SOME STUDY MATERIALS ON QUANTUM MECHANICS PART OF PAPER C9T (VU PHYSICS HONS. CBCS 4TH SEMESTER)

DEBASISH AICH VU CBCS Semester-IV 2019: Planck’s quantum, Planck’s constant and light as a collection of photons; Blackbody Radiation: Quantum theory of Light; Photo-electric effect and Compton scattering. De Broglie wavelength and matter waves; Davisson-Germer experiment. Wave description of particles by wave packets. Group and Phase velocities and relation between them. Two-Slit experiment with electrons. Probability. Wave amplitude and wave functions.

Historical Note 1. Situation towards the end of the 19th century and the beginning of the 20th century 1.1 Advancement in Physics:  Classical mechanics: Newtonian Mechanics (Principia 1687-1713-1726; Sir Isaac Newton, English, 1643-1727) > Lagrangian Formulation (1750s) (Joseph-Louis Lagrange, Italian-French, 1736-1813) > Hamiltonian Formulation (1833) (William Rowan Hamilton, Irish, 1805-1865)  Electrodynamics: Maxwell’s (James Clerk Maxwell, Scottish, 1831-1879) Equations of Electromagnetic waves [1861]. [In present form by Oliver Heaviside (English), Josiah W Gibbs (American), Heinrich Hertz (German, 1857-1894) in 1884] Lorentz (Dutch, 1853-1928) Force Equation [1861 Maxwell > 1881J. J. Thomson1 (English) > 1884 Heaviside > 1895 Lorentz]  Thermodynamics: Carnot Theorem (1824) [Nicolas Leonard Carnot, French 1796-1832], Maxwell-Boltzmann (Ludwig Eduard Boltzmann, German, 1844-1906) Statistics (1868).

1.2. Major unsolved Questions:  Energy Distribution [푢(휈)푑휈 표푟 푢(휆)푑휆] of Blackbody Radiation.  Photo electric effect: Experiment by Hertz in 1887.  Stability of Rutherford’s (New Zealand-born British) atom (1911 Gold Foil Expt.).  Existence of aether: Michelson (American) –Morley (American) Experiment (1887, at Western Reserve University, Ohio).  Atomic Spectra: Balmer (Swiss Mathematician) series:

Balmer formula (1885) 휆 = 퐵 = 퐵 Rydberg (Swedish Physicist) Formula (1888) 휈̅ = = − = 푅 − With 푅 = 1.09737309 × 10 푚 . Anomalous Zeeman (Dutch) effect, Fine structures and other observation in atomic spectroscopy.

1.3 End of the era of triumph of classical physics. 1900 Planck (German) distribution formula of Blackbody Radiation > Assumption of radiation quanta of energy 풉흂, where 휈 is the frequency of radiation and ℎ is a constant determined by Planck to fit the experimental distribution curve and is called Planck constant.

1 William Thomson is a different scientist having other name Lord Kelvin, Scots-Irish, 1824-1907.

1905 Einstein (German Jewish)> Photo electric effect > Particle nature of light/radiation > Photon. 1905 Einstein Special Theory of Relativity > Non existence of aether; dependence of mass, length and time on velocity. 1913 Niels Bohr (Danish) > Model of Hydrogen Atom> quantisation of angular momentum of atomic electron > explanation of atomic stability, Balmer formula, atomic spectroscopy. 1923 New observations: Compton (American) Effect > recoil of electron which scatters X-ray. X-ray photon has momentum ℎ휈⁄ 푐 > Radiation has particle nature. 1923 de Broglie (French) hypothesis: Electron and all matter have wave nature. 1925 Heisenberg (German): Matrix Formulation. 1926 Schrodinger (German): Schrodinger Equation > Wave mechanics. 1927 Heisenberg: Uncertainty Relation (Earle Hesse Kennard in late 1927 & Hermann Wey in 1928 gave the formal relation involving standard deviations as uncertainties: 휎휎 ≥ℏ⁄ 2). 1923-27 Davisson (American) and Germer (American) experiment and explanation > Diffraction of electrons > Confirmation of wave nature of electrons i.e. de Broglie hypothesis. 1927 Max Born (German Jewish) probabilistic interpretation of wave mechanics > 푃(푥, 푡)푑푥 = ∫ |휓(푥, 푡)| 푑푥, where 푥 and 푥 are the limits within which the particle exists. In 3D 푃(푟⃗, 푡)푑푥 = ∭|휓(푟⃗, 푡)|푑휏, where the integration is over the region of space in which the particle exists. 1928 Paul Dirac (English): Relativistic Quantum Mechanics > Prediction of Positron > Proof in 1932; 1939 bra ket notation by Dirac: Both Heisenberg’s matrix formulation and Schrodinger’s wave mechanics formulation can be handled with this. 1.3.1 Blackbody Radiation

Quantum Mechanics Concepts and Applications - Nouredine Zettili

1879 J. Stefan (Carinthian Slovene) established from Tyndall’s experimental results of IR emissions by platinum filament and its colour: Per unit area of the surface of a radiating solid at absolute temperature 푇 radiates normally (perpendicularly) a power (or energy per second)- 푃 = 푎휎푇 ……….. (1) −8 2 −4 where 휎 = 5.670367 × 10 Wm K is called Stefan’s constant; 푎 is a coefficient ≤1. For ideal blackbody 푎 =1. Equation (1) is called Stefan’s law or Stefan-Boltzmann Law. In 1884 a theoretical derivation of the law was done by Boltzmann (German). Up to a temperature 1535 K this law accurately matches experimental observations. But at higher temperature deviation from experimental results are observed. 1893 Wien ( Wilhelm Wien, 1864-1928, German Physicist) displacement law:

휆푇 = 푐표푛푠푡푎푛푡 ≈ 2900 휇푚. 퐾

1894 Wien energy density distribution: Wien proposed (from thermodynamic consideration) that, Stefan-Boltzmann law and Wien displacement law can be derived if the energy density of blackbody radiation at temperature 푇 per unit wavelength at 휆 i.e. 푢(휆, 푇) must be given by a relation: 푢(휆, 푇)푑휆 = 푓(휆푇)푑휆, where 푓(휆푇) is any function of 휆푇. From some arbitrary assumptions regarding mechanisms of emission he proposed that 푓(휆푇) = 푎푒⁄ and so 푢(휆, 푇)푑휆 = 푒⁄ 푑휆. In terms of frequency 푢(휈, 푇)푑휈 = 퐴휈푒⁄ 푑휈. Unit of 푢(휈, 푇) is 퐽푚퐻푧 or 퐽푚푠 and unit of 푢(휆, 푇) is 퐽푚. Constants 푎, 푏 or 퐴, 훽 were determined to fit these equations to experimental curves. Failure of Wien distribution: Wien’s distribution satisfies experimental curve at lower wavelengths or higher frequencies but fails to explain them at higher wavelengths or lower frequencies. [In those days producing radiation of higher frequencies or lower wavelengths was not easy.] Thus Wien’s distribution was insufficient to satisfy observations.

Quantum Mechanics Concepts and Applications - Nouredine Zettili

1900 Rayleigh’s (Lord Rayleigh, 1842-1919, British physicist) energy density distribution: Rayleigh assumed that in the cavity of a blackbody radiation exists in the form of electromagnetic standing waves with their nodes at the walls of the cavity. Density of states (or vibrational modes) of these standing waves i.e. number of states (or modes) per unit volume per frequency range of such standing waves is equal to . The electromagnetic standing waves are excited by the linear oscillation of the tiny electric dipoles of atomic or molecular dimension in the walls of the cavity. The energy of an oscillating dipole can have any value between 0 & ∞ i.e. the energy spectrum of an oscillator is continuous. At temperature 푇, the ⁄ number of electric dipoles having energy 퐸 is given by M-B statistics, i.e. 푁(퐸) = 푁푒 , where 푁 is the number of oscillators with zero energy and 푘 (= 1.38 퐽퐾 ) is Boltzmann constant. Then it can be shown that at temperature 푇, the average energy of the oscillators in the walls is 〈퐸〉 = 푘푇. In equilibrium the energy distribution of the standing waves is same as the energy distribution of the oscillators over the frequency range. Therefore the average energy of the vibrational modes of the standing waves will also be 〈퐸〉 = 푘푇. So, according to Rayleigh, the energy density distribution is given by:

푢(휈, 푇)푑휈 = 푛(휈, 푇)〈퐸〉푑휈 = 푘푇푑휈; Or, in terms of wavelength 푢(휆, 푇)푑휆 = 푘푇푑휆. Density of states of vibrations in a cubical cavity of side 푳 filled with a continuous elastic medium:

3D wave equation: + + = …………(A) Standing wave solution: 휓(푥, 푦, 푧, 푡) = 퐴푠푖푛 푠푖푛 푠푖푛 푐표푠(2휋휈푡). ……..(B)

푛, 푛, 푛 are integers ≥1. A vibrational mode is determined by the set of integers 푛, 푛, 푛. What is the number of such modes within frequency range 흂 풕풐 흂 + 풅흂 ? Consider a coordinate system in which coordinates are the positive integers and zero. This is the first octant of the 3D integer space. Each point in this space will be at unit distance from its nearest neighbours; in other words each point will share unit volume of this space.

