Postulates of Quantum Mechanics

1. The state of a quantum mechanical system is completely specified by a function (r,t) that depends on the coordinates of the particle(s) and on time. This function, called the wave function or state function, has the important property that *(r,t) (r,t)d is the probability that the particle lies in the volume element d located at r at time t

The wavefunction must satisfy certain mathematical conditions because of this probabilistic interpretation. For the case of a single particle, the probability of finding it somewhere is 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 eigenstate of . 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 ai is 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.

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

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 = h ν /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).

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 = h ν /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).

De Broglie wavelength of matter waves: in terms of kinetic energy and associated with particle in thermal equilibrium

De-Broglie wavelength expressed in term of kinetic energy 

If a particle has kinetic energy K.E., then

De-Broglie wavelength associated with particle in thermal equilibrium

If the particle is in thermal equilibrium at temperature T, then their kinetic energy is given by

Where K = 1.38 X 10-23 J/K

For an electron

m = 9.1X 10-31 Kg; e = 1.6 X 10-19 C; h = 6.62 X 10-34 J.s

Therefore

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.

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 J

E = 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

Properties of matter waves

• Matter waves are the waves associated with matter in motion. Wavelength of the matter waves is given by λ=h/p.
• Matter waves are not progressive waves. They are localized waves.
• Matter waves are not single waves. They are group of waves (wave packet) assumed formed due to the superposition of two or more progressive waves.
• Matter waves called pilot waves and they represent the direction of propagation of matter.
• Unlike electromagnetic waves, matter waves will not have constant speed. The speed of matter waves depends on the mass of the particle.
• Vg x Vp=C2 shows that Vp, phase velocity of matter waves is greater than velocity of light. This shows that matter waves not physical waves
• Lighter the particle, greater is the De-Broglie wave length.
• The faster the particle moves, the smaller is its De-Broglie wave length.
• The De-Broglie wave length of a particle in independent of the charge or nature of the particle.
• The matter waves are not electromagnetic in nature. The electromagnetic waves are produced only by charged particle.

1. λ =   =        =      
2. Where ℏ = h/2π. The expression known as the de Broglie relation

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.

Key Takeaways

1. Schrodinger wave equation, is the fundamental equation of quantum mechanics
2. The time-dependent Schrodinger equation is given by

⇒   = - + V(x) ψ 

3.     The time-independent Schrodinger equation is given by

   Eψ = - + V(x) ψ    

Let R be some region in space.

(Notation: the boundary of R—that is, the surface surrounding R—is often denoted ∂R.)

Suppose a particle has a wave function ψ(x,t). We are interested in the probability of finding the particle in the region R:

Prob(R) =∫ R|ψ|2d3x

Since there is some chance of the particle moving into or out of R, this probability can change in time. Its time derivative is (since|ψ|2=ψ∗ψ).

Prob(R) =∫ R(+ ψ∗)d3x

Now use the Schrodinger equation and its complex conjugate:

Solve for and and insert into (1) to obtain

Check the last equality in equation (3)—do the derivatives!) Then by Stokes theorem (orGreen’s theorem), equation (3) becomes

(Depending on the place you learned this, you may have seen the last term written instead as−∫ J·dA or  −∫ J·dS.)

Equation (5) should look familiar from E&M: it’s the same form as the equation for charge conservation. It says that the probability I R can change only by an amount equal to the flux of the “probability current ”J through the surface surrounding R. The current J therefore describes the flow of probability, the nearest thing we have in quantum mechanics to a description of the motion of a 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 + E ψ =0        

Substituting   E = k2                             

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

+ k2 ψ =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 k2 Thus the plot of E as a function of k gives a parabola.

 Figure 1: Free-particle wave-function

The momentum is well defined and in this case given by

px ψ =

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

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.

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.

Figure 2: Wave Packet

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)

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. 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 no choice 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.

APPLICATION OF UNCERTAINTY PRINCIPLE

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

1. Non-existence of electrons in the nucleus or Non-confinement of electron 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 =

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 or Calculation 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 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

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 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.

Key Takeaways

1. 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).
2. The three-dimensional form of the uncertainty relations for position and momentum can be written as follows:

3.     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.

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

5.    

6.     Electron cannot exist inside the nucleus because experimental results show that no electron or particle in the atom possess energy greater than 4 MeV.

7.     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.

