GR | # Login | Register
  GR8677 #18
GREPhysics.NET Official Solution    Alternate Solutions
Verbatim question for GR8677 #18
Quantum Mechanics}Schrodinger Equation

The problem gives the wave function, wherein the hidden mysteries of the problem are contained. The potential is referenced as V(x), which means that it's time-independent. Thus, the Time-Independent Schrodinger Equation can be used: -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi+V(x)\psi=E\psi. The second derivative of the given \psi is \frac{d^2}{dx^2}\psi=(-b^2+(b^2x)^2)\psi. Plug that into the TISE, and one gets -\frac{\hbar^2}{2m}(-b^2+(b^2x)^2)=E-V(x)
Now, plugging in the potential condition V(0)=0, one gets, \frac{\hbar^2}{2m}((b^2x)^2)=E-V(0)=E. This implies that the term on the left that disappeared from that substitution is the V(x) term. Therefore, one deduces that V(x)=\frac{\hbar^2}{2m}(b^2x)^2.
The right answer would be (B).

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
2013-10-13 06:48:24
The answer should be E . Because \frac{d^2}{dx^2}\psi=(-b^2+(b^2x)^2)\psi now the kinetic energy part \frac{\hbar^2}{2m}(-b^2+(b^2x)^2)=E-V(x)=K now putting x=0 and V(0) = 0 we get \frac{\hbar^2}{2m}(b^2)=E-V(0)=E .So the right answer goes to choice E .Alternate Solution - Unverified
2011-11-08 18:36:51
I agree with Kabuto Yakushi more or less. Couldn't you just plug \psi into the TDSE? Then the solution falls out.Alternate Solution - Unverified
2016-10-28 17:37:14
That doesn\'t give answer E. Answer E has a b^4 in it, your method (which is the same thing that I did originally) has a b^2.NEC
2013-10-13 06:48:24
The answer should be E . Because \frac{d^2}{dx^2}\psi=(-b^2+(b^2x)^2)\psi now the kinetic energy part \frac{\hbar^2}{2m}(-b^2+(b^2x)^2)=E-V(x)=K now putting x=0 and V(0) = 0 we get \frac{\hbar^2}{2m}(b^2)=E-V(0)=E .So the right answer goes to choice E .
2013-10-13 07:09:42
The 2nd EQ will be -\frac{\hbar^2}{2m}(-b^2+(b^2x)^2)=E-V(x)=K . By the way the choice B was a good candidate for the answer but it requires a minus sign .
2013-10-14 07:56:05
Sorry I was wrong . Choice B is the right answer .
2014-08-25 13:22:22
Choice E has b^\4, not b^\2.
2014-09-02 13:32:47
Remember that the term in front of the kinetic is NEGATIVE: \frac{-\hbar ^2}{2M}
2016-10-28 17:37:58
That doesn\'t give answer E. Answer E has a b^4 in it, your method (which is the same thing that I did originally) has a b^2.
Alternate Solution - Unverified
2013-08-15 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?
2013-08-15 18:08:12
This should be classified as HELP!, haha
2013-08-15 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!
2012-04-15 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.
2013-01-20 20:26:18
C is fine too by dimensional analysis.
2011-11-08 18:36:51
I agree with Kabuto Yakushi more or less. Couldn't you just plug \psi into the TDSE? Then the solution falls out.
2012-11-01 16:13:05
You could use the time-dependent SE, but this only adds to the already long problem. Kabuto Yakushi has stated that \frac{\partial\psi}{\partial t} = 0, but this is misleading since the time derivative of \Psi (\psi muliplied by the time component of the wavefunction) isn't zero and we must use it if we want to apply the time-dependent SE. In fact, since \Psi is an eigenfunction, it must be of the form \psi e^{\frac{Et}{i\hbar}}, which would make (Time dependent SE) H\Psi = i\hbar\frac{\partial \Psi}{\partial \t}=E\Psi. The hamiltonian operator for this system, H, has no time dependence, and so this reduces to the time-independent SE, H\psi=E\psi, and we're back at square one.

Using H\psi=0, as Kabuto suggests, would give me an incorrect result...maybe the fixed-typo some people are talking about has something to do with it.
Alternate Solution - Unverified
2011-06-23 05:56:14
In the solution, why is the term -b^2 omitted?
2012-10-23 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
Answered Question!
2011-04-06 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 \frac{\hbar^2b^2}{2m} 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 b^2 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 x is enough for you to choose the right answer, after you plug and chug of course.Common Pitfalls
2010-11-11 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
2011-09-23 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.
2010-09-20 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 wave-packet 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
2010-09-02 09:49:11
I just plugged the wave function into the schrodingers equation.

\frac{d\psi}{dt} is obviously zero so V(x)\psi
equals the second time derivative of psi time the
\frac{-\hbar^2}{2m}. Solving one gets B).

Thanks for the great site Yosun.
2013-10-13 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 \Psi (x,t) = \psi (x) \phi (t) we work with the time independent part \psi(x) . The SE we have here is a time independent SE . So your calculation is wrong .
2010-04-06 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.
2010-04-09 06:49:32
After you get E - V(0) = E as Yosun describes you have to calculate

\hat{p}^2/2m - E = V(x) which will give answer B).
2010-10-25 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
2009-12-24 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) the correct answer is b.
2009-08-24 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 x, the only way for V(0)=0 is for answer B to be true?NEC
2008-06-18 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] = [\hbar] [\omega]. Then, check if [\hbar b^4 / M] has units of 1/s.

Therefore it must be (B).
2008-09-05 03:07:22
Just to add to the "Elimination" time

For (C) just think about applying the \hat{p} to see that you can never get b^6

For (D) you just have to look at the units after the V(x) = 0 condition is applied. You get \hbar b^2 doesn't look like an energy to me... \frac{1}{s}^2

For (E)... Why do we need to know about the potential for this one? ETS and useless information... kick it
pam d
2011-09-23 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.
2006-11-14 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?
Typo Alert!
2006-05-05 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).NEC
2006-02-20 14:00:04
There is a typo in the solution. plugging in v(0)=0 gives you h^2*b^2/2m = ENEC
2005-11-30 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.
2007-10-29 10:29:11
Watch out! The given wave function is NOT a Gaussian wavepacket. And a constant potential (A) can yield "free-particle solutions": plane waves.
2008-11-05 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).
2010-09-20 07:24:28
the Gaussian wave packet isn't an eigenstate of the Schrodinger equation

Post A Comment!
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...