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Chapter 9: Q19P (page 365)
We have encountered stimulated emission, (stimulated) absorption, and spontaneous emission. How come there is no such thing as spontaneous absorption?
Spontaneous absorption would involve taking energy (a photon) from the ground state of the electromagnetic field
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Suppose you don’t assume
(a) Find and in first-order perturbation theory, for the case
.show that , to first order in .
(b) There is a nicer way to handle this problem. Let
.
Show that
where
So the equations for are identical in structure to Equation 11.17 (with an extra
(c) Use the method in part (b) to obtain in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.
As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody radiation is the source). Show that at room temperature (T = 300 K) thermal stimulation dominates for frequencies well below , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?
Solve Equation 9.13 for the case of a time-independent perturbation, assumingthatandcheck that
. Comment: Ostensibly, this system oscillates between “” Doesn’t this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this are not, and never were, Eigen states of the Hamiltonian—a measurement of the energy never yields. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end,are Eigen states of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T —this doesn’t affect the calculations, but it allows for a more sensible interpretation of the result.
…(9.13).
A hydrogen atom is placed in a (time-dependent) electric between the ground state (n = 1 ) the (quadruply degenerate) first excited states (n = 2 ) . Also show for all five states. Note: There is only one integral to be done here, if you exploit oddness with respect to z; only one of the n = 2 states is “accessible” from the ground state by a perturbation of this form, and therefore the system functions as a two-state configuration—assuming transitions to higher excited states can be ignored.
A particle starts out (at time t=0 ) in the Nth state of the infinite square well. Now the “floor” of the well rises temporarily (maybe water leaks in, and then drains out again), so that the potential inside is uniform but time dependent:.
(a) Solve for the exact , using Equation 11.116, and show that the wave function changes phase, but no transitions occur. Find the phase change, role="math" localid="1658378247097" , in terms of the function
(b) Analyze the same problem in first-order perturbation theory, and compare your answers. Compare your answers.
Comment: The same result holds whenever the perturbation simply adds a constant (constant in x, that is, not in to the potential; it has nothing to do with the infinite square well, as such. Compare Problem 1.8.
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