8.     Heisenberg uncertainty principle proves proton, neutron and alpha particles exists within the nucleus.

9.     Size of Elementary cell in Phase space is given by Heisenberg uncertainty principle

10. Frequency of light is uncertain by

∆ ν = = Hz 0.8 x 107 Hz

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

On the contrary, the first measurement (whose outcome is indeterminate) will On the contrary, the first measurement (whose outcome is indeterminate) will collapse 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 state ψ, which means that either you must find some way of returning the particle to its original state after each measurement, or else you prepare a whole 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 wavefunction over space:

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 3: 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 + E ψ =0    …………….(2)    

Substituting   E = k2                                       …………….(3)

writing the SWE for 1-D we get

+ k2 ψ =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  or  ϕ=0    …………….(6)

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

0 = Asin kL

As A then sinkL =0  or  kL=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.

Figure 4: The energy values for an electron in a potential box

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

ψ = ψn = Asin   …………….(9)

The probability density

|ψ(x,t)|2 = ψ ψ*

|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 = sin    n=1,2,3,4…..    …………….(12)

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

F

Figure 5: Wave function for Particle

4.8.2 SIMPLE HARMONIC OSCILLATOR (ONE DIMENSIONAL)

It’s time to study another example of solving the Schrodinger equation for a particular potential energy function V(x). This example is the harmonic oscillator, for which V(x) is quadratic.

For a Classical harmonic oscillator a mass m attached to spring of spring constant k. The motion of classical harmonic oscillator is governed by Hooke’s law

      …………(1)

 And the solution is given by

     …………(2)

       …………(3)

is the (angular) frequency of oscillation. The potential energy is

       …………(4)

its graph is a parabola. Of course, there's no such thing as  perfect simple harmonic oscillator-if you stretch it too far the spring is going to break, and typically Hooke's law fails long before that point is reached. But practically any potential is approximately parabolic, in the neighbourhood of a local minimum (Figure 6). Formally, if we expand V(x) in a Taylor series about the minimum:

…………(5)

Subtract V (xo) [you can add a constant to V(x) with impunity, since that doesn't change the force], recognize that V'(xo) = 0 (since xo is a minimum), and drop the higher-order terms [which are negligible as long as (x - xo) stays small], the potential becomes

     …………(6)

which describes simple harmonic oscillation (about the point xo), with an effective spring constant k = V” (xo). That's why the simple harmonic oscillator is so important: Virtually any oscillatory motion is approximately simple harmonic, as long as the amplitude is small.

Figure 6: Parabolic approximation (dashed curve) to an arbitrary potential, in the neighbourhood of a local minimum.

The quantum problem is to solve the Schrodinger equation for the potential

      …………(7)

(it is customary to eliminate the spring constant in favour of the classical frequency, using Equation (3). As we have seen, it suffices to solve the time-independent Schrodinger equation:

   …………(8)

In the literature you will find two entirely different approaches to this problem. The first is a straightforward "brute force" solution to the differential equation, using the method of power series expansion; it has the virtue that the same strategy can be applied to many other potentials. The second is a diabolically clever algebraic technique, using so-called ladder operators also known as algebraic method. But we will solve this with the analytic method for now.

Analytic Method

The Schrodinger equation for the harmonic oscillator (Equation 8):

Things look a little cleaner if we introduce the dimensionless variable

     …………(9)

in terms of ,the Schrodinger equation reads

       …………(10)

where K is the energy, in units of (1/2) ℏω:

        …………(11)

We have to solve Equation (10), and in the process obtain the "allowed" values of K (and hence of E).

To begin with, note that at very large  (which is to say, at very large x), 2 completely dominates over the constant K, so in this regime

        …………(12)

Which has the approximate solution

      …………(13)

The B term is clearly not normalizable (it blows up as |x| the physically acceptable solutions, then, have the asymptotic form

     …………(14)

This suggests that we “peel off" the exponential part,

       …………(15)

in hopes that what remains [h()] has a simpler functional form than ()itself. Differentiating Equation (15), we have

and

so the Schrodinger equation (Equation 10) becomes

     …………(16)

A solution to Equation (16) in the form of a power series in :

   …………(17)

Differentiating the series term by term,

and

Putting these into Equation 16, we find

  …………(18)

It follows (from the uniqueness of power series expansions) that the coefficient of each power of  must vanish,

and hence that

      …………(19)

This recursion formula is entirely equivalent to the Schrodinger equation itself. Given ao it enables us (in principle) to generate a2, a4, a6,..., and given a1 it generates a3, a5, a7,.... Let us write

      …………(20)

where

is an even function of  (since it involves only even powers), built on a0, and

is an odd function, built on a1. Thus Equation (19) determines h() in terms of two arbitrary constants (a0 and a1) –which is just what we would expect, for a second – order differential equation.

