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If you do any problems in sections 2-1 and 2-2 and if you have issue with them, please let us know. If it is judged general enough, we will address them in class. Otherwise, someone will help you by answering your questions.

Otherwise, we will be spending most of Friday on SHO.

On pages 52-53 of Griffiths: I understand why we drop the B term from Equation 2.75, but then I can't figure out why we need to throw on <math>h(\xi)</math> in Equation 2.76, and then proceed with all of the math on the following page.

From what I understand about the material, in Footnote 23 on page 52 it says that stripping off the asymptotic behavior is merely to provide an alternative method for solving the differential equations of the wave. Of course this is based on estimations that <math>h(\xi)</math> will provide a simpler wave function to work with since we want to be able to analyze the wave at all values of <math>\xi</math>.

On today's lecture, I noticed when Yuichi was talking about stationary state how Hamiltonian is time independent, he wrote <math> <\hat{H}>=\int\Psi^\ast E_n \Psi dx=E_n</math> here <math>E_n</math> is an eigenvalue instead of an operator <math> ih\frac{\partial }{\partial t} </math> . Does that mean when the subscript shows up, it is always an eigenvalue?

<math>E_n</math> is associated with the eigenfunction (or eigenstate if you will) of the Hamiltonian, and since energy is quantized there can only be n eigenvalues corresponding to specific eigenstates (note that the eigenfunctions have subscripts as well). I'm sure there's a more lucid, mathematical way to express this but I tried.

I still have a question about stationary states and how to use and apply Schrodinger's equation: Since the solution to Schro's equation is a linear combination of stationary states, doesn't that mean that a particle that is described by the equation can have any amount of energy? How does this connect to the fact that the energy we measure from, say, an electron in hydrogen, is quantized?

I am not sure whether I am correct, but according to my understanding, once you make a measurement to a particle, you can get only the eigenvalues of energy. Although the state is a combination of several stationary states, the result of measurement is among the different energy eigenvalues E1,E2,E3,… The probability of each result is related with the coefficient before the steady states. For example, if the state of particle is combined with two steady states, <math>\Psi=c_1\Psi_1+c_2\Psi_2</math> their eigenvalue of energy E1 and E2, the measurement can never get a result between E1 and E2, the result may be E1, or may be E2, and the probability of E1 result is <math>\|c_1|^2</math> and probability of E2 result is <math>\|c_2|^2</math> So the energy is quantized.

Okay, thanks, that makes some sense, in that it explains why the energy level you measure will always be quantized, but isn't it true that particles will only absorb certain amounts of energy based on what energy levels are available for them to jump up to? Your answer seems to imply that my hydrogen electron could absorb any amount of energy, not just specific amounts.

Yuichi

Daniel Faraday, you ask a tough question. I think you are saying that if an electron is in the “mixed state of the ground and 1st excited state, it should be able to absort photon of some energy appropriate for the “average energy” of this state - since no one is measuring the electron energy before the photon is absorbed, it does not have to decide if it is in the E1 or E2 state, you are thinking, right? - to go up to the 2nd excited state with E3 energy. I think what's missing at this point in our study is that the hamiltonian is incomplete because it does not contain the term corresponding to the interaction between the electron and the field of photons, or electromagnetic field. This part is not properly treated in our book until section 9.2. So at this point, we have to wiggle our way out with hand-waving argument. When you consider this interaction, the mix state is not very stable, and in a rather short time, emits a photon and will be found in the ground state.

I suppose that you can then ask why would the energy of the photon emitted in this situation be the difference between E1 and the average energy of the mixed state. Then I have to think about how the electron can be put in this mixed state in the first place. I would say that you have to start from some “stable state” and do something to this electron to take this mixed state. For example, if you start with this electron in the ground state. Then shine light of energy = E2-E1 to try to excite it to the first excited state, but you are not sure if this happened. This state of the electron may be described by the mixed state that we have been talking about. Then really, this state is the mixed state of (1) an electron with E=E1 with a photon and (2) an electron with E=E2 with no additional photon. If we wait long enough, this excited electron (component) emit a photon (of energy E2-E1) and decays to the ground state. If at that point, if you see a photon, you will conclude that the electron decided to have been in the excited state, but by emitting the photon, it made a transition to the ground state. “Observation of the photon” consists of “measuring the energy of the electron” which was E2 before the measurement, but after the measurement, it has already changed to E1. Meanwhile, between the two states mixed together (ground state electron+photon vs. excited electron), by making a measurement, you discovered that by seeing a photon, the system was in the latter, or by no observing a photon, you discovered that the system was in the former state. Well, getting very confusing, but I hope I have illustrated an interesting aspect of QM which reveal itself when you think about things deeply. And I am not sure I am “right” in my illustration 100%.