Substituting (B) in (A) and simplifying: 푛 + 푛 + 푛 = = 푅 (푠푎푦). Above equation represents the portion of a sphere of radius 푅 = in the first octant of the integer space. In this space a spherical shell between radii 푅 and 푅 + 푑푅 corresponds to the frequency range 휈 to 휈 + 푑휈. Volume of such a shell in integer space is: 1 1 1 4퐿휈 2퐿 4휋퐿휈 ×4휋푅푑푅 = 휋푅푑푅 = 휋 × × 푑휈 = 푑휈 8 2 2 푐 푐 푐 The number of coordinate points 푛 , 푛 , 푛 in this shell will also be 휋푅푑푅, since each point shares unit volume in integer space. But this is equal to the number of vibrational modes in the frequency range 휈 to 휈 + 푑휈. Thus the number of modes in the frequency range 휈 to 휈 + 푑휈 per unit volume of the cavity will be 1 4휋퐿휈 4휋휈 × 푑휈 = 푑휈 퐿 푐 푐 Now unpolarised electromagnetic waves contains two types of circularly polarised waves with the plane of polarisation rotating in clockwise and anticlockwise sense. Now two modes of the electromagnetic standing waves with plane of polarisation rotating in opposite sense but identical in all other respect will have same set of 푛, 푛, 푛, i.e. each point in the integer space represents two states. Therefore number of states per unit volume of the cavity in the frequency range 휈 to 휈 + 푑휈 will be

4휋휈 8휋휈 푛(휈)푑휈 =2× 푑휈 = 푑휈. 푐 푐 8휋휈 푛(휈) = ………….(퐶) 푐 is called the density of states. Though for simplicity here we derive this result for a cubical medium, it is applicable to any shape. Average energy per vibrational mode: According to M-B statics the number of oscillators (or vibrational states in this case) having ⁄ energy 퐸 at temperature 푇 is 푁 = 푁푒 , where 푁 is the number of oscillators in the state of zero energy (ground state) and 푘 is Boltzmann constant (푘 = 1.38×10퐽퐾). Therefore for a continuous energy distribution: ⁄ ⁄ ∫ 퐸푁푒 푑퐸 (푘푇) ∫ (퐸⁄ 푘푇 )푒 푑(퐸⁄ 푘푇 ) Γ(2) 〈퐸〉 = = = 푘푇 = 푘푇 ….. (퐷). ⁄ ( ) ⁄ ( ⁄ ) Γ(1) ∫ 푁푒 푑퐸 푘푇 ∫ 푒 푑 퐸 푘푇 Rayleigh formula satisfies experimental curves at higher wavelengths or lower frequencies but deviates badly from the experimental curves towards lower wavelengths or higher frequencies i.e. towards ultraviolet region of the spectrum. This failure is known as ultraviolet catastrophe. Particle nature of wave 1905 Planck blackbody radiation formula: Planck’s quantisation rule / Planck’s quantum hypothesis / Planck’s postulate: According to classical mechanics, a harmonic oscillator of frequency 휈 can have any amount of energy 퐸 [ = 4휋푚푎휈], which is proportional to the square of its amplitude 푎. And it can have any energy between 0 & ∞. But to explain blackbody radiation Planck made the following revolutionary assumptions: i) An oscillator frequency 휈 in the wall of the blackbody can have only discrete energies given by 휀 = 푛ℎ휈, where 푛 = 0,1,2….. and ℎ is a constant, which was determined by him to fit his formula with the experimental distribution curves of blackbody radiation. ii) When an oscillator of frequency 휈 absorbs or emits energy in the form of radiation its energy can change only in the steps of ℎ휈. Since the radiation absorbed or emitted by an oscillator have same frequency as that of the oscillator therefore it follows from Planck assumptions that an oscillator of frequency 휈 can absorb or emit radiation of frequency 휈 and this emitted or absorbed radiation can have only an amount of energy ℎ휈, no less no more. Regarding the nature and density of states of the radiation inside the cavity of blackbody, Planck’s assumption was same as that of Rayleigh.

Average energy of the oscillators of frequency 흂 in the walls of the blackbody:

Oscillators of frequency 휈 have discrete energies 휀 = 푛ℎ휈, 푛 = 0,1,2… . According to M-B statics the ⁄ ⁄ number of such oscillators at temperature 푇 is 푁 = 푁푒 = 푁푒 . Therefore the average energy of the oscillators of frequency 휈 will be:

∑ 휀 푁 푒⁄ ∑ 휀 푒⁄ 〈 〉 휀 = ⁄ = ⁄ . ∑ 푁푒 ∑ 푒 Note that the integrations of equation (D) have been replaced here by summations since in this case discrete energies are assumed for the oscillators in place of continuous energies of the oscillators in Rayleigh theory. Now

⁄ ⁄ ⁄ ⁄ ⁄ ∑ ∑ ⋯ 〈휀〉 = ⁄ = ⁄ = ⁄ ⁄ ⁄ ∑ ∑ ⋯ … = (where 푥 = 푒⁄ ) … () =ℎ휈푥 =ℎ휈푥 = = . () ⁄ Since the radiation in the cavity of the blackbody is in equilibrium with the oscillators in the wall so this above expression will also give the average energy of the vibrational modes of the standing waves in the cavity. The number of vibrational modes or states per unit volume of the cavity in the frequency range 휈 to 휈 + 푑휈 can be determined as before and is given by:

푛(휈)푑휈 = 푑휈. ℎ휈 Thus the energy distribution of the radiation is given by: 푢(휈)푑휈 = 〈휀 〉푛(휈)푑휈 = ∙ 푑휈. 휈 푒ℎ휈⁄ 푘푇−1 8휋ℎ휈 1 푢(휈)푑휈 = ∙ 푑휈 …………(1) 푐 푒ℎ휈⁄ 푘푇 −1 In terms of wavelength: 8휋ℎ푐 1 푢(휆)푑휆 = ∙ 푑휆 ……………(2) 휆 푒ℎ푐⁄푘푇 −1

Note that the discreteness of vibrational modes indexed by 푛, 푛, 푛 arises here from purely classical considerations. But the discreteness of possible energies of an oscillator is due to the assumptions of a new type, called Planck’s quantum conditions.

Fitting his equation with experimental curves Planck determined the value of ℎ. Its value is 6.626 × 10 퐽. 푠 and it is a universal constant of immense importance as was revealed later years with the advancement of quantum mechanics. Derivations from Planck’s law:

Stefan-Boltzmann law: Total energy (in all wavelength range) per unit volume of the cavity of a black body is

8휋ℎ 휈 8휋ℎ 푘푇 푥 푢 = 푢(휈)푑휈 = 푑휈 = 푑푥 [푤ℎ푒푟푒 푥 = ℎ휈⁄ 푘푇] 푐 푒ℎ휈⁄ 푘푇 −1 푐 ℎ 푒푥 −1

8휋ℎ 푘푇 8휋ℎ 푘푇 = 푥푒−푥(1− 푒−푥)푑푥 = 푥푒−푥1+ 푒−푥 + 푒−2푥 + 푒−3푥 +⋯ 푑푥 푐 ℎ 푐 ℎ

8휋ℎ 푘푇 8휋ℎ 푘푇 = 푥푒−푥 + 푒−2푥 + 푒−3푥 + 푒−4푥 +⋯ 푑푥 = 푥푒−푝푥푑푥 푐 ℎ 푐 ℎ

8휋ℎ 푘푇 1 8휋ℎ 푘푇 1 8휋ℎ 푘푇 3! = (푝푥)푒−푝푥푑(푝푥) = Γ(4) = 푐 ℎ 푝 푐 ℎ 푝 푐 ℎ 푝

48휋ℎ 푘푇 휋 8휋푘 = = 푇 푐 ℎ 90 15푐ℎ

It can be shown that the energy radiated normally per unit area from a blackbody is 퐸 = 푢 . Thus 퐸 = ∙ 푇 = 푇 or 퐸 ∞ 푇.

Wien displacement law:

1 푢(휆) = ∙ = , say. Where 푧(휆) = 휆푒ℎ푐⁄푘푇 −1. 푒ℎ푐⁄푘푇−1 ()

The value of 휆 for which 푢(휆) is maximum is obtained from the condition:

휕푢(휆) =0 휕휆 This is equivalent to: 휕푧(휆) =0 휕휆 ℎ푐 ℎ푐⁄푘푇 ℎ푐⁄푘푇 Or, 5휆 ∙ 푒 −1 − 휆 ∙ 푒 =0; 푘푇

Or, 1− 푒⁄휆푚푎푥 = ; 5휆푚푎푥

Or, with = 푥, this equation can be written as: 1− 푒 = 휆푚푎푥 5

This equation can not be solved analytically, but can be solved numerically. Or by writing as a pair of 풙 equation: 풚 = ퟏ − 풆풙; and 풚 = it can also be solved graphically. The curves represented by these ퟓ equations intersect for 푥 ≈ 4.965.

.× ×× Thus ≈ 4.965; 휆푇 = = ≈ 0.0029 푚. 퐾 = 푐표푛푠푡푎푛푡 . .×.× Wien and Rayleigh-Jeans distribution law:

Planck’s law: 푢(휈)푑휈 = ∙ 푑휈 ⁄ At high frequencies 푒⁄ −1≈푒⁄

So 푢(휈)푑휈 = ∙푒⁄ 푑휈 = 퐴휈푒⁄ 푑휈 where 퐴= and 훽=ℎ⁄ 푘 are constants. [Wien’s distribution law.]

(⁄ ) (⁄ ) At low frequencies 푒⁄ −1=(1+ℎ휈⁄ 푘푇 + + …)−1≈ℎ휈⁄ 푘푇 ! ! So 푢(휈)푑휈 = ∙ 푑휈 = 푑휈 [Rayleigh-Jeans law]. ⁄ Problems: JAM 2017

1 Ans.: 푢(휈) = ∙ 푒ℎ휈⁄ 푘푇−1

() () 1 푒ℎ휈⁄ 푘푇−1 푒ℎ휈⁄ 푘푇−1 푒ℎ휈⁄ 푘푇−1 = ∙ ∙ ∙ = 8∙ = 8∙ () 푒2ℎ휈⁄ 푘푇−1 1 푒2ℎ휈⁄ 푘푇−1 푒ℎ휈⁄ 푘푇−1푒ℎ휈⁄ 푘푇+1

= 8∙ = 8∙ 푒⁄ +1 ⁄

() −1 ∞푒ℎ휈⁄ 푘푇 +1 ⇒ (퐵). () JAM 2014

Ans.: 푢 (휆) =0

So, from 푢 (휆) = 푒⁄ , we have − 푒⁄ + 푒⁄ =0

⇒ −5+ =0 ⇒휷=5휆푇=5×2.9×10 = ퟎ.ퟎퟏퟒퟓ 풎.푲 휎푇 =퐸

= = ∫ 푢 (휆)푑휆 = ∫ 푒⁄ 푑휆

Let 훽⁄ 휆푇 =푥. Then 푑푥 =−(훽⁄ 휆푇 )푑휆 ⇒ 푑휆 = − 푑푥.