However, not all the solutions so obtained are normalizable. For at very large j, the recursion formula becomes (approximately)

with the (approximate) solution

For some constant C, and this yields (at large , where the higher powers dominate)

Now, if h goes like exp(2),then  (remember ? - that's what we're trying to calculate) goes like exp (2/2) (Equation 15), which is precisely the asymptotic behaviour we don't want. There is only one way to wiggle out of this: For normalizable solutions the power series must terminate. There must occur some "highest" j (call it n) such that the recursion formula spits out an+2 = 0 (this will truncate either the series heven or the series hodd; the other one must be zero from the start). For physically acceptable solutions, then, we must have

for some positive integer n, which is to say (referring to Equation 11) that the energy must be of the form

   …………(21)

For the allowed values of K, the recursion formula reads

       …………(22)

If n = 0, there is only one term in the series (we must pick a1= 0 to kill hodd, and j = 0 in Equation 22 yields a2 = 0):

and hence

(For n = 1 we pick ao = 0, and Equation (22) with j = 1 yields a3 = 0, so

and hence

For n = 2, j = 0 yields a2 = -2ao, and j = 2 gives a4 = 0, so

and

and so on.

In general, hn() will be a polynomial of degree n in , involving even powers only, if n is an even integer, and odd powers only, if n is an odd integer. Apart from the overall factor (a0 or a1) they are the so-called Hermite polynomials, Hn (). The first few of them are listed in Table 1. By tradition, the arbitrary multiplicative factor is chosen so that the coefficient of the highest power of  is 2n. With this convention, the normalized stationary states for the harmonic oscillator are

   …………(23)

In Figure 7 a; We have plotted n(x) for the first few n's.

The quantum oscillator is strikingly different from its classical counterpart-not only are the energies quantized, but the position distributions have some bizarre features. For instance, the probability of finding the particle outside the classically allowed range (that is, with x greater than the classical amplitude for the energy in question) is not zero and in all odd states the probability of finding the particle at the center of the potential well is zero, Only at relatively large n do we begin to see some resemblance to the classical case.

Table 1: The first few Hermite polynomials, Hn (x).    

Figure 7: (a) The first four stationary states of the harmonic oscillator.

(b) Graph of |100|2, with the classical distribution (dashed curve) superimposed.

In Figure 7(b) We have superimposed the classical position distribution on the quantum one (for n = 100); if you smoothed out the bumps in the latter, the two would fit pretty well, however, in the classical case we are talking about the distribution of positions over time for one oscillator, whereas in the quantum case we are talking about the distribution over an ensemble of identically-prepared systems.

4.8.3 CONCEPT OF SCATTERING FROM A POTENTIAL BARRIER AND TUNNELING

Consider the following piecewise continuous, finite potential energy:

U=U0     x <0      (1)

U= 0  0x L      (2)

U=U0  L < x.       (3)

We want to solve Schroedinger’s Equation for this potential to get the wave functions and allowed energies for E < U0. We will refer to the three regions as regions 0, 1, and 2 with associated wave functions ψ0,ψ1,ψ2.

Figure 8: Finite Square Well Potential Energy

The time-invariant, non-relativistic Schroedinger’s equation is

         (4)

that can be rearranged to give

         (5)

It is convenient to define two new variables (both positive), one for regions 0 and 2, and one for region 1—they are wavenumbers:

         (6)

            (7)

and Schrödinger’s equation becomes

In regions 0 and 2 the general solution is a linear combination of exponentials with the same form, but with different constants, namely

In region 1 we have the same general solution that we had for the infinite square well,

Equations (10) to (12) have 7 unknowns—A,B,C,D,F,G and the energy E that is im-plicitly contained in the variables κ0,k1. Therefore we need to get 7 equations to be ableto solve for the unknowns.

We will first use the requirement that the wavefunction remain finite everywhere. Consider ψ2 as x→∞. For this to remain finite we must require G= 0. Similarly, as x→−∞, we require A= 0. Our solutions become

The next step is to require that the wavefunction and its first derivative be continuous everywhere, and in our case we look at the boundaries, x= 0 and x=L.

Hence ψ0=D exp(+κ0x).Take derivatives of the wave functions,

There remain 3 unknowns, D,F, and E. Finding them is a bit messier! Consider the boundary conditions at x=L,

We are not going much farther, but if we divided Equation (22) by Equation (23), we can see that the constants D and F cancel leaving us with one rather difficult equation to solve for energy E (remember this is implicitly included in the values of k1and κ0.)

