Easy integration by parts but getting wrong answer. Help

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Homework Help Overview

The original poster is investigating the expectation value of x for a given wavefunction, specifically \(\Psi (x) = \sqrt{\frac{2}{L}} \sin(\frac{\pi x}{L})\) over the interval \(0 < x < L\). They are attempting to evaluate the integral \(f = \int_0^L \! \Psi^* x \Psi \, dx\) and have encountered difficulties using integration by parts.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to solve the integral using integration by parts but arrives at an incorrect result of zero. Some participants question the method of finding the antiderivative and suggest checking the power reduction of the sine squared term.

Discussion Status

Participants are actively engaging with the original poster's approach, offering guidance on the integration process and pointing out potential errors in the setup. There is a focus on clarifying the correct application of integration techniques without reaching a consensus on the final solution.

Contextual Notes

The discussion includes references to the original poster's use of Mathematica for verification, indicating a reliance on computational tools alongside manual calculations. There is also a light-hearted exchange about the curriculum in quantum mechanics, suggesting a shared context among participants.

Bacat
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Homework Statement



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

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

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


The Attempt at a Solution



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

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

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

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

Then [tex]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)[/tex]

[tex]v = \frac{L}{2} - Sin^2(\pi) = \frac{L}{2}[/tex]

Then [tex]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.[/tex]

Where have I made an error?
 
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[itex]v[/itex] is supposed to be the antiderivative of [itex]dv[/itex], not the definite integral from 0 to L of [itex]dv[/itex]!
 
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 (Schrödinger) cats are projected with speed [tex]\nu[/tex] 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 [tex]10^{10}[/tex] years. Assume the cats each weight 1 kg.
 

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