휎푇 =퐸=− ∫ 푒푑푥 = ∫ 푒푑푥 = ∫ 푥푒푑푥

= ∙6=

⇒ = 휎푇 ⇒ =휎 ⇒훼 =

×× ×.××(ퟎ.ퟎퟏퟒퟓ) ××.×(ퟏ.ퟒퟓ) ⇒훼= = = 150.3856. JAM 2013

() Ans.: 휆푇 = 푐표푛푠푡푎푛푡 ⇒ (휆)푇 = (휆)푇 ⇒ (휆) = = = ⇒ (퐷). JAM 2012

1 1 Ans.: 푢(휆) = ∙ = ∙ 푒ℎ푐⁄푘푇−1 푒ℎ푐⁄푇−1

1 For 휆=휆 we have 푢(휆) = ∙ ℎ푐⁄ 푇 . 푒 −1

[] Also 휆푇 = 푐표푛푠푡푎푛푡 ⇒ [휆]푇 = [휆]푇 ⇒ = []

([ ] ) 1 [ ] 푒ℎ푐⁄[]−1 [ ] = ∙ ∙ ∙ = = ℎ푐⁄[] ([]) [] 푒 −1 1 []

=2 = 32 ⇒ (퐷).

JAM 2007:

Ans.: Clearly the temperatures of the two blackbody will be different. If 푇 and 푇 are the temperatures then:

[ ] = = =3. []

= = =3 = 81.

Photoelectric effect: In 1888 Hertz and afterwards other scientists observed that when the surface of metals like zinc is irradiated with ultraviolet light the metal gets positively charged i.e. the metal loses negative charge. 1n 1899 P. Lenard (Philipp Lenard, German Physicist, 1862-1947, Supporter of Hitler) showed that the loss of negative charge is due to emission of negatively charged electrons from the metal surface. The following laws were discovered experimentally prior to 1905:  If the frequency of the incident radiation is smaller than the metal’s threshold frequency - a frequency that depends on the properties of the metal—no electron can be emitted regardless of the radiation’s intensity (Philipp Lenard, 1902).  No matter how low the intensity of the incident radiation, electrons will be ejected instantly the moment the frequency of the radiation exceeds the threshold frequency 휈.

1< 2<3 i I3 > I2 > I1 eV0 1> 2>3 i I3 I2 I1 1 2 3 -V01 -V02 -V03 O V -V0 O V 0  light intensity  constant  or   constant -W (a) (b) 0 (c)

Fig.X

 At any frequency above 휈, the photo current and hence the number of electrons ejected per second increases with the intensity of the light but does not depend on the light’s frequency.

 The stopping potential (푉) and so the kinetic energy of the ejected electrons depends on the frequency but not on the intensity of the beam; 푉 and so the maximum kinetic energy of the ejected electrons 푚푣 = 푒푉 increases linearly with the incident frequency.

Einstein showed that the plot of maximum kinetic energy of the electrons or of 푒푉 with frequency (휈) of the incident light is a straight line (Fig.-X(c)) which can be given by:

푒푉 =ℎ휈 − 푊 The slope of the straight line graph does not depend on the metal and as determined from experimental results, it is equal to Planck’s constant ℎ. If the electron emitting metal surface is clean and oxide free then the intercept 푊 is characteristic of the metal and it is equal to ℎ휈. Einstein, extending Planck’s quantum condition to radiation, proposed that light is made up of discrete energy packets or quanta – photons – each of which have energy ℎ휈, where 휈 is the frequency of light. In photo electric effect a photon, incident on an electron, is completely absorbed by it. At normal temperatures the maximum energy of an electron inside the metal is less than the minimum energy required by an electron to come out of the metal surface by an amount equal to 푊 which is called work function. So if ℎ휈 < 푊 the photon absorbing electron cannot come out of the metal surface. But if ℎ휈 > 푊 the electron can emit from the metal surface. And the emitted electron possesses a kinetic energy ℎ휈 − 푊 with which it can reach to the anode even if the anode is given no positive potential with respect to the cathode. Moreover to stop such an emitted electron from leaving the metal surface a negative potential (say −푉) should be applied to the anode with respect to the cathode. 푉 is called stopping potential. Clearly 푒푉 = ℎ휈− 푊 and it is equal to the maximum kinetic energy of the electrons. The equation: 푒푉 = ℎ휈− 푊 is Einstein’s photoelectric equation.

Problems: JAM 2014

Ans. 푒푉 =ℎ휈−푊 = −푊 = − 2.1 electron Volt = 1.775 electron Volt

. ⇒푉 = electron Volt = 1.775 Volt ⇒ (퐴). JAM 2011

Ans.: 퐸=퐸 sin휔푡 represents the electric field vector of light having angular frequency 휔 i.e. frequency 휈=휔⁄ 2휋 where 퐸 is the amplitude of the electric field vector. So 퐸 = (90 푉⁄ 푚 )[sin(6.28 × 10푠)푡+sin(12.56 × 10푠)푡]

= (90 푉⁄ 푚 ) sin(6.28 × 10푠)푡+ (90 푉⁄ 푚 ) sin(12.56 × 10푠)푡

.× represents two light waves of frequencies 휈 = 푠 = 10푠 = 10퐻푧 and

.× 휈 = 푠 = 2×10푠 = 2×10퐻푧 Clearly the maximum kinetic energy will be determined by the larger frequency. So, in this problem the maximum kinetic energy will be:

ℎ휈 −푊 퐽표푢푙푒 = − 푒푉

.××× = − 2.0 푒푉 = 8.28 푒푉 − 2.0 푒푉 = 6.28 푒푉 ⇒ (퐶) .× JAM 2007:

Ans. 푃 = 1.55×10 × 푊 = 1.55×10 푊 Number electrons emitted from the cathode per second is equal to the number of photons participating in photoemission per second:

.× .× = = = ⁄

.× .× = = [⁄ ]×.× .×

.× Current = ×1.6×10 = 0.05 푚퐴. .×

(퐾. 퐸) = 푒푉 =ℎ휈 − 푊 = − 푊 ⇒ 푒푉 = − 푊; 푒푉 = − 푊

푒푉 − 푒푉 = − = − electron Volt = 3.1 electron Volt

⇒5푒푉 − 푒푉 = 3.1 electron Volt ⇒ 4푒푉 = 3.1 electron Volt ⇒ 푉 =3.1⁄ 4 V

⇒ 푉 = 0.7525 V; 푉 = 5 × 0.775 eV = 3.875 V

Problem of estimation of Plank Constant (Example 1.2, Zettili):

Compton Effect:

# Before starting to study this topic, solve the following problems:

1. An electron has energy 퐸 푒푉. Obtain the expression of de Broglie wavelength (휆=ℎ⁄ 푝) of the electron, considering the problem to be (i) relativistic and (ii) non relativistic. Find the value of 휆 for (a) 퐸 = 10 푒푉 and (b) 퐸 = 10 푀푒푉 for both relativistic and nonrelativistic treatments.

2. An electron is accelerated through a potential difference 푉. Obtain the expression of de Broglie wavelength (휆=ℎ⁄ 푝) of the electron after acceleration, considering the problem to be (i) relativistic and (ii) non relativistic. Find the value of 휆 for (a) 푉 = 10 V and (b) 퐸 = 100 kV for both relativistic and nonrelativistic treatments.

*Photons require high energy to produce Compton Effect. Such photons are X-ray and 훾-ray photons. *Electron participating in Compton Effect is assumed to be free. *Collision is considered to be elastic, i.e. both the total kinetic energy and total linear momentum of the colliding particles remain unchanged through the collision.

Conservation of energy:

ℎ휈 = ℎ휈 +푇 = ℎ휈 + 푚푐 −푚푐

− +푚 푐 = 푚푐 ⇒ − +푚 푐 =푚푐.

Squaring

+ −2 + 2푚푐ℎ − +푚 푐 =푚 푐

⇒ + −2 + (휆 −휆) =푚 푐 −푚 푐 ⇒ + −2 + (휆 − 휆) = − 푚 푐 = = [Where 훽 = 푣⁄ 푐]

⇒ + −2 + (휆 − 휆) = 푚 푣 …………(A)

Conservation of momentum: Relativistic energy relation ⇒ 퐸 = 푝 푐 + 푚 푐 .

For photon, 퐸 =ℎ휈 and rest mass 푚 =0. Therefore ℎ휈 = 푝푐.

⇒ 푝ℎ표푡표푛 푚표푚푒푛푡푢푚 푝 = = .

