Unit 1 Quantum Mechanics PDF

Summary

This document provides an introduction to quantum mechanics, covering basic concepts, theories, and applications. It discusses foundational ideas like wave-particle duality and introduces problems and solutions related to the topic.

Full Transcript

Unit 1 Quantum Mechanics Basics of Quantum Mechanics Why Quantum Physics? The three theories, namely Newtonian mechanics, thermodynamics and Maxwell's electromagnetic theory together constitute the classical physics. The classical physics is based on: 1) The physical quantities can take...

Unit 1 Quantum Mechanics Basics of Quantum Mechanics Why Quantum Physics? The three theories, namely Newtonian mechanics, thermodynamics and Maxwell's electromagnetic theory together constitute the classical physics. The classical physics is based on: 1) The physical quantities can take any value. 2. An object can be isolated from its environment and treated as an independent quantity for the investigation. Classical mechanics (Newton's mechanics) and Maxwell's equations (electromagnetics theory) can explain MACROSCOPIC phenomena such as motion of billiard balls or rockets. Basics of Quantum Mechanics There are a few phenomena which classical mechanics failed to explain such as Stability of atom, spectral series of hydrogen atom (line spectra), black body radiation, photoelectric effect etc. Failures of Classical mechanics led to the Need of Quantum mechanics Quantum mechanics is used to explain microscopic phenomena such as photon-atom scattering and flow of the electrons in a semiconductor. QUANTUM MECHANICS is a collection of postulates based on a huge number of experimental observations. Basics of Quantum Mechanics 1) According to Planck's hypothesis in 1900, the radiating body consists of an enormous number of atomic oscillators vibrating at all possible frequencies and that each oscillator emits or absorb energy in discrete portions. 2) This discrete portions cannot be further subdivided. Einstein extended Planck’s hypothesis.. An electromagnetic wave having a frequency ν contains identical photons, each photon having an energy h ν Photoelectric effect: h ν = K.E + W K.E = h ν –W In 1921 A. Einstein received Nobel Prize De BROGLIE HYPOTHESIS Light can exhibit both kind of nature of waves and particles so the light shows wave particle duality nature. In some cases like interference ,diffraction and polarization it behaves as a wave while in other cases like photoelectric effect and Compton effect it behaves as a particle (photon) Not only light but every materialistic particle such as electron, proton or even heavier object exhibits wave particle duality nature. De Broglie proposed that a moving particle whatever its nature has waves associated with it these waves are called matter waves. De BROGLIE HYPOTHESIS De BROGLIE HYPOTHESIS De BROGLIE HYPOTHESIS Properties of matter waves Properties of matter waves Problems Obtain an expression for the de Broglie wavelength associated with an electron accelerated through V volts. Also find the wavelength for 100v and 54V Solution: The k.E. Acquired by the electron accelerated through V volts is (½)mv2 = eV mv2 = 2eV m2v2=2meV mv= 2meV or p = 2meV ------(1) = h/p or  = h/2meV (h= 6.625 x 10-34 J-s, m=9.1 x 10-31kg, e= 1.6 x 10-19 C Using all these values  = (12.27 /V) AU For v=100V ,  = 1.227 AU For v=54V  = 1.67AU Problems Calculate the de Broglie wavelength of an electron moving with speed 1/10 th of the velocity of light. m = 9.1 x 10-31kg, h= 6.625 x 10-34 J-s, c = 3x108m/s V = c/10 = 3 x 107 m/s P = mv = 2.73 x 10-23kg-m/s = h/p = 2.43 x 10-11 m  = 0.243 AU Find the De broglie wavelength of neutron whose K.E. is 1eV. Given: mass of neutron = 1.676 x 10-27kg k.E. of neutron = 1ev k.E. of neutron = 1 x 1.6 x 10-19 J (½) mv2 = 1.6 x 10-19 J V2 = (2 x 1.6 x 10-19 J) / m = 1.9093 x 108 V = 1.38 x 104m/s  = h / mv = 0.286AU Physical Significance of Wave Function and Probability Density Interpretation  We know there is a wave associated with a moving particle and the motion of a particle is guided by the wave group.  The mathematical function which describes wave group is the wave function ψ.  