Quantam Mechanic - Particle in a rigid one-d box (PDF)

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SUMMARY

The discussion focuses on calculating the probability distribution for the second excited state (n=3) of a particle in a rigid box of length 'a'. The probability density function is given by |ψ(x)|^2 = (2/L)sin^2(n∏x/L). Users clarify the calculation of probabilities in specific intervals, [0.50a, 0.51a] and [0.75a, 0.76a], emphasizing the use of approximations for the sine squared term. The first interval approximates sin^2(3π/2) as 1, while the second requires the exact evaluation of sin^2(9π/4).

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Homework Statement



Write down and sketch the probability distribution for the second excited state (n=3) of a particle in a rigid box of length a.

What are the probabilities of finding a particle in the intervals [0.50a, 0.51a] and [0.75a, 0.76a]?

Homework Equations



|ψ(x)|^2 = (2/L)sin^2(n∏x/L)

The Attempt at a Solution



I've found the solution online: http://www.uic.edu/classes/phys/phys244ma/p244hw7.html

However, I'm not understanding how they are getting the answers.

Multiply the wave-function squared times the distance between the two points, but I'm not sure why the sine squared term in the first part is one, while in the second part you are to use sin^2(9∏/4)?

Thank you.
 
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In the first part, the author is making an approximation. In the range [itex]x \in [ 0.5a,0.51a][/itex] :

[tex]\sin^2{\frac{3\pi}{a}x} \approx \sin^2{\frac{3\pi}{2}} = 1[/tex]

In the second part, that approximation does not hold, so the author uses a more general expression.
 

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