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Quantum Mechanics Lecture 2 Dr. Mauro Ferreira E-mail: [email protected] Room 2.49, Lloyd Institute Postulates of Quantum Mechanics What’s th e differenc e to classica P1: l mechanic s ? The state of a dynamic system is completely specified by a state (wave) function ... whose physical meaning is of a probability density, i.e., ! |ψ(x)|2 dx Probability of finding the particle in the range between x and x+dx Mean values and deviations are direct consequences of the probabilistic nature of the state function { !x" = ! −∞ ∞ ! dx ψ ∗ (x) ψ(x) = 1 square-integrable functions dx ψ ∗ (x) x ψ(x) = −∞ (∆x)2 = ∞ ∞ −∞ ! ∞ −∞ dx ψ ∗ (x) (x − "x#)2 ψ(x) dx x |ψ(x)|2 What does it mean to have a probabilistic description of the position x of a particle ? { !x" = ! ∞ dx ψ ∗ (x) x ψ(x) = −∞ (∆x)2 = ! ∞ −∞ ! ∞ −∞ dx ψ ∗ (x) (x − "x#)2 ψ(x) dx x |ψ(x)|2 (Mean) (Mean square deviation) (∆x)2 = (∆x)2 = ! ! ∞ −∞ ∞ −∞ dx ψ ∗ (x) (x − "x#)2 ψ(x) dx ψ ∗ (x) [x2 + !x"2 − 2x!x"] ψ(x) (∆x)2 = !x2 " − !x"2 e h t t u o ? b a at tum h W men mo (Mean square deviation) ! ∆x = !x2 " − !x"2 (Uncertainty in the position) Following the same formalism, the momentum is given by !p" = ! ∞ dx ψ ∗ (x) p ψ(x) −∞ To solve the integral above, we should be able to express the momentum p in terms of the position x. However, unlike classical mechanics, p and x are not simultaneously known, which requires an alternative procedure. Let’s start with a wave that describes the motion of a free particle: ψ(x, t) = A ei(kx−ωt) ∂ψ(x, t) = ikψ(x, t) ∂x ! ∂ψ(x, t) p ψ(x, t) = i ∂x Bearing in mind that p = !k ! !p" = i ! ∞ ∂ ψ(x) dx ψ (x) (Mean) ∂x −∞ ∗ (∆p)2 = !p2 " − !p"2 ∆p = where ! (Mean square deviation) !p2 " − !p"2 (Uncertainty in the momentum) !p2 " = −!2 ! ∞ 2 ∂ ψ(x) ∗ dx ψ (x) ∂x2 −∞ Example: A particle is described by the following wave function ψ(x) defined in the region -2<x<2 (see graph below). Calculate the average and the mean square deviations for both the position and the momentum of the particle. πx ψ(x) = A cos( ) 4 2.5 2 1.5 1 0.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 x 2 Example: Normalization condition: A 2 ! x + 2 A= ! πx "+2 sin( 2 ) π 1 2 −2 A 2 ! +2 −2 =1 πx dx cos ( ) = 1 4 2 Example: Average position: Integrand is odd !x" = A 2 ! +2 πx dx x cos ( ) 4 2 −2 ⇒ "x# = 0 The same conclusion could have been obtained from the graph ψ(x) = A cos( 2.5 πx ) 4 2 1.5 1 0.5 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 x 2 Example: ! !p" = i ! ∞ ∂ ψ(x) dx ψ (x) ∂x −∞ πx ψ(x) = A cos( ) 4 { ∗ ∂ψ(x) Aπ πx =− sin( ) ∂x 4 4 −A ! π !p" = 4i 2 ! Integrand is odd 2 πx πx dx cos( ) sin( ) 4 4 −2 ⇒ "p# = 0 Example: !x2 " = ! !x " = A 2 dx ψ ∗ (x) x2 ψ(x) −∞ 2 !x " = 2 A 2 ∞ ! +2 −2 2 πx dx x cos ( ) 4 2 ! +2 0 4 8 !x " = − 2 3 π 2 Integrand is even ? 0 ≠ > 2 x < d πx dx x cos ( ) 4 2 2 n a 0 2 n a C = > <x Example: !p2 " = −!2 ! ∞ 2 ∂ ψ(x) ∗ dx ψ (x) ∂x2 −∞ ∂ 2 ψ(x) −A π 2 πx −π 2 = cos( )= ψ(x) 2 ∂x 16 4 16 ! π !p " = 16 2 2 2 2 2 ! π 2 !p " = 16 ! 2 −2 dx ψ ∗ (x) ψ(x) Normalization Example: !x" = 0 !p" = 0 4 8 2 !x " = − 2 3 π 2 2 ! π 2 !p " = 16 It is instructive to calculate the product Δx.Δp : ∆x ∆p ≈ 0.568 ! ... which obviously satisfies the uncertainty principle. ! ∆x ∆p ≥ 2