As the particle moves under the action of external forces the wave function changes with time. Thus, the variable quantity characterizing the De Broglie wave is called as wave function and it is denoted by the symbol ψ ( x,y,z,t ).  ψ describes the wave as a function of position and time.  The wave function usually contains all the information which the uncertainty principle allows us to know about the associated particle. But the wave function ψ itself has no physical interpretation, as it may be positive, negative or complex. Physical Significance of Wave Function and Probability Density Interpretation  ψ has no direct physical significance, as it is not an observable quantity. In general, ψ is a complex-valued function.  According to Heisenberg uncertainty principle, we can only know the probable value in a measurement.  The probability cannot be negative. Hence ψ cannot be a measure of the presence of the particle at the location (x, y, z).  This difficulty is solved by taking IψI2 = ψ ψ*. This quantity is called probability density.  The product of ψ and its complex conjugate is having physical significance and it represents the probability of finding the particle in a volume dxdydz.  A probability interpretation of the wave function was given by Max Born in 1926.  He suggested that the square of the magnitude of the wave function |ψ|2 evaluated in a particular region represents the probability of finding the particle in that region. Physical Significance of Wave Function and Probability Density Interpretation Probability, P, of finding the particle in an infinitesimal volume dV (= dx dy dz) is proportional to |ψ (x, y, z)|2 dx dy dz at time t. |ψ|2 is called the probability density and ψ is the probability amplitude. Since the particle is certainly somewhere in the space, the probability P = 1 and the integral of |ψ|2dV over the entire space must be equal to unity. That is The wave function ψ is in general a complex function. But the probability must be real. Physical Significance of Wave Function and Probability Density Interpretation Therefore to make probability a real quantity, ψ is to be multiplied by its complex conjugate ψ* Thus, ψ has no physical significance but |ψ|2 gives the probability of finding the particle in a particular region.. The above condition on ψ is known as normalisation condition. the wave function that satisfies above condition is called normalised wave function. Physical Significance of Wave Function and Probability Density Interpretation How to normalise a wave function? If the wave function is not normalised in order to normalise the wave function it is multiplied by some arbitrary constant and then the integral is evaluated over the entire space. The normalisation procedure is as follows: If ψ is not normalised wave function then multiply it by some constant A. Then evaluate the integral and equate it to unity and calculate the constant A called normalisation constant i.e. ∫ A ψ (A ψ)* dV = 1 Or AA*∫ ψ ψ* dV = 1 As A is real constant, we get IAI2∫ ψ ψ* dV = 1 This gives normalisation constant as IAI2 = 1 / (∫ψ ψ* dV ) The normalisation constant can be positive square root of above equation. Requirements of wave function Conditions to be satisfied by ψ – function 1) ψ function must be finite: The wave function must be finite everywhere. It must remain finite for all values of x, y, z. If ψ is infinite, it would imply an infinitely large probability of finding the particle at that point. Not finite at x=a Requirements of wave function 2) ψ function must be single–valued: The function related to a physical quantity cannot have more than one value at that point. If it has more than one value at a point,it means that there is more than one value of probability of finding the particle at that point. Not single valued Requirements of wave function 3) ψ function must be continuous: ψ function should be continuous across any boundary. Since ψ is related to a physical quantity, it cannot have a discontinuity at any point. Therefore, the wave function ψ and its space derivatives should be continuous across any boundary. Since ψ is related to a real particle, it cannot have a