timfinnigan 20180520 03:18:29  Plug into the TDSE and then set x=0. You will end up with (should be hbar). Put that into the TDSE and then cancel like terms. You will end up with V(x)=choice B.
timfinnigan 20180520 03:19:45 
This E value will allow you to eliminate choices D&E

 
GrahamS 20161028 17:37:14  That doesn\'t give answer E. Answer E has a in it, your method (which is the same thing that I did originally) has a .  
arifmujib 20131013 06:48:24  The answer should be E . Because now the kinetic energy part now putting x=0 and V(0) = 0 we get .So the right answer goes to choice E .
arifmujib 20131013 07:09:42 
The 2nd EQ will be . By the way the choice B was a good candidate for the answer but it requires a minus sign .

arifmujib 20131014 07:56:05 
Sorry I was wrong . Choice B is the right answer .

tjurra 20140825 13:22:22 
Choice E has , not .

QuantumCat 20140902 13:32:47 
Remember that the term in front of the kinetic is NEGATIVE:

GrahamS 20161028 17:37:58 
That doesn\'t give answer E. Answer E has a in it, your method (which is the same thing that I did originally) has a .

 
natec 20130815 18:07:19  why do I get E = (hbar^2)*(b^4)/2M when I plug in V(x)=0 and x=0 (Which is answer E)? Why are the units wrong?
natec 20130815 18:08:12 
This should be classified as HELP!, haha

natec 20130815 18:16:13 
NEVER MIND. found my mistake; i made a silly mistake with exponents. I actually get E=hbar^2 b^2/2M which is NOT answer E. Sorry for these three useless comments... Thank you for the great website though!

 
mpdude8 20120415 20:19:31  You really don't have to know much about QM to get this one. If you know that the terms in the exponent must be dimensionless, the rest is just elimination.
b must have units of 1/length to cancel with x^2. V and E must be in units of Joule, or, using base units, (kg * m^2 / s^2). Knowing that b ~ 1/length quickly tells you the answer must be B without ever invoking the TISE.
Eliminate C immediately because you cannot add or subtract terms with different units.
Also, note that both V and E have the same units  logically, A and E must both be eliminated, then.
Jovensky 20130120 20:26:18 
C is fine too by dimensional analysis.

 
deafmutemouse 20111108 18:36:51  I agree with Kabuto Yakushi more or less. Couldn't you just plug into the TDSE? Then the solution falls out.
oliTUTilo 20121101 16:13:05 
You could use the timedependent SE, but this only adds to the already long problem. Kabuto Yakushi has stated that , but this is misleading since the time derivative of (\psi muliplied by the time component of the wavefunction) isn't zero and we must use it if we want to apply the timedependent SE. In fact, since is an eigenfunction, it must be of the form , which would make (Time dependent SE) . The hamiltonian operator for this system, H, has no time dependence, and so this reduces to the timeindependent SE, , and we're back at square one.
Using , as Kabuto suggests, would give me an incorrect result...maybe the fixedtypo some people are talking about has something to do with it.

 
ali8 20110623 05:56:14  In the solution, why is the term b^2 omitted?
drizzo01 20121023 11:53:50 
it was a typo. There when the B.C. was applied, all x=0, so really all that should have remained was the b^2 term. The rest follows

 
walczyk 20110406 17:07:23  I understand the source of the confusion I think. It isn't necessarily a typo but a confusion in the accepted solution. E cannot depend on x, so the quadratic term generated by the kinetic energy term in the SE must be eliminated by the potential energy term. So folks, the constant term is IN FACT the eigenvalue of the energy of the system. The answer makes it seem that plugging in the given conditions for the potential makes disappear, but no, that's not it. Check again. The conditions given in the problem lets us not worry about any constant potential term that would modify eigenvalues at the end. Its not really that helpful. Just the knowledge that the energy must not depend on is enough for you to choose the right answer, after you plug and chug of course.  
Almno10 20101111 13:53:44  One way to look at this is that
i) E cannot depend on x
ii) V(0) = 0
B is the only option which satisfies i and ii.
pam d 20110923 20:21:08 
Actually this line of reasoning only eliminates (A) and (D). It certainly helps to make these eliminations, but there are more steps.

