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## GR0877 Problem 85

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (B) As one can check for himself/herself the constant $A$ is equal to $\displaystyle \sqrt{\frac{2}{L}}$ (this is from normalization condition $\displaystyle \int_0^L \psi^2(x) dx = 1$). The probability in question is: $\displaystyle \frac{2}{L} \int_{L/3}^{2L/3} \sin^2 \left(\frac{3 \pi x}{L}\right)dx = \frac{1}{L} \left[ \int_{L/3}^{2L/3} dx - \int_{L/3}^{2L/3} \cos \left( \frac{6 \pi x}{L} \right)dx \right] = \frac{1}{L} \left[ \frac{L}{3} -0 \right] =$ $= \displaystyle \frac{1}{3}$.

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1. 14.10.2011 at 01:55

$A= \sqrt{2/L}$, not just $2/L$. Using $2/L$ results in the probability being equal to $2/3L$, which is wrong. To get $A$, one needs to remember to square $A$, resulting in $A^2=2/L$, so therefore $A=\sqrt{2/L}$.

• 14.10.2011 at 12:16

Thank you, Oliver. This typo has been fixed.

2. 07.11.2011 at 05:25

Easier: if you recall that the wavefunctions for the infinite square well go as $A\sin(n \pi x/L)$, then one can see that $n=3$. The probability distribution for the n=3 state is three humps. The given range, $x=L/3$ to $x=2L/3$, bounds the middle hump. Thus the probability is 1/3!

• 11.11.2011 at 00:07

That’s the way to do it in 1.7 minutes. I don’t think ETS expects you do that integral in the given time.

• 20.10.2015 at 22:11

I agree! There has got to be a way to be able to do this in 1.7 minutes. It seems a rather long problem to do it conventionally.