There is one remaining condition, normalization, that for this problem is

Even without solving the entire problem we can make some conclusions about the wave-function and the allowed energy levels. Recall that for an infinite square well potential of width L the allowed energies are quantized and

Here is Equation(22)/Equation(23)

Now put in the values ofκ0andk1from Equations (6) and (7) and do some algebra to get

This equation is a single equation in a single unknown, E < U0. Once we have the details of our particle (its mass) and the potential energy (depth U0 and width L), we can solve it. There is no analytic solution, only a numerical one.

From equation (25) With n being any positive integer. Outside the well the wavefunction is 0. We are certain that the particle is somewhere inside the box, so ∆x ∞ =L.

With the finite well, the wavefunction is not zero outside the well,

Figure 9: Wave function in Finite well

So ∆x finite> L, hence from the uncertainty principle, ∆pfinite x<∆p∞x. This suggests that the average value of momentum is less for the finite well, and therefore that the kinetic energy inside the well is less for the finite well than for the infinite well. Indeed this is borne out with detailed analysis. In addition, the number of allowed energy levels is finite, and there is a possibility that a well may be sufficiently narrow or sufficiently shallow that no energy levels are allowed. Also note that the non-zero wave functions in regions 0 and 2 mean that there is a non-zero probability of finding the particle in a region that is classically forbidden, a region where the total energy is less than the potential energy so that the kinetic energy is negative.

TUNNELING

The so-called tunnel effect of quantum mechanics can be derived from a special case of the potential well, by changing −V0 into +V0, thus creating a potential barrier, as seen in Figure 10.

Figure 10: Tunnel effect: For a given potential barrier with height V0 the solutions of the Schrodinger equation with energy E < V0 still have a non vanishing  probability density in region III, which allows them to ”tunnel” through the barrier although this would classically be forbidden.

Classically, a particle with less energy than the potential barrier could only be reflected. But in quantum mechanics, due to continuity the wave function decreases exponentially in the forbidden region II, resulting in a non-vanishing probability density in region III.

It allows the particle to pass the barrier as if it was through a tunnel, this linguistic illustration gives rise to the name tunnel effect.

Mathematically, we can use the solutions

   …………(1)

Where

…………(2)

Since the exponent of the solution in region II is not imaginary anymore, So we have solution in the form their hyperbolic counterparts

Tunneling – Transmission Coefficient

For the transmission amplitude we get the result

  …………(3)

which, using the identity

     …………(4)

gives the transmission coefficient

…………(5)

It can be simplified under the condition that qL1, which is a good approximationin most cases. Then

    …………(6)

and we can rewrite Equation (5) as

 …………(7)

We then use Equation (2) to express the transmission coefficient by the energy and potential strength to get

    …………(8)

Using exey=ex+y we can write the whole coefficient as an exponential function

  …………(9)

Since we required that qL be much bigger than one and the logarithm increases only very slowly we can conclude that the first term in the exponent outweighs the second one, which we therefore neglect to obtain.

      …………(10)

We now have a good approximation for the transmission probability of a single potential step, constant in certain interval and vanishing outside. This potential is of course

Figure 11: Calculation of Gamow factor: The generalization of the transmission coefficient, from a single constant potential barrier with width 2L, to the incorporation of a function V(x) is done straightforward by integration of infinitesimal potential barriers.

a very crude approximation of real life potentials, which usually are more complicated functions of x. To meet these concerns we can generalize the transmission coefficient of Equation (10) to the so called Gamow factor by ”chopping” a given potential in infinitesimal potential steps with constant values and integrating over a reasonable range [x1,x2]for which the potential stays above a certain value, see Figure 11.

Physical Examples of Tunneling

We will now briefly present some examples, where the tunnel effect explains the observed phenomena.

Tunneling between conductors:

Imagine two conducting materials, separated by a thin insulating material, sketched in Figure 12. The tunnel effect then allows the electrons to tunnel through that barrier, thus creating a current. This effect is also observed for super conducting materials, where it is named Josephson effect.

Cold emission:

Electrons can be emitted from metals at very low temperatures, even without incident light if an exterior electric field is applied. Assuming that the electrons have a very low energy compared to the potential height, it is not very probable that an electron can tunnel through the potential barrier. Only if an electric field raises the energy of the electron, the transmission coefficient increases and the electron emission can be observed.

Figure 12: Tunneling through insulator: Between two conducting materials, separated by a thin insulating barrier, the tunnel effect creates a current.