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Assignment 7: Solution Quantum Mechanics I / Quantum Theory I. 14 December, 2015 Assignment 7: Solution 1. In class, we derived that the ground state of the harmonic oscillator j0 is given in the position space as 1=4 m! 2 −m!x =2~ 0(x) = xj0 = e : π~ (a) This question deals with finding the state in the momentum basis. Let's start off by finding the position operatorx ^ in the momentum basis. Show that @ pjx^j = i pj : ~@p This is analogous to the expression of the momentum in the position basis, @ xjp^j = −i~ @x xj . You will use [Beck Eq. 10.56] expressing the momentum eigenstate in the position basis as well as the definition of the Fourier transform. @ (b) Sincea ^j0 = 0 and the position operator pjx^ = i~ @p , construct a differential equation to find the wavefunction in momentum space pj0 = f0(p). Solve this equation to find f0(p). Plot f0(p) in the momentum space. (c) Verify that your f0(p) = pj0 derived above is the Fourier transform of xj0 = 0(x). Answer (a) Lets find the position operator in momentum space pjx^j = pjx^1^j Z = dx pjx^jx xj Z = dx pjx x xj * x^jx = xjx : Now the normalized wavefunction of the momentum eigenstate in the position representation is 1 xjp = p eipx=~ Eq 10.56 in Beck 2π~ Z 1 −ipx= ) pjx^j = p dxe ~x xj : (1) 2π~ Due Date: December. 14, 2015, 05:00 pm 1 Quantum Mechanics I / Quantum Theory I. 14 December, 2015 Now @ −ix e−ipx=~ = e−ipx=~ @p ~ @ i e−ipx=~ = xe−ipx=~: ~ @p Inserting into Eq (1): i Z @ pjx^j = p ~ dx e−ipx=~ xj 2π~ @p Z @ 1 −ipx= = i~ p dxe ~ xj @p 2π~ @ = i pj ; ~@p where e(p) is the Fourier transform of (x). So we have @ pjx^j = i pj and ~@p @ pjx^ = i pj; ~@p which is the required form. (b) We need to solve the lowering operatora ^ in momentum space as rm! i pja^j0 = pj x^ + p^ j0 2~ m! rm! i = pjx^j0 + pjp^j0 : 2~ m! @ Sincea ^j0 and the position operator pjx^ = i~ @p , r m! d i pja^j0 = i~ pj0 + p pj0 2~ dp m! r m! d i 0 = i~ pj0 + p pj0 2~ dp m! Let f0(p) = pj0 be the wavefunction in momentum space, rm! d i 0 = i~ f0(p) + p f0(p) 2~ dp m! d i i f0(p) = − p f0(p) ) ~dp m! d f0(p) 1 = − p dp f0(p) ~m! Due Date: December. 14, 2015, 05:00 pm 2 Quantum Mechanics I / Quantum Theory I. 14 December, 2015 Z Z d f0(p) 1 = − p dp f0(p) ~m! p2 ln f0(p) = − + ln A 2~m! 2 − p f0(p) = Ae 2~m! : Now Z Z 2 p 2 2 − p 2 f0(p) dp = jAj e ~m! dp = jAj π~m! = 1 1 1=4 A = π~m! 1=4 2 1 − p ) f0(p) = e 2~m! : π~m! (c) We need to find the Fourier transform of 0(x) and compare it with f0(p) derived in (b). 1 Z 1 −ipx=~ f0(p) = p dxe 0(x) 2π~ −∞ 1=4 1 1 m! Z 2 = p dxe−ipx=~e−m!x =2~ by using Eq (1) 2π~ π~ −∞ 1=4 1 1 m! Z 2 = p dxe−(ipx=~+m!x =2~): 2π~ π~ −∞ Consider m!x2 ipx rm! 2 rm! ip ip 2 ip 2 + = x +2 x p + p − p 2~ ~ 2~ 2~ 2m!~ 2m!~ 2m!~ rm! ip 2 ip 2 = x + p − p 2~ 2m!~ 2m!~ rm! ip 2 p2 = x + p + · 2~ 2m!~ 2m!~ Due Date: December. 14, 2015, 05:00 pm 3 Quantum Mechanics I / Quantum Theory I. 14 December, 2015 Let rm! ip y = x + p 2~ 2m!~ r rm! 2 ) dy = dx or dx = ~ dy 2~ m! So the integral becomes 1 r 1 Z 2 2 Z 2 2 dxe−ipx=~em!x =2~ = ~ dye−y e−p =2m!~ −∞ m! −∞ r 1 2 2 Z 2 = ~ e−p =2m!~ dye−y m! −∞ r r 2 2 p 2π 2 = ~ e−p =2m!~ π = ~e−p =2m!~: m! m! Hence 1=4r 1 m! 2π 2 ~ −p =2m!~ f0(p) = p e 2π~ π~ m! 1=4 1 2 = e−p =2m!~; m!π~ which is the required result. 2. A particle of mass m in a one-dimensional harmonic oscillator is in a state for which a measurement of the energy yields the values ~!=2 or 3~!=2, each with a probability 1=2 of one-half. The average value of momentum px at time t = 0 is (m!~=2) . This information specifies the state of the particle completely. What is this state and what is px at time t? Answer The initial state is a superposition of two adjacent states j0 and j1 , with eigenvalues E0 = ~!