UNIT 5

QUANTUM PHYSICS

If the systems of quantum theory are thought of as elementary information carriers in the first place, rather than elementary constituents of matter, and their connections are logical connections within a given algorithm, rather than space-time relations, then we need to find the origin of mechanical concepts that characterise quantum mechanics as a theory of physical systems. To this end, we will illustrate how physical laws can be viewed as algorithms for the update of memory registers that make a physical system.

The term "quantum mechanics" was first coined by Max Born in 1924.

Quantum mechanics is the branch of physics dealing with the behaviour of matter and light on the atomic and subatomic level. . It attempts to describe and account for the properties of molecules and atoms and their constituents—electrons, protons, neutrons, and other more particles such as quarks and gluons.

At the scale of atoms and electrons, many of the equations of classical mechanics which describe how things move at everyday sizes and speeds cease to be useful.

In classical mechanics objects exist in a specific place at a specific time. However, in quantum mechanics, objects instead exist in a haze of probability they have a certain chance of being at point A another chance of being at point B and so on. It results some very strange conclusions about the physical world.

At the turn of the twentieth century, however, classical physics not able to explain some of the phenomena.

Relativistic domain: Einstein’s 1905 theory of relativity showed that the validity of

Newtonian mechanics ceases at very high speeds i.e. Speeds comparable to that of light.

Microscopic domain: As soon as new experimental techniques were developed to the point of searching atomic and subatomic structures, it turned out that classical physics fails miserably in providing the proper explanation for several newly discovered phenomena. It thus became evident that the validity of classical physics ceases at the microscopic level and that new concept had needed to describe.

The classical physics fails to explain several microscopic phenomena such as blackbody radiation, the photoelectric effect, atomic stability and atomic spectroscopy

In 1900 Max Planck introduced the concept of the quantum of energy. He successfully explained the phenomenon of blackbody radiation .He introduced the concept of discrete or quantized Energy. He also explained that the energy exchange between an electromagnetic wave of frequency ν and matter occurs only in integer multiples of hν, which he called the energy of a quantum, where h is called Planck’s constant.

In 1905 Einstein provided a powerful consolidation to Planck’s quantum concept. In tryingto understand the photoelectric effect, Einstein recognized that Planck’s idea of the quantizationof the electromagnetic waves must be valid for light as well.

So, He Gave light itself is made of discrete bits of energy or tiny particles, called photons,each of energy hν, νbeing the frequency of the light. The introduction of the photon conceptenabled Einstein to give an elegantly accurate explanation to the photoelectric problem.

Bohr introduced in 1913 his model of the hydrogen atom. He explained that atoms can be found only in discrete states of energy and the emission or absorption of radiation by atoms, takes place only in discrete amounts of hν.This successfully explained to problems such as atomic stability and atomic spectroscopy

Then in 1923 Compton made an important discovery of scattering X-rays with electrons, he confirmedthat the X-ray photons behave like particles with momenta hν/c, ν is the frequency of theX-rays.

Planck, Einstein, Bohr, and Compton gave the theoretical and experimental confirmation for the particle aspect of wave that means that waves exhibit particle behaviour at the microscopicscale.

Some of the most prominent scientists to subsequently contribute in the mid-1920s to what is now called the "new quantum mechanics" or "new physics" were Max Born, Paul Dirac, Werner Heisenberg, Wolfgang Pauli, and Erwin Schrödinger.

Through a century of experimentation and applied science, quantum mechanical theory has proven to be very successful and practical.

Historically, there were two independent formulations of quantum mechanics. The firstformulation, called matrix mechanics, was developed by Heisenberg (1925) to describe atomic structure starting from the observed spectral lines.

The second formulation, called wave mechanics, was due to Schrödinger (1926).

It is a generalization of the de Broglie postulate. This method describes the dynamics of microscopic matter by means of a wave equation, called the Schrodinger equation.

Dirac derived in 1928 an equation which describes the motion of electrons. This equation, known as Dirac’s equation, predicted the existence of an antiparticle.

In short, quantum mechanics is the founding basis of all modern physics:solid state, molecular, atomic, nuclear, and particle physics, optics, thermodynamics, statisticalmechanics, and so on. Not only that, it is also considered to be the foundation of chemistry andbiology.

Postulates of Quantum Mechanics

The wavefunction must satisfy certain mathematical conditions because of this probabilistic interpretation. For the case of a single particle, the probability of finding it somewhereis 1, so that we have the normalization condition

It is customary to also normalize many-particle wavefunctions to 1. The wavefunction must also be single-valued, continuous, and finite.

2. To every observable in classical mechanics, there corresponds a linear, Hermitian operator in quantum mechanics. For example, in coordinate space, the momentum operator corresponding to momentum px in the direction for a single particle is -iℏ .

