Solutions to Exam 1 - Quantum Mechanics I | PHYS 3810, Exams of Quantum Mechanics

Download Solutions to Exam 1 - Quantum Mechanics I | PHYS 3810 and more Exams Quantum Mechanics in PDF only on Docsity! Phys 3810, Spring 2009 Exam #1 1. What is the physical meaning of the expectation value of a particular physical quantity Q? The expectation value ofQ (for a particular quantum state Ψ) is the result one gets if one makes many repeated measurements of Q (at the same time t) and then takes the average. 2. Show that the operator for p is Hermitian. Explain all the steps! When the p̂ operator sits in an integral between two wave functions, using integration by parts we can do the steps: ∫ ∞ −∞ φ∗p̂ψ dx = ∫ ∞ −∞ φ∗ h̄ i ∂ ∂x ψ dx = h̄ i ∫ ∞ −∞ ( −∂φ ∗ ∂x ) ψ dx But now the factors in front can be brought inside = ∫ ∞ −∞ ( h̄ (−i) )( ∂φ∗ ∂x ) ψ dx = ∫ ∞ −∞ ( h̄ i ∂φ ∂x )∗ ψ dx But the last is the inner product of (p̂φ) with ψ, whereas we started with the inner product of φ with (p̂ψ). The fact that we can switch the position of the p̂ operator in this way shows that p̂ is Hermitian. 3. What is a stationary state? A stationary state is a state Ψ(x, t) which can be expressed as a separated function of of x and t. With this condition, the Schrdinger equation requires the wave function to be of the form Ψ(x, t) = ψ(x)e−iEt/h̄ Such a state has the nice property that expectation values don't depend on time. A general solution to the Schrdinger equation can be built up from all of the stationary state solutions. 4. A particle in the infinite square well has the initial wave function Ψ(x, 0) = A sin(πx/a)(1 + cos(πx/a)) (0 ≤ x ≤ a) Determine A, find Ψ(x, t). What is the expectation value of the energy? Hint: The decom- position into the stationary states can be done by inspection! As the stationary states are given by ψn(x) = √ 2 sin(nπx/a) 1 the given state can be decomposed easily: Ψ(x, 0) = A(sin(πx/a) + sin(πx/a) cos(πx/a)) = A(sin(πx/a) + 1 2 sin(2πx/a)) = A √ a 2 ( √ 2 a sin(πx/a) + 1 2 √ 2 a sin(2πx/a) ) = A √ a 2 (ψ1(x) + 1 2 ψ2(x)) and in this form we can use the orthonormality of the ψn(x)'s. Normalizing Ψ(x, 0) gives ∫ ∞ −∞ |Ψ(x, 0)|2 dx = A2a 2 (1 + 1 4 ) = 5A2a 8 = 1 This gives A2 = 8 5a =⇒ A = 2 √ 2 5a Then the decomposition is Ψ(x, 0) = 2√ 5 (ψ1(x) + 1 2 ψ2(x)) With E1 = π2h̄2 2ma2 E2 = 4π2h̄2 2ma2 then putting in the time dependence gives Ψ(t) = 2√ 5 (ψ1(x)e −iE1t/h̄ + 1 2 ψ2(x)e −iE2t/h̄) The expectation of value of the energy at t = 0 is (using orthonormality) 〈E〉 = ∫ Ψ(x, 0)ĤΨ(x, 0) dx = 4 5 (E1 + 1 4 E2) = 4 5 E1 + 1 5 E2 which is, pulling out the common factor, π2h̄2 2ma2 (1 · 4 5 + 4 · 1 5 ) = 8 5 π2h̄2 2ma2 = 8π2h̄2 10ma2 This is same as the 〈E〉 at all other times, from the orthogonality of the stationary states. 