Therefore

= cos 휑 + 푚푣 cos 휃 ⇒ − cos 휑 = 푚푣 cos 휃 …………(B) and

0= sin 휑 − 푚푣 sin 휃 ⇒ sin 휑 = 푚푣 sin 휃 . …………(C)

Squaring and adding (B) and (C):

+ −2 = 푚 푣 …………(D)

(A) – (D)

−2 (1−cos 휑) + (휆 − 휆) =0 Wavelength shift:

⇒ 휆 − 휆 = (1−cos 휑) = 휆(1−cos 휑) ……….. (E) ⇒ 휆 = 휆 + 휆(1−cos 휑). ……….. (F)

.× 휆 = = Compton wavelength = = 0.02424 Å. .× ××





Recoil angle of electron: We have − cos 휑 = 푚푣 cos 휃 …………(B) and sin 휑 = 푚푣 sin 휃 . …………(C)

tan 휃 = =

From (F), =1+ (1−cos 휑) =1+ 훼(1−cos 휑). Then tan 휃 = = = = () ()( ) () () cot 휃 = (1+ 훼) tan ………….. (G)

Recoil energy of electron:

푇 =ℎ휈 −ℎ휈 =ℎ휈 1− =ℎ휈 1− = ℎ휈 1− ()

( ) = ℎ휈 . ( )

Useful way to calculate 푻풆 if 흀 (or 흂) and 흋 are given: First find 휆 = 휆 + 휆(1−cos 휑) Now 푇 =ℎ휈 −ℎ휈 = − Use ℎ푐 = 1240 푛푚. 푒푉

Then: 푇 = − =ℎ푐 − = 1240 − 푒푉 풉풄 휶(ퟏ퐜퐨퐬흋) Plot 퐓 = 풗풔. 흋. Given 흀 = ퟎ. ퟕퟎퟗ Å. 풆 흀 ퟏ휶(ퟏ퐜퐨퐬흋)

.× Ans.: 훼 = = = 34.2328 × 10 .× ×× ×.×

.××× = = 28.036671 × 10 퐽 .×

( ) .×( ) T = = . ( ) .×( )

.( ) We plot 푣푠. 휑 .( )

Te

휑

() (T ) = = . ; (T ) = . . ( ) ⁄

Comparison between Photoelectric Effect and Compton Effect:

Photoelectric Effect Compton Effect Participating Visible or UV light X-Rays and 훾-Rays Radiation 휆 ~ 10 푛푚 휆~ 10 − 10 푛푚 (photon) 퐸~ 푒푉 퐸~ 푘푒푉 − 푀푒푉 Energy partly transferred to electron, Completely absorbed by the electron Direction changes up to 180°, What happens to (or other scattering particle) Wavelength increases, Energy completely transferred to the the photon For electrons ∆휆 푚푎푦 푏푒 푢푝 푡표 electron 2휆 = 2 × 0.002426 푛푚 Participating Conduction electrons or electrons Loosely bound atomic electrons of non-metals electron loosely bound the atoms of a metal e.g. graphite Electrons have negative energy of several 푒푉 In magnitude equal or slightly greater Several 푒푉. Magnitude may be greater than that than the work function. of electrons of Photoelectric effect. Electrons free or Work function of 퐴푔 and 푁푎 are Ionisation potential of Carbon is 11.26 푒푉 bound 4.54 푒푉 & 2.28 푒푉 Magnitude of electron energy is small Magnitude of electron energy is compared to the photon energy (푘푒푉). So comparable to the energy of photon electrons are considered as free. (푒푉). So electrons are considered as bound. What happens to Recoils. If recoil energy is high then emits from Emits from the metal the electron the material. Energy of Several 푒푉. Several 푒푉 to 푘푒푉. emitted / Non-relativistic treatment is Relativistic treatment is required recoiled electron allowable

Problems: JAM 2017

Ans.: (A) 휆 − 휆 = (1−cos 휑) ⇒ 푐 − = (1−cos 휑) 2 1 ℎ ⇒ 푐 − = (1−cos 휑) 휈 휈 푚푐 [푆푖푛푐푒 푙표푠푠 표푓 푒푛푒푟푔푦 =ℎ휈 −ℎ휈 =ℎ휈⁄2 ⇒ 휈 = 휈⁄2] ⇒ = (1−cos 휑) ⇒ (1−cos 휑) = = ⇒ cos 휑 =1− ⇒⇒ (퐴) 푖푠 푐표푟푟푒푐푡.

퐿표푠푠 표푓 푒푛푒푟푔푦 표푓 퐶표푚푝푡표푛 푠푐푎푡푡푒푟푒푑 푝ℎ표푡표푛 = 푘푖푛푒푡푖푐 푒푛푒푟푔푦 푔푎푖푛푒푑 푏푦 푡ℎ푒 푟푒푐표푖푙푒푑 푒푙푒푐푡푟표푛

() ⁄ ⁄ =푇 = ℎ휈 , where 훼 = = = = . ( ) ⁄ () Here 푇 = = ℎ휈 ⇒ 2훼(1−cos휑) =1+훼(1−cos휑); ()

⇒1−cos휑= ⇒ cos휑 = 1− =1− ⇒⇒ (푨) 풊풔 풄풐풓풓풆풄풕.

(B) tan휃 = = = = = = √ ()( ) ()⁄ ()⁄

sin휃 = = () = =

Given relation in (B) : sin휃 =1− =1− ⇒⇒ (푩) 풊풔 풘풓풐풏품. (C) We have sin휑 = 푚푣sin휃. ⇒ = ⇒⇒ (푪) 풊풔 풄풐풓풓풆풄풕. ⁄ (D) Change in photon wavelength 휆 −휆= (1−cos휑) ⇒⇒ (푫) 풊풔 풘풓풐풏품. JAM 2016

××.× ×. Ans.: Given: ℎ휈 = 20푘푒푉; Also 훼= = = =0.039072 .× ×(× ) .×

() . 푇 = ℎ휈 ⇒ 푇 = 20푘푒푉 × ≈ 0.75푘푒푉 ( ) .

Alternately: ℎ휈 = 20푘푒푉 ⇒ 휆 = = 0.062 푛푚 ×

.× 휆 =휆+ (1−cos휑) = 0.062 푛푚 + = 0.062 푛푚 + 0.002427푛푚 .× ×× = 0.064427푛푚

ℎ휈 = 푒푉 = 19.25 푘푒푉 . 푇 =ℎ휈−ℎ휈 = 0.75 푘푒푉 JAM 2015: Section C

Ans.: 휆 −휆= (1−cos휑) =휆(1−cos휑) = 0.00243 × (1− cos60°) = 0.00243⁄ 2 = 0.001215 푛푚 휆 = 0.241215 푛푚 Loss of energy of the X-ray is the gain in kinetic energy of the electron. So:

푇 = − 푒푉 = (5166.67 − 5140.64) 푒푉 = 26.03 푒푉 . . JAM 2013

Ans.: 휆 −휆= (1−cos휑) =휆(1−cos휑) = 0.002426 × (1 − cos180°) = 0.004856 푛푚 휆 = 0.204856 푛푚 % loss of energy of the X-rays:

− % = (6200 − 6053.03)⁄6200 % = 2.37 %. . . . JAM 2010

Ans.: (a) In case of head on collision the photon will bounce back by 180°.

Then ∆휆 = 2푐휆 sin = 2푐휆 sin = 2푐휆 If the electron is initially at rest then: 푃=0.

⁄ Then ∆휆 = 2푐휆 sin = 2푐휆 sin = 2푐휆 sin

(b) Resolving power = = = = = 82440.23 ∆ ( ) .×(°) . JAM 2008

Ans.: = (1−cos 휑) = (1−cos 휋) = . JAM 2006:

Ans.: (a) 0.05 = = =1− =1− 휆 = 0.95휆 = 0.95∆휆 + 0.95휆 ⇒ 0.05휆 = 0.95∆휆 ⇒ 휆 = 19∆휆

휆 = 19∆휆 = 19휆 (1−cos 휑) = 19휆 (1− cos60°) = 휆 = 9.5휆 .

. . (b) 퐸 = 푇 + 푚푐 = 0.05ℎ휈 + 푚푐 = + 푚푐 = + 푚푐 .

. . = + 푚 푐 = + 푚 푐 = + 푚 푐. . .

= 퐽 = 푖푛 푡ℎ푒 푢푛푖푡푠 표푓 푟푒푠푡 푚푎푠푠 Zettili: Page-21

Wave nature of particle

Davisson-Germer Experiment (1923-27): Bombarding the surface of nickel single crystal normally with collimated beam of electrons and detecting the number of electrons scattered in different angles with the incident beam. It was observed that scattering was maximum for an angle 휑 = 50° when the electron energy was 54 푒푉.

휑 휑 2 휃 2

휃 휃 푑

퐷 = 2.15Å, 푑 = 퐷 sin 25° = 0.91Å

To explain their result Davisson and Germer used the concept of wave nature of particle, here electrons, postulated by de Broglie. So the de Broglie wavelength of the electrons: 풉 풉 ퟔ.ퟔퟐퟔ × ퟏퟎퟑퟒ = = 풎 =ퟏ.ퟔퟕÅ ퟑퟏ ퟏퟗ 풑 ퟐ풎푬풌 √ퟐ ×ퟗ × ퟏퟎ × ퟓퟒ ×ퟏ.ퟔ × ퟏퟎ Verify for relativistic relations Again when waves, like X-rays, are incident on the atomic planes of a crystal at different angles they suffers intense reflection like scattering for some particular angles given by Bragg’s equation: ퟐ풅퐬퐢퐧휽 = 풏흀 where 휃 is the angle of the incident beam with the set of parallel atomic planes (glancing angle) responsible for scattering, 푑 is the inter-planar spacing between these planes and 푛 is a positive integer, called order number. It is to be noted that intensity of scattered beam is small for large 푛. In the Davisson-Germer experiment angle of incident 휑⁄ 2 =50Å⁄ 2 = 25° and so 휃=90°−휑⁄ 2 = 65°. Thus the atomic planes responsible for the Bragg like scattering are oriented at an angle 25° with the surface of the crystal. Inter planer spacing of these planes was 푑 = 0.91Å. Thus taking 푛=1, since a single scattering maximum was observed, 흀 = ퟐ풅퐬퐢퐧휽 = ퟐ×ퟎ.ퟗퟏ×퐬퐢퐧ퟔퟓ° Å = ퟏ.ퟔퟓ Å which matches quite well with the de Broglie wavelength of the electrons (1.67 Å) within experimental errors. Thus the results of Davisson-Germer experiment confirms de Broglie relation 휆=ℎ⁄ 푝.

