# Probability Density

1. Oct 3, 2014

### terp.asessed

1. The problem statement, all variables and given/known data
There is an equal probability density for finding a particle anywhere in the box. Assume that the box is of length L.
What would the root-mean-square fluctation in position be?

2. Relevant equations
root-mean-square fluctation2 = <x2> - <x>2

3. The attempt at a solution
since there is an equal probability density anywhere in the box, I assumed normalization:
1 = integral (x=0 to L) p(x)dx = integral (x=0 to L) C dx
1 = C integral (x=0 to L) dx = CL
C = 1/L
probability density = p(x) = 1/L

So....from here, I got:

<x> = integral (x=0 to L) p(x) x p(x) dx = 1/L2 integral (x=0 to L) x dx = 1/2
<x2> = integral (x=0 to L) p(x) x2 p(x) dx = 1/L2 integral (x=0 to L) x2 dx = 1/2 = L/3

therefore:
root-mean-square fluctation2 = <x2> - <x>2 = L/3 - 1/2...which does not make sense at ALL--could someone point out the error I've made? I have no idea where I made a mistake because I believe the solution to be something like of one value, like 3/12 or something, not what I got....

2. Oct 3, 2014

### Orodruin

Staff Emeritus
It is $\langle x\rangle^2$, not $\langle x\rangle$ ... Also, either your normalisation is wrong or your expectation values are. You have to decide whether p(x) is the probability density or the wave-function. You have used it as the probability density when normalising but as a wave function when computing the expectation values. This results in very strange units on your resulting expectation values.

3. Oct 3, 2014

### terp.asessed

root-mean-square fluctation2 = <x2> - <x>2 = L/3 - 1/4.
And, honestly, I am using p(x) as probability density--hence normalization...teacher hinted that the problem is NOT a wave...just a matter of probability density and expectation values....

4. Oct 3, 2014

### Orodruin

Staff Emeritus
Then your computations of the expectation values are wrong as you are including $p(x)^2$. The expectation value of $f(x)$ should be
$$\langle f(x) \rangle = \int_0^L f(x) p(x) dx.$$

5. Oct 3, 2014

### terp.asessed

Ok, wait--what is exactly f(x)? if p(x) is a probability density?

6. Oct 3, 2014

### Orodruin

Staff Emeritus
$f(x)$ is a function of $x$ that you want to know the expectation value of. In your case $f(x) = x$ and $x^2$, respectively.

7. Oct 3, 2014

### terp.asessed

wait, so, to make a loooooong story short:

<x> = integral (x=0 to L) x p(x) dx = 1/L integral (x=0 to L) x dx = L/2
<x2> = integral (x=0 to L) x2 p(x) dx = 1/L integral (x=0 to L) x2 dx = L2/3

Also, I am sorry to keep asking, but shouldn't I use normalization in this case? It seems I got wrong p(x) value?

8. Oct 3, 2014

### Orodruin

Staff Emeritus
Yes, as you can see, those expressions also have the correct dimensions and the expectation for x is L/2, which is reasonable. Yes, you should use normalisation, the normalisation is p(x) = 1/L. The problem was that you squared p(x) when computing the expectation values in your first post.

9. Oct 3, 2014

### terp.asessed

..ok, so for <x2> = integral (x=0 to L) x2 p(x) dx = 1/L integral (x=0 to L) x2 dx = L2/3 ?

10. Oct 3, 2014

### Orodruin

Staff Emeritus
Yes, so in the end, what do you get for <x^2> - <x>^2?

11. Oct 3, 2014

### terp.asessed

(L2/3 - L/2)1/2 for root-mean square fluctation? Does this solution make any sense for the problem? I am sorry but I thought I wouldn't get any squared values in the solution.

12. Oct 3, 2014

### Orodruin

Staff Emeritus
You are now missing the square of the expectation value of x (i.e., <x>^2). You can see that your expression does not make sense since you are trying to subtract something with units length from something with units length squared. Once you have this sorted out, you can always pull out L^2 as an L outside of the square root.

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