# Simple expectation value calculation

1. Jan 15, 2013

### Syrus

1. The problem statement, all variables and given/known data

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

Given the Gaussian Distribution: p(x) = Aec(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?

2. Relevant equations

3. The attempt at a solution

2. Jan 15, 2013

### haruspex

∫Axec(x-a)2.dx
subst x+a for x:
∫A(x+a)ecx2.dx = ∫Axecx2.dx + a∫Aecx2.dx
= (1/2)∫Aecx2.dx2 + a

3. Jan 15, 2013

### Syrus

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

Last edited: Jan 15, 2013
4. Jan 15, 2013

### Syrus

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

Last edited: Jan 15, 2013
5. Jan 15, 2013

### 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)

6. Jan 15, 2013

### haruspex

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.

7. Jan 15, 2013

### Syrus

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

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8. Jan 15, 2013

### haruspex

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

9. Jan 15, 2013

### 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.

10. Jan 15, 2013

### haruspex

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

11. Jan 15, 2013

### Syrus

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

12. Jan 16, 2013

### haruspex

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

Last edited: Jan 16, 2013
13. Jan 16, 2013

### Syrus

There's an x multiplied by that too however, i.e. xecx2, so at each extreme it becomes +/- infinity * e^-infinity

Last edited: Jan 16, 2013