***Story (Wikipedia): Davisson began work in 1921 to study electron bombardment and secondary electron emissions. A series of experiments continued through 1925. Davisson and Germer's actual objective was to study the surface of a piece of nickel by directing a beam of electrons at the surface and observing how many electrons bounced off at various angles. They expected that because of the small size of electrons, even the smoothest crystal surface would be too rough and thus the electron beam would experience diffused reflection. The experiment consisted of firing an electron beam at a nickel crystal, perpendicular to the surface of the crystal, and measuring how the number of reflected electrons varied as the angle between the detector and the nickel surface varied. To measure the number of electrons that were scattered at different angles, an electron detector that could be moved on an arc path about the crystal was used. The detector was designed to accept only elastically scattered electrons. During the experiment, air accidentally entered the chamber, producing an oxide film on the nickel surface. To remove the oxide, Davisson and Germer heated the specimen in a high temperature oven, not knowing that this caused the formerly polycrystalline structure of the nickel to form large single crystal areas with crystal planes continuous over the width of the electron beam. When they started the experiment again and the electrons hit the surface, they were scattered by nickel atoms in crystal planes (so the atoms were regularly spaced) of the crystal. This, in 1925, generated a diffraction pattern with unexpected peaks. On a break, Davisson attended the Oxford meeting of the British Association for the Advancement of Science in summer 1926. At this meeting, he learned of the recent advances in quantum mechanics. To Davisson's surprise, Max Born gave a lecture that used diffraction curves from Davisson's 1923 research which he had published in Science that year, using the data as confirmation of the de Broglie hypothesis. He learned that in prior years, other scientists – Walter Elsasser, E. G. Dymond, and Blackett, James Chadwick, and Charles Ellis – had attempted similar diffraction experiments, but were unable to generate low enough vacuums or detect the low-intensity beams needed. Returning to the United States, Davisson made modifications to the tube design and detector mounting, adding azimuth in addition to colatitude. Following experiments generated a strong signal peak at 65 V and an angle θ = 45°. He published a note to Nature titled, "The Scattering of Electrons by a Single Crystal of Nickel". Questions still needed to be answered and experimentation continued through 1927. By varying the applied voltage to the electron gun, the maximum intensity of electrons diffracted by the atomic surface was found at different angles. The highest intensity was observed at an angle θ = 50° with a voltage of 54 V, giving the electrons a kinetic energy of 54 eV. According to the de Broglie relation, electrons with kinetic energy of 54 eV have a wavelength of 0.167 nm. The experimental outcome was 0.165 nm via Bragg's law, which closely matched the predictions. Davisson and Germer's accidental discovery of the diffraction of electrons was the first direct evidence confirming de Broglie's hypothesis that particles can have wave properties as well.***

Double slit experiment with light (photons) and with electrons:

Double slit experiment with parallel monochromatic beam of light:

b

a

Path difference: 훿 =(푎 + 푏)sin 휃, where 푎 & 푏 are respectively the widths of the slits and the separation between the slits. 휃 is the angular position of a point on the screen with respect to the central line between the slits. Condition of maximum intensity: (푎 + 푏)sin 휃 = 푛휆 ⇒ sin 휃 = ; Now for small 휃 , sin 휃 ≈ tan 휃 =

So = ; 푦 = 푛휆

퐹푟푖푛푔푒 푤푖푑푡ℎ= 훽 = 푦 − 푦 = 휆 = 휆 = 푐표푛푠푡푎푛푡. where 푑 = 푎 + 푏 is the distance between the centres of the slits.

Resultant intensity on the screen (considering the slits to be extremely narrow):

( ) (()) 휓 = 퐴푒 ; 휓 = 퐴푒

휓 = 휓 + 휓 ;

∗ ∗ 퐼 = |휓| = (휓 + 휓)(휓 + 휓 ) = |휓| + |휓| + |휓||휓|푒 + 푒

= |휓| + |휓| +2|휓||휓| cos(푘훿)

퐼 = 퐼 + 퐼 +2퐼퐼 cos[푘(푎 + 푏)sin 휃]

For 퐼 = 퐼 = 퐼, we have:

() 퐼 =2퐼 +2퐼 cos[푘(푎 + 푏)sin 휃] =4퐼 cos 훽, where 훽 = sin 휃.

Again the intensity distribution on the screen, for a single slit of finite width 푎, is given by:

퐼 = 퐼 , where 훼 = sin 휃 = sin 휃 .

It can be shown that in case of the double slit experiment if the finite width of the slits is considered, then intensity distribution will be given by:

퐼 =4퐼 cos 훽.

Double slit experiment with macroscopic particles:

퐼=퐼 +퐼

Quantum Mechanics Concepts and Applications - Nouredine Zettili

Double slit experiment with parallel mono-energetic beam of electrons:

1. 퐼≠퐼 +퐼

2. Count distribution pattern on the screen is similar to the double slit diffraction pattern of light of wavelength 휆=ℎ⁄ 푝, where 푝 is the linear momentum of the electrons and ℎ is Planck constant.

3. The count distribution pattern remains same if even one electron at a time is fired from the electron gun.

Quantum Mechanics Concepts and Applications - Nouredine Zettili

Single slit diffraction of light:

Y

A p a=y px Δpy  X

B

D

The intensity of light on the screen is given by:

퐼=퐼 , where 훼= sin휃 = sin휃, 휃 being the angular position on the slit with respect to the centre of the slit. Condition of minima is = 0, = +푣푒 ⇒ sin훼 = 0 ⇒ 훼 = sin휃 = 푛휋 . ⇒ 푎sin휃 = 푛휆; 푛 = ±1,±2,±3 …;푛 ≠ 0,as 푛 = 0 corresponds to the central maximum. For the first minimum on the both sides of the central maximum |sin휃| = . Before passing through the slit, the beam of light is parallel and so have no angular spread. But after passing through the narrow slit, the beam of light gets some angular spread. In the figure the angular spread of the central maximum is 2휃 ≈ 2sin휃 (푠푖푛푐푒 휃 푖푠 푠푚푎푙푙) = . Thus the angular spread of the central maximum increases if the width of the slit (푎) decreases.

Single slit diffraction of electrons and Heisenberg uncertainty principle: Now let in the place of light/photons electrons are used for the diffraction experiment. Let detectors are placed at every point on the screen which can count the number of electrons hitting that point. The distribution pattern of electron counts on the screen will be similar to that for light of wavelength 휆 = ℎ⁄ 푝, where 푝 is the momentum of the mono-energetic electrons incident on the slits.

The electrons, emitting from the source, move in positive X direction. So we can write 푝=푝. But after passing through the slit the electrons gain a little momentum in Y direction, which causes angular spread of the diffracted electron beam. Let ±∆푝 are the 푦-components of momentum gained by the electrons reaching to the first minima on the both sides of the central maximum. Then we can say that, the electrons reaching the screen within the central peak 퐴퐵 have gained at the slit an uncertainty ∆푝 in 푦-component of momentum in either positive or negative 푦-direction. These electrons have passed through any point in the slit of width 푎. So, at the slit, the uncertainty in their 푦-coordinate is ∆푦=푎. ∆ Now: =tan휃 ≈휃 ≈sin휃 (푠푖푛푐푒 휃 푖푠 푠푚푎푙푙) = = ∆ So, at the slit, ∆푦∆푝 ≈ 휆푝 ⇒ ∆푦∆푝 ≈휆 ⇒ ∆푦∆푝 ≈ℎ. ℏ i.e. the product of the uncertainties in 푦-component of momentum and position is ℎ> . Thus Heisenberg uncertainty principle is obeyed. From uncertainty principle show that, in the double slit diffraction of electrons, it is not possible to determine the slit, through which the electron passes and to produce the diffraction pattern in a single experiment.

Quantum Mechanics Concepts and Applications - Nouredine Zettili

Y

X  d C d/2 훿푝 푝 D

Ans.: Fringe width of the interference fringes in the double slit diffraction experiment is given by: 훽=푦 −푦 = 휆 . For the fringe system to be stationary the uncertainty in measurement of the position (푦- coordinate) of the slit plate must be small compared to the fringe width 훽. i.e. ∆푦 ≪ 훽 ⇒ ∆푦 ≪ 휆. ….. ……. (A) Now let we want to determine through which the electron passes to reach a point 퐶 on the screen. If the electron passes through the lower slit, it would have to gain a momentum, say 훿푝, in +푣푒 푌 direction and if it passes through the upper slit, it would have to gain a momentum, say 훿푝, in −푣푒 푌 direction. In the two cases the slit plate will recoil in opposite directions and measuring the recoil momentum of the slit plate it will be possible to determine through which slit the electron passes. But to measure the recoil momenta of the slit plate in these two cases distinctly the uncertainty in measurement of the momentum of the slit plate must be much small compared to the difference of these two recoil momenta.

i.e. ∆푝 ≪ 훿푝 − −훿푝, or, ∆푝 ≪ 2훿푝. Now from figure:

⁄ = tan훼 = = . Therefore 훿푝 =푝 = . So, ∆푝 ≪2 , or, ∆푝 ≪ ……. (B) Thus to measure through which slit the electron passes and also to have a diffraction pattern at the same time, we must have both the conditions (A) and (B) satisfied at the same time. Or in other words: ∆푝 ∆푦 ≪ 휆;

Or, ∆푝∆푦≪ℎ…..(C). Equation (C) is clear violation of uncertainty principle and so impossible. Therefore measuring through which the electron passes and having a diffraction pattern at the same time is impossible. Or in other words, any successful trial of ‘seeing’ through which slit the electrons pass will destroy the diffraction pattern.

Problems of Waveparticle Duality, Uncertainty Principle, Two Slit Experiment: JAM 2017

Ans.: Rest mass energy of the electron is 0.511 MeV Kinetic energy gained by the electron is 200 푘푒푉 = 0.2 MeV, which is of the order of the rest mass energy. Therefore relativistic treatment is required. Kinetic energy gained by the electron is 200 푘푒푉 = 3.2×10퐽

퐾. 퐸 = 퐸 − 푚푐 = 푝 푐 + 푚 푐 − 푚푐

푝 푐 + 푚 푐 = 퐾. 퐸 + 푚푐

. 푝 =

(. ) .(. ) 푝 = =

.×(.××.×) 푝 = .× . = × 10 .× . = × 10 = 2.64×10 푘푔. 푚⁄ 푠 .×

.× 휆 = = = 2.5×10푚 .× Best possible resolution = 휆 = 2.5×10 푚 JAM 2016

Ans.: Rest mass energy of electron is 0.511 MeV. Given kinetic energy of the electron is 10 eV is much less than the rest mass energy of the electron. Therefore nonrelativistic treatment can be done.