discontinuity at any boundary where potential changes. Discontinuous at x=a Well-behaved Wave Functions 4) ψ function must be normalised: Well behaved wave function Schrodingers time dependent wave equation Schrodingers time dependent wave equation Schrodingers time dependent wave equation. Schrodingers time dependent wave equation. Schrodingers time dependent wave equation. Schrodingers time independent wave equation. Schrodingers time independent wave equation. Schrodingers time independent wave equation. Schrodingers time independent wave equation. Operators in quantum mechanics. Operators in quantum mechanics. Eigen Functions and Eigen Values. Expectations Values. Problems.... Numericals. Numericals Problem : A wave function of the free particle in the range -∞ to +∞ is given by ψ = A. e –αx2/2 Motion of a free particle Suppose the particle is moving with constant potential i.e V(x) = constant. Then the force acting on the particle F = -dV/dx = 0 Because force is called as negative gradient of potential. Thus the particle is free particle so to predict the quantum mechanical behaviour of a free particle, we have to solve the schrodingers time independent wave equation Motion of a free particle. The solution of equation 4.2 are eigen functions. The possible solutions of equation 4.2 are e-ikx and eikx The solution of the time part is given by  (t) = e-iEt/ħ Since we have E = ħω There fore  (t) = e-iωt The wave function is Ψ (x,t) = ψ (x)  (t) Ψ (x,t) = eikx e-iωt = ei(kx-ωt) --------4.3 Ψ (x,t) = e-ikx e-iωt = e-i(kx+ωt) --------4.4 Equation 4.3 represents wave propagating along +x direction and Equation 4.4 represents wave propagating along –ve x direction Motion of a free particle. Let us consider equation 4.3 and Let A is the normalization constant Ψ (x,t) = A ei(kx-ωt) Ψ* (x,t) = A* e -i(kx-ωt) Ψ Ψ* = A A* = ei(kx-ωt) e -i(kx-ωt) Ψ Ψ* = A A* = Constant ( This probability density) There is difficulty in the normalisation of free partcile wave function. According to normalisation condition A A* [x] ∞ =1 -∞ A A* ∞ = 1 A A* is the amplitude of the wave. There fore the probability of finding the particle will be very small everywhere. Particle in an infinite deep potential well (Rigid box). A partcile moving under the influence of infinite square well potential is often called a particle in one dimensional rigid box Particle in an infinite deep potential well (Rigid box). The general solution of this equation may be written as Particle in an infinite deep potential well (Rigid box)... Particle in a finite deep potential well (Non rigid box). If the particle has total energy E < V0, Then according to classical mechanics particle may be in the region II only ( - a < x < a ). Because in the region I and III the momentum P = (2m(E-V0) will be imaginary and it is not possible. Thus the particle is permanently in the region II with momentum of constant magnitude P = (2mE). If E > V0 then only particle can go in the region I and III Particle in a finite deep potential well (Non rigid box. Particle in a finite deep potential well (Non rigid box. Numericals.. Numericals. 1 ) An electron is trapped in a rigid box of width 1 Å. Find its lowest energy level and momentum. Hence find energy of 5th level. As P= ħk= (h/2π)x(nπ/L) and n= 1  P= h/2L Numericals.2) Lowest energy of an electron trapped in a infinite potential well is 38 electron Volt. Calculate the width of the well. Numericals. 3) A small object of mass 1 g is confined to move between two rigid walls separated by a distance of 1mm. (a) Calculate the minimum speed of the object (b) If the speed is 3 x108 m/s, find the corresponding value of n. Sol: (a) Numericals. Numericals 4) Determine the probability that a particle trapped in a rigid box of length. be found between 0.45L and 0.55L for ground and first excited L can state. Given : ψ = (2/L) ( sin(nπx/L) ) Numericals. 5) Ruby laser emits light of wavelength 693.4 nm. If the light is due to transition from n=2 to n=1 state of an electron in one dimensional rigid box, find the width of the box. Numerical. Numerical. Numerical. Numerical. Numerical. Numerical. Numerical..

Use Quizgecko on...
Browser
Browser