Probabilty of Particle in Box Between 0.4a and 0.6a: Solution Attempt

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SUMMARY

The discussion focuses on calculating the probability of a particle in the ground state of an infinite one-dimensional square well, specifically between the positions 0.4a and 0.6a. The normalized wave function is given as sqrt((2/a)sin(pi*x/a). The probability calculation involves integrating the function (2/a)Int(sin^2(kx)dx) from 0.4 to 0.6, where k is defined as pi/a. Participants suggest using the trigonometric identity sin^2(θ) = (1 - cos(2θ))/2 to simplify the integral for easier computation.

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Homework Statement



A particle is trapped in an infinite one-dimensional square well which extends between x=0 and x=a. If the particle is in the ground state, calculate the probability that it is between 0.4a and 0.6a.

2. The attempt at a solution

The normalised of the ground state is sqrt((2/a)sin(pi*x/a)

Setting k = pi/a, I get the probability

(2/a)Int(sin^2(kx)dx for 0.4 - 0.6. How can I solve this integral?
 
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If you've taken first-year integral calculus, you have seen this type of integral. Do you know any trig identities that involve sin^2 \theta?
 
sin2x = (1-cos2x)/2 and then integrate.

May be someone said that:
"Physics is too tough for Physicists."
... may be he was Schrödinger.
 

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