GR8677 #90



Alternate Solutions 
BillNyeTheRussianSpy 20180522 03:06:49  I think, notationally, it is easier to have . But it still comes out the same.\r\n\r\n implies   BillNyeTheRussianSpy 20180522 03:06:49  I think, notationally, it is easier to have . But it still comes out the same.\r\n\r\n implies  

Comments 
BillNyeTheRussianSpy 20180522 03:06:49  I think, notationally, it is easier to have . But it still comes out the same.\r\n\r\n implies   BillNyeTheRussianSpy 20180522 03:06:49  I think, notationally, it is easier to have . But it still comes out the same.\r\n\r\n implies   Moush 20100919 06:20:35  Yosun, in the 2nd paragraph you have "load" instead of "node". Richard mentioned it but didn't classify it as a typo so you probably didn't see it.
I just noticed that the preview button resets the classification to NEC.
Thanks for this site!   neon37 20081106 14:49:13  I solved this slightly differently,
We know,
and De Broglie wavelength,
Plug this into the E and we find out,
So wavelength is doubled going from this state to ground state so the Energy has to be a quarter so .
wittensdog 20090727 13:10:26 
That works in this case, but only because it makes sense to talk about a specific wavelength. The infinite square well is weird in the sense that because there are definite energy states in a region where there is no potential energy, you have E = p^2/2m for a set of definite energies, you end up with states that have a definite magnitude of momentum. So then these definite momentum magnitude states correspond to a precise wavelength. Normally there would be a spread in p, but the square well has this nice feature. I didn't even really think about this until a semester after taking my QM course, when I read it in the appendix of my statmech book. There was actually kind of an interesting derivation of the infinite square well properties, just making reference to the idea of de Broglie waves being fit inside the well. But if you were to try to apply the method to another potential, I imagine it would not be so successful. It makes a lot of sense thinking about the fourier transform. Inside the well you have pure sine waves, and the fourier transform of a pure sine wave with some frequency w just has two delta spikes at +/ w, each normalized with a factor of 1/2 to make the total probability equal to 1. Of course there would be mixing in from other wavelengths if you were to consider the entire range of x values, outside of a and a, because then you have a wave pulse.

wittensdog 20090727 13:11:30 
But I guess my point was, I always thought that was a really interesting property of the square well.

  FortranMan 20081030 07:41:56  Try not to confuse this with the harmonic oscillator where the ground state is n=0.   jesford 20080403 17:57:00  Even if you can't remember the infinitesquarewell energies, the problem asks for the LOWEST possible energy... so it can't be more than the 2 eV energy given for the diagram. This rules out choices D and E. Also, recall that the particle can't have zero energy, so the only options left are B and C.
jesford 20080403 17:58:41 
oops I guess 1 eV is less than 2, eh?

  Richard 20071101 12:37:50  You wrote "load," but you should have written "node."
:)
realcomfy 20081101 09:50:32 
I also believe that it should state that the particle has nodes thus giving us

realcomfy 20081101 09:51:31 
Check here and correct me if I am wrong
http://en.wikipedia.org/wiki/Image:Particle_in_a_box_wavefunctions.png

gt2009 20090625 17:17:22 
2 antinodes, not nodes

wittensdog 20091008 11:37:54 
This isn't very scientific, but instead of fretting over nodes vs. antinodes, I always just remember that the state number is the number of "bumps."

IRFAN 20110818 02:25:34 
Actually you ppl confuse node i thimk it might be loops

  grae313 20071007 19:01:27  Alternately, knowing we can write
Divide the first equation by the second and solve for x.  

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