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classes:2009:fall:phys4101.001:q_a_1023 [2009/10/23 12:26] – x500_razi0001classes:2009:fall:phys4101.001:q_a_1023 [2009/11/12 13:49] (current) – x500_chap0326
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 ===Devlin 10/23 830a=== ===Devlin 10/23 830a===
 As far as I can tell, determinate states are just eigenfunctions of certain operators.  For example, the stationary states we've been studying are determinate states of the Hamiltonian.  For the stationary states, every measurement of the particle in a stationary state gives a corresponding energy.  It really is just eigenfunctions and eigenvalues. As far as I can tell, determinate states are just eigenfunctions of certain operators.  For example, the stationary states we've been studying are determinate states of the Hamiltonian.  For the stationary states, every measurement of the particle in a stationary state gives a corresponding energy.  It really is just eigenfunctions and eigenvalues.
 +
 +===chap0326 11/12===
 +I think indeterminacy is the phenomenon in QM when you have a bunch of identical systems all in the same state and you don't get the same result each time you measure the observable. So a determinate state is the idea that you could prepare some state where you get the same value for every measurement of the system (Griffith calls it 'Q'). Its in a state that is 'determinate'.
  
 ====Daniel Faraday 10/23 12:30 pm==== ====Daniel Faraday 10/23 12:30 pm====
 What did y'all think of the test? I thought there was less time pressure, which meant I could take the time I needed to do the best solutions I could. What did y'all think of the test? I thought there was less time pressure, which meant I could take the time I needed to do the best solutions I could.
  
---------------------------+===Dark Helmet 10/25 10:53 pm=== 
 +I definitely thought i had an appropriate amount of time-that didn't change the fact that since i didn't now how to solve prob 3, i didn't get to solve problem 3-but i still thought the problems where of good length and difficulty.   
 + 
 +=== Zeno 10/26 9:45AM === 
 +I thought it was pretty perfect. Great length, great difficulty, and representative of what we were expected to know. I got tripped up toward the end of the third problem, which is my own fault for not fully understanding the strategy of solving the delta function potential outlined in detail in the book, but the other two were completely manageable (as the third one should have been). I hope the rest of the exams and final are very similar. 
 + 
 +===Captain America 10/27 10:08=== 
 +I thought it was good as well.  I too got tripped up on the last problem, and a bit on the first problem (at least I got an answer that should be right, I just don't know how well I explained myself).  I liked the length and it would have been very doable had I read through the derivation of the energy for the Kronecker Delta function a bit more.  Again I got some work down and I think the right energy, but it may be too scattered to show how I came up with it. 
 + 
 +===Esquire 10/26 10:33am=== 
 +The test was fairly straightforward. After the first test, I made a more concentrated effort at studying for this one and it seems to have paid off.  
 + 
 +===Blackbox 10/26 5:59 pm=== 
 +I'm not quite sure I've got all correct answers for the test but it was reasonable length and also appropriate amount of time.  
 + 
 +====Schrodinger's Dog 10/23 1:09 pm==== 
 +When developing the differentiability boundary for the delta potential, why does the integral -epsilon to epsilon when integrating E(Psi) go to 0? 
 + 
 +===Daniel Faraday 10/23 2pm=== 
 +Because the integral of any function at just one point is zero; it's the area of a line. Since you're taking the limit of the integral as epsilon goes to zero, the integral of psi goes to zero. 
 + 
 +Unless it's the magic delta function, of course. Which is part of what makes the delta function so useful, and also what makes it feel like math legerdemain*. 
 + 
 +* sorry, too many GRE flash cards. I meant math //magic//. 
 + 
 +====Blackbox 10/23 2:40pm==== 
 +Griffiths says "degenerate" is two or more linearly independent eigenfunctions share the same eigenvalue. Could you explain more about this with physical concepts or examples? Thanks,  
 + 
 +===chavez 10/23 3:16PM=== 
 +Think of it like degenerate energy levels for an atom. Different configurations of elections (eigenfunctions) can cause the particle to have the same energy (eigenvalue). 
 + 
 +=== Zeno 10/26 10am === 
 +A completely watered-down description of degeneracy is: multiple ways to achieve the same result. In QM this corresponds to multiple electron configurations, as Chavez said above, to achieve the same energy. Note that the degree of degeneracy is usually also important; you can have double degeneracy, triple, all the way up to infinite degeneracy, depending on how many (independent) ways you can achieve the same result. 
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 **Return to Q&A main page: [[Q_A]]**\\ **Return to Q&A main page: [[Q_A]]**\\
 **Q&A for the previous lecture: [[Q_A_1021]]**\\ **Q&A for the previous lecture: [[Q_A_1021]]**\\
 **Q&A for the next lecture: [[Q_A_1026]]** **Q&A for the next lecture: [[Q_A_1026]]**
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classes/2009/fall/phys4101.001/q_a_1023.1256318774.txt.gz · Last modified: 2009/10/23 12:26 by x500_razi0001

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