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classes:2009:fall:phys4101.001:q_a_0916 [2009/09/16 10:39] – x500_sohnx020classes:2009:fall:phys4101.001:q_a_0916 [2009/09/26 23:47] (current) – yk
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 ==== Malmx026 8:00 pm 9/15 ==== ==== Malmx026 8:00 pm 9/15 ====
 Page 65 Griffiths shows how to calculate the group velocity for wave function made up of many frequencies.  It seems to me that the shape of the wave would change (different freq. would travel at different velocities) if the higher order derivatives of the frequency were non zero, am I interpreting this correctly? Page 65 Griffiths shows how to calculate the group velocity for wave function made up of many frequencies.  It seems to me that the shape of the wave would change (different freq. would travel at different velocities) if the higher order derivatives of the frequency were non zero, am I interpreting this correctly?
 +
 +===Yuichi 14:10 9/16/09===
 +
 +That's right.  Since wave packet contains components with different (usually slightly) frequencies, and they travel at slightly different speed, the shape of the wave packet will change over time.  
 +
 +If the original wave is Gaussian (both in space and in frequency domains), it will stay Gaussian, but the width will grow over time. as indicated in problem 2.22 and 2.43.  When the initial width of the Gaussian (in space) is very small, it will spread very quickly.  This is because small <math>\Delta x</math> suggest that <math>\Delta p</math> which suggests that the momentum (and speed) of the particle can be either + or - large values, which make the wave packet spread out very quickly.
  
 ==== Esquire 10:25pm 9/15 ==== ==== Esquire 10:25pm 9/15 ====
 One issue concerning quantum mechanics has always bothered me. Why was is a wave model assumed for quantum events? Is this solely in attempt to match experimental observation? Wave mechanics is still classical in foundation quantum happenings seem to be in a completely scheme than classical events, which is the source of my discontent. Have other models been successfully applied to quantum mechanics?  One issue concerning quantum mechanics has always bothered me. Why was is a wave model assumed for quantum events? Is this solely in attempt to match experimental observation? Wave mechanics is still classical in foundation quantum happenings seem to be in a completely scheme than classical events, which is the source of my discontent. Have other models been successfully applied to quantum mechanics? 
  
-==Spherical Chicken==+===Spherical Chicken23:59 9/15===
 It is my understanding that the wave model was derived and created before very much experimental observation had been made at all (at least on small scales...  we've still not got very good small scale observational skills, instruments limited as they are) .  But I believe this is born in part from Planck's discovery of light quanta and Einstein's wave-particle duality description of light.  As well as an attempt at explaining the behavior of electrons staying in orbital -- so inherently there was a need for quantized energy levels.   It is my understanding that the wave model was derived and created before very much experimental observation had been made at all (at least on small scales...  we've still not got very good small scale observational skills, instruments limited as they are) .  But I believe this is born in part from Planck's discovery of light quanta and Einstein's wave-particle duality description of light.  As well as an attempt at explaining the behavior of electrons staying in orbital -- so inherently there was a need for quantized energy levels.  
  
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 I think you should come up with a new model.  Something Einstein would like better! I think you should come up with a new model.  Something Einstein would like better!
  
-==John Galt==+===John Galt 9:57 9/16===
 This question is similar to the one I posed earlier and I would still like to ask it. If we could find the emission time of a photon with greater precision, would that narrow its probability function? (For all time, because its probability function does not change over time?) This question is similar to the one I posed earlier and I would still like to ask it. If we could find the emission time of a photon with greater precision, would that narrow its probability function? (For all time, because its probability function does not change over time?)
 +
 +==vinc0053 9/20 17:45==
 +I have the same question.  I haven't been given compelling reason to accept waves, and aspects of QM like the uncertainty principle (or duality) don't make it better.  Thus far, waves have pulled through in explaining everything we have based on the assumption that particles move in waves.  I think new models should be posed, but I think it would be most helpful to know exactly what experiments and measurements in labs have shown.
  
 ==== Chap0326 10:41pm 9/15 ==== ==== Chap0326 10:41pm 9/15 ====
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 What exactly is a stationary state? They talk a little bit about it on page 26 of Griffiths...Is it just every solution of the time independent solution to the Sch. Eqn.? What exactly is a stationary state? They talk a little bit about it on page 26 of Griffiths...Is it just every solution of the time independent solution to the Sch. Eqn.?
  
-== Spherical Chicken ==+=== Spherical Chicken 0:41 9/16 ===
 A stationary state is synonymous with Eigen state.  Basically, a state that doesn't radiate energy. The probability density of these states has no time dependence. Small perturbations can be ignored.  Hence... stationary... in time.  It should be noted that the only true stationary state is the ground state.  With enough perturbation ... a stationary state can be made un-stationary. yes? Like granny in her rocker.  Small movement doesn't tip her out... but come hurricane season... she's not really stationary till she's on the ground.   A stationary state is synonymous with Eigen state.  Basically, a state that doesn't radiate energy. The probability density of these states has no time dependence. Small perturbations can be ignored.  Hence... stationary... in time.  It should be noted that the only true stationary state is the ground state.  With enough perturbation ... a stationary state can be made un-stationary. yes? Like granny in her rocker.  Small movement doesn't tip her out... but come hurricane season... she's not really stationary till she's on the ground.  
 +
 +=== Yuichi 14:26 9/16/09 ===
 +Yes, to the last part of the question.
 +
 +In addition, we discussed why stationary states are special in today's class.  Did the discussion make sense?  Do you have any additional questions, Joh04684?
 +
 +=== joh04684 10:45 9/16/09 === 
 +Yes it did, thank you!
  
 ==== joh04684 12:05 am 9/16 ==== ==== joh04684 12:05 am 9/16 ====
 On page 54 of Griffiths, I got a little bit lost in their jump at the top of the page from the sum of (1/j!)Squiggle^2j to turning it into an exponential, as well as their deduction that you can only have either even or odd components, but not both.  Could someone try to elaborate on how you can do this? On page 54 of Griffiths, I got a little bit lost in their jump at the top of the page from the sum of (1/j!)Squiggle^2j to turning it into an exponential, as well as their deduction that you can only have either even or odd components, but not both.  Could someone try to elaborate on how you can do this?
  
-== Blackbox ==+=== liux0756 02:51 pm 9/16 === 
 +About the sum of (1/j!)Squiggle^2j to turning it into an exponential, it is the taylor expansion that e^x=1+x/1!+x^2/2!+x^3/3!+..., here x=Squiggle^2 
 + 
 +==== Blackbox 10:30 am 9/16====
 On page 41 of Griffiths, could you explain why V(xo) is subtracted from V(x) in a Taylor series and the physical meaning of this, "(you can add a constant to V(x) with impunity, since that doesn't change the force)"? On page 41 of Griffiths, could you explain why V(xo) is subtracted from V(x) in a Taylor series and the physical meaning of this, "(you can add a constant to V(x) with impunity, since that doesn't change the force)"?
  
 +=== liux0756 02:56 pm 9/16===
 +I think it is because the abstract value of potential is not important, what we concern about is the value relative to another value, then we can set the zero potential point as we like. The force is the derivative of potential, and if we add a constant to V(x), the force will not change, so here we let V(x0)=0.
 +
 +==== time to move on ====
 +
 +It's time to move on to the next Q_A: [[Q_A_0918]]
classes/2009/fall/phys4101.001/q_a_0916.1253115581.txt.gz · Last modified: 2009/09/16 10:39 by x500_sohnx020

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