3. The average value of the observable corresponding to operator is given by

4. The wavefunction evolves in time according to the time-dependent Schrödinger equation

5. In any measurement of the observable associated with operator , the only values that will ever be observed are the eigenvalues a which satisfy = a. Although measurements must always yield an eigenvalue, the state does not originally have to be in an eigenstateof . An arbitrary state can be expanded in the complete set of eigenvectors of ( = a) as, = where the sum can run to infinity in principle. The probability of observing eigenvalue aiis given by ci*ci.

6. The total wavefunction must be antisymmetric with respect to the interchange of all coordinates of one fermion with those of another. Electronic spin must be included in this set of coordinates. The Pauli Exclusion Principle is a direct result of this antisymmetry principle.

Einstein Equation

Quantum mechanics began with two deceptively simple formulas

E= and p= h/λ

These are the Einstein and de Broglie relations, respectively. Youmay already be familiar with both of these equations, but let’s review where theycame from and what they tell us.

The Einstein relation, E=, says that a particle’s energyEis proportionalto its frequency. Albert Einstein proposed this relation for light, introducing theradical idea that light comes in discrete lumps, now called photons, each with anenergy determined by the light’s frequency. So a blue photon (high frequency) hasmore energy than a red photon (low frequency), while a gamma-ray photon has farmore energy than either, and a radio-wave photon has much less.

The first directexperimental evidence for the Einstein relation came from the photoelectric effect,in which high-frequency light, aimed at a metal surface, ejects electrons with moreenergy than low-frequency light. Each electron, it turns out, absorbs the energyof just one photon from the light. If you plot the energy vs. the frequency you get a straight line, whose slope is the constant of proportionality,

h= 6.63×10−34J s = 4.14×10−15eV s

called Planck’s constant.

But the Einstein relation doesn’t just apply to particles of light; it applies equallywell to electrons, protons, quarks, neutrinos, and also baseballs.

Defining what we mean by the “frequency” of one of these particles is trickier thanfor light, so here we usually stick to photons.

Example:What is the frequency of a photon with energy of 4.5 eV?

Solution: E = (4.5 eV) x (1.60 x 10-19 J/eV) = 7.2 x 10-19 JE = hf

h = 6.63 x 10-34 J x s

f = E / h = (7.2 x 10-19 J) / (6.63 x 10-34 J x s)

f = 1.1 x 1015 Hz

De Broglie Hypothesis of Matter Waves

As we know in the Photoelectric Effect, the Compton Effect, and the pair production effect—radiation exhibits particle-like characteristics in addition to its wave nature. In 1923 de Broglie took things even further by suggesting that this wave–particle duality is not restricted to radiation, but must be universal.

In 1923, the French physicist Louis Victor de Broglie (1892-1987) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions.

All material particles should also display dual wave–particlebehaviour. That is, the wave–particle duality present in light must also occur in matter.

So, starting from the momentum of a photon p = hν/c = h/λ.

We can generalize this relation to any material particle with nonzero rest mass. Each material particle of momentum behaves as a group of waves(matter waves) whose wavelength λand wave vector aregoverned by the speed and mass of the particle. De Broglie proposed that the wave length λ associated with a particle of momentum p is given as where m is the mass of the particle and v its speed.

λ = = …….(1) = …….(2)

Where ℏ = h/2π. The expression known as the deBroglie relation connects the momentum of a particle with the wavelength and wave vector of the wave corresponding to this particle.The wavelength λ of the matter wave is called de Broglie wavelength. The dual aspect of matter is evident in the de Broglie relation.

λ is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. Planck’s constant h relates the two attributes. Equation (1) for a material particle is basically a hypothesis whose validity can be tested only by experiment.

However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen, p = hν/c.

Therefore

= = λ

Matter waves: According to De-Broglie, a wave is associated with each moving particle which is called matter waves.

That is, the wave–particle duality present in light must also occur in matter. So, starting from the momentum of a photon p= hν /c =h/λ, we can generalize this relation to any material particle with nonzero rest mass: each material particle of momentum behaves as a group of waves (matter waves) whose wavelength λ and wave vector are governed by the speed and mass of the particle.

λ = =

Wave has wavelength λ here h is Planck's constant and p is the momentum of the moving particle.

We have seen that microscopic particles, such as electrons, display wave behaviour. What about macroscopic objects? Do they also display wave features? They surely do. Although macroscopic material particles display wave properties, the corresponding wavelengths are too small to detect; being very massive, macroscopic objects have extremely small wavelengths.

At the microscopic level, however, the waves associated with material particles are of the same size or exceed the size of the system. Microscopic particles therefore exhibit clearly noticeable wave-like aspects.

The general rule is: whenever the de Broglie wavelength of an object is in the range of, or exceeds, its size, the wave nature of the object is detectable and hence cannot be neglected.

But if its de Broglie wavelength is much too small compared to its size, the wave behaviour of this object is undetectable.

For a quantitative illustration of this general rule, let us calculate in the following example the wavelengths corresponding to two particles, one microscopic (electron) and the other macroscopic (ball).

Example: What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4×106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?

