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I work at a help/tutoring center at my university. Today a kid came in with this problem. I've only studied math and haven't drifted into physics, but he had this problem:

Let $P(x)=Ne^{-\frac{|x|}{a}}$. Then:

(a) Find $N$ such that $P(x)$ is properly normalized.

(b) Find $\langle x \rangle$ and $\langle x^2 \rangle $

Attempt at solution:

$1=\displaystyle\int_{-\infty}^{\infty}Ne^{-\frac{|x|}{a}}=2\int_{0}^{\infty}Ne^{-\frac{x}{a}}=2\int_{0}^{\infty}Ne^{-u}a\,du=2Na\int_{0}^\infty e^{-u}du=2Na.$

So $N=\frac{1}{2a}$

I'm not sure how to do part (b). The text, Robinett's Quantum Mechanics, was a bit beyond me. Could I get some guidance on how to solve this seems really interesting.

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migrated from Jun 20 '13 at 2:05

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up vote 1 down vote accepted

For part a) you should check that the symmetry of $P(x)$ with respect to $x$ gives

$$\int_{-\infty}^{\infty}Ne^{-|x|/a} dx=2\int_{0}^{\infty}Ne^{-x/a}dx$$

and then do a change of variable $\displaystyle u=\frac{x}{a}$

For part b) note that by definition $\displaystyle\langle f(x) \rangle=\int_{-\infty}^{\infty}f(x)Ne^{-|x|/a}dx$ and that both $x$, $x^2$ and $P(x)$ have defined symmetry with respect to $x$ and you can exploit that symmetry to know the value of some of them (in particular $\langle x \rangle$ without actually computing the integrals

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Is my $N$ now correct? – emka Jun 19 '13 at 10:20
@Mark when you do the change $u=x/a$ the integral is that of $e^{-u}$ which happens to be $1$ for $u=0\to \infty$. Otherwise $N$ is correct – Jorge Jun 19 '13 at 10:44
I'm not sure I follow your hint for part (b). So $<x>=\int\limits_{-\infty}^{\infty}xNe^{-|x|/a}\,dx=2\int\limits_{0}^{\infty}xNe‌​^{-x/a}\,dx$. After it is set up it looks like a pretty ugly integration by parts. – emka Jun 19 '13 at 19:24
Using integration by parts I got $a^2N$ and $-2a^3N$. – emka Jun 19 '13 at 19:46
Note that $P(x)$ is even on $x$ and $x$ is odd on $x$ so $P(x)x$ is odd on $x$, and the integral is on a symmetric interval with respect to $x$ so $\langle x \rangle = 0$ – Jorge Jun 19 '13 at 19:53

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