Q1) Explain dual nature of matter?

A1)

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

Q2) 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?

A2)

(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.

Q3) What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?

A3)

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.

Q4) What is Heisenberg’s Uncertainty Principle?

A4)

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.

Q5) 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.

A5)

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

Q6) Prove non-existence of electrons in the nucleus?

A6)

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.

Q7) Discuss applications of Heisenberg uncertainty principle?

A7)

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

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.     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 in

Fluence 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.

Q8) Derive the Schrodinger wave equation?

A8)

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.

Q9) 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?

A9)

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

Q10) 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?

A10)

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

Q11) The mass of a bullet is 10gm & uncertainty in its velocity is 5.25×10–26 cm/sec. Calculate the uncertainty in its position?

A11)

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

Q12) Discuss Particle in a One-Dimensional Deep Potential Well?

A12)

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: 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: 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

Figure: Wave function for Particle

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

Q13) A 45 kW broadcasting antenna emits radio waves at a frequency of 4MHz.

(a) How many photons are emitted per second?

(b) Is the quantum nature of the electromagnetic radiation important in analysing the radiation emitted from this antenna?

A13)

(a) The electromagnetic energy emitted by the antenna in one second is E = 45 000 J. Thus, the number of photons emitted in one second is

(b) Since the antenna emits a huge number of photons every second, 1.71031, the quantum nature of this radiation is unimportant. As a result, this radiation can be treated fairly accurately by the classical theory of electromagnetism.

Q14) Consider a one-dimensional particle which is confined within the region 0   x a and whose wave function is (x, t) =sin(πx/a) exp-iωt.

(a) Find the potential V(x).

(b) Calculate the probability of finding the particle in the interval a/4 x 3a/4.

A14)

Since the first time derivative and the second x derivative of  (x, t), are given by =-iω (x, t) and =- (π2/a2)  (x, t) the Schrödinger equation yields

Hence V(x,t) is time independent and given by

(b) The probability of finding the particle in the interval a/4 x 3a/4.can be obtained from

Q15) Calculate the de Broglie wavelength of an electron having a kinetic energy of 1000 eV. Compare the result with the wavelength of x-rays having the same energy?

A15)

The kinetic energy

For x-rays

Q16) Determine the de Broglie wavelength of an electron that has been accelerated through a potential difference of (i) 100 V, (ii) 200 V.

A16)

(i)The energy gained by the electron = 100 eV. Then

(ii)

Q17) The wave function of a particle of mass m moving in a potential V (x) is

Where A and k are constants. Find the explicit form of the potential V (x).

A17)

Substituting these values in the time dependence Schrödinger equation, we have

Q18) Define phase and group velocity?

Or

Define term vp and vg?

A18)

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: 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

2A cos [ 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 =          ….….(7)

Q19) Write a note on quantum computing?

A19)

Quantum computing is an area of computing focused on developing computer technology based on the principles of quantum theory, which explains the behaviour of energy and material on the atomic and subatomic levels. Quantum computing is the study of how to use phenomena in quantum physics to create new ways of computing. The basis of quantum computing is the Qubit. Unlike a normal computer bit, which can be 0 or 1, a Qubit can be either of those or a superposition of both 0 and 1.

Classical computers that we use today can only encode information in bits that take the value of 1 or 0. This restricts their ability. Quantum computing, on the other hand, uses quantum bits or qubits. It harnesses the unique ability of subatomic participles that allows them to exist in more than one state i.e. a1 and a0 at the same time. Superposition and entanglement are two features of quantum physics on which these supercomputers are based. This empowers quantum computers to handle operations at speeds exponentially higher than conventional computers and at much lesser energy consumption.

Three quantum mechanical properties are used in quantum computing to manipulate the state of a qubit.

1)     superposition,

2)     entanglement

3)     interference

Superposition refers to a combination of states we would ordinarily describe independently. To make a classical analogy, if you play two musical notes at once, what you will hear is a superposition of the two notes. Understanding superposition makes it possible to understand the basic component of information in quantum computing, the qubit. In classical computing, bits are transistors that can be off or on, corresponding to the states 0 and 1. In qubits such as electrons, 0 and 1 simply correspond to states like the lower and upper energy levels discussed above. Qubits are distinguished from classical bits, which must always be in the 0 or 1 state, by their ability to be in superpositions with varying probabilities that can be manipulated by quantum operations during computations.

Entanglement is a famously counter-intuitive quantum phenomenon describing behaviour we never see in the classical world. Entangled particles behave together as a system in ways that cannot be explained using classical logic. Entanglement is a phenomenon in which quantum entities are created and/or manipulated such that none of them can be described without referencing the others. Individual identities are lost. This concept is exceedingly difficult to conceptualize when one considers how entanglement can persist over long distances. A measurement on one member of an entangled pair will immediately determine measurements on its partner, making it appear as if information can travel faster than the speed of light.

Finally, quantum states can undergo interference due to a phenomenon known as phase. Quantum interference can be understood similarly to wave interference; when two waves are in phase, their amplitudes add, and when they are out of phase, their amplitudes cancel.

The degrees of freedom required to hold data and perform computation should be available as dimensions of the Hilbert space of a quantum system. This quantum system should be more-or-less isolated from its environment.

Another requirement of quantum computation is that it must be possible to place the quantum system in a fiducial starting quantum state.

A quantum computer is any device for computation that makes direct use of distinctively quantum mechanical phenomena, such as superposition and entanglement, to perform operations on data.

In a classical (or conventional) computer, information is stored as bits; in a quantum computer, it is stored as qubits (quantum bits).

The basic principle of quantum computation is that the quantum properties can be used to represent and structure data, and that quantum mechanisms can be devised and built to perform operations with this data.

Quantum computing could contribute greatly in the fields of finance, military affairs, intelligence, drug design and discovery, aerospace designing, utilities (nuclear fusion), polymer design, Artificial Intelligence (AI) and Big Data search, and digital manufacturing. If large-scale quantum computers can be built, they will be able to solve certain problems exponentially faster than any of our current classical computers. Quantum computers are different from other computers such as DNA computers and traditional computers based on transistors.

Q20) List the conditions wave function should satisfy?

A20)

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. 

dx =1

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