Solution:

(a)For the electron:

Mass m = 9.11×10–31 kg, speed v = 5.4×106 m/s.

Then, momentum

p = m v = 9.11×10–31 kg × 5.4 × 106 (m/s)

p = 4.92 × 10–24 kg m/s

de Broglie wavelength, λ = h/p = 6.63 x 10-34Js/ 4.92 × 10–24 kg m/s

λ= 0.135 nm

(b)For the ball:

Mass m’ = 0.150 kg,

Speed v ’= 30.0 m/s.

Then momentum p’ = m’ v’= 0.150 (kg) × 30.0 (m/s)

p’= 4.50 kg m/s

de Broglie wavelength λ’ = h/p’ =6.63 x 10-34Js/ 4.50 kg m/s =1.47 ×10–34 m

The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of the proton, quite beyond experimental measurement.

In his 1923 experiment, Compton provided the most conclusive confirmation of the particle aspect of radiation.

By scattering X-rays off free electrons, he found that the wavelength of the scattered radiation is larger than the wavelength of the incident radiation. This can be explained only by assuming that the X-ray photons behave like particles.

According to classical physics, the incident and scattered radiation should have the same wavelength.

Also we know that the energy of the X-ray radiation is too high to be absorbed by a free electron therefore the incident X-ray would then provide an oscillatory electric field which sets the electron into oscillatory motion, hence making it radiate light with the same wavelength but with an intensity I that depends on the intensity of the incident radiation I0

But neither of these two predictions of classical physics is compatible with experiment.

Figure 1: Elastic scattering of a photon from a free electron

By experiment Compton reveal that the wavelength of the scattered X-radiation increases by an amount , called the wavelength shift, and that depends not on the intensity of the incident radiation, but only on the scattering angle.

Compton succeeded in explaining his experimental results only after treating the incident radiation as a stream of particles—photons—colliding elastically with individual electrons.

Here we will discuss elastic scattering of a photon from a free electron as shown in figure. Consider that the incident photon, of energy E =hν and momentum p = hν /c, collides with an electron that is initially at rest. If the photon scatters with a momentum at an angle θ while the electron recoils with a momentum the conservation of linear momentum yields

= + …………(1)

Which leads

= (- )2= + -2pp’cosθ = ( + -2νν’cosθ) ………..(2)

Energy Conservation

The energies of the electron before and after the collision are given, respectively, by

E0 =mec2 ………..(3)

Since the energies of the incident and scattered photons are given by E = hν and E0 = hν’, respectively, conservation of energy dictates that

E + E0 = E’ + Ee ………..(4)………..(5)

Which leads to

………..(6)

Squaring both sides of (5) and simplifying, we get

Hence wavelength shift is given by

Where λC = h/mec =2.426 x 10-12 m is called the Compton wavelength of the electron.

This relation connects the initial and final wavelengths to the scattering angle.

Compton’s experimental observation:

Example: High energy photons are scattered from electrons initially at rest. Assume the photonsare backscattered and their energies are much larger than the electron’s rest-mass energy, E = mec2.

(a) Calculate the wavelength shift.

(b) Show that the energy of the scattered photons is half the rest mass energy of the electron, regardless of the energy of the incident photons.

(c) Calculate the electron’s recoil kinetic energy if the energy of the incident photons is150 MeV.

Solution:

(a) In the case where the photons backscatter i.e.θ = π

The wavelength shift becomes

= - = 2csin2θ = 2csin2π = 2c =2 x2.426 x 10-12 m =4.86 x 10-12 m

(b) Since the energy of the scattered photons E’ is related to the wavelength by E =hc /

Where E = hc/ is the energy of the incident photons. If E = mec2 we can approximate

(c) If E = 150 MeV, the kinetic energy of the recoiling electrons can be obtained from conservation of energy

Ke = E – E’ 150 – 0.25 =14.75 MeV

Example: X-rays with energy of 300 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 30° relative to the incident X-rays, determine the Compton shift at this angle, the energy of the scattered X-ray, and the energy of the recoiling electron.

Solution:

The Compton shift

When it is scattered through an angle θ by an electron:λ′−λ=λe(1−cosθ)

We know Compton wavelength of the electron λe=h/mec= 2.43 pm

me mass of the electron = 511 keV/c2

θ = 30°

Compton shift is

λ′−λ=λe(1−cosθ) = 2.426 x 10-12 m (1−cos30◦.) = 0.325 pm

The energy E′of the scattered photon is E′=hc/ λ′

And λ=hc/E

E=300 keV is the wavelength of the incoming photon. It follows that E′= 278 ke V. By conservation of energy, the energy lost by the photon in the collision is converted into kinetic energy K of the recoiling electrons K= 22 keV

We deal here with another physical process which confirms that radiation (the photon) has corpuscular properties.

