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Probablities from wave function

by Badger
Tags: function, probablities, wave
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Dec5-07, 04:52 PM
P: 29
1. The problem statement, all variables and given/known data
[tex]P = \int_a^b \, \left| \psi(x) \right|^2 \, dx[/tex]

If the particle in the box is in the second excited state (i.e., n=3), what is the probability P that it is between x=L/3 and x=L? To find this probability, you will need to evaluate the integral:

[tex]\int_{L/3}^L \left(\sqrt{\frac{2}{L}} \sin \left( \frac{n \pi x}{L} \right) \right)^2 \, dx = \frac{2}{L} \int_{L/3}^L \sin^2 \left( \frac{n \pi x}{L} \right) \, dx .[/tex]

2. Relevant equations
This one I think.
* \int \sin^2(kx) dx = \frac{x}{2} - \frac{1}{4k} \sin(2 k x) +C, and

3. The attempt at a solution
Set k = n(pi)/L or np/L

[2/L][(x/2 - 1/4k)sin(2kx)]|L and L/3

put k back in as it is.

[2/L]{[L/2 - L/4np]sin(2*np/L*L)]-[L/3*2 - L/4np]sin(2 * np/L * L/3)]}

cancel out the L's and plug-in n = 3

2{[(1/2 - 1/12p)sin(6p)] - [(1/6) - (1/12p)]sin(2p)]}

well the answer isn't 0. but sin6p and sin2p both equal 0
2[0 - 0] = 0

Where did I go wrong?
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Shooting Star
Dec5-07, 05:37 PM
HW Helper
P: 1,982
sin^2(kx) = [1-cos(2kx)]/2. Now can you integrate directly?
Dec5-07, 07:10 PM
P: 29
doesn't seem like it... integrating the cos(2kx) part just gives me a sin function again and again I get an integer times pi within leading to more 0. right?

Shooting Star
Dec6-07, 02:20 AM
HW Helper
P: 1,982
Probablities from wave function

(I had merely showed you how to evaluate the integral sin^2(kx), for which you have already written the formula.)

Anyway, you have done this mistake:
2{[(1/2 - 1/12p)sin(6p)] - [(1/6) - (1/12p)]sin(2p)]} = 2{(1/2) - (1/6)} = 2/3.
Feb10-09, 08:26 AM
P: 1
What is mean probability is zero?
Feb10-09, 09:22 AM
Sci Advisor
PF Gold
jambaugh's Avatar
P: 1,776
It looks like you mistakenly are grouping the sine function of your integral outside the difference from your integral formula: (u-v)sin(w) should be (u - v sin(w) ).

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