5. An electron is confined to a one–dimensional box of length 2.0 × 10−10 m. Find the difference in energies between ground state and first excited state. As the energy levels of the confined particle are given by En = n2π2h̄2 2ma2 , the energy of the ground state (above the bottom of the well) is E1 = π2(1.055 × 10−34 J · s)2 2(9.11 × 10−31 kg)(2.0 × 10−10 m)2 = 1.51 × 10 −18 J 2 At least I think I did that right. Anyway, it could be worked out. c) Construct Ψ(x, t) in the form of an integral (which you don’t need to evaluate). Having φ(k), the full time--dependent Ψ(x, t) is constructed via Ψ(x, t) = 1√ 2π ∫ ∞ −∞ φ(k)ei(kx− h̄k 2 2m t) dk using the φ(k) that we got in (b); but since this integral will be very hard to work we'll leave it at that. 10. The finite square well was a potential given by: V (x) = { −V0 for − a < x < a 0 for |x| > a What boundary conditions were imposed upon ψ for a proper solution? At the boundaries of the well (x = ±a) we required that the wavefunction ψ(x) be continuous and that its derivative, dψ dx be continuous. (We require the derivative condition because the potential only makes a finite jump at the boundary.) 11. What was the physical meaning of the transmission coefficient T derived for the finite square well potential? T represents the probability that an incident particle with nearly definite energy E will keep moving freely past the well, rather than being reflected backwards in the direction from whence it came. 5 Useful Equations Math ∫ ∞ 0 xne−x/a = n! an+1 ∫ ∞ 0 x2ne−x 2/a2 dx = √ π (2n)! n! (a 2 )2n+1 ∫ ∞ 0 x2n+1e−x 2/a2 dx = n! 2 a2n+2 ∫ b a f dg dx dx = − ∫ b a df dx g dx+ fg ∣ ∣ ∣ ∣ ∣ b a Numbers h̄ = 1.05457 × 10−34 J · s me = 9.10938 × 10−31 kg mp = 1.67262 × 10−27 kg e = 1.60218 × 10−19 C c = 2.99792 × 108 m s Physics ih̄ ∂Ψ ∂t = − h̄ 2 2m ∂2Ψ ∂x2 + VΨ Pab = ∫ b a |Ψ(x, t)|2 dx p→ h̄ i d dx ∫ ∞ −∞ |Ψ(x, t)|2 dx = 1 〈x〉 = ∫ ∞ −∞ x|Ψ(x, t)|2 dx 〈p〉 = ∫ ∞ −∞ Ψ∗ ( h̄ i ∂ ∂x ) Ψ dx σ = √ 〈j2〉 − 〈j〉2 σxσp ≥ h̄ 2 − h̄ 2 2m d2ψ dx2 + V ψ = Eψ φ(t) = e−iEt/h̄ Ψ(x, t) = ∞ ∑ n=1 cnψn(x)e −iEnt/h̄ = ∞ ∑ n=1 Ψn(x, t) ∞ Square Well : En = n2π2h̄2 2ma2 ψn(x) = √ 2 a sin (nπ a x ) ∫ ψm(x) ∗ψn(x) dx = δmn cn = ∫ ψn(x) ∗ f(x) dx ∞ ∑ n=1 |cn|2 = 1 〈H〉 = ∞ ∑ n=1 |cn|2En Harmonic Oscillator : V (x) = 1 2 mω2x2 1 2m [p2 + (mωx)2]ψ = Eψ a± ≡ 1√ 2h̄mω (∓ip+mωx) [A,B] = AB −BA [x, p] = ih̄ H(a+ψ) = (E + h̄ω)(a+ψ) H(a−ψ) = (E − h̄ω)(a+ψ) a−ψ0 = 0 ψ0(x) = (mω πh̄ )1/4 e− mω 2h̄ x2 ψ1(x) = (mω πh̄ )1/4 √ 2mω h̄ xe− mω 2h̄ x2 Free particle : Ψk(x) = Ae i(kx− h̄k 2 2m )t vphase = ω k vgroup = dω dk 6 Ψ(x, t) = 1√ 2π ∫ ∞ −∞ φ(k)ei(kx− h̄k 2 2m t) dk φ(k) = 1√ 2π ∫ ∞ −∞ Ψ(x, 0)e−ikx dx Delta Fn Potl : ψ(x) = √ mα h̄ e−mα|x|/h̄ 2 E = −mα 2 2h̄2 7