The theory of quantum mechanics that Schrödinger and Heisenberg proposed works only for non relativistic phenomena. This theory, which is called non relativistic quantum mechanics, was immensely successful in explaining a wide range of such phenomena. Combining the theory of special relativity with quantum mechanics, Dirac succeeded (1928) in extending quantum mechanics to the realm of relativistic phenomena. The new theory, called relativistic quantum mechanics, predicted the existence of a new particle, the positron. This particle, defined as the antiparticle of the electron, was predicted to have the same mass as the electron and an equal but opposite (positive) charge.

Four years after its prediction by Dirac’s relativistic quantum mechanics, the positron was discovered by Anderson in 1932 while studying the trails left by cosmic rays in a cloud chamber.

When high-frequency electromagnetic radiation passes through a foil, individual photons of this radiation disappear by producing a pair of particles consisting of an electron, e-, and a positron, e+: photone- + e+. This process is called pair production.

Figure 2: Pair production

a highly energetic photon, interacting with a nucleus, disappears and produces an electron and a positron.

Anderson obtained such a process by exposing a lead foil to cosmic rays from outer space which contained highly energetic X-rays. It is useless to attempt to explain the pair production phenomenon by means of classical physics, because even non relativistic quantum mechanics fails utterly to account for it.

Due to charge, momentum, and energy conservation, pair production cannot occur in empty space. For the process photone- + e+ to occur, the photon must interact with an external field such as the Coulomb field of an atomic nucleus to absorb some of its momentum. In there action depicted in Figure 2, an electron–positron pair is produced when the photon comes near (interacts with) a nucleus at rest; energy conservation dictates that

Where ℏω= the energy of the incident photon,

2mec2= the sum of the rest masses of the electron and positron,

ke- and ke+ are the kinetic energies of the electron and positron, respectively.

As for EN=KN, it represents the recoil energy of the nucleus which is purely kinetic. Since the nucleus is very massive compared to the electron and the positron, KN can be neglected to a good approximation.

Note

The inverse of pair production, called pair annihilation, also occurs. For instance, when an electron and a positron collide, they annihilate each other and give rise to electromagnetic radiation

e- + e+photon

This process explains why positrons do not last long in nature.

When a positron is generated in a pair production process, its passage through matter will make it lose some of its energy and it eventually gets annihilated after colliding with an electron.

The collision of a positron with an electron produces a hydrogen-like atom, called positronium. Mean lifetime of positronium is about 10-10s. Positronium is like the hydrogen atom where the proton is replaced by the positron.

Note

The pair production process is a direct consequence of the mass–energy equation of EinsteinE =mc2, which states that pure energy, can be converted into mass and vice versa.

Conversely, pair annihilation occurs as a result of mass being converted into pure energy. All subatomic particles also have antiparticles (e.g., antiproton).

Even neutral particles have antiparticles; for instance, the antineutron is the neutron’s antiparticle. Pair production and pair annihilation, which are relativistic processes, merely to illustrate how radiation interacts with matter, and also to underscore the fact that the quantum theory of Schrödinger and Heisenberg is limited to non relativistic phenomena only.

According to classical physics, given the initial conditions and the forces acting on a system, the future behaviour (unique path) of this physical system can be determined exactly. That is, if the initial coordinates,velocity, and all the forces acting on the particle are known, the position, and velocityare uniquely determined by means of Newton’s second law. So by Classical physicsit can be easily derived.

Does this hold for the microphysical world?

Since a particle is represented within the context of quantum mechanics by means of a wave function corresponding to the particle’s wave, and since wave functions cannot be localized, then a microscopic particle is somewhat spread over space and, unlike classical particles, cannot be localized in space. In addition, we have seen in the double-slit experiment that it is impossible to determine the slit that the electron went through without disturbing it. The classical concepts of exact position, exact momentum, and unique path of a particle therefore make no sense at the microscopic scale. This is the essence of Heisenberg’s uncertainty principle.

In its original form, Heisenberg’s uncertainty principle states that: If the x-component of the momentum of a particle is measured with an uncertainty ∆px, then its x-position cannot, at the same time, be measured more accurately than ∆x =ℏ/(2∆px). The three-dimensional form of the uncertainty relations for position and momentum can be written as follows:

This principle indicates that, although it is possible to measure the momentum or position of a particle accurately, it is not possible to measure these two observables simultaneously to an arbitrary accuracy. That is, we cannot localize a microscopic particle without giving to it a rather large momentum.

We cannot measure the position without disturbing it; there is no way to carry out such a measurement passively as it is bound to change the momentum.

To understand this, consider measuring the position of a macroscopic object (you can consider a car) and the position of a microscopic system (you can consider an electron in an atom). On the one hand, to locate the position of a macroscopic object, you need simply to observe it; the light that strikes it and gets reflected to the detector (your eyes or a measuring device) can in no measurable way affect the motion of the object.

On the other hand, to measure the position of an electron in an atom, youmust use radiation of very short wavelength (the size of the atom). The energy of this radiation is high enough to change tremendously the momentum of the electron; the mere observation of the electron affects its motion so much that it can knock it entirely out of its orbit.

