(Quantum Mechanics) Gaussian Distributions, Expected Values, and Sketches

[QuantumBunnii]

Homework Statement

Consider the gaussian distribution

ρ(x) = Aexp[(-λ^2)(x-a)^2] ,

where A, a, and λ are positive real constants.

(a) Find A such that the gaussian distribution function is normalized to 1.

(b) Find <x> (average; expected value) , <x^2>, and σ (standard deviation).

(c) Sketch the graph of ρ(x)

Homework Equations

Gaussian integrals (integrated from 0 to infinity):

∫x^(2n)exp[(-x^2)/a^2]dx = $\sqrt{\pi}$$\frac{(2n)!}{n!}$($\frac{a}{2}$)^(2n+1)

∫x^(2n+1)exp[(-x^2)/a^2]dx = $\frac{n!}{2}$a^(2n+2)

The Attempt at a Solution

(a) This part didn't give me any problems (I think), but I would like to make sure the general methodology is correct.
In order to normalize the distribution function to 1, you would merely set the integration over all space equal to 1:

∫Aexp[(-λ^2)(x-a)^2]dx = 1 (all space)

Since the only variable that differs from the given gaussian integral is the (x-a) term in the exponent, we can change the variable to, say, y = x-a and get dy = dx. Plugging everything in straight from the gaussian integral and multiplying everything by 2 (the given integrals are from 0 to inf., whereas this is from negative inf. to positive inf.), A is readily obtained as $\frac{2λ}{\sqrt{\pi}}$ .

(b) This is where I start to have trouble.

The average, <x> is given as follows:

∫xρ(x) dx = ∫xAexp[(-λ^2)(x-a)^2]dx (all space)

and this is as far as I can get. My problem is essentially one of mathematics: I don't know how to solve this integral. Just as in the previous problem, it is extremely close to the given gaussian integral, expect that-- in this case-- we can't employ a change of variables because we have the 'x' sitting in front of everything. The same will, of course, apply to <x^2>, and-- since σ = <x^2> - <x>^2-- this inhibits me from moving any further.

(c) is again a problem which arises from my lack of mathematical competence. I'm simply not sure how to sketch gaussian distribution functions. Could it be that the average (or, 'expected') value represents the peak of this function? (If so, I don't understand why-- as the average value is not necessarily the most probable value). And perhaps the standard deviation σ represents something akin to the full-width-at-half-maximum?
Or maybe I simply need to plug in a variety of points to get a general idea?

Thanks for any help~

P.S. Sorry about the formatting of the equations. I'm not sure how to make them look any better. :p

 

[Mute]

Homework Statement

Consider the gaussian distribution

ρ(x) = Aexp[(-λ^2)(x-a)^2] ,

where A, a, and λ are positive real constants.

(a) Find A such that the gaussian distribution function is normalized to 1.

(b) Find <x> (average; expected value) , <x^2>, and σ (standard deviation).

(c) Sketch the graph of ρ(x)

Homework Equations

Gaussian integrals (integrated from 0 to infinity):

∫x^(2n)exp[(-x^2)/a^2]dx = $\sqrt{\pi}$$\frac{(2n)!}{n!}$($\frac{a}{2}$)^(2n+1)

∫x^(2n+1)exp[(-x^2)/a^2]dx = $\frac{n!}{2}$a^(2n+2)

The Attempt at a Solution

(a) This part didn't give me any problems (I think), but I would like to make sure the general methodology is correct.
In order to normalize the distribution function to 1, you would merely set the integration over all space equal to 1:

∫Aexp[(-λ^2)(x-a)^2]dx = 1 (all space)

Since the only variable that differs from the given gaussian integral is the (x-a) term in the exponent, we can change the variable to, say, y = x-a and get dy = dx. Plugging everything in straight from the gaussian integral and multiplying everything by 2 (the given integrals are from 0 to inf., whereas this is from negative inf. to positive inf.), A is readily obtained as $\frac{2λ}{\sqrt{\pi}}$ .

[strike]Your post has some conflicting information here: you say you want "∫Aexp[(-λ^2)(x-a)^2]dx = 1 (all space)", which to me would imply $-\infty < x < \infty$, but then you say that x is from 0 to infinity? Could you clarify which is the correct case?

I would normally assume that for some reason in this problem "all space" is just 0 to infinity, but I would to double-check because if that is really the case you run into trouble when you change variables: if the integral over x is from 0 to infinity, then the integral over y = x-a is from -a to infinity. This is not a simple integral, and the resulting normalization constant would involve the error function (which is defined in terms of a Gaussian integral).

So, I would suggest you double-check your limits. If they are indeed from x=0 to infinity, then you will need to know the error function to express the normalization constant.[/strike]

Ok, re-reading your post it looks like your integral limits are from $x=-\infty$ to $\infty$, you're just using the symmetry to do the integral from 0 to $\infty$, correct? That's fine as long as you do your change of variables before you double the integral and change the limits to 0 to $\infty$.

(b) This is where I start to have trouble.

The average, <x> is given as follows:

∫xρ(x) dx = ∫xAexp[(-λ^2)(x-a)^2]dx (all space)

and this is as far as I can get. My problem is essentially one of mathematics: I don't know how to solve this integral. Just as in the previous problem, it is extremely close to the given gaussian integral, expect that-- in this case-- we can't employ a change of variables because we have the 'x' sitting in front of everything. The same will, of course, apply to <x^2>, and-- since σ = <x^2> - <x>^2-- this inhibits me from moving any further.

You can certainly employ a change of variables here. The factor of x in the integrand would just mean you have to replace it by (y+a), so you would end up splitting your integral into two terms.

(c) is again a problem which arises from my lack of mathematical competence. I'm simply not sure how to sketch gaussian distribution functions. Could it be that the average (or, 'expected') value represents the peak of this function? (If so, I don't understand why-- as the average value is not necessarily the most probable value). And perhaps the standard deviation σ represents something akin to the full-width-at-half-maximum?
Or maybe I simply need to plug in a variety of points to get a general idea?

The average value is not necessarily the most probable value, but that doesn't mean it can't be in some cases. You can compute both in this case to see if they match.

To learn how to do LaTeX formatting of equations, see this post.

 

[QuantumBunnii]

Your post has some conflicting information here: you say you want "∫Aexp[(-λ^2)(x-a)^2]dx = 1 (all space)", which to me would imply $-\infty < x < \infty$, but then you say that x is from 0 to infinity? Could you clarify which is the correct case?

Sure. The given guassian integrals are from 0 to inf., but the problem requires us to integrate from negative inf. to inf. (this is also why I said we would need to multiply by two).
Perhaps I should have simply listed the integrals from negative inf. to inf. in the outset, so that they agree. I guess I was really trying to ask if multiplying by a factor of two does in fact correct the difference in the lower limit from 0 to negative inf. in the two integrals.

You can certainly employ a change of variables here. The factor of x in the integrand would just mean you have to replace it by (y+a), so you would end up splitting your integral into two pieces.

You're right-- I don't know what I was thinking. :)
I intuitively dismissed the change of variable technique here and ended up trying to find another method for hours. Winter break was much too long x_x

Thanks for the help!