Probability with a normalized wave function

AI Thread Summary
The discussion revolves around a normalized wave function, psi = sqrt(2/a) * sin^2(pi*x/a), and the challenge of calculating the probability for specific intervals. The integral for probability, P(a,b) = int(psi^2)dx, is causing confusion, particularly regarding normalization, as the wave function appears to diverge outside the defined limits. Participants suggest that the wave function may not be normalized correctly, indicating that it must equal one when integrated over its range. There are recommendations to use substitutions to simplify the integral and to carefully verify the wave function provided in the assignment. The conversation highlights the importance of understanding normalization in quantum mechanics.
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Homework Statement


So I have this normalized wave function

psi=sqrt(2/a) * sin^2(pi*x/a)

with limits of 0 and a.

I'm supposed to find the probability for a bunch of points if this form:

p(x=0.00a,x=0.002a)


Homework Equations



P(a,b)=int(psi^2)dx

The Attempt at a Solution



So, partially solving the integral, I'm stuck here:

(2/a) * [ (x/2) - (a/4pi) * sin(2*pi*x/a) ] between 0 and a

My first question. If I continue to solve the definate integral P=1. This is appearently not the correct answer. I know my calculus is right and I've verified it in wolfram alpha and octave.

Once I figure out how to proceed I need to plug in these values which have a in them, not really sure how that is going to work out either.

Any suggestions?
 
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$$P(\alpha<x<\beta)=\int_\alpha ^\beta \psi^\star(x)\psi(x)\; dx$$

You have been given ##\psi(x)= \sqrt{\frac{2}{a}}\sin^2(\pi x/a)##

- there has to be more to the function than that because this is not normalized - there must be part where it says ##\psi = 0## for some values of x.

You need to be able to find probabilities of form:
$$P(0<x<ra)=\frac{2}{a}\int_0 ^{ra} \sin^4(\pi x/a)\; dx\; :\; 0<r\leq 1$$

(eg. in your example, post #1, r=0.002)

this will be easier with a substitution like: ##\theta=\pi x/a##

$$P(0<x<ra)=P(0<\theta <\pi r)=\cdots$$

... and you can look up the indefinite integral for sin4(x)
 
Ok, I'm pretty new to this stuff but if by normalized you mean is equal to one than it is normalized.
If that's not what you mean then maybe I'm misunderstanding the meaning of the word normalized.

I've already taken it to my professor once, because I thought there was a mistake. He insists there is not.

Here is a link to the actual assignment
https://dl.dropboxusercontent.com/u/22548767/CHEC%20352%20Exam%20II%20Assignment.pdf BTW: is that Latex you are using? can you just type that into a message or does it have to be encapsulated by something?
 
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Ok, I'm pretty new to this stuff but if by normalized you mean is equal to one than it is normalized. If that's not what you mean then maybe I'm misunderstanding the meaning of the word normalized.
By "normalized" I mean that:
$$\int_{-\infty}^\infty \psi^\star\psi\; dx = 1$$

... the wavefunction given in post #1 is:
psi=sqrt(2/a) * sin^2(pi*x/a)
... that wavefunction is periodic and positive everywhere - so the integral is going to diverge. If we restrict the range to a single wavelength, [0,a], assuming zero elsewhere, then the integral does not come to 1 (check). Therefore, it is not normalized. Therefore there is missing information =)

Here is a link to the actual assignment
Excellent <reads>

The wavefunction given in your assignment sheet has some important differences from the one in post #1 - it's much simpler for a start. Please go over it more carefully.

Anyway - even with the differences - the same advise will hold as of post #2 but with the actual wavefunction this time.

BTW: is that Latex you are using? can you just type that into a message or does it have to be encapsulated by something?
Well spotted - yes it is ##\LaTeX##
... below my post, to the right, there is a button marked "quote" - click on it and you will see how I did it ;)
 
got it.
thanks for your help.
 
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