INTRODUCTION TO QUANTUM MECHANICS | unit 2 introduction to quantum mechanics

Engineering Physics

INTRODUCTION TO QUANTUM MECHANICS

UNIT 2

QUESTION BANK

1 Question: Explain dual nature of matter?

Solution:

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–particle behaviour. 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. 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 de Broglie 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

2 Question: 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.

3 Question: What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?

Solution: Accelerating potential V = 100 V. The de Broglie wavelength λ is

λ= h /p = 1 227/ nm

λ.1 227/ nm = 0.123 nm

The de Broglie wavelength associated with an electron in this case is of the order of x ray wavelengths.

4 Question: What is wave packet or wave group?

Solution:

We know that in classical physics a particle is well localized in space. We can calculate its position and velocity simultaneously.

But for quantum mechanics, a particle is not well localized in space. We cannot calculate its position and velocity simultaneously.

In Quantum mechanics it is describe that the matter wave associated with the particle. Wave functions depend on the whole space. Hence they cannot be localized.

If the wave function is made to vanish everywhere except in the neighbourhood of the particle or the neighbourhood of the classical trajectory, it can then be used to describe the dynamics of the particle. That is a particle which is localized within a certain region of space can be described by a matter wave whose amplitude is large in that region and zero outside it. This matter wave must then be localized around the region of space within which the particle is confined.

A localized wave function is called a Wave Packet or Wave Group. A wave packet or wave group therefore consists of a group of waves of slightly different wavelengths with phases and amplitudes so chosen that they interfere constructively over a small region of space and destructively elsewhere.

Wave packets are Not only useful in the explanation of isolated particles that are confined to a certain spatial region but they also play a key role in understanding the connection between quantum mechanics and classical mechanics. The wave packet concept also represents a uniting mathematical tool that describes particle-like behaviour and also its wave-like behaviour.

5 Question: Define phase and group velocity?

Define term vp and vg?

Solution:

Phase and group velocity are two important and related concepts in wave mechanics. They arise in quantum mechanics in the time development of the state function for the continuous case, i.e. wave packets.

According to the theory of relativity particle velocity (v) is always less than the speed of light c. But according to the De-Broglie wave velocity must be greater than c. This is an unexpected result. According to this the de-Broglie wave associated with the particle would travel faster than the particle itself thus leaving the particle far behind.

The difficulty was recovered by Schrodinger. He proposed that a material particle in motion is equivalent to a wave packet rather than a single wave. Wave packet comprises of a group of waves, each with slightly different velocity and wavelength.
Such a wave packet moves with its own velocity vg called the group velocity.

The individual wave forming the wave packet possesses an average velocity vp called the phase velocity.

It can be shown that the velocity of the material particle v is the same as group velocity

Let us assume two wave trains have same amplitude but different frequency and phase velocities

u (x, t) =A sin(ωt - kx)…….(1)

u’(x, t) =A sin[(ω+∆ω)t-(k+∆k)x]…….(2)

Where ω and (ω+Δω) are angular frequencies and k and (k +Δk) are propagation constants

The superposition of two waves is of the form

ψ(x,t) = u +u’ = A sin(ωt - kx)sin[(ω+∆ω)t - (k+∆k)x] …….(3)

As Δ ω and Δk are small therefore (ω+Δω) and (k + Δk) k, equation (3) reduces to

ψ(x,t) = 2Acos[ t -x] sin(ωt - kx).…….(4)

This equation represents a vibration of amplitude

2Acos[ t -x]…….(5)

The phase of the resultant wave moves with the velocity known as phase velocity

vp= ……..(6)

And the amplitude moves with the velocity known as group velocity

vg = =

vg = ….….(7)

6 Question: Explain in detail Davisson and Germer experiment?

Solution:

In this experiment, Davisson and Germer scattered a 54 eV monoenergetic beam of electrons from a nickel (Ni) crystal. The electron source and detector were symmetrically located with respect to the crystal’s normal, as shown in figure this is similar to the Bragg setup for X-ray diffraction by a grating. What Davisson and Germer found was that, although the electrons are scattered in all directions from the crystal, the intensity was a minimum at θ = 35°

Wave Aspect Of Particles

Davisson–Germer experiment: electrons strike the crystal’s surface at an angle ϕ

The detector, symmetrically located from the electron source, measures the number of electrons scattered at an angle θ, where θ is the angle between the incident and scattered electron beams.

It is maximum at θ = 50°; that is, the bulk of the electrons scatter only in well-specified directions. They showed that the pattern continued even when the intensity of the beam was so low that the incident electrons were sent one at a time. This can only result from a constructive interference of the scattered electrons. The reflected electrons formed diffraction patterns that were identical

With Bragg’s X-ray diffraction by a grating instead of the diffuse distribution pattern.

