Probablities from wave function

In summary, the conversation discusses the probability P that a particle in a box in the second excited state is between two specific points. To find this probability, an integral is evaluated using a formula for integrating sin^2(kx). The attempt at a solution involves a mistake in grouping the sine function of the integral outside of the difference from the integral formula, resulting in a probability of 2/3 instead of 0.
  • #1
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Homework Statement


[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]

Homework Equations


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

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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  • #2
sin^2(kx) = [1-cos(2kx)]/2. Now can you integrate directly?
 
  • #3
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
(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
What is mean probability is zero?
 
  • #6
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) ).
 

1. What is a wave function?

A wave function is a mathematical representation of the quantum state of a physical system. It describes the probability amplitude of finding a particle in a certain position or state.

2. How is probability calculated from a wave function?

In quantum mechanics, probability is calculated by taking the squared magnitude of the wave function. This gives the probability density of finding a particle in a particular state or location.

3. Can the wave function predict the exact outcome of a measurement?

No, the wave function only gives the probability of finding a particle in a particular state or location. The exact outcome of a measurement is determined by the collapse of the wave function when the measurement is made.

4. What is the uncertainty principle and how does it relate to the wave function?

The uncertainty principle states that it is impossible to know both the exact position and momentum of a particle at the same time. This relates to the wave function because the more precisely we measure one of these properties, the less certain we are about the other, as determined by the wave function.

5. Can the wave function be used to describe macroscopic objects?

No, the wave function is only applicable to describe the behavior of microscopic particles in the quantum realm. Macroscopic objects do not exhibit wave-like behavior and are governed by classical mechanics rather than quantum mechanics.

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