(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] "Particle in a Box" Measurement scenario

Problem statement:

Measurement of the position of a particle in a 1D well with walls at x=0 and x=L finds the value at x=L/2. Show that in a subsequent measurement, the particle will be in any odd eigenstate with equal probability.

Relevant equations/attempt at solution:

Well, we know that if the energy is not sharp, then it is realized as an expectation value, so

[tex]<E>=\int_0^L\Psi^*(x)\hat{H}\Psi(x)dx[/tex]

But, we know that any function, [tex]\Psi(x)[/tex] can be described as being a sum of "base states:"

[tex]\Psi(x)=\Sigma_{k=0}^\infty a_k\Psi_k(x)[/tex]

where

[tex]<\Psi(x),\Psi_n(x)>=a_k \delta_{kn}=a_n[/tex]

so, we can rearrange this to say that

[tex]a_n=<\Psi(x),\Psi_n(x)>[/tex]

where even solutions take the form

[tex]\Psi(x)=\sqrt{\frac{2}{L}}Sin\left(\frac{(2n)\pi x}{L}\right)[/tex]

and odd solutions

[tex]\Psi(x)=\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)[/tex]

So, if we wanted to find the probability of detecting the particle in an odd eigenstate, we would calculate:

[tex]\left|a_{2n+1}\right|^2=<\Psi(x),\Psi_{2n+1}(x)>^2=\left<\sqrt{\frac{2}{L}}Sin\left(\frac{n\pi x}{L}\right),\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right >^2

=\left [\int_0^L \left(\sqrt{\frac{2}{L}}Sin\left(\frac{n\pi x}{L}\right)\right)^*\left(\sqrt{\frac{2}{L}}Sin\left(\frac{(2n+1)\pi x}{L}\right)\right)dx\right]^2[/tex]

To my understanding, [tex]\left|a_n\right|^2[/tex] is the probability of finding a particle with a energy value of n, is this right? Or am I at least on the right track? Thanks all (as always...) :)

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# "Particle in a Box" Measurement scenario

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