Answers for Assigned Problems

2.11 (a)
n=0:
      <x> = 0
      <p> = 0
      <x²> = hbar/(2mw)
      <p²> = m hbar w / 2

n=1: [not required]
      <x> = 0
      <p> = 0
      <x²> = 3 hbar/(2mw)
      <p²> = 3 m hbar w / 2

        (c)
n=0:
      <T> = hbar w / 4
      <V> = hbar w / 4

n=1: [not required]
      <T> = 3 hbar w / 4
      <V> = 3 hbar w / 4

In each case, <T> + <V> is the energy of the state.

2.12 <x> = 0
        <p> = 0
        <x²> = (n + 1/2) hbar /(m w)
        <p²> = (n + 1/2) m hbar w
        <T> = (1/2)(n + 1/2) hbar w
        <V(x)> = (1/2)(n + 1/2) hbar w
        s_x² = (n + 1/2) hbar / (m w)
        s_p² = (n + 1/2) m hbar w

2.16 H₅ is given in the text in Table 2.1.
        H₆(x) = -120 + 720x² - 480x⁴ + 64x⁶

2.21 (a) A = Öa

        (b) f(k) = Ö[a / (2 p)] [2 a / ( k² + a²) ]

        (c) y(x,t) = [a^3/2 / p ] ¤ _-¥ ^¥ {(k² + a²)^-1 exp [ikx - i hbar k² t / (2m)]} dk

        (d) For large a the particle is poorly confined, so y(x) is broad; f(k) is sharply peaked. For small a the particle is confined to a small region so y(x) is sharply peaked; f is broad. These two relationships are as predicted by the Uncertainty Principle.

2.24 Proofs.

2.42 E_n = (n + 1/2) hbar w, n odd only

Last revised 2005/03/03