|
GR0177 #44
|
|
|
|
|
Alternate Solutions |
ericimo
2007-11-01 15:05:07 | All you need to know is  . The rest is given by ETS to try and make you overthink it and skip it. All you see is a scary quantum question when it's just monkey work.
It's apparent from the given that  =   = 4 and equally apparent that  = 9 and so on.
If it's not obvious to you...plug and chug n=1,2,3..., and you see that D works for n=3 which is reason enough to choose it. |  |
|
|
Comments |
anmuhich
2009-03-14 13:49:30 | As others have said but I'll state differently...the key word is MEASUREMENT. This indicates that if you were able to open the box and "see" the energy of the particle then you would have to see one of the energies corresponding to an energy eigenvalue of the system. The expectation value is a statistical analysis of the system and is not a real energy value of the system. An analogous situation would be like pulling three numbers out of a bag let's say 1, 3 and 6. Their average (aka expectation value) is 3.33, but you can't pull 3.33 out of the bag. Pulling a number out of the bag is analogous to taking a measurement, you can only pull out 1, 3 or 6. This is a fundamental idea in quantum mechanics but it can be tricky because we spend a lot of time computing expectation values for a composition of eigenstates. |  | ericimo
2007-11-01 15:05:07 | All you need to know is . The rest is given by ETS to try and make you overthink it and skip it. All you see is a scary quantum question when it's just monkey work.
It's apparent from the given that = = 4 and equally apparent that = 9 and so on.
If it's not obvious to you...plug and chug n=1,2,3..., and you see that D works for n=3 which is reason enough to choose it.
ericimo
2007-11-01 15:12:26 |
probably what yosun was getting at, but people didn't seem to be getting it.
|
| | YeonPil
2007-10-16 05:19:40 | I think, We should suppose that Hy(Psi) = E_n y is correct. If y is not eigenstate, we can not get a E_n.
Therefore, I can get values only such as E1 , 4E1, 9E1, 16E1.........
I agree with author. | | kevglynn
2006-11-03 08:25:01 | The above comments point out that people commonly mix up concepts within phrases such as "possible energy measurements"; "expectation values"; and "probability of finding where A is some eigenvalue." Remember what they mean physically and then the math makes sense. It's simple stuff, too, so everyone should pick up these easy points. |  | Healeyx76
2006-11-02 16:32:05 | Um, the answer is indeed (D), 9E_1 according to the testbook.
a
I get the same conclusion due to the E measurement being dependent on n^2. The only possible solution as a multiple of the ground stte E1 is 9E1 (for state #3)
It seems like for this one they give you a lot of unnecessary info. | | cailh
2006-10-28 06:46:49 | the answer is c
==1/14*
=1/14*(1+16+81)E1=7E1
beddi
2007-06-13 14:10:10 |
I also think so. Is the solution wrong?
|
deimos
2007-08-27 15:16:10 |
This is the expectation value of the energy. If you were to make a large number a measurements, they would average to that value. In this case we are only concerned with one measurement.
|
| | lolo
2006-10-28 01:45:27 | Or may be you use the difference between "measurement" and "mean value"? | | lolo
2006-10-28 01:43:15 |  =1/14(+4etc) | | lolo
2006-10-28 01:37:57 | This solution is absolutely wrong. Three eigenstates, corresponding to different eigenvalues are mixed. Of course, this function is still an eigenfunction, but the square dependence on n doesn't exist anymore. For example, take a square spin operator S^2. Take two eigenfunctions: singlet and triplet ones. Their linear combination is still an eigenfunction of the operator, but the eigenvalue is not S(S+1) anymore. This happens because we mix functions from the subspaces, corresponding to different eigenvalues.
Moreover, a given function here, is also normalized to unity, so following the elementary quantum mecanics, we get
=1/14{+4etc}
Blake7
2007-09-19 06:33:58 |
The point is not to argue with ETS.
The point is to understand what they are looking for, which (in all my experience) is eventually explicable, and get a good score.
|
gryphia
2008-11-03 20:52:41 |
The point is:rnThe wavefunction is constructed out of a set of eigenstates of the hamiltonian. When you measure the particle, the wavefunction you measure will be one of those eigenfunctions. That means that the energy you measure will be the energy associated with one of those eigenstates. Those energies go as  , so the only possible energies you can measure are  . The only perfect square in the choices we are given is answer d. Thus d is the correct answer. That it corresponds to one of our eigenstates is essential, of course, and if we look at the original function we notice that we have a  , so everything is good.
|
| | lolo
2006-10-28 01:29:17 | | |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
