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| classes:2009:fall:phys4101.001:q_a_0918 [2009/09/17 22:20] – kuehler | classes:2009:fall:phys4101.001:q_a_0918 [2009/09/26 23:43] (current) – yk |
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| //**Yuichi**// | //**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>. | 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>. |
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| ==== Daniel Faraday 9/17 2pm==== | ==== Daniel Faraday 9/17 2pm==== |
| 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 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? |
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| 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%. | 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%. |
| | ====spillane 7:00 9-18==== |
| | 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. |
| ====Pluto 4ever 10:11pm 9/17/09==== | ====Pluto 4ever 10:11pm 9/17/09==== |
| 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)\ħ\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. | 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. |
| | ====spillane 8ish 9/18==== |
| | Im also fuzzy on this. fig. 2.6???? |
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| | === Mercury 8:11 09/18/2009 === |
| | I'm also having problems understanding this concept and the plots in fig. 2.6. |
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| | === Zeno 9:10 9/18 === |
| | 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? |
| | ===spillane 9-18==== |
| | 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? |
| | ===John Galt 10:27 9/18=== |
| | 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. |
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| | ==== time to move on ==== |
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| | It's time to move on to the next Q_A: [[Q_A_0921]] |