푃 = √2푚퐸 = √2×9.1×10 ×1.6×10 × 10 = 1.7065× 10 푘푔. 푚푠 .

.× 휆 = = = 3.88×10 푚 = 0.388 푛푚 .× 휆 = = 124 푛푚

.× 푃 = = = 5.34×10 푘푔. 푚푠 ×

×.× 푃 = = = 5.33×10 푘푔. 푚푠 ×

Thus 휆 < 휆 & 푃 > 푃 JAM 2016

Ans.: (A) If (A) would be true then the momentum of the electrons in 푥-direction would be completely indeterminable. Then one could not say that they are accelerated in 푦- direction. (B) From Heisenberg uncertainty relation, if the position in 푥-direction has an uncertainty 푑, ℏ ℏ then uncertainty in momentum in 푥-direction will be ∆푝 ≈ ≈ . So (B) is correct. (C) Uncertainty in momentum along 푦-direction: Energy of the electron is 퐸 = 100±0.1 푒푉 = (100 ± 0.1) ×1.6×10 푒푉 Nonrelativistic treatment:

∆푝 = 2푚푒(푉 +∆푉) − 2푚푒(푉 −∆푉)

= √2×9.1×10 ×1.6×10 (푉 +∆푉) − (푉 −∆푉)

∆ ∆ = 5.3963× 10 √푉 1+ − √푉 1−

∆ . = 5.3963× 10 × √푉 × = 5.3963 × 10 ×10× 푘푔. 푚푠 = 5.3963× 10 푘푔. 푚푠. Relativistic treatment:

(∆)((∆) ) (∆)((∆) ) ∆푝 = −

(∆)( ∆) (∆)( ∆) = −

∆ ∆ ∆ ∆ ( ) ( ) = −

( ) ∆ ∆ ∆ ∆ = 1+ 1+ − 1− 1− ( ) ( )

( ) ∆ ∆ ∆ ∆ = 1+ + − 1− − ( ) ( ) ( ) ∆ ∆ = + ( )

(× ) . . = + , (×)

(since 푚푐 = 0.511 푀eV = 511000 × 푒 퐽표푢푙푒)

× = √ × ×0.1 = 5.39247× 10 푘푔.푚⁄ 푠 ×

ℏ .× ∆푦 ≈ = = 9.776 푛푚 ≠ 0 ∆ ××.× (D) The presence of the slit has no influence on the 푦-direction motion of the electrons. So it does not effect the momentum of the electrons in 푦-direction. JAM 2016

Ans.: 퐸 =푝푐 +푚 푐 ⇒ 푝 = +푚 푐

(×.×) ⇒푝= +푚 푐 = + (9.1 × 10 ×3×10) (×)

= (0.53 × 10) + (27.3 × 10 ) = 10 × (0.53) + (0.273) = 0.59×10 푘푔 푚 푠

.× 휆= = = 11.23×10 푚 = 1.123×10 푚 . .× JAM 2015: Section A

Ans.: 푀푖푛푖푚푢푚 푢푛푐푒푟푡푎푖푛푡푦 푖푛 푚표푚푒푛푡푢푚 ≈ (ℏ⁄ 2 )⁄푀푎푥푖푚푢푚 푢푛푐푒푟푡푎푖푛푡푦 푖푛 푝표푠푖푡푖표푛

ℏ .× ∆푝 ≈ = = ≈ ≈ 푘푔.푚푠. ∆ ×∆ ×∆ × Minimum momentum of the electron is:

푝 =∆푝≈ 푘푔.푚푠. If non relativistic treatment is done then the velocity of the electron comes to be:

푣=푝 ⁄푚 ≈ ≈ 10 푚⁄ 푠 ≫ 푐 푎푛푑 푖푠 푖푚푝표푠푠푖푏푙푒. ×.× Therefore relativistic treatment is to be done.

퐸 = 푝 푐 +푚 푐

퐸 =푐푝 +푚 푐 =3×10 × + (9.1 × 10)(3×10) =3×10 × 10 √0.25+27.3× 10 ≈3×10 × 10 ×0.5 = 1.5×10퐽 .× = 푒푉 = 10푒푉 = 10푀푒푉 .× JAM 2007:

Ans.: 휆 = Compton wavelength = 휆 =

푝= =푚 푐 푝=푚푣= = ⁄ ⁄

=푚푐 ⇒ 푣 ⁄푐 =1−푣 ⁄푐 ⇒ 2푣 ⁄푐 = 1 ⇒ 푣 = . ⁄ √

QX. Zero point energy of harmonic oscillator from uncertainty principle: If ∆푝 is the standard deviation of measurement of 푝 then: (∆푝) = 〈푝 〉 − 〈푝〉 = 〈푝 〉 Similarly if ∆푥 is the standard deviation of measurement of 푥 then: (∆푥) = 〈푥〉 − 〈푥〉 = 〈푥〉 Now 퐸 = + 푘푥 〈 〉 (∆ ) (∆ ) So 〈퐸〉 = + 푘〈푥〉 = + 푘(∆푥) = + (∆푥) ℏ Uncertainty Principle: (∆푥)(∆푝 )≥ ℏ ⇒ (∆푥)(∆푝 ) ≥ ℏ ⇒ (∆푝 ) ≥ (∆) ℏ Thus 〈퐸〉 ≥ + (∆푥) (∆) (..) ℏ The r.h.s will be minimum if = 0; 푖.푒. 푖푓 − + = 0; 푖.푒. 푖푓 (∆푥) = [(∆)] (∆) ℏ

ℏ ℏ So 〈퐸〉 = + ℏ ퟏ = ℏ흎 = 퐳퐞퐫퐨 퐩퐨퐢퐧퐭 퐞퐧퐞퐫퐠퐲 퐨퐟 퐪퐮퐚퐧퐭퐮퐦 퐡퐚퐫퐦퐨퐧퐢퐜 퐨퐬퐜퐢퐥퐥퐚퐭퐨퐫. ퟐ

SOME PROBLEMS ON WAVE PACKET Eigen functions of particle in a one dimensional infinite potential well:

ℏ ℏ 훹(푥. 푡) = ∑ 푐 휓 (푥, 푡) = ∑ 푐 휓 (푥)푒⁄ℏ = ∑ 푐 sin 푒

Where 휓 (푥) = sin are the eigen functions or stationary states.

2⁄푎

−2⁄푎 Plot of 휓 (푥) = sin for 푎 = 1, 푛 = 1,2,3

Plot of 푃 (푥) = sin for 푎 = 1, 푛 = 1,2,3

ℏ ℏ The functions 휓 (푥, 푡) = sin 푒 , are not of the form 푓(푥 ± 푣푡) or 푓(푘푥 ± 휔푡). So they do not represent propagating waves. They actually represent standing waves with amplitude sin and wavelength . The probability density is 푃 (푥) = sin for = 1, 푛 = 1,2,3… . ********* Free particle:

ℏ (,) (,) = 푖ℏ Using separation of variables: 훹(푥, 푡) = 휓(푥)휂(푡) 풊푬 ℏ () 풕 = 퐸 ⇒ =− 푑푡 ⇒ 휂(풕) = 풆 ℏ () ℏ Since 휂(푡) must not blow up at 푡 =∞, so 퐸 must have to be real.

ℏ () = 퐸 ⇒ + 휓 = 0 ⇒ + 푘휓 = 0; 푘 = ⇒ 휓(푥) = 퐴푒 + () ℏ ℏ 퐵푒

⇒ 휓 (푥) = 퐴 푒 푤푖푡ℎ 푘 =± for waves travelling to right and to left ℏ respectively… (A)

Since 휓(푥) must not blow up at 푥 = ±∞, so 푘 must have to be real.

ퟐ풎푬 푬 풊± 풙 풕 ℏퟐ ℏ 흍풌(풙, 풕) = 퐴푒 ℏ = 푨풌풆 . …… (B)

ℏ For free particle there are no boundary conditions. Therefore values of 푘 or 퐸 = are not discrete. Rather they have continuous values.

Comparing the equation 휓(푥, 푡) = 퐴푒 ℏ = 퐴푒 ℏ with the standard wave equation 푓 = 푓(푥 − 푣푡) we see that the velocity of the wave is:

ℏퟐ 푬 푣 = = = ……… (C). ℏ ℏ ퟐ풎푬 ퟐ풎

ퟐ푬 But the particle velocity 푣 = …… (D) 퐹푟표푚 퐸 = 푚푣 . 풎

Therefore, = × = . The ‘eigen function’ (A) [or (B)] do not represent the stationary states of a free particle since:

(i) 푃 = |휓(푥, 푡)| = |퐴| is independent of 푥 and the probability of finding the particle at anywhere between 푥 = −∞ to +∞ is the same (see Fig.). Therefore the eigen function does not represent a localised particle.

푃(푥)

−∞ 0 푥 +∞

( ) | | (ii) ∫ 푃 푥 푑푥 = 퐴 ∫ 푑푥 =∞. i.e the ‘eigen functions’ 휓 are not square integrable and total probability becomes infinite which is impossible. (iii) The wave, representing the particle, moves with a velocity which is half of the velocity of the particle.

