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Latest comments |
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| Latest 25 Comments | | 0 | Click here to jump to the problem! | evanb
2008-07-03 14:25:31 | p --absolutely must-- be self-adjoint, because it is observable. The only boundary conditions that we need to apply is that  is square-integrable (goes to zero quick enough that the boundary term is ignored), which holds true for a harmonic potential. From this, you can deduce that a is not hermitian, eliminated II.
For the last piece, "doesn't the fact that a and the hamiltonian have different eigenvalues in this way mean that they don't commute?" That's a weirdly-phrased question, mainly because you don't explain what "this way" is supposed to mean. However, if you know that the eigenstates of the ladder operators are coherent states and the eigenstates of the hamiltonian are NOT coherent states (they are stationary states), that is enough to say that H and a do not commute, eliminating I and leaving (C) as the answer.
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| 2 | Click here to jump to the problem! | evanb
2008-07-02 14:55:19 | They're trying to lure you into computing how much power is emitted by the oven, then focus the light into a smaller area, so you have a larger power flux, then absorb that power into the sample.
Of course, we're to smart for that trap! |
| 3 | Click here to jump to the problem! | evanb
2008-07-02 13:59:18 | At r = 0, the field must be zero, because an infinitely-small amperian loop contains no current.
At r=c, the field must be zero, because the inner current and the outer current cancel out.
The only graph that satisfies these two contraints is B. |
| 4 | Click here to jump to the problem! | evanb
2008-07-02 13:50:36 | See Griffith's page 358.
However, the stored momentum is dependent on  , so the stored momentum could be dependent on d, through  . |
| 5 | Click here to jump to the problem! | evanb
2008-07-01 17:22:24 | Unfortunately, the way would know it works is by doing the general relativity. You can deduce that in the Schwarzschild metric for a point mass, any timelike or null (lightlike) trajectory (ie. any trajectory that a particle might take) that starts within R = 2M will ever wind up outside of R = 2M ( G = c = 1).
Un-unfortunately, GR reproduces lots of "Newtonian" results. You can even do most cosmology with Newtonian math. Why does that work? Because someone did it fully-general-relativistically and found that the math wound up the same... |
| 6 | Click here to jump to the problem! | evanb
2008-06-30 16:38:53 | Call each spherical harmonic | l, m >.
We know, because spherical harmonics are orthonormal, that
<  ,  | l, m > = 
The operator we want to apply is
{ 1 if l = 5, 0 if l not = 5 }
Applying that operator to any ket | l m > yields zero if l isn't 5, and yields | 5 m > if l IS 5.
Taking < | operator | > gives us the following:
(5 <1 1| + 3 < 5 1 | + 2 <5 -1| ) /  * ( 3 | 5 1> + 2 |5 -1>)/$\sqrt{38}
Contracting, that yields, ( 3^2 <5 1| 5 1> + 2^2 <5 -1|5 -1> ) / 38
Simplifying, (9+4)/38 = 13/38, which is answer (C). |
| 7 | Click here to jump to the problem! | evanb
2008-06-30 13:38:24 | Yes, a safer choice is to track seconds.
[G] is m^3 / kg s^2
[h] is J s which is kg m^2 / s
[c] is m/s.
We MUST eliminate seconds, so there must be a division in our answer.
Thus, the answer is E. |
| 10 | Click here to jump to the problem! | HariSeldon
2008-06-29 11:30:45 | You can also eliminate E with the knowledge that electrons have a relatively short interaction length, while muons travel much farther before interacting. |
| 12 | Click here to jump to the problem! | evanb
2008-06-27 14:46:04 | The wave function of 1s is even, the perturbative operator = - q * Efield * Z is odd.
Thus the first-order matrix element must be zero. |
| 16 | Click here to jump to the problem! | sirius
2008-06-25 13:05:18 | the potential is 0 from 0 to a. it says so in the problem. this is a particle in an infinite square well, and answer (E) is the only answer that makes sense. |
| 17 | Click here to jump to the problem! | evanb
2008-06-25 12:56:18 | It turns out that it doesn't matter what  is equal to... it adds a factor of  to the numerator and to all the factors in the denominator. Better off setting it to zero, in my opinion, but in terms of the physics it doesn't make one lick of difference: it's like setting your h in mgh potential energy to 0 on the ground.
See radicaltyro's comment on moonrazor's solution for evidence. |
| 18 | Click here to jump to the problem! | evanb
2008-06-25 12:47:47 | A way to do Andresito's method without math.
With a  of 100, the thing is definitely ultra-relativistic, meaning that it could effectively be treated as an almost-zero-mass particle.
Photons have p = E/c, so this particle should have approximately p = E / c = 100 m c |
| 19 | Click here to jump to the problem! | sirius
2008-06-24 14:27:26 | i have never heard of conducting cavities, what topic is this from? is there a text i should look at for this? |
| 20 | Click here to jump to the problem! | evanb
2008-06-24 11:56:52 | How about a uniform-density infinite-plane slab. So, it would be thick and the region of interest would be from the middle of the slab to the edge of the slab. |
| 22 | Click here to jump to the problem! | evanb
2008-06-23 19:11:02 | Dammit, the preview de-TYPO-ified my entry.
Yes, (B) should be >= and not <=
and it should be the answer, not (E) |
| 24 | Click here to jump to the problem! | evanb
2008-06-23 18:57:17 | A useful fact is that c is just about 1 foot per nanosecond.
3 meters is about 10 feet, so the flight takes about 10 nanoseconds, so we have to be able to resolve on the order of 1 nanosecond to say for sure. |
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