GREPhysics.NET: GR8677 #18

GR8677 #18

Problem

GREPhysics.NET Official Solution

Quantum Mechanics $\Rightarrow$ }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 $-\frac{\hbar^2}{2m}(-b^2+(b^2x)^2)=E-V(x)$
\par
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 $V(x)=\frac{\hbar^2}{2m}(b^2x)^2$ .
\par
The right answer would be (B).

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Comments

evanb
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).

ssp
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

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barefoot0
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?

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transoid
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).

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marshiesbudda
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 = E

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tarlen
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.

kicksp
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.

sirius
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).

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Post A Comment!

You are replying to:

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?

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LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
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$\partial$	$\partial$
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$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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