It is therefore impossible to determine the position and the momentum simultaneously to arbitrary accuracy. If a particle were localized, its wave function would become zero everywhere else and its wave would then have a very short wavelength. According to de Broglie’s relation p =ℏ /λ,

Time Energy Uncertainty Relation

The momentum of this particle will be rather high. Formally, this means that if a particle is accurately localized (i.e., ∆x 0), there will be total uncertainty about its momentum (i.e., ∆px∞).

Since all quantum phenomena are described by waves, we have nochoice but to accept limits on our ability to measure simultaneously any two complementary variables.

Heisenberg’s uncertainty principle can be generalized to any pair of complementary, or canonically conjugate, dynamical variables: it is impossible to devise an experiment that can measure simultaneously two complementary variables to arbitrary accuracy .If this were ever achieved, the theory of quantum mechanics would collapse.

Energy and time, for instance, form a pair of complementary variables. Their simultaneous measurement must obey the time–energy uncertainty relation:

This relation states that if we make two measurements of the energy of a system and if these measurements are separated by a time interval ∆t, the measured energies will differ by an amount ∆E which can in no way be smaller than ℏ /∆t. If the time interval between the two measurements is large, the energy difference will be small. This can be attributed to the fact that, when the first measurement is carried out, the system becomes perturbed and it takes it a long time to return to its initial, unperturbed state. This expression is particularly useful in the study of decay processes, for it specifies the relationship between the mean lifetime and the energy width of the excited states.

In contrast to classical physics, quantum mechanics is a completely indeterministic theory. Asking about the position or momentum of an electron, one cannot get a definite answer; only a probabilistic answer is possible.

According to the uncertainty principle, if the position of a quantum system is well defined its momentum will be totally undefined.

1 Example: The uncertainty in the momentum of a ball travelling at 20m/s is 1×10−6 of its momentum. Calculate the uncertainty in position? Mass of the ball is given as 0.5kg.

Solution:

Given

v = 20m/s,

m = 0.5kg,

h = 6.626 × 10-34 m2 kg / s

Δp =p×1×10−6

As we know that,

P = m×v = 0.5×20 = 10kgm/s

Δp = 10×1×10−6

Δp = 10-5

Heisenberg Uncertainty principle formula is given as,

∆x∆p

∆x

∆x

∆x =0.527 x 10-29 m

2 Example: .The mass of a ball is 0.15 kg & its uncertainty in position to 10–10m. What is the value of uncertainty in its velocity?

Solution:

Given

m=0.15 kg.

h=6.6×10-34 Joule-Sec.

Δx = 10 –10 m

Δv=?

Δx.Δv ≥h/4πm

Δv≥h/4πmΔx

≥6.6×10-34/4×3.14×0.15×10–10

≥ 3.50×10–24m

The Heisenberg uncertainty principle based on quantum physics explains a number of facts which could not be explained by classical physics.

One of the applications is to prove that electron cannot exist inside the nucleus.

But to prove it, let us assume that electrons exist in the nucleus.

As the radius of the nucleus in approximately 10-14m. If electron is to exist inside the nucleus, then uncertainty in the position of the electron is given by

According to uncertainty principle

∆x∆p =h/2π

Thus ∆p=h/2π∆x

Or ∆p=6.62 x10-34/2 x 3.14 x 10-14

Or ∆p=1.05 x 10-20 kg m/ sec

If this is p the uncertainty in the momentum of electron then the momentum of electron should be at least of this order that is p=1.05*10-20 kg m/sec.

An electron having this much high momentum must have a velocity comparable to the velocity of light. Thus, its energy should be calculated by the following relativistic formula

E =

Therefore, if the electron exists in the nucleus, it should have an energy of the order of 19.6 MeV.

However, it is observed that beta-particles (electrons) ejected from the nucleus during b –decay have energies of approximately 3 Me V, which is quite different from the calculated value of 19.6 MeV.

Another reason that electron cannot exist inside the nucleus is that experimental results show that no electron or particle in the atom possess energy greater than 4 MeV.

Therefore, it is confirmed that electrons do not exist inside the nucleus.

2. Ground State Energy of A Harmonic Oscillator orCalculation of zero point energy

Zero-point energy (ZPE) is the lowest possible energy that a quantum mechanical system may have. Unlike in classical mechanics, quantum systems constantly fluctuate in their lowest energy state as described by the Heisenberg uncertainty principle. As well as atoms and molecules, the empty space of the vacuum has these properties. According to quantum field theory, the universe can be thought of not as isolated particles but continuous fluctuating fields: matter fields, whose quanta are fermions and force fields, whose quanta are bosons. All these fields have zero-point energy.

The minimum energy of a system at o k (zero kelvin) is called zero point energy.

Consider a particlewhichis confined to move under the influence of potential of dimensions a. Since the particlemay be present anywhere in these dimensions, so uncertainty in position

∆x = a/2 (half thediameter).