Thus these ‘eigen functions’ [eqn. (A) or (B)] “do not represent the physically realisable states”[Griffiths]. In other words “a free particle cannot exist in a stationary state” or “there is no such thing as a free particle with a definite energy”[Griffiths]. Now the general solution will be the linear combination of all the eigen functions or of all the ℏ solutions exp 푖 푘푥 − 푡. Since 푘 or 퐸 are continuous so the summation in the linear combination will be replaced by integration: ℏ 훹(푥, 푡) = ∫ 휑(푘)푒 푑푘 , √ where 휑(푘) play the role of the coefficients of the linear combination. √ 휑(푘) can be determined from the initial wave function 훹(푥,0) = ∫ 휑(푘)푒풊풌풙 푑푘 with √ the help of Fourier Transform: 1 휑(푘) = 훹(푥,0)푒풊풌풙 푑푥 √2휋

Fourier Transform:

If 푓(푥) = ∫ 퐹(푘)푒푑푘 ……… (푖) then 퐹(푘) = ∫ 푓(푥)푒푑푥 ……… (푖푖) √ √ and vice versa. (ii) is called the Fourier Transform of (i) and (i) is called inverse Fourier transform of (ii). Thus the free particle wave function at 풕 = ퟎ and at 풕 = 풕 are:

훹(푥,0) = ∫ 휑(푘)푒풊풌풙 푑푘 √

ퟐ 푬 ℏ풌 풊풌풙 풕 풊풌풙 풕 and 훹(푥, 푡) = ∫ 휑(푘)푒 ℏ 푑푘 = ∫ 휑(푘)푒 ퟐ풎 푑푘 √ √ with 휑(푘) = ∫ 훹(푥,0)푒풊풌풙 푑푥. √ Now we see that |훹(푥,0)| = 푃(푥,0) = ∫ |휑(푘)| 푑푘 and ∫ 푃(푥,0)푑푥 = ∫ ∫ |휑(푘)| 푑푘푑푥 is not infinite essentially.

ퟏ What does 휳(풙, ퟎ) = ∫ 흋(풌)풆풊풌풙 풅풌 represent? √ퟐ흅 The type or shape of the wave represented by 훹(푥,0) = ∫ 휑(푘)푒 푑푘 depends on √ 휑(푘). To have an idea about what is represented by 훹(푥,0), let us first solve the following problems (A) to (D). Problem A. (From: Int. to Q. M. 2nd edn. D. J. Griffiths Problem 2.20 (a)): Dirichlet’s theorem says that ‘any’ function 푓(푥)on the interval [−푎, 푎]can be expanded as a Fourier series:

푓(푥) = ∑ 푎 sin + 푏 cos

Show that this can be written equivalently as: 푓(푥) = ∑ 푐푒 ….. ……(A) where 푐 = 푏 for 푛 = 0; 푐 = (−푖푎 + 푏 ) for 푛 = 1,2,… ∞ and 푐 = (푖푎 + 푏 ) for 푛 = −1,−2,… −∞.

Ans.: 푓(푥) = ∑ 푎 sin + 푏 cos = ∑ 푎 + 푏

= ∑ + 푒 + − + 푒 = ∑ (−푖푎 + 푏 )푒 + ∑ (푖푎 + 푏 )푒

= (−푖푎 + 푏 ) + (푖푎 + 푏 ) + ∑ (−푖푎 + 푏 )푒 + ∑ (푖푎 + 푏 )푒

.. = ∑ (−푖푎 + 푏 )푒 + 푏 푒 + ∑ (푖푎 + 푏 )푒

.. = ∑ (푖푎 + 푏 )푒 + 푏 푒 + ∑ (−푖푎 + 푏 )푒

= ∑ 푐푒 Problem B. (From: Int. to Q. M. 2nd edn. D. J. Griffiths Problem 2.20 (b)):

Show that: 푐 = ∫ 푓(푥)푒 푑푥 ….. ……(B)

Ans.: 푓(푥)= ∑ 푐푒

() ∑ ∑ ∫ 푓(푥)푒 푑푥 = ∫ 푐푒 푒 푑푥 = ∫ 푐푒 푑푥 …… (X) For 푛 = 푚,

() ∫ 푐푒 푑푥 = ∫ 푐푒 푑푥 =2푎푐 For 푛 ≠ 푚,

() () () ( ) ∫ 푐 푒 푑푥 = 푐 푒 − 푒 = 푐 푒 − () () 푒()

= 푐 (−1)() − (−1)() (Since 푒 = −1) () =0 Therefore all the terms of the summation of the r.h.s of (X) vanish except for 푛 = 푚 and:

() ∑ ∫ 푓(푥)푒 푑푥 = ∫ 푐푒 푑푥 =2푎푐.

Hence 푐 = ∫ 푓(푥)푒 푑푥. Problem C. (From: Int. to Q. M. 2nd edn. D. J. Griffiths Problem 2.20 (c)):

Let 푘 = , ∆푘 is 푡ℎ푒 increment in 푘 for increment in 푛 by 1 and 푐 = 퐹(푘). Show that (A) and (B) reduce to:

푓(푥) = ∑ 퐹(푘)푒∆푘 ……. (C) and 퐹(푘) = ∫ 푓(푥)푒푑푥. ……. (D) √ √

Ans.: We have 푓(푥) = ∑ 푐푒

With the substitutions 푘 = , and 푐 = 퐹(푘) we have:

푓(푥) = ∑ 퐹(푘)푒 = ∑ √2휋 퐹(푘)푒 = ∑ 퐹(푘)푒 √ √

Now ∆푘 is 푡ℎ푒 increment in 푘 for increment in 푛 by 1. i.e. ∆푘 = . Then: 푓(푥) = ∑ 퐹(푘)푒 = ∑ 퐹(푘)푒∆푘. ……. (C) √ √

Also we have: 푐 = ∫ 푓(푥)푒 푑푥 ⇒ 퐹(푘) = ∫ 푓(푥)푒푑푥

⇒ 퐹(푘) = 푎 ∫ 푓(푥)푒푑푥 = ∫ 푓(푥)푒푑푥 ……. (D) √ Problem D. (From: Int. to Q. M. 2nd edn. D. J. Griffiths Problem 2.20 (d)): Show that for 푎 →∞ (C) and (D) reduce to:

푓(푥) = ∫ 퐹(푘)푒푑푘 ……. (E) and 퐹(푘) = ∫ 푓(푥)푒푑푥. ……. (F) √ √ Ans.: For 푎 →∞, from (D) we have: 퐹(푘) = ∫ 푓(푥)푒푑푥 = ∫ 푓(푥)푒푑푥 √ √ From (C) 푓(푥) = ∑ 퐹(푘)푒∆푘. Now for 푎 →∞, ∆푘 = →0. Therefore the √ summation is to be replaced by integration: 푓(푥) = ∫ 퐹(푘)푒푑푘. √ Wave Packet: Now let us plot the wave obtained by adding sine waves having uniformly varying propagation constants within a finite range [푓표푟 푒푥푎푚푝푙푒, 푤푒 푚푎푦 푡푎푘푒 1 ≤ 푘 ≤ 1.1] i.e. ∑ sin 푘푥, where 1≤ 푘 ≤ 1.1. We see that [Fig.1-7] the resultant wave is a series of wave groups. As the number of component waves within the finite range of propagation constants increases, the resultant wave groups increase in height, decrease in width and the separation between the groups increases. The small wave groups in between can be neglected in the present discussion since they decrease in height and thus become insignificant with increase in the number of component waves. If the number of component waves tends to infinity then the separation between the groups will increase to infinity and we will get a single wave group or wave packet. The height of the group or packet will be very large and width will be very small. If the resultant wave represents the wave function of a particle, then the probability density of the particle, i.e. the square of the resultant wave function will be very high at a very narrow region i.e. within the width of the wave group and the particle will be well localised.

푦 = sin(푥) 푦 = sin(푥) + sin(1.1푥)

푦 = sin(푥) + sin(1.05푥) + sin(1.1푥) 푦 = sin(푥) + sin(1.025푥) + sin(1.05푥) + sin(1.075푥)+ sin(1.1푥)

푦 = sin(푥) + sin(1.0125푥) + sin(1.025푥) + sin(1.0375푥) + sin(1.05푥) + sin(1.0625푥) + sin(1.075푥) + sin(1.0875푥) + sin(1.1푥)

푦 = sin(1 + 0.001푛)푥

We see that the wave packet is composed of waves of smaller wavelength modulated by a profile or envelope which gives the shape of the group (Fig.-8).

QQ: Let two superposing sine waves have wave lengths 휆 and 휆. Let at any particular instant of time their peaks coincide at 푥 =0. Show that at the same instant their peaks will also coincide at:

푥 = 푛 × LCM(휆, 휆), where 푛 is an integer. From the problems (A) to (D) it is clear that the set of sine waves forming the above wave groups can be obtained as special case from 푓(푥) = ∫ 퐹(푘)푒푑푘 in proceeding back √ from problem (D) to (A) i.e. from 푓(푥) = ∫ 퐹(푘)푒푑푘 to 푓(푥) = √ ∑ 푎 sin + 푏 cos and by suitably choosing 푛 and the coefficients 푎 and 푏 . Therefore 훹(푥,0) = ∫ 휑(푘)푒풊풌풙 푑푘 represents a localised wave packet or group, whose √ shape depends on 휑(푘) and may be different from those given in the above figures. This will be clear from the following problems (E) & (F). However since the wave packet Problem E. (From: Int. to Q. M. 2nd edn. D. J. Griffiths Problem 2.21): A free particle has initial wave function 훹(푥,0) = 퐴푒풂|풙|, where 퐴 and 푎 are constants with 푎 real and positive. (i) Normalise 훹(푥,0). (ii) Find 훹(푥, 푡) in integral form. (iii) Discuss the cases for 푎 very large and 푎 very small.

