Consider a spin ½ particle whose Hamiltonian is H0 = (A/ħ2)S2
+ (B/ħ)Sz. For t < 0 the particle is in the |+> eigenstate of H0.
At t = 0 the Hamiltonian changes to H = H0 + (C/ħ)Sy.
Let A, B, C be positive constants and let C << B.
Use time-dependent perturbation theory to find the probability that the particle
is found in the |-> eigenstate of H0 states for times t > 0.
A particle of mass m is in the ground state of an infinite square well
(U = 0 for 0 < x < a and U = ∞ otherwise).
At time t = 0,
the right "wall" (i.e. at x = a) shifts to x = 3a. A measurement of
the energy of the particle is made just after the wall shifts. (Assume that all
this happens so quickly so the spatial wave function of the particle does not
change). What is the probability that the energy measurement yields a value
EXACTLY the same as the energy of the ground state of original well?
In one dimension, consider a spinless particle trapped in a delta-function potential U(x) = -Cδ(x), C > 0.
At t = 0, a time dependent perturbation W(t) = Wcosωt is turned on.
Assume ω >> mC2/(2ħ3) so that the particle can be ejected
from the trap. Use perturbation theory to find the transition rate.
You can assume that the free particles will be in a box of size L, L =
very large.
A hydrogen atom in its ground state [(n, l ,m) = (1, 0, 0)] is placed between
the plates of a capacitor. A time dependent but spatially uniform electric field
(not potential!) is applied as follows:
E = 0 for t < 0,
E = E0exp(-t/τ) for t > 0.
E0 =
E0 k.
Using first-order time-dependent perturbation theory compute the probability for
the atom to be found at t >> τ in each of the three p-states n = 2, l
= 1, m = ±1 or 0).
Useful information: <210|z|100> = (128*√2/243)a0 and E2 - E1 = 3e2/8a0.
Hydrogen atom wave functions:
R10(r) = 2 a0-3/2 exp(-r/a0),
R20(r) = (2a0)-3/2 (2 - r/a0)
exp(-r/(2a0)),
R21(r) = 3-½(2a0)-3/2(r/a0)
exp(r/(2a0)),
Y00 = (4π)-½, Y1±1 = ∓(3/(8π))½sinθ
exp(±iφ), Y10 = (3/(4π))½cosθ.