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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>.

Yuichi When one tries to find a well-behaving asymptotic solution at infinity, the implicit hope is the the remaining function, which in our case, <math>h(\xi)</math> can be more easily expressed by a polynomial. The additional twist is the if the polynomial turns out to be an infinite series (meaning that K does not satisfy the condition given by the equation just above 2.83, the asymptotic behavior will be so bad that it override the well behaving exponential part of the solution we had found in the previous step. To avoid this disaster, we have to impose the condition on K, which is mentioned above. This still fit in the general senario that solving the Schrodinger equation is not give a complete answer, but rather we always have to think about the normalizability of the solution (in this case the behavior at the <math>\pm</math> infinities), which determines the eigenvalue of the original equation, K.

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.

Yuichi When Can says <math> <\hat{H}>=\int\Psi^\ast E_n \Psi dx=E_n</math>, I would write <math> <\hat{H}>_{\Psi_n}=\int\Psi_n^\ast E_n \Psi_n dx=E_n</math> to be sure we have a common ground, i.e. we are not talking about the expectation value of the Hamiltonian for an arbitrary wave function. Rather, hopefully it's clear that we are calculating the expectation value of the Hamiltonian for a stationary state, <math>\Psi_n</math>, which is equal to <math>E_n</math>. Then, <math>E_n</math> is clearly(?) associated with the stationary state, <math>\Psi_n</math>.

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%.

In response to Mr. Faradays first Q, The time independent schr. eq. is a linear eq but,in contrast to the schr eq., it contains explicitly the total energy E. Therefore, an arbitrary linear combination of diff. solutions will satisfy the equation only if they all correspond to the same value of E. Definite total energy follows from the fact that separation of variables is the tool used to get schr. eq. It is one of the answers to “Whats so great about separable solutions”pg 26 Griffiths.

On pg. 54 it talks about how the Schrodinger equation for the harmonic oscillator can give solutions for any value of E but only the values for E defined by <math>E_n=(n+1/2)\hbar\omega</math> can produce normalizable solutions. Meaning any value in between them will blow up. Maybe I'm reading this wrong but I just want to know how that works.

Im also fuzzy on this. fig. 2.6????

I'm also having problems understanding this concept and the plots in fig. 2.6.

This is a very interesting concept. I also don't understand it entirely either. In reference to the The Schrodinger Equation for the Harmonic Oscillator [2.70], I understand that the equation is “unstable” for many solutions. The general power series solution developed is essentially a sum of weighted decaying exponentials, and a finite series is needed to integrate a finite probability density (which is later normalized); and I see proof in figure [2.6] that eqn [2.83] must hold for all normalizable (physically real) solutions, so my corollary question is: do non-normalizable solutions (those that blow up to +/- <math>\infty</math>) to the Schrodinger Equation have any physical significance? Or are they born out of the thoroughness of mathematics and completely thrown away?

Correct me if im wrong but, isnt a fundemental constraint on Hookes law: that the pertubations of x most relatively small. That being said how is it in the analytical method between 2.73 and 2.74 we now change this condition by allowing x to be very large in order to simplify the expression. Isnt this a contradiction? How is this valid? Is it related to the graph 2.6 and the fact that eq. 2.70 has linearly independent solutions for any value of E but, almost all of this solutions blow up exponentially at large x? WHATS GOING ON?

I am also not completely positive, but isn't the reason you need to have a E=.5h(bar)w value due to the fact that the power series in EQ. 2.79 blows up at other values? If so, it is just how the math works. I am not sure how to describe it in a qualitative or visual sense, I guess. I'm guessing that the math tools were chosen to follow the experimentally observed events, so a proper function had to be determined which would only allow probability functions to exist in places where the particle could actually be, so 2.6 is probably just showing that, yes, as a visual confirmation, En = (n+1/2)h(bar)w.

It's time to move on to the next Q_A: Q_A_0921