QUANTUM MECHANICS/PHYSICS, Lecture notes of Quantum Mechanics

Download QUANTUM MECHANICS/PHYSICS and more Lecture notes Quantum Mechanics in PDF only on Docsity! Quantum Mechanics 1 Introduction In 1924, French scientist Louis de Broglie proposed that all moving objects exhibit wave like properties under suitable conditions. The waves associated with a particle in motion are called matter waves. The idea originated by analogy with radiation which has dual nature of wave and particle. The momentum of a photon (particle of light) is given by p = h λ where h is the Planck’s constant and λ is the wavelength of light. From this λ = h p . De Broglie assumed that this equation is very general and can be applied to other particles as well. Hence the wavelength of a particle of mass m moving with velocity v is given by λ = h p = h mv If an electron of charge e and mass m is accelerated by a potential difference of V volts, its kinetic energy is K.E. = eV 1 2 mv2 = eV v = √√√√2eV m Then the wavelength of the electron is λ = h mv = h m √ m 2eV λ = h√ 2meV = 1.23× 10−9 √ V m Remember that this formula is applicable only for electrons Problem : Find the De Broglie wavelength of a)a ball of mass 500g moving with a speed 50km/hr and b) an electron of mass 9.1 x 10−31kg accelerated through 1 100V In the first part of the problem, we find that the wavelength of the ball 10−34 m is so small compared to its size and hence we cannot find any wave aspect in its behavior. In the second part we find that the wavelength 0.123 x 10−10m is much larger compared to the classical electron radius of ' 10−15 m. Hence it cannot be neglected. De Broglie had no experimental evidence to support his hypothesis. How- ever he was able to show that this leads to the stationary orbits and energy quantization postulated in the Bohr model of hydrogen atom in 1913. Later in 1927, De Broglie’s hypothesis was experimentally proved by electron diffraction experiments conducted by Davission and Germer and also by G P Thomson. 2 Wave function and Probability In water waves, the quality that varies periodically with space and time is the relative height of the water surface. In sound waves, it is pressure. In light waves, electric and magnetic fields vary. The quantity whose variations make up matter waves is called wave function represented by Ψ (Greek letter pronounced as psi). In general Ψ is a function of all space coordinates and time ie., Ψ = Ψ(x, y, z, t). Ψ can be expressed as a mathematical function but is not physically observable or measurable. The mathematical theory of quantum mechanics describes how to use Ψ(x, y, z, t) to determine values of particle’s position, velocity, momentum, energy, angular momentum etc. Even though Ψ(x, y, z, t) does not have a physical meaning, square of the absolute value of Ψ(x, y, z, t) ie., |Ψ|2 represents the probability of finding the particle at the point (x,y,z) at the time t. |Ψ|2 is called probability density or probability distribution function. Larger the value of |Ψ|2, larger is the probability and smaller the value of |Ψ|2, smaller is the probability. In general Ψ can be a complex function.|Ψ|2 = ( √ (Ψ∗Ψ))2 is always positive.Wave function Ψ is also called probability amplitude because of this reason. |Ψ|2dV is the probability of finding the particle in a small volume dV around the point (x,y,z) at time t. 3 Wave packets The wave function of the De Broglie waves corresponding to a moving body reflects the probability that it will be found at a particular place at a particular 2 Now the probability of finding the particle |Ψ|2 is zero everywhere except in the region of width ∆x. Hence the uncertainty in the position of the particle is only ∆x. This type of a wave is formed by interference of a large number of waves of slightly different wavelengths. Since wavelength of the combination can have any values in the range, momentum of the particle p = h λ can also have different values. Thus there is an uncertainty in the momentum of the particle represented by ∆p and the momentum cannot be exactly determined. For narrower wave packets the number of component waves is larger and hence ∆p will be larger. It is impossible to