 
alemsalem 20100920 07:42:47  from the Schrodinger equation it is obvious that the units of energy are same as the units of h^2/(2m dx^2) all the answers should have a unit of energy b has units of inverse length ,, this leaves B and C.
a Gaussian wavepacket has a shape like this (at t=0) but it's not an eigenstate of the equation, and it must be a harmonic potential anyway (ground state).
also if you derive the wave function twice you can easily tell that at most you will get x to the power of 2 and because energy should be a constant for all x the potential cannot turn out to have x^4.  
Kabuto Yakushi 20100902 09:49:11  I just plugged the wave function into the schrodingers equation.
is obviously zero so V(x)
equals the second time derivative of psi time the
. Solving one gets B).
Thanks for the great site Yosun.
arifmujib 20131013 06:57:50 
Usually the simple form of SE(schrodingers equation) we work with is the space part of the real SE . The actual wave equation we work with the time independent part . The SE we have here is a time independent SE . So your calculation is wrong .

 
Plantis 20100406 05:42:52  I don't agree with solution. When we are plugging the potential conditions V(x)=0 and x = 0. It's mean that we will recieve (A) correct answer.! I don't clearly understand the solution.
apr2010 20100409 06:49:32 
After you get as Yosun describes you have to calculate
which will give answer B).

faith 20101025 19:48:53 
there is a typo in yosun;s solution.rnrnplugging in V(0) would require E to be just h^2*b^2/2m = E (which was commented by marshiesbudda). solving that you'll arrive at the answer of B

 
physics_gre 20091224 02:33:42  the given vave function is in the usual form of a one dimensional linear harmonic oscillator.we know the expression for the potential of a lho is
v(x)=(1/2)kx.x.so the correct answer is b.
 
solarclathrate 20090824 00:28:17  I don't quite understand the final implication in the solution. Is he trying to say that because the energy is independent of , the only way for is for answer B to be true?  
evanb 20080618 17:43:24  (A) can be eliminated based on given knowledge that V(x=0) = 0, except in the pathological b = 0 case.
(C) can be eliminated based on knowledge that the potential is harmonic.
(D) can be eliminated because an eigenfunction of the SE has a constant E, not one that varies based on position.
(E) can be eliminated based on units: just assume [E] = [] []. Then, check if [] has units of 1/s.
Therefore it must be (B).
ssp 20080905 03:07:22 
Just to add to the "Elimination" time
For (C) just think about applying the to see that you can never get
For (D) you just have to look at the units after the V(x) = 0 condition is applied. You get doesn't look like an energy to me...
For (E)... Why do we need to know about the potential for this one? ETS and useless information... kick it

pam d 20110923 20:23:47 
ssp, you cannot eliminate (E) like that. They are asking for what statement is correct. Choice (E) is certainly incorrect but not for the reason you are using. evanb's reasoning is much more sound.

 
barefoot0 20061114 10:32:18  Would you not plug 0 in for x and thus V(0)> 0 and then whatever remains on the left side must be E. Then by subtracting E when x =/ 0 gives us the V term?
In other words h^2*b^2/2m = E + V(0) = E,
E + V(x)  E = h^2*(b^2 x)^2/2m
or am I wrong?  
transoid 20060505 10:14:05  No, there is no typo. The solution did say "This implies that the term on the left that disappeared from that substitution is the V(x) term". Thus you have to put E back in to find out V(x).  
marshiesbudda 20060220 14:00:04  There is a typo in the solution. plugging in v(0)=0 gives you h^2*b^2/2m = E  
tarlen 20051130 21:17:28  While this solution is very illuminating, a very quick way of solving this one is to realize that this wavefunction is a gaussian wave packet. There are only 2 potentials that give a Gaussian wave packet: one is the free particle and another is the ground state wavefunction of the harmonic oscillator potential. Since the free particle potential isn't there, we know the answer must be b), the harmonic oscillator potential.
kicksp 20071029 10:29:11 
Watch out! The given wave function is NOT a Gaussian wavepacket. And a constant potential (A) can yield "freeparticle solutions": plane waves.

sirius 20081105 20:17:00 
tarlen is mostly correct. It is not a wave packet however, since it is not in motion as in the free particle. It is simply a gaussian, which is characteristic of a harmonic potential. So (B).

alemsalem 20100920 07:24:28 
the Gaussian wave packet isn't an eigenstate of the Schrodinger equation

 