Quantum Theory 2 O Chapter 1- Variational Methods Lecture 1 1.1 - Brief Summary of quantum theory A particle in Rc (d=1, 2, 3) in a potential V Schrodinger Eq: - h2 74(x,+)+ V(x)+(x,+)= it d4 (2,+) 2m dt where 1.14(x, +) 12dxc = 1 (normalisation) Rd · Seperation of time variable: 4(x,t)= 4(x) e = = + - Time dependent -> plug this into the TOSE to get the TISE. TISE: -th2 4(x)+v(x)+(x) = = +(x) 2m il. 4 is an eigenfunction of the Hamiltonian operator H= h2 O+V(x) 2m with eigenvalue E. This leads to the concept of Hilbert Spaces and operators. -> Hilbert Space: L2(Ra) > space of functions 4: RC with , 14(x)12 doc < 00. Rd This space is a complex vector space , with inner product : <4, 4> = 14(x) +(x) dx VY,YELERd Properties of an inner product . < 4, 4 + x> = < 4,4) + <4, x> . < 4, 04) = c <4,4> . < 4, 4 > = < 4, 4> VY, 4, 7 € L' c = complex constant VY, Y EL2, VOER This complex vector space with inner product is a Hilbert space. (L' (IRd) is a Hilbert Space ->[To prove this we must show that every (auchy series converges] Then: · H is an operator, i.e, a linear map on It H(4+4) = HY+HY, H (C4) = CHY V 4,4 € Il , VIEL op. A+ => <A4,4> = < 4,A+4> · Adjoint operator: operator A, adjoint · Physical observables are represented by Hermitian operator A = A+ V4,4€Je
Lecture 2 > List of facts ·L2 (Rd) = { 4: Rd - C ; 5 14 (x ) 12 dxc < 00 } is a Hilbert Space Il , with inner product Rd <4,6) = { 4(x) 4(x) dx IRd norm: 114112 = < 4,4> = {14(x)12 dxc · Linear Operators : A : · Adjoint Operators A+ : <4,A4> =< A+4,4> · Hermitian A= A+ · A hermitian operator , with discrete spectrum: => Then: (1) eigenvalues In are real (ii) AYn = xn In , In EL2 (Rd), In + 0 eigenfunctions form an orthonormal basis in Il, i.e. (x) <41,4m> = Som ={1} n=m 10, n+m (B) 4€+1 => +={cnYn, CHEC · In MT3260, the stationary Schrödinger eq. was solved in 3 basic examples: 1 Particle in a box T 1 Here: Il= L2([0,L]) the limits H = - t2 d2 with boundary conditions: 4' (0)=0 = 4 (L) 0 L 1 × - 2m doc? Particles can not be in OCCO, x>L. -> Dirichlet b.c Probability current must vanish at boundary conditions We have the normalisation condition 5 14(x)13 doc = 1 2 Harmonic Oscillator Il = L 2 (R), H = - t2 d2 + mw2x2 2m doc? 2 with normalisation condition 1. 14(x) \2 doc = Then we can solve HYn = EnUn where En = hw (n+1/2) 3 Hydrogen atom d=3, V(x) = - e2 1001 with normalisation