know simultaneously and exactly the position and momentum of a particle. This statement is known as Heisen- berg’s uncertainty principle. If ∆x is the uncertainty in position and ∆p is the uncertainty in the corresponding momentum (x component of momentum) then it can be shown that ∆x∆p ≥ 1 2 h̄ where h̄ = h 2π . Similar uncertainty relations occur in the case of pairs of variables angular displacement (θ) and angular momentum (L), and energy (E) and time (t). ∆L∆θ ≥ 1 2 h̄ ∆E∆t ≥ 1 2 h̄ It should be noted that there is no uncertainty relationship between ∆x and component of momentum other than x component. Why uncertainty principle is not relevant in the macroscopic world? A bullet of mass 25g is moving with a speed 400m/s. Speed is measured accurate up to 0.02%. Calculate the certainty with which the position of the bullet can be located. h = 6.625× 10−34Js 5 Ans. The uncertainty in position is ∆x ≥ 2.626× 10−32m. This uncertainty in position is much less than the size of an atom which can be neglected in this case. 6 Applications of uncertainty principle Planck’s constant h is so small that the limitations imposed by the uncertainty principle are significant only in the subatomic level. On such a scale, this principle is of great help in understanding many phenomena. 1. Non existence of electrons inside the atomic nucleus In earlier days it was thought that the electrons coming out of nucleus as β rays in radioactivity actually exist within the nucleus. Uncertainty principle showed that this is impossible. The diameter of the nucleus is typically 10−14m. Hence if the electron exists inside the nucleus the maximum uncertainty in its position is ∆x = 10−14m. The minimum uncertainty in the momentum of the electron is ∆p = h̄ 2∆x = 6.625× 10−34 4× 3.14× 10−14 = 0.5275× 10−20kgm/s If this is the uncertainty in momentum the minimum momentum itself should be at least this much.ie; pmin = 0.5275× 10−20kgm/s Minimum kinetic energy = pc = 0.5275×10−20×3×108 = 1.5825×10−12J = 1.5825×10−12 1.602×10−19 eV ≈ 10MeV The kinetic energy of the electron must be greater than 10MeV if it exists inside the nucleus. Electrons emitted in β decay have energies which are only a fraction of this value. Thus it is concluded that electrons from the nuclei are not those which previously exist inside. 2. Width of spectral lines When excited atoms return to the ground level, the radiation emitted by them do not have a sharply defined frequency. This can be proved using uncertainty principle. The average time between the excitation of an atom and the time it radiates (also called the life time of an atom in an energy state) is of the 6 order of 10−8s.Since ∆E∆t ≈ h̄ 2 the uncertainty in the energy of the emitted photon is ∆E ≈ h̄ 2∆t = 6.625× 10−34 2× 3.14× 2× 10−8 ≈ 5.3× 10−27J Therefore the minimum uncertainty in the frequency of the emitted photon is ∆ν = ∆E h = 8× 106Hz Hence the emitted frequency is not sharp, or the spectral lines have a finite width. 7 How can we determine Ψ for a particle? - The Schrodinger Equation We assume that the wave function of a particle moving freely along the +ve direction of X-axis Ψ = Ae−iω(t−xv ) (1) Here ω = 2πν and v = λν. Therefore Ψ = Ae−i2πν(t− x λν ) = Ae−2πi(νt−xλ ) (2) We know frequency and wave length of the wave are related to energy and momentum of the associated particle by the relations E = hν and λ = h p . Using these relations, the equation for Ψ becomes Ψ = Ae−2πi(Eth − px h ) (3) Let h 2π = h̄. Therefore 2π h = 1 h̄ . Substituting this,equation 3 becomes Ψ = Ae− i h̄ (Et−px) (4) Differentiating equation 4 partially with respect to t ∂Ψ ∂t = Ae− i h̄ (Et−px). −iE h̄ = −iE h̄ Ψ or EΨ = −h̄ i ∂Ψ ∂t = ih̄ ∂Ψ ∂t (5) Differentiating equation 4 partially with respect to x ∂Ψ ∂x = Ae− i h̄ (Et−px). ip h̄ = ip h̄ Ψ or pΨ = h̄ i ∂Ψ ∂x = −ih̄∂Ψ ∂x (6) 7 operation is to be carried out on the quantity that follows it. Symboli- cally an operator is represented by a cap (ˆ) above the usual symbol of the variable.eg. x̂, p̂, Ê etc. Some dynamic variables and their quantum mechanical operators are given below. Variable Symbol Operator Position x̂ x Linear momentum p̂ −ih̄∇ Kinetic energy Ê − h̄2 2m∇ 2 Hamiltonian Ĥ − h̄2 2m∇ 2 + V (b) When the operator acts on a wave function, in general we will get a new wave function as shown below. ÂΨ = φ If φ = aΨ, the above equation takes the form ÂΨ = aΨ This type of an equation is known as Eigen value equation. Here Ψ is called the Eigen function and a is called the corresponding Eigen value. For quantum mechanical operators, eigen value is real number. Eigen value is the result that we obtain when we make a measurement of the variable corresponding to  in a particular trial. (c) In general when a measurement of a dynamic quantity A is made on a system whose wave function is Ψ, we get different values during different trials. the value that can be expected from a particular trial is given by <  >= ∫ Ψ∗ÂΨdV∫ Ψ∗ΨdV <  > is called the expectation value. It is the value of A that we would obtain if we measured the quantity on a large number of particles described by the same wave function at some instant t and then averaged the results. If the wave function is normalized <  >= ∫ Ψ∗ÂΨdV 10 Applications of Schrodinger equation 10.1 Particle moving inside a box (particle moving in an one dimensional infinite square well potential) Consider a particle moving inside a box along the x-direction, bouncing back and forth between the walls of the box. L is the length of the box. The box has insurmountable potential barriers at x=0 and x=L ie; the box is supposed to have walls of infinite height at x=0 and x=L 10 The variation of potential energy is as follows V = 0, 0 < x < L V =∞, for x ≤ 0, x ≥ L The particle has a mass m and its position at any instant is 0 < x < L. Since the particle cannot exist outside the box, its wave function Ψ is 0 for x≤0 and x≥L. Our task is to find what Ψis within the box.ie., between x=0 and x=L. Since the potential energy does not depend on time, we can use time independent Schrodinger equation. d2ψ dx2 + 2m h̄2 (E − V )ψ = 0 In the region 0 < x < L; V = 0 d2ψ dx2 + 2m h̄2 Eψ = 0 put 2mE h̄2 = k2, then d2ψ dx2 + k2ψ = 0 The general solution of this differential equation is ψ = Aeikx +Be−ikx To evaluate the constants we apply the boundary conditions. Since the particle cannot jump over the walls the wave function does not exist outside the box. ψ = 0 at x = 0 and ψ = 0 at x = L 11 Applying the first condition we get A+B = 0 or B = −A ψ = A(eikx − e−ikx) ψ = 2iA sin kx Applying the second condition we get 2iA sin kL = 0 Since 2iA cannot be zero, this implies sin kL = 0 or kL = nπ or k = nπ L where n = 1, 2, 3, ... Thus ψ = 2iA sin nπx L This equation shows that the the wave function has several expressions depending on the value of n. The nthwave function is given by ψn = C sin nπx L where C = 2iA The value of n depends on the energy of the particle in the box E as follows. We have kL = nπ but k2 = 2mE h̄2 ; k = √ 2mE h̄2√ 2mE h̄2 L = nπ since n should be an integer, the energy of the particle E can have only certain values. Squaring the above expression 2mE h̄2 L2 = n2π2 E = En = n2π2h̄2 2mL2 We find that the energy of the particle in the potential well can have certain values only or the energy is quantized . The minimum energy is π2h̄2 2mL2 . For each value of n, there is an energy level En and a corresponding wave function ψn. Each value of En is called energy eigen value and the corre- sponding ψn is called the energy eigen function. Thus inside the potential well the particle can have only discrete energy values. The particle cannot have zero energy. The energy levels are not equally spaced. 12 Region I ψ1 = Aeiαx +Be−iαx Region II ψ2 = Ceβx +De−βx Region III ψ3 = Feiαx +Me−iαx Here Aeiαxis the wave incident on the barrier, Be−iαx is the reflected wave from the barrier(Region I). On the other side of the barrier (x>L, Region III) Feiαxrepresents the transmitted wave. Since there are no particles coming from the right side in this region, the coefficient M in the second term of ψ3 must be zero. In region II, De−βxrepresents the wave penetrating the barrier and Ceβxrepresents the reflected wave. Square of the absolute value of the wave function gives the probability of the associated process (like reflection, penetration, transmission etc.). Since the probability is non zero