According to uncertainty principle

∆x∆px = ℏ/2

or

Thus ∆px=ℏ/2∆x

∆px=ℏ/2a/2 = ℏ/a

Uncertainty in momentum of particle along x-axis is

∆px=ℏ/a

Assuming the momentum of particle to be at least equal to uncertainty in it, the lowest possible value of K.E. of particle is given by

K.E. == =

Where mis the mass of the particle.

Energy even at OK is given by the above equation. This minimum energy is called the zero-point energy.

This implies that even at zero kelvin, the particle is never at rest. If it is so, then ∆pX = 0, which is not possible. [It gives ∆x =∞]

3. Existence of proton, neutron and alpha particles within the nucleus.

We know that the rest mass of the protons and neutron is of the order of

1.67x 10-27 kg. Hence, the value of momentum 5.27 x 10-21kg.m/sec from calculation and also the value of v come out to be 3x 105m/sec.

The corresponding value of kinetic energy of a neutron or a proton is

E= = =8.33 x 10-15J =eV

52.05 keV

Since the rest mass of the a-particle is nearly four times the proton mass, therefore the alpha particle should have a minimum kinetic energy of one fourth of 52.05 keV, or about 13 keV. Since the energy carried by the protons or neutrons emitted by the nuclei are greater than 52 keV and for a-particle more than 13 keV, these particles can exist in the nuclei.

4. Size of Elementary cell in Phase space

We have studied in our previous class that state of a microsystem is defined by six variables – three are due to position and three due to momentum.

Hence, a system of N particles needs 6 N variables

If ∆ x and ∆px be the uncertainly in position and in momentum measurements then

∆ x ∆ px= ;

Similarly ∆ y ∆ py= ,

And ∆ z ∆ pz= ,

Multiplying these three equations, we get

∆ x ∆ y ∆ z ∆ px∆ py∆ pz= ( )3 in the units (J3 S3).

The above product is called the volume of elementary cell in phase space.

So, volume of an elementary cell in phase space 10-101 units, (for quantum statistics) being .

5. Accurate limit of frequency of radiation emitted by an atom

Consider the radiation emitted from an excited atom. The energy of this atom will decrease when it emits one or more photons of characteristic frequency.

The average period between excitation of the atom and the release of energy is about 10-8seconds.

Thus, uncertainly in energy is

∆ E

Or ∆ E J

Or ∆ E 5.3 x 10-27 J

Frequency of light is uncertain by

∆ν = = Hz 0.8 x 107 Hz

As a result, the radiation from an excited atom does not have the noted precise frequency new ν - ∆ν andν + ∆ν.

Quantum theory and determinism usually do not go together. A natural combination is quantum theory and randomness. Indeed, when in the end of 19th century physics seemed to be close to provide a very good deterministic explanation of all observed phenomena, Lord Kelvin identified “two clouds” on “the beauty and clear-ness of the dynamical theory”. One of this “clouds” was the quantum theory which brought a consensus that there is randomness in physics. Recently we even “certify” randomness using quantum experiments

The quantum theory of the wave function of the Universe is a very successful deterministic theory fully consistent with our experimental evidence. However, it requires accepting that the world we experience is only part of the reality and there are numerous parallel worlds. The existence of parallel worlds allows us to have a clear deterministic and local physical theory.

Wave Function

The wave function, at a particular time, contains all the information that anybody at that time can have about the particle. But the wave function itself has no physical interpretation. It is not measurable. However, the square of the absolute value of the wave function has a physical interpretation. We interpret |ψ(x,t)|2 as a probability density, a probability per unit length of finding the particle at a time t at position x.

The wave function ψ associated with a moving particle is not an observable quantity and does not have any direct physical meaning. It is a complex quantity. The complex wave function can be represented as

ψ(x, y, z, t) = a + ib

and its complex conjugate as

ψ*(x, y, z, t) = a – ib.

The product of wave function and its complex conjugate is

ψ(x, y, z, t)ψ*(x, y, z, t) = (a + ib) (a – ib) = a2 + b2

a2 + b2is a real quantity.

However, this can represent the probability density of locating the particle at a place in a given instant of time.

The positive square root of ψ(x, y, z, t) ψ*(x, y, z, t) is represented as |ψ(x, y, z, t)|, called the modulus of ψ. The quantity |ψ(x, y, z, t)|2 is called the probability.This interpretation is possible because the product of a complex number with its complex conjugate is a real, non-negative number.

We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one.

For the probability interpretation to make sense, the wave function must satisfy certain conditions.

dx =1

Only wave function with all these properties can yield physically meaningful result.

Physical significance of wave function

The Schrodinger equationalso known as Schrodinger’s wave equation is a partial differential equation that describes the dynamics of quantum mechanical systems by the wave function. The trajectory, the positioning, and the energy of these systems can be retrieved by solving the Schrodinger equation.

