(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle of mass m is in the lowest energy (ground) state of the infinite potential energy well (V(x)=0 for 0<x<L, and infinite elsewhere)

At time t=0 the wall located at x=L is suddenly pulled back to a position at x=2L. This change occurs so rapidly that instantaneously the wave function does not change. Calculate the probability that a measurement of the energy will yield the ground-state energy of the new well. What is the probability that a measurement of the energy will yield the first excited energy of the new well?

2. Relevant equations

So far I am using the following relevant equations/formulas:

(1) [tex]

\Psi (x) = \sum c_n \sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} \: \: for \: 0<x<L

[/tex]

(2) [tex]

\int {\psi_{n}}^{\ast} \Psi dx = c_n

[/tex]

and

(3) [tex]

E_n = \frac{n^2 h^2 \pi^2}{2 m L^2}

[/tex]

3. The attempt at a solution

This problem is tripping me up a little bit... I am not really sure what is implied/asserted by the phrase "instantaneously the wave function does not change"

So far, I have come up with the following solution....

Obviously, in the new system E will be able to take on 2 values instead of just the ground state since [tex]E'_1=\frac{h^2 \pi^2}{8 m L^2}[/tex] and [tex]E'_2 = E_1 = \frac{h^2 \pi^2}{2 m L^2}[/tex]

I try to use (1) and (2) to calculate [tex]c_1^'[/tex] and [tex]c_2^'[/tex] using :

[tex]

c'_2 = \frac{\sqrt{2}}{L} \int \sin{\frac{\pi x}{L}} \sin{\frac{\pi x}{L}}

[/tex]

from 0 to L which comes out as 1/sqrt(2) which seems reasonable and is nice (P = 1/2) so i left it...

the reasoning i used behind this calculation is that i am multiplying the psi_2 prime with the first wave equation, which has a c_1 of 1, and 0 elsewhere.... which makes the limits 0 to L...

next, i find c'_1 by:

[tex]

c'_1 = \frac{\sqrt{2}}{L} \int \sin{\frac{\pi x}{L}} \sin{\frac{\pi x}{2 L}}

[/tex]

from 0 to L which comes out as

[tex]\frac{4 \sqrt{2}}{3 \pi}[/tex]

in which case P=.3602, and thus this must be wrong unless the particle can take the 2nd excited energy state, which doesn't seem to make sense to me (plus that answer doesn't look as clean or nice)

Can someone help me out or explain to me where it is I am going wrong????

also... if anyone can qualitatively explain to me how it is the wave function "does not change"???

any help appreciated!!!

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# Infinite Well Potential : going from 0<x<L to 0<x<2L

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