=2 and E1 = 3=2~!, 1 j (0) = p j0 +eiφj1 ; 2 where eiφ is the phase factor and its magnitude can be find as p^x = (0)jp^xj (0) r m! = (0)j −i ~(^a − a^y) j (0) 2 r m! = −i ~ (0)j(^a − a^y)j (0) 2 Due Date: December. 14, 2015, 05:00 pm 4 Quantum Mechanics I / Quantum Theory I. 14 December, 2015 r m! 1 1 = −i ~ p 0j + e−iφ 1j (^a − a^y) p j0 +eiφj1 : 2 2 2 Herea ^y anda ^ are raising and lowering operators: p a^jn = njn − 1 p a^yjn = n + 1jn + 1 : Therefore, r i m! p^ = − ~ 0j + e−iφ 1j eiφa^j1 −a^yj0 x 2 2 r i m! = − ~ 0j + e−iφ 1j eiφj0 −|1 2 2 r i m! = − ~(eiφ − e−iφ) 2 2 r r m! (eiφ − e−iφ) m! = ~ = ~ sin φ. 2 2i 2 q m!~ Given that p^x at t = 0 = 2 . So we have r r m! m! ~ = ~ sin φ 2 2 ) sin φ = 1 or φ = π=2 1 ) j (0) = p j0 +ij1 ; 2 is the required state. If j (0) = p1 (j0 +ij1 ), then after time t, the state evolves to 2 j (t) = e−iHt=~j (0) : For the harmonic oscillator, the Hamiltonian is expressed as 1 1 H = !(^aya^ + ) = !(N^ + ) (as derived in class) ~ 2 ~ 2 with eigenvalues En = (n + 1=2)~! for n = 0; 1. Therefore, −i !(N^+ 1 )t= j (t) = e ~ 2 ~j (0) = e−i!t=2e−i!tN^ j (0) 1 = e−i!t=2e−i!tN^ p j0 +ij1 2 −i!t=2 e −i!t ^ = p j0 +ie j1 * Njn = njn : 2 Due Date: December. 14, 2015, 05:00 pm 5 Quantum Mechanics I / Quantum Theory I. 14 December, 2015 So the expectation value of momentum operatorp ^x at time t is p^x = (t)jp^xj (t) r ei!t=2 m! e−i!t=2 = p 0j − iei!t 1j −i ~(^a − a^y) p j0 +ie−i!tj1 2 2 2 r −i m! = ~ 0j − iei!t 1ja^ − a^yj0 +ie−i!tj1 2 2 r −i m! = ~ 0j − iei!t 1j ie−i!tj0 −|1 2 2 r −i m! = ~ie−i!t + iei!t 2 2 r r m! e−i!t + ei!t m! = ~ = ~ cos !t: 2 2 2 3. Calculate the probability that a particle in the ground state of the harmonic oscillator is located in a classically disallowed region, namely, where V (x) > E. Obtain a numerical value for the probability. Suggestion: Express your integral in terms of a dimensionless variable and compare with the tabulated values of the error function. Answer The ground state of the harmonic oscillator (or the minimum uncertainty state) 1=4 m! 2 −m!x =2~ 0(x) = e π~ is a Gaussian function as drawn in figure. Here ±xc are classical turning points (or critical points), at these points the energy of Due Date: December. 14, 2015, 05:00 pm 6 Quantum Mechanics I / Quantum Theory I. 14 December, 2015 the particle equals to its potential energy, 1 1 ! = m!2x2 2~ 2 c r ) x = ~ · c m! In one of the classically forbidden regions (x > xc), where V (x) > E, the probability of finding a particle in ground state of the harmonic oscillator is 1 r 1 Z m! Z 2 2 −m!x =~ dx 0(x) = e dx: xc π~ xc Let rm! 1 a = ; ) xc = · ~ a By these substitutions, integral becomes 1 p a Z 2 2 a π 1 1 p e−a x dx = p − Erf π 1=a π 2 a a 1 = (1 − Erf(1)) ≈ 7:9%: 2 But this is only half of the desired result since we have two classically forbidden regions, xc > a and xc < a. Hence the required probability is 16%. 4. Calculate E for the coherent state jα . Answer The energy eigenvalue corresponds to hamiltonian for harmonic oscillator is p2 1 E = + m!2x2 2m 2 1 1 E = p^2 + m!2 x^2 : (2) ) 2m 2 The expectation values ofp ^2 andx ^2 for the coherent state jα as derived in class m! p^2 = ~ 4jαj2 sin2(!t − δ) + 1 2 x^2 = ~ 4jαj2 cos2(!t − δ) + 1 : 2m! Plugging these values in Eq(2), 1 m! 1 E = ~ 4jαj2 sin2(!t − δ) + 1 + m!2 ~ 4jαj2 cos2(!t − δ) + 1 2m 2 2 2m! ! ! = ~ 4jαj2 sin2(!t − δ) + 1 + ~ 4jαj2 cos2(!t − δ) + 1 4 4 ! 1 = ~ 4jαj2 + 2 = ! jαj2 + ; 4 ~ 2 Due Date: December. 14, 2015, 05:00 pm 7 Quantum Mechanics I / Quantum Theory I.
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