Ans.: (i) Normalise 훹(푥,0): | ( )| |풙| ∫ 훹 푥,0 = 1 ⇒ 퐴 ∫ 푒 푑푥 = 1 ⇒ 퐴 ∫ 푒풙푑푥 + ∫ 푒풙푑푥 = 1 ⇒ 퐴 ∫ 푒푑푦 + ∫ 푒푑푦 = 1 ⇒ 퐴 ([푒] − [푒]) =1 ⇒ 퐴 (푒 − 푒 − 푒 + 푒) =1 ⇒ =1⇒ 퐴 = 푎 . √ So 훹(푥,0) = √푎푒풂|풙|. (ii) 훹(푥,0) = √푎푒풂|풙| 휑(푘) = ∫ 훹(푥,0)푒 푑푥 = √푎 ∫ 푒풂|풙|푒 푑푥 √ √ = ∫ 푒(풂풙) 푑푥 + ∫ 푒(풂풙) 푑푥 = 푒풂풙 + 푒(풂풙) () = − () = + = 휑(푘) =

ℏ풌ퟐ ℏ풌ퟐ 풊풌풙 풕 ⁄ 풊풌풙 풕 훹(푥, 푡) = ∫ 휑(푘)푒 ퟐ풎 푑푘 = ∫ 푒 ퟐ풎 푑푘 √ (iii) For very large 푎, 훹(푥,0) = 퐴푒풂|풙|is a very narrow spike at 푥 =0. But for small 푎, 훹(푥,0) = 퐴푒풂|풙| is flat. 훹(푥, 0) = √푎푒풂|풙| , 푤푖푡ℎ 푎 = 1

x

풂 ퟐ풂 흋(풌) = 풘풊풕풉 풂 = ퟏ ퟐ흅 풂ퟐ + 풌ퟐ

k

Problem F. (From: Int. to Q. M. 2nd edn. D. J. Griffiths Problem 2.22, with modification):

A free particle has initial wave function 훹(푥,0) = ⁄ 푒 푒 = 퐴푒 푒 , ( ) where 퐴 and 푎 are constants with 푎 real and positive.

(i) Show that 훹(푥,0) = ⁄ 푒 푒 is normalised. ( ) (ii) Normalise 훹(푥,0) = 퐴푒 푒 to show that 퐴 = . (iii) Plot 퐼푚훹(푥,0) and |훹(푥,0)| for 푎 = 1, 푘 = 100. (iv) Show that the expression of 휑(푘) in the Fourier expansion of 훹(푥,0) is: ( ) 휑(푘) = 푒 . √ (v) Plot 휑(푘) vs. 푘 for 푎 =1. (vi) Find the expression of 훹(푥, 푡). (vii) Find the expression of |훹(푥, 푡)|. Comment on it. Plot |훹(푥, 푡)| vs. 푡. What happens to |훹(푥, 푡)| as time goes on. (viii) At what time does the system come closest to uncertainty limit?

Ans.: (i) (ii) Normalise 훹(푥,0) = 퐴푒 푒: | ( )| ∫ 훹 푥,0 = 1 ⇒ 퐴 ∫ 푒 푑푥 = 1 ⁄ Let 2푎푥 = 푦 ⇒ 4푎푥푑푥 = 푑푦 ⇒ 푑푥 = = √ = √ √ ⇒ ∫ 푦⁄ 푒푑푦 =1⇒ Γ =1⇒ Γ ⇒ 퐴 = √ √ √ (iii) 훹(푥,0) = 푒 푒 퐼푚훹(푥,0) = 푒 sin(푘 푥) = 푒 sin(100푥). |훹(푥,0)| = 푒 = 푒 .

(iv) 훹(푥,0) = 푒 푒 휑(푘) = ∫ 훹(푥,0)푒 푑푥 = ∫ 푒 푒() 푑푥 √ √ = ∫ 푒 () 푑푥 √

ퟐ ퟐ 풃 ퟐ 풃 풃 풃 ퟐ 풂풙ퟐ 풙 풂풙 ퟐ풙 ퟐ 풆풂풙 풃풙 풅풙 = 풆 풂 풅풙 = 풆 ퟐ풂 ퟒ풂 ퟒ풂 풅풙

풃ퟐ 풃 ퟐ 풃ퟐ 풃ퟐ 풃ퟐ 풂풙 ퟐ ퟐ ퟐ √흅 흅 = 풆ퟒ풂 풆 ퟐ풂 풅풙 = 풆ퟒ풂 풆풚 풅풚 = 풆ퟒ풂 = 풆ퟒ풂 √풂 ퟎ √풂 ퟐ 풂

( ) ( ) 휑(푘) = 푒 = 푒 . √ √

ퟏ ퟒ 풌ퟐ ퟏ ퟐ풂 흋(풌) = 풆 ퟒ풂 풘풊풕풉 풂 = ퟏ √ퟐ풂 흅

k ℏ ( ) ℏ 훹(푥,푡) = ∫ 휑(푘)푒 푑푘 = ∫ 푒 푒 푑푘 √ √ √

ℏ = ∫ 푒 푑푘 √ √

ℏ = 푒 ∫ 푒 푑푘 √ √

ퟐ ℏ 1 1 2푎 흅 ퟒ = 푒 풆 √2휋 √2푎 휋 푖ℏ푡 1 2푚 + 4푎 ퟐ 2푎 1 1 ퟒ풂흅 풂 (ℏ⁄ ) = 푒 풆 휋 √2휋 √2푎 1+2푎푖ℏ푡⁄ 푚

( ⁄ ) 2푎 ퟏ 풂 ℏ = 푒 풆 휋 1+2푎푖ℏ푡⁄ 푚 풂 (ℏ⁄ ) 2푎 ퟏ = 풆 (ℏ⁄ ) 휋 1+2푎푖ℏ푡⁄ 푚

ℏ 풂 2푎 ퟏ = 풆 (ℏ⁄ ) 휋 1+2푎푖ℏ푡⁄ 푚

ℏ ⁄ (ℏ⁄ ) = ℏ 푒 = . √ √ ℏ⁄

⁄ (ℏ⁄ ) (ℏ⁄ ) ⁄ ℏ ⁄ (iii) |훹(푥, 푡)| = = . (ℏ⁄ )(ℏ⁄ ) ℏ⁄

ퟐ풂 ퟏ⁄ퟐ ퟐ |휳(풙, ퟎ)|ퟐ = 풆ퟐ풂풙 흅 ퟐ풂풙ퟐ ퟐ풂 풆 ퟏퟒ풂ퟐℏퟐ풕ퟐ⁄풎ퟐ |휳(풙, 풕)|ퟐ = 흅 ퟏ + ퟒ풂ퟐℏퟐ풕ퟐ⁄풎ퟐ

x |휳(풙, 풕)|ퟐ at 풕 = ퟎ & 풕 = 풎⁄ퟐ풂ℏ

Phase velocity, group velocity and particle velocity:

Phase velocity: ℏ Eigen functions of free particle: 휓 (푥, 푡) = 퐴 푒(); where 푘 = , 휔 = , 퐸 = = . ℏ ℏ Let the wave form, which was at position 푥 at time 푡, appears at position 푥 + 푑푥 at time 푡 + 푑푡. In other words:

휓(푥, 푡) = 휓(푥 + 푑푥, 푡 + 푑푡),

() () ℏ Or, 퐴푒 ℏ = 퐴푒 Or, 푘(푥 + 푑푥) − (푡 + 푑푡) = 푘푥 − 푡; ℏ ℏ

ℏ⁄ ℏ Or, 푣 ======. ℏ ℏ Thus the velocity of the wave form, or phase velocity, is equal to half of the particle velocity. We see that the phase velocity of the monochromatic (particular 푘) eigen functions of the free particle is half of the particle velocity. Now let us calculate the group velocity i.e. the velocity of the envelope of the wave group or packet in the following cases: Group velocity: 1. Wave packet composed of two plane waves: Consider two sine waves of equal amplitudes but slightly different propagation constants and frequencies –

휓(푥, 푡) = 퐴 sin(푘푥 + 휔푡) and 휓(푥, 푡) = 퐴 sin(푘 + 푑푘)푥 + (휔 + 푑휔)푡

The wave group formed by these waves will be: 휓(푥, 푡) = sin(푘푥 + 휔푡) and

휓 = 휓(푥, 푡) + 휓(푥, 푡) = 퐴 sin(푘푥 + 휔푡) + 퐴 sin(푘 + 푑푘)푥 + (휔 + 푑휔)푡

=2퐴 sin 푘 + 푥 + 휔 + 푡 cos 푥 + 푡 ≈2퐴 sin(푘푥 + 휔푡) cos 푥 + 푡 =2퐴 cos 푥 + 푡 sin(푘푥 + 휔푡) This is not a plane wave, but a sine wave [sin(푘푥 + 휔푡)] of propagation constant 푘 and frequency 휔 whose amplitude is modulated or enveloped by the function 2퐴 cos 푥 + 푡. The sin(푘푥 + 휔푡) = sin 푘 푥 + 푡 part oscillates with frequency 휔 and moves with velocity 휔⁄ 푘 as the component waves but its amplitude or envelope 2퐴 cos 푥 + 푡 = 2퐴 cos 푥 + 푡 is slowly varying with small frequency 푑휔⁄2 moves with velocity . 푣 = 휔⁄ 푘 represents the velocity of the phase of the part sin 푘 푥 + 푡 and is called the phase velocity and 푣 = represents the velocity of the envelope of the wave and is called the group velocity of the wave. 2. Fourier Packet:

풑 훹(푥, 푡) = ∫ 휑(푘)푒풊(풌풙흎풕) 푑푘, where propagation vector 풌 = √ ℏ

풑 푬 풊 풙 풕 = ∫ 휑(푘)푒 ℏ ℏ 푑푘 √

ℏ풌ퟐ 풊풌풙 풕 = ∫ 휑(푘)푒 ퟐ풎 푑푘 √ Given:

ퟐ The Gaussian wave packet 훹(푥,0) = 퐴푒풂풙 can be written in terms of Fourier transform as:

ퟐ 훹(푥,0) = 퐴푒풂풙 = ∫ 휑(푘)푒 푑푘, with 휑(푘) = 푒 . √ √

Where 휓(푥) = 푒 are plane wave solutions of the Schrodinger eqn. for a free particle. Question:

ℏ ∗ ℏ ∗ Let: 퐽 = 휓 − 휓 ∗ and 퐽 = 훹 − 훹∗ and in a problem of step || || || || potential 푅 = , 푇 = and 푅 = , 푇 = . || || || ||

Verify whether: 푅 = 푅, 푇 = 푇.