# General Physics I Principles of Quantum Mechanics Physics 123 02/28/20 Lecture XV 1 Concepts De Broigle waves Heisenbergs uncertainty principle Schrdingers equation Particle in a box Boundary conditions Unitarity condition 02/28/20 Lecture XV 2 Wave Particle duality If light exhibits both wave and particle properties

then particles (e.g. electrons) must also exhibit wave properties e.g. interference. Matter (de Broglie) waves =h/ph/p p=h/pmv 02/28/20 Lecture XV 3 Interference of electrons Send electron beam (a lot of electrons) on crystal structure Interference pattern is determined by h/p Double slits distance d~1nm Interference pattern Maxima (more e): Minima (no e): 02/28/20 d sin =h/p m m=h/p0,1,2,3,. d sin =h/p (m+ ) m+ ) Lecture XV 4

Matter waves Particle position in space cannot be predicted with infinite precision Heisenberg uncertainty principle x p h / 2 h t E h / 2 h (Wave function of matter wave)2 dV=probability to find particle in volume dV. Laws of quantum mechanics predict for a given system Given one can estimate probability for certain outcomes of experiment 02/28/20 Lecture XV 5 Schrdingers equation Equivalent of energy conservation equation in classical mechanics. Predicts the shape of the wave function. System is defined by potential energy, boundary conditions 2 2 h d ( x)

U ( x ) ( x ) E ( x ) 2 2m dx 02/28/20 Lecture XV 6 Particle in a box Infinite potential well Particle mass m in a box length L U(m+ ) x)=h/p0, if 0L Boundary conditions on (m+ ) 0)=h/p0=h/p(m+ ) L) h 2 d 2

E 2 2m dx d 2 2 Em 2 2 dx h 02/28/20 Lecture XV 7 Particle in a box Second derivative proportional to the function with - sign d 2 2 Em 2 2 dx h Possible solutions: sin(m+ ) kx) and cos(m+ ) kx) 2 Em k 2

h 2 Em p 2 k h h 2 02/28/20 ( x) A sin(kx) B cos(kx) k-wave vector Lecture XV 8 Particle in a box ( x) A sin(kx) B cos(kx) Lets satisfy boundary conditions (0) A sin(k 0) B cos(k 0) B cos(k 0) 0 B 0 ( x) A sin( kx) ( L) A sin(kL) 0 kL n n kn L 02/28/20

Wave vector is quantized! Lecture XV 9 Particle in a box n ( x) A sin(k n x) Quantum number n 2 Em k h n kn L n 2 Em L h 2 n 2 2 Em 2 2 L h 2 2

2 2 h n hn En 2 2 2mL 8mL 02/28/20 Energy is quantized! 2 We are not done yet, We dont know A Lecture XV 10 Particle in a box n ( x) A sin(k n x) We know for sure that the particles is somewhere in the box Probability to find the particle in 0

L 2 | ( x ) | dx 1 0 L L 2 kL A 1 | ( x) | dx A sin (kx)dx k 0 0 2 02/28/20 2 2 Lecture XV

2 sin ( y)dy 0 11 Particle in a box n ( x) A sin(k n x) L 2 | ( x ) | dx 1 0 kL kL kn L 1 sin ( y )dy (1 cos( 2 y )) dy 20 2

0 2 2 kL A 1 k 2 2 A kL A L sin ( y )dy 2k 2 0 2 2 A L 02/28/20 Lecture XV 12 2

n ( x) sin(k n x) L n kn L Count knots n 1 1 ( x) 2 sin( x) 0 L L n 2 2 x m L m x L ,0 x L 2 0 knots in the box x 0, L / 2, L 1 knot in the box x m

L x mL,0 x L x 0, L 02/28/20 Lecture XV n 3 2 knots in the box (N-1) knots in the box 3 x m L m x L ,0 x L 3 13 x 0, L / 3,2 L / 3, L n ( x) Wave function Probability 2 L sin(k n x) n kn L

Probability to find particle between x and x+dx 2 dP( x to x dx) ( x) dx Probability to find particle between x1 and x2 x2 2 P ( x1 to x2 ) ( x) dx x1 02/28/20 Lecture XV Probability density dP 2 2 2 ( x) sin k n x dx L 14 Symmetry considerations The observable: probability density must respect the symmetry of the system System is symmetric around x=h/pL/2

This means that the probability density function does not change if I replace x with (m+ ) L-x): P ( x ) P ( L x ) 2 ( x) ( L x) 2 Wave function is either symmetric or Odd harmonics have even number of knots and are antisymmetric symmetric ( x) ( L x) Even harmonics have odd number of knots and are 02/28/20 Lecture XV 15 antisymmetric