Simple expectation value calculation

[Syrus]

Homework Statement

This problem comes from the second edition of Griffiths's, Introduction to Quantum Mechanics.

Given the Gaussian Distribution: p(x) = Ae^c(x-a)2

find <x>, that is, the expectation (or mean) value of x.

Clearly, to do this you evaluate the following integral: ∫xp(x)dx on (-∞,∞). I've tried performing this integration analytically, and it seems to be more bothersome than a rectal exam. The solution should be x = a since, by definition, p is centered at x = a (this is evident graphically as well). Is a qualitative solution sufficient here? Another method (which obviously does not work in general) is to set p'(x) = 0 and solve for x since we know the Gaussian distrubtion contains only one extremum. Any advice on the best means to proceed?

Homework Equations

The Attempt at a Solution

 

[haruspex]

∫Axe^c(x-a)2.dx
subst x+a for x:
∫A(x+a)e^cx2.dx = ∫Axe^cx2.dx + a∫Ae^cx2.dx
= (1/2)∫Ae^cx2.dx² + a

 

[Syrus]

I tried this already. Wouldn't the differential be required to change also? That is, dx becomes d(x+a).

 

[Syrus]

EDIT* sorry, sorry, the differential is dx'/dx = d/dx[x+a], so dx' = dx.

 

[Syrus]

Now let's talk about <x²>, I seem to encounter an indeterminate form upon evaluating the integral at bounds +/- ∞ (see the integral of x²e^-cx2)

 

[haruspex]

Now let's talk about <x²>, I seem to encounter an indeterminate form upon evaluating the integral at bounds +/- ∞ (see the integral of x²e^-cx2)

Same substitution gets rid the -a. ∫x²e^-cx2.dx (with a suitable multiplier to get total prob of 1) is just var(X), a standard result. To work it out for yourself, use integration by parts.

 

[Syrus]

I obtain the integral in the attached photo. Obviously there is a problem evaluating it at +/- ∞ because of the second (subtracted) term, no?

 

[haruspex]

The attachment is illegible for me. Much too small. Did you try integration by parts?
$\int_{-\infty}^{+\infty} x^2e^{-cx^2}.dx = [x \int^x ye^{-cy^2}.dy]_{-\infty}^{+\infty} - \int_{-\infty}^{+\infty} \left(\int^x ye^{-cy^2}.dy\right) dx$

 

[Syrus]

I think you're misunderstanding me. Look here for the integral of interest (the one you're attempting to solve by parts): http://integral-table.com/ and you'll see the "problem" I'm encountering with evaluating the result at +/- infinity.

 

[haruspex]

I think you're misunderstanding me. Look here for the integral of interest (the one you're attempting to solve by parts): http://integral-table.com/ and you'll see the "problem" I'm encountering with evaluating the result at +/- infinity.

Number (70)? erf(-∞) = -1, erf(+∞) = +1. The other term vanishes at both extremes, leaving √(π/a)/(2a)

 

[Syrus]

Why does the other term vanish at both extremes? Isn't 0*infinity an indeterminate form?

 

[haruspex]

e^-x2 tends to zero much faster than x tends to infinity.

 

[Syrus]

There's an x multiplied by that too however, i.e. xe^cx2, so at each extreme it becomes +/- infinity * e^-infinity