All of the information for a subatomic particle is encoded within a wave function. The wave function will satisfy and can be solved by using the Schrodinger equation. The Schrodinger equation is one of the fundamental axioms that are introduced in undergraduate physics.

Statistical Interpretation

It is not possible to measure all properties of a quantum system precisely. Max Born suggested that the wave function was related to the probability that an observable has a specific value.

In any physical wave if ‘A’ is the amplitude of the wave, then the energy density i.e., energy per unit volume is equal to ‘A2’. Similar interpretation can be made in case of mater wave also. In matter wave, if ‘Ψ‘is the wave function of matter waves at any point in space, then the particle density at that point may be taken as proportional to ‘Ψ2’ . Thus Ψ2 is a measure of particle density.

According to Max Born ΨΨ*=Ψ2gives the probability of finding the particle in the state ‘Ψ’. i.e., ‘Ψ2’ is a measure of probability density. The probability of finding the particle in a volume (dv=dxdydz) is given by

=

Since the particle has to be present somewhere, total probability of finding the particle somewhere is unity i.e., particle is certainly to be found somewhere in space. i.e.

=1

Or =1

This condition is called Normalization condition. A wave function which satisfies this condition is known as normalized wave function.

The wave function, at a particular time, contains all the information that anybody at that time can have about the particle. But the wave function itself has no physical interpretation. It is not measurable. However, the square of the absolute value of the wave function has a physical interpretation. We interpret |ψ(x,t)|2 as a probability density, a probability per unit length of finding the particle at a time t at position x.

With every physical observable there is associated a mathematical operator which is used in conjunction with the wavefunction. Suppose the wavefunction associated with a definite quantized value (eigenvalue) of the observable is denoted by Ψn (an eigenfunction) and the operator is denoted by Q. The action of the operator is given by

The mathematical operator Q extracts the observable value qn by operating upon the wavefunction which represents that particular state of the system. This process has implications about the nature of measurement in a quantum mechanical system. Any wave function for the system can be represented as a linear combination of the eigen functions Ψn, so the operator Q can be used to extract a linear combination of eigen values multiplied by coefficients related to the probability of their being observed.

With each measurable parameter in a physical system there is an associated quantum mechanical operator.

Need of operators arise because in quantum mechanics you are describing nature with waves (the wave function) rather than with discrete particles whose motion and dymamics can be described with the deterministic equations of Newtonian physics.

The operators associated with the parameters needed to describe the system. Some of those operators are listed below.

It is part of the basic structure of quantum mechanics that functions of position are unchanged in the Schrodinger equation, while momenta take the form of spatial derivatives. The Hamiltonian operator contains both time and space derivatives.

The quantum mechanical operator Q associated with a measurable property q must be Hermitian. Mathematically this property is defined by

whereΨa and Ψb are arbitrary normalizable functions and the integration is over all of space. Physically, the Hermitian property is necessary in order for the measured values (eigenvalues) to be constrained to real numbers.

We are looking for expectation values of position and momentum knowing the state of the particle, i.e., the wave function ψ(x,t).

Position expectation:

=

What exactly does this mean?

It does not mean that if one measures the position of one particle over and over again, the average of the results will be given by

Onthecontrary,thefirstmeasurement(whoseoutcomeisindeterminate)willOnthecontrary,thefirstmeasurement(whoseoutcomeisindeterminate)willcollapse the wave function to a spike at the value actually obtained, and the subsequent measurements (if they're performed quickly) will simply repeat that same result.

Rather, <x> is the average of measurements performed on particles all in the stateψ, which means that either you must find some way of returning the particle to its original state after each measurement, or else you prepare awhole ensemble of particles, each in the same state ψ, and measure the positions of all of them: <x> is the average of these results. The position expectation may also be written as:

=

To relate a quantum mechanical calculation to something you can observe in the laboratory, the "expectation value" of the measurable parameter is calculated. For the position x, the expectation value is defined as

This integral can be interpreted as the average value of x that we would expect to obtain from a large number of measurements.

While the expectation value of a function of position has the appearance of an average of the function, the expectation value of momentum involves the representation of momentum as a quantum mechanical operator.

where

is the operator for the x component of momentum.

Since the energy of a free particle is given by

and the expectation value for energy becomes

for a particle in one dimension.

In general, the expectation value for any observable quantity is found by putting the quantum mechanical operator for that observable in the integral of the wave function over space:

The Schrodinger Equation

Schrodinger wave equation, is the fundamental equation of quantum mechanics, same as the second law of motion is the fundamental equation of classical mechanics. This equation has been derived by Schrodinger in 1925 using the concept of wave function on the basis of de-Broglie wave and plank’s quantum theory.

Let us consider a particle of mass m and classically the energy of a particleis the sum of the kinetic and potential energies. We will assume that the potential is a function of only x.

So We have

E=K+V=mv2+V(x) = +V(x) ……….. (1)

By de Broglie’s relation we know that all particles can be represented as waves with frequency ω and wave number k, and that E= ℏω and p= ℏk.

