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CYK\2010\PH405+PH213\Tutorial 5 Quantum Mechanics 1. Find Allowed Energies of the Half Harmonic Oscillator 2. a Charged Particle

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CYK\2010\PH405+PH213\Tutorial 5 Quantum Mechanics 1. Find Allowed Energies of the Half Harmonic Oscillator 2. a Charged Particle CYK\2010\PH405+PH213\Tutorial 5 Quantum Mechanics 1. Find allowed energies of the half harmonic oscillator ( 1 m!2x2; x > 0; V (x) = 2 1; x < 0: 2. A charged particle (mass m, charge q) is moving in a simple harmonic potential (frequency !=2π). In addition, an external electric eld E0 is also present. Write down the hamiltonian of this particle. Find the energy eigenvalues, eigenfunctions. Find the average position of the particle, when it is in one of the stationary states. 3. Assume that the atoms in a CO molecule are held together by a spring. The spacing between the lines of the spectrum of CO molecule is 2170 cm−1. Estimate the spring constant. 4. If the hermite polynomials Hn(x) are dened using the generating function G(x; s) = exp −s2 + 2xs, that is X Hn(x) exp −s2 + 2xs = sn; n! n (a) Show that the Hermite polynomials obey the dierential equation 00 0 Hn(x) − 2xHn(x) + 2nHn(x) = 0 and the recurrence relation Hn+1(x) = 2xHn(x) − 2nHn−1(x): (b) Derive Rodrigues' formula n 2 d 2 H (x) = (−1)n ex e−x : n dxn 5. Let φn be the nth stationary state of a particle in harmonic oscillator potential. Given that the lowering operator is 1 a^ = p m!X^ + iP^ : 2~m! p and ξ = m!=~x, (a) show that 1 d a^ = p ξ + : 2 dξ (b) Show that p aφ^ n(ξ) = nφn−1(ξ) 6. Let B = fφn j n = 0; 1;:::g be the set of energy eigenfunctions of the harmonic oscillator. Find the matrix elements of X^ and P^ wrt to basis B th 7. Suppose that a harmonic oscillator is in its n stationary state. 1 (a) Compute uncertainties σx and σP in position and momentum. [Hint: To calculate expectation values, rst write X^ and P^ in terms of the lowering operator a^ and its adjoint.] (b) Show that the average kinetic energy is equal to the average potential energy (Virial Theorem). 8. A particle of mass m in the harmonic oscillator potential, starts out at t = 0, in the state 2 Ψ(x; 0) = A (1 − 2ξ)2 e−ξ p where A is a constant and ξ = m!=~x: (a) What is the average value of energy? (b) After time T , the wave function is 2 Ψ(x; T ) = B (1 + 2ξ)2 e−ξ for some constant B. What is the smallest value of T ? 9. Let φn be eigenstates of the harmonic oscillator. For a given complex number µ, let 1 n − jµj X µ χµ = e 2 p φn: n=0 n! Such states are called coherent states. (a) Show that aχ^ µ = µχµ that is χµ is an eigenstate of a^. (b) If the state of the oscillator is χµ, then show that σxσp = ~=2. (c) The state of the oscillator Ψ(t = 0) = χµ; then show that Ψ(t) = χµ0 where µ0 = e−i!tµ. That means, if the state of the system, at an instant is a coherent state, then it is a coherent state at all times. 2 (d) Optional: If you choose the hilbert space to be L2(R); then show that jΨ(x; t)j is a gaussian wave packet and the wave packet performs a harmonic oscillations without changing the shape. Solutions: 1. Since V (x) = 1 for x ≤ 0; (x) = 0 for x ≤ 0. The Schrodinger time-independent equation is then 2 1 − ~ 00 + m!2x2 = E x ≥ 0 2m 2 (0) = 0 and must be square integrable. This problem is same as usual harmonic oscillator except that we must choose only those eigenfunction which satisfy the bc of the half harmonic 2 oscillator, that is (0) = 0. If φn(x) = Hn(ξ) exp(−ξ =2), then