in all the three regions we can conclude that the particles in region I can actually penetrate the barrier and get transmitted into region III. This phenomenon is called quantum mechanical tunnel- ing. The transmission coefficient is defined as T = Probability density of the transmittedwave Probability density of the incidentwave This is a finite non zero quantity in quantum mechanics. Its value depends on the width of the potential barrier (Larger the width smaller is the value of T) and the height of the barrier i.e., V0 (Larger the value of V0 compared to E, smaller is the value of T). Thus there is a non-zero probability of transmission even though E<V0 i.e., the energy of the object is less than the height of the potential barrier. This is in contrast to classical situation. It is purely a quantum effect signifying the wave property of quantum objects. The probability of tunneling depends on the height and width of the barrier. 15 Applications of tunneling (a) Tunnel diode (b) Explanation of α-decay (c) Scanning tunneling microscope (STM) Problems (a) Find the expectation value of position of a particle in a one dimensional infinite potential well of width L (Ans L 2 ) (b) Find the expectation value of momentum of a particle in a one dimen- sional infinite potential well of width L ( Ans. 0) (c) Estimate the De Broglie wavelength of an electron moving with a kinetic energy of 100eV (d) A certain excited state of helium atom is known to have an average lifetime of 2.11× 10−8s. What is the minimum uncertainty with which the frequency of the emitted radiation can be measured? (e) An electron and a proton are moving with the same kinetic energy. Which one has shorter wavelength? (f) Calculate the uncertainty in frequency of the emitted radiation if the uncertainty in time of an excited atom is 5× 10−8s. (g) The time gap between the excitation of an atom and emission of ra- diation is ∆t = 10−8s. Find the uncertainty in the frequency of the radiation. (h) Find the De Broglie wavelength of an electron whose kinetic energy is 10keV. (i) Find the energy of the ground state of an electron moving in an infinitely high potential box of width 1A0. (j) A particle limited to the x axis has a wave function ψ = ax between x=0 and x=1. ψ = 0 elsewhere. Find (a) The probability that the particle can be found between x=0.45 and x=0.55. (b) The expectation value of the particle’s position. (0.0251a2, 3 4) (k) An electron confined in a one dimensional box of width L is known to be in its first excited state. Determine the probability density of electron in the central half. (Jan 2017) Short answer type questions (a) What is tunnel effect? What is quantum mechanical tunneling? (May 2019, Apr 2018, Dec 2017, July 2016) 16 (b) State and explain Heisenberg’s uncertainty principle. (July 2018) (c) What are the characteristics of a well-defined wave function? (Dec 2018) (d) Obtain energy and momentum operators (Dec 2018) (e) A well behaved wave function is normalized. Why?(Apr 2018, Aug 2016) (f) Give the probability interpretation of the wave function. (May 2017, Jan 2016) (g) Explain the absence of electrons in in the nucleus on the basis of the uncertainty principle.(May 2017) (h) What is meant by wave function? Write its normalization condition. (Jan 2017) (i) What is the significance of operators in Quantum mechanics? What is Hamiltonian operator? Write its expression. (Aug 2016) (j) How do you account for the natural line broadening on the basis of Heisenberg’s uncertainty principle? (May 2016) Essay type questions (a) Write the Schrodinger equation for a particle trapped in a one dimen- sional box of width L and solve it to obtain the energy eigen values and wave function. (May 2019, July 2018, May 2017, Aug 2016, July 2016) (b) State uncertainty principle. Explain the absence of electron inside the nucleus using this principle, (Dec 2018, April 2018, Jan 2017) (c) Starting from the time dependent equation, derive Schrodinger’s time independent wave equation. (Dec 2017) (d) Formulate Schrodinger’s time dependent equation starting from a plane wave equation using De Broglie’s formula and Einstein’s relation for photon energy. (Jan 2016) 17