Griffiths Chapter 1. Dan Wysocki. February 12, = A exp( λu 2 ) d u. e cx2 d x = 1 = λ = A = π. λ exp( λ(x a)2 ) λ exp( λ(x a)2 ) = πa2.

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1 Griffiths Chapter 1 Dan Wysocki February 12, 215 Problem 1. Consider the gaussian distribution ρ) = A ep λ a) 2 ), where A, a, and λ are positive real constants. a. Use Equation 1.16 to determine A. 1 = A ep λ a) 2 ) d Pull out constant. Let u = a, d u = d From integral table: Substitute: Now we have 1 = A ep λ a) 2 ) d 1 = A ep λu 2 ) d u e c2 d = π c π λ 1 = A λ = A = π ρ) = π λ ep λ a)2 ) b. Find, 2, and σ. = ρ) d = 2 = 2 ρ) d = σ = σ 2 = c. Sketch the graph of ρ). 2 2 = π λ ep λ a)2 ) = πa λ 2 π λ ep λ a)2 ) = πa2 λ + π πa 2 λ + π πa ) 2 π 2a2 λ 2πa 2 + 1) = λ 1

2 ρ) a λ a a + λ Problem 1.4 At time t = a particle is represented by the wave function A, if a, a Ψ, ) = A b, if a b, b a, otherwise, where A, a, and b are constants. a. Normalize Ψ that is, find A, in terms of a and b). 2

3 , if a, a2 ρ, ) = Ψ, ) 2 = Ψ 2, ) = A 2 b ) 2, if a b, b a) 2, otherwise 1 = ρ) d = A 2 a 2 2 a 2 d + A2 b a b ) 2 b a) 2 d u := b = d u = d ; = a = u = b a; = b = u = a 2 u 2 a 2 a 2 d + b a u 2 b a) 2 d u A 2 = a 2 d b a b a) 2 d u = = 1 ] a a u ] b a b a) 2 = 1 a + b a)] = b = A = b, if a, a Ψ, ) = b b, if a b, b a, otherwise, b. Sketch Ψ, ) as a function of Ψ, ) ρ, ) a b c. Where is the particle most likely to be found, at t =? The particle s most likely position is given by argma ρ, ). To the left of a, ρ is positive and increasing, to the right of it, it is positive and decreasing, and outside the interval, b], it is zero, therefore the most likely position is at = a.

4 d. What is the probability of finding the particle to the left of a? Check your result in the limiting cases b = a and b = 2a. P < a] = b a 2 a 2 d = b 1 a 2 ] a = a b In the limiting case of b = a, this gives a probability of 1, which is to be epected as the probability is 1 over the interval, b], which is now the same as, a]. In the limiting case of b = 2a, the probability is 1/2, which is also epected, as P, ) is symmetric about a when both intervals have equal size, distributing half of the probability on, a] and half on a, b]. e. What is the epectation value of? Problem 1.5 Consider the wave function = ρ) d = a b a 2 d + = ] b2a + b) = 2a + b b 12 4 where A, λ, and ω are positive real constants. a. Normalize Ψ b Ψ, t) = A ep λ ) ep ıωt), a ] b )2 b a) 2 d ρ) = Ψ, t) 2 = Ψ Ψ Ψ = A ep λ ıωt) Ψ = A ep λ + ıωt) ρ) = A 2 ep 2λ ) 1 = A 2 = ρ) d ep 2λ ) d = 1 λ A = λ ρ) = λ ep 2λ ) Ψ, t) = λ ep λ ıωt) b. Determine the epectation values of and 2. = ρ) d = 2 = 2 ρ) d = λ ep 2λ ) = 2 λ ep 2λ ) = 1 4

5 c. Finhe standard deviation of. Sketch the graph of Ψ 2, as a function of, and mark the points + σ) and σ), to illustrate the sense in which σ represents the spread in. What is the probability that the particle would be found outside this range? σ = 2 2 = 2 = 2 = ρ) σ + σ The probability of finding the particle outside this range is the complement of the probability of finding the particle inside this range, which is given by ρ σ < < + σ] = 1 ρ σ < < + σ] = 1 +σ σ ) 1 ρ) d = 1 λ ep 2λ ) d ) 1 = 1 1 ep λ 1 ) ) = ep λ 1 ) Problem 1.7 Calculate /. p d Ψ d Ψ d Ψ ] t + d t 5

6 From Schrödinger equation: t = ı 2 Ψ 2 ı V Ψ; Ψ t = ı 2 Ψ t 2 + ı V Ψ ı 2 Ψ 2 + ı ) V Ψ + Note that the wavefunction has continuous second partial derivatives, anherefore the partial derivatives are commutative by Schwarz theorem) ] d = Ψ Ψ { Ψ ı t + ı 2 Ψ ı 2 Ψ 2 + ı ) ] V Ψ d + ı 2 Ψ 2 ı ) V Ψ 2 Ψ ı V Ψ + V )] + Ψ ı V + ı ) V ı 2 Ψ 2 { { ı Ψ ı V Ψ V 2 Ψ 2 + Ψ ) Ψ d ı ) Ψ d + 2 ı Ψ ı + ı V Ψ ) ] d ı 2 Ψ 2 + ı V Ψ ] ı + Ψ V Ψ Ψ ı Ψ Ψ Ψ d ]} d ı Ψ ı V Ψ 2 Ψ 2 2 Ψ ] 2 d ) d ) ]} d Let u = Ψ, then d u = d. Let d v = Ψ d, then v = 2 Ψ 2. Let µ =, then d µ = 2 Ψ 2 d. Let d ν = 2 Ψ 2 d, then ν =. Using integration by parts, that gives: = Ψ V ) { ) Ψ d + 2 Ψ 2 Ψ 2 Ψ 2 2 d = Ψ V ) ) Ψ d + 2 Ψ 2 Ψ ) ] 2 Ψ, t) must go to zero as goes to ±) infinity, so the entire 2 / term is zero. = Ψ V ) Ψ d = V We have now arrived at Equation 1.8. } d ) ]} 2 Ψ 2 d 6