Using this equation (1) for the energy will become

ℏω = +V(x) ……….. (2)

A wave with frequency ω and wave number k can be written as usual as

ψ(x, t) =Aei(kx−ωt) ……….. (3)

the above equation is for one dimensional and for three dimensional we can write it as

ψ(r, t) =Aei(k·r−ωt) ……….. (4)

But here we will stick to one dimension only.

=−iωψ⇒ ωψ= ……….. (5)

=−k2ψ ⇒k2ψ = - ……….. (6)

If we multiply the energy equation in Eq. (2) byψ, and using(5) and (6) , weobtain

ℏ(ωψ) = ψ+V(x)ψ⇒ = - + V(x) ψ……….. (7)

This is the time-dependent Schrodinger equation.

If we put the x and t in above equation then equation (7) takes the form as given below

= - + V(x) ψ(x,t) ……….. (8)

In 3-D, the x dependence turns into dependence on all three coordinates (x, y, z) and the term becomes∇2ψ.

The term |ψ(x)|2gives the probability of finding the particle at position x.

Let us again take it as simply a mathematical equation, then it’s just another wave equation. However We already know the solution as we used this function ψ(x, t) =Aei(kx−ωt) to produce Equations (5), (6) and (7)

But let’s pretend that we don’t know this, and let’s solve the Schrodinger equation as if we were given to us. As always, we will guess an exponential solution by looking at exponential behaviour in the time coordinate, our guess isψ(x, t) =e−iωtf(x) putting this into Equation (7) and cancelling thee−iωt yields

= - + V(x)f(x) ……….. (9)

We already know that E=. However ψ(x, t) is general convention to also use the letter ψto denote the spatial part. So we will now replace f(x) withψ(x)

Eψ = - + V(x) ψ ……….. (10)

This is called the time-independent Schrodinger equation.

We will now apply Schrodinger's wave equation in several examples using various potential functions. These examples will demonstrate the techniques used in the solution of Schrodinger's differential equation and the results of these examples will provide an indication of the electron behaviour under these various potentials.

A Free Particle

A particle is said to be free when no external force is acting on during its motion in the given region of space, and its potential energy V is constant.

Let us consider an electro is freely moving in space in positive x direction and not acted by any force, there potential will be zero. The Schrodinger wave equation reduces to

ψ + Eψ =0

Substituting E = k2

As the electron is moving in one direction (say x axis), then the above equation can be written as

+k2ψ =0

The general solution of the equation (2) is of the form ψ = ψ0e-iωt

The electron is not bounded and hence there are no restrictions on k. This implies that all the values of energy are allowed. The allowed energy values form a continuum and are given by

E =

The wave vector k describes the wave properties of the electron.

It is seen from the relation that E k2Thus the plot of E as a function of k gives a parabola.

Figure3

The momentum is well defined and in this case given by

pxψ =

Therefore, according to uncertainty principle it is difficult to assign a position to the electron.

Particle in a One Dimensional Deep Potential Well

Let us consider a particle of mass ‘m’ in a deep well restricted to move in a one dimension (say x). Let us assume that the particle is free inside the well except during collision with walls from which it rebounds elastically.

The potential function is expressed as

V= 0 for 0 ………. (1)

V= for x <0, x>L ………. (2)

Figure 4: Particle in deep potential well

The probability of finding the particle outside the well is zero (i.e.Ѱ =0)

Inside the well, the Schrödinger wave equation is written as

ψ + Eψ =0 …………….(2)

Substituting E = k2 …………….(3)

writing the SWE for 1-D we get

+k2ψ =0 …………….(4)

The general equation of above equation may be expressed as

ψ = Asin (kx + ϕ) …………….(5)

Where A and ϕ are constants to be determined by boundary conditions

Condition I: We have ψ = 0 at x = 0, therefore from equation

0= A sinϕ

As A then sinϕ =0 orϕ=0 …………….(6)

Condition II:Further ψ = 0 at x = L, andϕ=0,therefore from equation(5)

0= AsinkL

As A then sinkL =0 orkL=nπ

k = …………….(7)

where n= 1,2,3,4………

Substituting the value of k from (7) to (3)

)2 = E

This gives energy of level

En = n=1,2,3,4…so on …………….(8)

From equation En is the energy value (Eigen Value) of the particle in a well.

It is clear that the energy values of the particle in well are discrete not continuous.

Figure5

Using (6) and (7) equation (5) becomes, the corresponding wave functions will be

ψ =ψn= Asin …………….(9)

The probability density

|ψ(x,t)|2 =ψψ*

|ψ(x,t)|2=A2sin2 …………….(10)

The probability density is zero at x = 0 and x = L. since the particle is always within the well

…………….(11)

=1

A =

Substituting A in equation (9) we get

ψ =ψn= sinn=1,2,3,4….. …………….(12)

The above equation (12) is normalized wave function orEigen function belonging to energy value En

F

Figure6: Wave function for Particle