Problem 2.3 Show that there is no acceptable solution to the (time-independent) Schrödinger equation for the infinite square well with E = 0 or E < 0. (This is a special case of the general theorem in Problem 2.2, but this time do it by explicitly solving the Schrödinger equation, and showing that you cannot satisfy the boundary conditions.)
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- (a_)*(a_V)ax- J-o Problem 2.11 Show that the lowering operator cannot generate a state of infinite norm (i.e., f la.-²dx < oo, if y itself is a normalized solution to the Schrödinger equation). What does this tell you in the case y = vo? Hint: Use integration by parts to show that y*(a,a_) dx. = -00 Then invoke the Schrödinger equation (Equation 2.46) to obtain la-yl² dx E - hw, -0- where E is the energy of the state y. **Problem 2.12 (a) The raising and lowering operators generate new solutions to the Schrödinger equation, but these new solutions are not correctly normalized. Thus a Vn is proportional to yn+1, and a n is proportional to yn-1, but we'd like to know the precise proportionality constants. Use integration by parts and the Schrödinger equation (Equations 2.43 and 2.46) to show that roo | la+ Vl² dx = (n+ 1)hw, la- Vnl? dx = nhw, -00 -00 and hence (with i's to keep the wavefunctions real) a+ Vn = iv(n + 1)hw yn+1, [2.52] a_n = -ivnhw n-1. [2.53] Sec. 2.3: The Harmonic…Problem 2.21 Suppose a free particle, which is initially localized in the range -a < x < a, is released at time t = 0: А, if -a < х <а, otherwise, (x, 0) = where A and a are positive real constants. 50 Chap. 2 The Time-Independent Schrödinger Equation (a) Determine A, by normalizing V. (b) Determine (k) (Equation 2.86). (c) Comment on the behavior of (k) for very small and very large values of a. How does this relate to the uncertainty principle? *Problem 2.22 A free particle has the initial wave function (x, 0) = Ae ax where A and a are constants (a is real and positive). (a) Normalize (x, 0). (b) Find V(x, t). Hint: Integrals of the form e-(ax?+bx) dx can be handled by "completing the square." Let y = Ja[x+(b/2a)], and note that (ax? + bx) = y? – (b²/4a). Answer: 1/4 e-ax?/[1+(2ihat/m)] 2a Y (x, t) = VI+ (2iħat/m) (c) Find |4(x, t)2. Express your answer in terms of the quantity w Va/[1+ (2hat/m)²]. Sketch |V|? (as a function of x) at t = 0, and again for some very large t.…Find the corresponding Schrödinger equation and wave function, The energy for the infinite-walled well problem of size L, The expected value of x (<x>) on the interval [0,a/4], The expected value of p (<p>) for the same interval and the probability of finding at least one particle in the same interval. Do not forget the normalization, nor the conditions at the border.
- In this question we will consider a finite potential well in which V = −V0 in the interval −L/2 ≤ x ≤ L/2, and V = 0 everywhere else (where V0 is a positive real number). For a particle with in the range −V0 < E < 0, write and solve the time-independent Schrodinger equation in the classically allowed and classically forbidden regions. Remember to keep the wavenumbers and exponential factors in your solutions real!2.4. A particle moves in an infinite cubic potential well described by: V (x1, x2) = {00 12= if 0 ≤ x1, x2 a otherwise 1/2(+1) (a) Write down the exact energy and wave-function of the ground state. (2) (b) Write down the exact energy and wavefunction of the first excited states and specify their degeneracies. Now add the following perturbation to the infinite cubic well: H' = 18(x₁-x2) (c) Calculate the ground state energy to the first order correction. (5) (d) Calculate the energy of the first order correction to the first excited degenerated state. (3) (e) Calculate the energy of the first order correction to the second non-degenerate excited state. (3) (f) Use degenerate perturbation theory to determine the first-order correction to the two initially degenerate eigenvalues (energies). (3)Consider a classical particle moving in a one-dimensional potential well u(x), as shown in Figure 6.10 (attached). The particle is in thermal equilibrium with a reservoir at temperature so the probabilities of its various states are determined by Boltzmann statistics.