we know that φn satises the above de and bc if n is odd. Thus, the energy eigenvalues of the half harmonic oscillator are 1 E = n + ! n = 1; 3; 5;::: n 2 ~ 2 2. The potential energy can be written as 1 V (x) = m!2x2 − qE x 2 0 1 qE 2 q2E2 = m!2 x − 0 − 0 2 m!2 m!2 Let 2 and 2 2 2. Let and Then the x0 = qE0=m! H0 = −q E0 =m! x − x0 = z H1 = H − H0: hamiltonian P 2 1 H = + m!2z2: 1 2m 2 The eigenvalues of are 1 , then the eigenvalues of are . Note H1 En = n + 2 ~! H En + H0 H0 is just a number. −1 −4 2 3. Then ~! = 2170 cm = 2170 × 1:24 × 10 eV. Calculate force constant K = m! . 4. See Arfken. 5. Prove this by using the recurrence relations given in problem 4. 6. Note r X^ = ~ a + ay 2m! r m! P^ = 1 ~ a − ay i 2 The matrix elements are D E X^mn = φm; Xφ^ n r D E = ~ φ ; a + ay φ 2m! m n r p p = ~ φ ; nφ + n + 1φ 2m! m n−1 n+1 r p p = = ~ nδ + n + 1δ 2m! m;n−1 m;n+1 Simillarly D E P^m;n = φm; P^ φn r m! p p = ~(−i) nδ − n + 1δ 2 m;n−1 m;n+1 7. Note r X^ = ~ a + ay 2m! r m! P^ = 1 ~ a − ay : i 2 Then D E X^ = X^n;n = 0 and D E P^ = P^n;n = 0: 3 Now D E D E X^ 2 = ~ φ ; a + ay a + ay φ 2m! n n D E = ~ φ ; a2 + ay2 + aay + aya φ 2m! n n = ~ (0 + 0 + (n + 1) + n) = ~ (2n + 1): 2m! 2m! Simillarly D E m! P^2 = ~ (2n + 1) 2 (a) Thus, 1 σ σ = n + X P 2 ~ (b) Note: 1 D E ! 1 hKi = P^2 = ~ (n + ) 2m 2 2 and 1 D E ! 1 hV i = m!2 X^ 2 = ~ (n + ) 2 2 2 8. Now, 1 p p (x; 0) = 3φ − 2 2φ + 2 2φ 5 0 1 2 where φn is the nth eigenfunction of energy operator. (a) The average energy 1 hEi = (9E + 8E + 8E ) 25 0 1 2 1 24 = + !: 2 25 ~ (b) After time T 2 (x; T ) = B 1 + 2ξ2 e−ξ 1 p p 1 p p 3φ e−i!T=2 − 2 2φ e−i!3T=2 + 2 2φ e−i!5T=2 = 3φ + 2 2φ + 2 2φ 5 0 1 2 5 0 1 2 Must nd T such that e−i!T=2 = e−i5!T=2 = 1 and e−i3!T=2 = −1. when, !T = π; exp (−i!T=2) = −i exp (−i3!T=2) = i exp (−i!T=2) = −i This will do. 9. Given: 2 1 n − jµj X µ χµ = e 2 p φn n=0 n! 4 (a) Now 2 1 n − jµj X µ aχ^ µ = e 2 p aφ^ n n=0 n! 2 1 n − jµj X µ = e 2 p nφn−1 n=1 n! 2 1 n−1 − jµj X µ = µe 2 p φn−1 n=1 (n − 1)! = µχµ y Interestingly, χµ is not an eigenstate of a^ : (b) First note: 1 n m −|µj2 X (¯µ) µ hχµ; χµi = e p p hφn; φmi n=0 n! m! 1 n m −|µj2 X (¯µ) µ = e p p δm;n n=0 n! m! 1 2n 2 X jµj = e−|µj = 1 n! n=0 Then, hχµ; aχ^ µi = µ D y E χµ; a^ χµ = haχ^ µ; χµi =µ. ¯ q Now, ^ ~ y, X = 2m! a^ +a ^ D E r D E χ ; Xχ^ = ~ χ ; a^ +a ^y χ µ µ 2m! µ µ r r 2 = ~ (µ +µ ¯) = ~ (Reµ) 2m! m! 2 2 And, ^2 ~ y ~ 2 y y , so X = 2m! a^ +a ^ = 2m! a^ + a^ + 2^a a^ + 1 D E X^ 2 = ~ µ2 +µ ¯2 + 2jµj2 + 1 2m! = ~ (2Reµ)2 + 1 2m! Finally, 2 σ2 = ~ (2Reµ)2 + 1 − ~ (Reµ)2 x 2m! m! = ~ 2m! 2 Now, simillary, σ2 = ~ m! and p 2 ~ σ σ = ~: x p 2 Here the product of uncertainties is as minimum as it can get! 5 (c) If Ψ(0) = χµ, then 2 1 n − jµj X µ −i!(n+ 1 )t Ψ(t) = e 2 p e 2 φn n=0 n! 2 1 −i!tn −i!t=2 − jµj X µe = e e 2 p φn n=0 n! −i!t=2 = e χµ0 where µ0 = µe−i!t. (d) Now if we write χµ(x) in space represenetation, we need to substitute 1=2 α 1 −ξ2=2 φn(x) = p p e Hn(ξ) π 2nn! 1=2 n n α (−1) 2 d 2 = = p p eξ =2 e−ξ π 2nn! dxn using Rodrigue's formular for Hermite polynomials. n 1=2 2 1 n −i!t n α −i!t=2 − jµj ξ2=2 X (−1) µe d −ξ2 Ψ(x; t) = p e e 2 e p p e π n dxn n=0 2 n! n! 1=2 2 2 α −i!t=2 − jµj −ξ −(ξ−η)2 = p e e 2 e π p Use Taylor expansion for the last step and η = jµj exp(−i(!t − x))= 2. Now α 2 2 2 2 jΨ(x; t)j2 = p e−|µj +ξ e−(ξ−η) e−(ξ−η¯) π α p Arg 2 = p e−(ξ− 2jµj cos(!t− µ)) π This is a gaussian wave packet performing simple harmonic motion with frequency ! p and amplitude A = 2~=m!jµj .
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