, it's just for utility).

( ) ( )8

17 8t tc 200 eV nm

E(r) 3 10 eV E2 2c 2 3 10 nm/s 10 s

−

−∆ ∆

⋅∆ ≈ × ∆

⋅ × ⋅≈ = = ≈

E(r)=Enuclear(r)+ECoulomb(r) E(r)=1/r−1/r4

F(r)=−ddr(Enuclear(r)+ECoulomb(r))=Fnuclear(r)+FCoulomb(r) F(r)=1/r2−1/r5

My intent is that this answer allows you to explicitly answer to yourself what the potential and force contributions are, and give some very rudimentary example for how the "edge"

6.18

can appear. Beyond that I hope it's clear how someone may take the shape of the above curve and lump it into a "wall".

As you state the Coulomb force is repulsive, this is why the potential increases with smaller separation. But at small enough distances the strong force takes over which has an attracting nature. The strong force is stronger (hence the name) than the Coulomb force, which is why in total the alpha particle in the nucleus sits at a lower potential than the Coulomb force at the edge.

The strong nuclear force can be modeled as a rectangular potential well to first order

Example 6.3 Many excited states of atoms are unstable and decay by emission of a photon in a time of order ∆t ≈ 10-8 s. What is the minimum uncertainty in the energy of such an atomic state?

Solution According to ∆t ∆E ≥ ½ , the minimum uncertainty in energy is

( ) ( )8

17 8t tc 200 eV nm

E 3 10 eV E2 2c 2 3 10 nm/s 10 s

−

−∆ ∆

⋅∆ ≈ × ∆

⋅ × ⋅≈ = = ≈

Compared to the several eV between typical atomic energy levels, this uncertainty ∆E is very small. Nevertheless, the resulting spread in the energy, and hence frequency, of the ejected photon is easily measurable with a modern spectrometer.

Nowadays, the frequencies of photons ejected in atomic transitions are used as standards for the definition and calibration of frequency and time. Because of the uncertainty principle, the frequency of any such photon is uncertain by an amount

E 1t2

∆∆

∆ω = ≈

where ∆t is the lifetime of the emitting state. Therefore, it is important to choose atomic states with very long lifetime’s ∆t to use as standards. Example 6.4 (P5.44) Suppose that a particle of mass m is constrained to move in a one-dimensional space between two infinitely high barriers located A apart. a. Compute the minimum energy of an electron for A = 10−10 m and A = 10−2 m. Interpret

the results. b. Calculate the minimum energy for a 100-mg bead moving on a thin wire between two

stops located 2 cm apart. Interpret the results.

Solution The zero point energy of the particle has already been derived:

2 2

min 2 2E2mL 2mA

≥ =

It means that a particle must have some minimum energy according to the uncertainty principle; otherwise it would violate the uncertainty principle and one could know both the position and momentum of the particle simultaneously. If we confine a particle to a smaller box, then the minimum energy will automatically be greater than when confined to a larger box.

a. For an electron with A = 10−10 m

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2 2 210

min min2 2 6 1e

c (197.3eV nm)E 3.81 eV E (A 10 m)2m c A 2 (0.511 10 eV)(10 nm)

−−

⋅= = = = =

⋅ ×

For A = 10−2 m, the minimum energy is 2 16

minE (A 10 m) 3.81 10 eV− −= = ×

This makes physical sense; in the example of the quantum billiards game as soon as the ball was enclosed with the rack, the uncertainty in its momentum was extremely larger. As soon as one tries to confine the electron into a smaller box, its zero-point energy goes up just like the confined quantum billiard ball and the momentum or energy has a high uncertainty. On the other hand, a larger box allows the zero-point energy to be easily measured at the cost of not knowing its location.

b. For the bead, it is more convenient to work with units of Joules: 2 34 2

min 2 2bead

61 43min

(1.055 10 J s)E2m A 2 (0.100 kg)(2 10 m)

1.39 10 eV 8.7 10 eV E (bead)

−

−

− −

× ⋅= =

⋅ ×

= × = × =

This macroscopic example is extremely small due to two factors: the macroscopic size of the box and the mass of the bead. This zero-point energy implies that the bead’s energy or linear momentum are unaffected by the zero-point energy as far as measurements are concerned.

Example 6.5 (P5.48) It is possible for some fundamental particles to “violate” conservation of energy by creating and quickly reabsorbing another particle. For example, a proton can emit a π+ according to p → n + π+ where n represents a neutron. The π+ has a mass of 140 MeV/c2. The reabsorption must occur within a time ∆t consistent with the uncertainty principle. a. By how much ∆E is energy conservation violated? (Ignore KE) b. For how long ∆t can the π+ exist? c. Assuming that the π+ is moving at nearly the speed of light, how far from the nucleus

could it get in the time ∆t?

Solution a. The minimum amount of energy required to produced this decay occurs with the proton

is not at rest since the rest of the neutron is greater than the proton. The threshold energy is given by the s-invariant:

2 20

2 21 1

S S

s (E ) (p )

=

= − 2 22 3 2 3(E E ) (p p )= + − +

2 2 2p nm c m c m cπ− −

b. d

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Bell’s Theorem

The shit has hit the fan! There is something called Bell’s theorem that makes a straightforward assumption about a particle being real and separable. These two things form the most fundamental ideas we have about what a real object is. Bell’s theorem sets up a simple inequality that can be tested using a quantum experiment. For example, suppose that the real part is defined by the parameter A and the separable part is defined by B. Bell’s theorem sets up the inequality

( )2 2 2 2 2A B A B 2AB A B+ = + + ≥ + If we find that

2 2 2 2A B 2AB A BA is not true the property of real objects is not correct

either B is not true the property of separable objects is not correctboth are not true

+ + ≤ +

→⇒ →

Either way, Bell’s theorem states that we live in a very weird world and that objects at the microscopic level have properties where real and separable characteristic of particles is NOT VALID.