In fact, the intensity maximum of the scattered electrons in the Davisson–Germer experiment corresponds to the first maximum (n = 1) of the Bragg formula,

nλ = 2d sin ϕ

Where d is the spacing between the Bragg planes

ϕ is the angle between the incident ray and the crystal’s reflecting planes

θ is the angle between the incident and scattered beams

d is given in terms of the separation D between successive atomic layers in the crystal by d = D sin θ

For an Ni crystal, we have d = 0.091 nm, since D = 0.215 nm. Since only one maximum is seen at θ = 50° for a mono-energetic beam of electrons of kinetic energy 54 eV, and since 2ϕ + = π and hence sin ϕ = cos(θ /2) from figure.

We can obtain the wavelength associated with the scattered electrons:

λ = sin ϕ = cos θ = cos 25° = 0.165 nm

Now using results from de Broglie’s relation we will calculate the numerical value of λ. Since the kinetic energy of the electrons is K = 54 eV, and the momentum is p = with mec2 = 0.511 MeV.

Where mec2 is the rest mass energy of the electron and hc 197.33 eV nm, we can show that the de Broglie wavelength is

λ = = = 0.167 nm

This is in excellent agreement with the experimental value.

We have seen that the scattered electrons in the Davisson–Germer experiment produced interference fringes that were identical to those of Bragg’s X-ray diffraction. Since the Bragg formula provided an accurate prediction of the electrons’ interference fringes, the motion of an electron of momentum must be described by means of a plane wave

ψ(r, t) =Aei(k·r−ωt) = Aei(p·r−Et)/

7 Question: What is Heisenberg’s Uncertainty Principle?

Solution:

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 velocity are uniquely determined by means of Newton’s second law. So by Classical physics it 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, you must 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.

8 Question: 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 =0.527 x 10-29 m

9 Question: Prove non-existence of electrons in the nucleus?

Solution:

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.

10 Question: Discuss applications of Heisenberg uncertainty principle?

Solution:

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

Non-existence of electrons in the nucleus

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 =

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Therefore, if the electron exists in the nucleus, it should have an energy of the order 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. Calculation of zero point energy

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

Consider a particle which is confined to move under the influence of potential of dimensions a. Since the particle may be present anywhere in these dimensions, so uncertainty in position

∆x = a/2 (half the diameter).

According to uncertainty principle

∆x∆px = ℏ/2

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 m is the mass of the particle.

Energy even at O K 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 10-21 kg.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 10-8 seconds.

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 and ν + ∆ ν.

11 Question: Derive the Schrodinger wave equation?

Solution:

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 particle is 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) by ψ, and using(5) and (6) , we obtain

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)|2 gives 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 is ψ(x, t) =e−iωtf(x) putting this into Equation (7) and cancelling the e−iωt yields

= - + V(x) f(x) ……….. (9)

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

Eψ = - + V(x) ψ ……….. (10)

This is called the time-independent Schrodinger equation.

The Schrodinger equation also 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.

12 Question: The mass of an electron is 9.1×10–31 kg. Its uncertainty in velocity is 5.7×105 m/sec. Calculate uncertainty in its position?

Solution:

Given m= 9×10–31

ΔV= 5.7×105 m/sec

Δx=?

h= 6.6×10–34 Joule-Sec.

Δx.Δv ≥h/4πm

Δx≥h/4πmΔv

≥6.6×10–34/9×3.14×9.1×10–31×5.7×105

≥ 0.010×10–8

≥ 1×10–10m

13 Question: 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

14 Question: The mass of a bullet is 10gm & uncertainty in its velocity is 5.25×10–26 cm/sec. Calculate the uncertainty in its position?

Solution:

m=10 gm. h=6.6×10–27 erg-sec

Δv= 5.25×10–26 cm

Δx=?

Δx.Δv ≥h/4πm

Δx≥h/4πmΔv

≥6.6×10-34/4×3.14×10×5.25×10–26

≥ 0.10×10–2m

≥ 1×10–3 cm

15 Question: List the conditions wave function should satisfy?

Solution:

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 + b2 is 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.

The wave function must be single valued at each point.
The probability of finding the particle at time t in an interval ∆x must be some number between 0 and 1.
ψ must be finite everywhere.
ψ must be continuous everywhere and must also be continuous everywhere except where V(x) is infinite.
ψ (x) must vanish ψ0 as x.
The wave function should satisfy the normalization condition. Normalization condition of a wave function ψ is mathematical statement of existence of the particle somewhere. so that if we sum up all possible values ∑|ψ(xi,t)|2∆xi we must obtain 1. The total probability of finding the particle anywhere must be one. Normalization condition is given as

dx =1

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

16 Question: What is Physical significance of wave function?

Solution:

The wave function ‘Ѱ’ has no physical meaning. It is a complex quantity representing the variation of a matter wave.
The wave function Ѱ(r,t) describes the position of particle with respect to time .
It can be considered as ‘probability amplitude’ since it is used to find the location of the particle.
The square of the wave function gives the probability density of the particle which is represented by the wave function itself.
More the value of probability density, more likely to find the particle in that region.

17 Question: Discuss Particle in a One Dimensional Deep Potential Well

Solution:

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)