Easy integration by parts but getting wrong answer. Help

Bacat
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Homework Statement



I am actually looking for the expectation of x for the wavefunction that is \Psi (x) = \sqrt{\frac{2}{L}}Sin(\frac{\pi x}{L}) for 0 < x < L.

To do this I need to find the solution to this integral:

f = \int_0 ^L \! \Psi^* x \Psi \, dx = \int_0^L \! x*(\sqrt{\frac{2}{L}}Sin(\frac{\pi x}{L}))^2 \, dx


The Attempt at a Solution



Using Mathematica, I know that the answer to the integral is \frac{L^2}{4}.

However, when I attempt the solution by integration by parts, I get 0. Help!

f = \frac{2}{L} \int_0^L \! xSin^2(\frac{\pi x}{L}) \, dx = \frac{2}{L} \left[ uv - \int v du \right]

Let u=x, du=dx, dv=Sin^2(\frac{\pi x}{L}) dx

Then v=\int dv = \int_0^L Sin^2(\frac{\pi x}{L}) dx = \frac{1}{2} \int_0^L (1-Cos(\frac{\pi x}{L}) dx)

v = \frac{L}{2} - Sin^2(\pi) = \frac{L}{2}

Then f = \frac{2}{L}\left[uv - \int v du\right] = \frac{2}{L}\left[x\frac{L}{2}\right|_{0}^{L} - \int_0^L \frac{L}{2} dx\right] = \frac{2}{L}\left[ \frac{L^2}{2} - \frac{L^2}{2}\right] = 0.

Where have I made an error?
 
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v is supposed to be the antiderivative of dv, not the definite integral from 0 to L of dv!
 
guess who goes to U of M and is in Prisca's class...am i right?

in addition to the reply above...

check your power reduction of the sine squared term again, and you should notice right away what you did wrong...hint, what happens to the angle when you go from the squared trig function to a first power trig function?
 
Ah, thank you gabbagabbahey!

And Dahaka14, no. I live in Houston. Oddly enough, the curriculum for quantum mechanics is pretty much the same wherever you go.
 
Bacat said:
Oddly enough, the curriculum for quantum mechanics is pretty much the same wherever you go.

Only once you measure it. Before measurement there is a non-zero probability that the curriculum consists entirely of exorcising the vengeful spirits of cats which are both alive and dead simultaneously.
 
Haha!

Just for fun, here is one of the other problems that were part of this homework (I solved it already):

Suppose a lecture hall is evacuated and (Schrodinger) cats are projected with speed \nu at the two doors leading out of the lecture hall in a double-slit experiment. Assume that in order for the interference fringes to be observed as the cats pile up against a distant wall the wavelength of each cat must be greater than 1 meter. Estimate the maximum speed of each cat. If the distance between the front of the lecture hall to the wall is 30 meters, how long will it take to carry out the experiment? Compare this time with the age of the universe, roughly 10^{10} years. Assume the cats each weight 1 kg.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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