Calculating Expectation Values for x, x^2 in 1D Box

In summary, the conversation discusses the calculation of expectation values for a particle in a one-dimensional box. The method used is to form an integral and then use integration by parts to solve for <x> and <x^2>. However, there is confusion about the correct method to use for <x^2>, with one person suggesting to replace x with x^2 in the integral equation. Another person suggests checking for computational errors and following the solution found on another page.
  • #1
thenewbosco
187
0

Homework Statement


Calculate the expectation values of x, [tex]x^2[/tex] for a particle in a one dimensional box in state [tex]\Psi_n[/tex]


Homework Equations


[tex]\Psi_n = \sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})[/tex]


The Attempt at a Solution


i formed the integral
[tex]\int_{-\infty}^{+\infty}\Psi ^2 x dx[/tex] as the expectation value of x. (Psi squared simply because this psi is not complex)
this gives [tex]<x>=\frac{2}{a} \int_{-\infty}^{+\infty}x sin^2(\frac{n\pi x}{a})dx[/tex]. The problem is i do not know a way to simplify this integral, similarly i have the problem when there is an x^2 in the integral.
any help would be appreciated
 
Physics news on Phys.org
  • #2
First, the limits of integration are the ends of the box, not infinity...(since psi is zero outside the box anyway)

To solve, try integration by parts. (BTW I just did the exact same problem!)
 
  • #3
I am working on this same problem for x^2. Since x is just a multiplicative operator, shouldn't you just be able to put in x^2 in front of the sin term? It doesn't seem to give me the correct result.
Thanks
 
  • #4
gardman007 said:
I am working on this same problem for x^2. Since x is just a multiplicative operator, shouldn't you just be able to put in x^2 in front of the sin term? It doesn't seem to give me the correct result.
Thanks

Can I see your work? You may be making a computational error. You should be able to do that integral by integration by parts as well, just like the previous one. It just takes one more step.
 
  • #5
I found your solution on another page G01, I'm still not sure what I was doing wrong though. It seems like my method should work. Any ideas?
 
  • #6
gardman007 said:
I found your solution on another page G01, I'm still not sure what I was doing wrong though. It seems like my method should work. Any ideas?

What was your method? If you mean putting the x^2 in front of the sine term that should be fine. How did the solution you find solve it?
 
Last edited:
  • #7
I am just calculating the actual values, so I'm letting my calculator due the integration. Could you explain the process to get to the correct integration?
 
  • #8
So, for <x^2>, I should just be able to replace x with x^2 in front of the sin term in the last equation of the first post?
 

1. How do you calculate the expectation value for x in a 1D box?

The expectation value for x in a 1D box is calculated by taking the integral of x multiplied by the probability density function (PDF) over the entire length of the box. This can be represented by the equation:
⟨x⟩ = ∫ x |Ψ(x)|^2 dx where Ψ(x) is the wavefunction.

2. What is the significance of calculating the expectation value for x in a 1D box?

The expectation value for x in a 1D box represents the average position of a particle within the box. It provides important information about the behavior and characteristics of the system, such as the spread or localization of the particle.

3. How does the expectation value for x^2 differ from the expectation value for x?

The expectation value for x^2 in a 1D box is calculated in a similar way as the expectation value for x, except that x is squared in the integral instead of just being multiplied by x. This value represents the average of the squared position of a particle within the box, and is related to the spread or variance of the particle's position.

4. Can the expectation value for x, x^2 in a 1D box be negative?

No, the expectation value for x, x^2 in a 1D box cannot be negative. This is because the position and squared position of a particle are always positive values, and the probability density function also cannot be negative. Therefore, the integral will always result in a positive value.

5. How does the length of the 1D box affect the calculation of the expectation value for x, x^2?

The length of the 1D box does not affect the calculation of the expectation value for x, x^2. This is because the integral is taken over the entire length of the box, so the overall scale of the box does not change the result. However, the shape of the probability density function and the spread of the particle's position may be affected by the length of the box.

Similar threads

  • Advanced Physics Homework Help
Replies
30
Views
1K
  • Advanced Physics Homework Help
Replies
24
Views
809
  • Advanced Physics Homework Help
Replies
4
Views
918
Replies
16
Views
549
  • Advanced Physics Homework Help
Replies
8
Views
3K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
24
Views
2K
Back
Top