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Probablities from wave function 
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#1
Dec507, 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. [tex] * \int \sin^2(kx) dx = \frac{x}{2}  \frac{1}{4k} \sin(2 k x) +C, and [/tex] 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 plugin 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? 


#2
Dec507, 05:37 PM

HW Helper
P: 1,979

sin^2(kx) = [1cos(2kx)]/2. Now can you integrate directly?



#3
Dec507, 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?



#4
Dec607, 02:20 AM

HW Helper
P: 1,979

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. 


#5
Feb1009, 08:26 AM

P: 1

What is mean probability is zero?



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