# Expectation value <x>

## Homework Statement

It's an old assignment for exam, but the solution manual gives little help:

Describing a particle of mass m moving in one dimension (x) the wave function at time t=0 is:

$\Psi(x,t=0) = A \frac{1}{\sqrt{(x-x_0)^4 + l^4}}$

$x_0$ and $l$ are positive constants describing some length ($x_0, l>0$).

Normalize the wave equation and find the expectation value of x.

These integrals can be helpful (given with the assignment):
(all integrals are from -∞ to ∞, I'm just a tad new at texing, and you'll get it right)

$\int \frac{1}{x^2+l^2} dx = \frac{\pi}{l}$

$\int \frac{1}{sqrt{x^4+l^4}} dx = \frac{3.7081}{l}$

$\int \frac{1}{x^4+l^4} dx = \frac{\pi}{\sqrt{2}l^3}$

I am having trouble mathematically or principally calculate the expectation value.

## Homework Equations

Normalization demands $1 = < \Psi | \Psi >$

Exceptation value is $<x> = < \Psi | x \Psi >$

## The Attempt at a Solution

Normalizing the function was all right; since the function is symmetrical around $x_0$, we can choose the function to be symmetric around zero, and so:

$1 = |A|^2 \int \frac{1}{x^4+l^4} dx = |A|^2 \frac{\pi}{\sqrt{2}l^3}$

So then $A = \frac{\sqrt{2}\sqrt{l^3}}{\sqrt{\pi}}$ or $A = \sqrt{2} \sqrt{\frac{l^3}{\pi}}$ if you like..

For the expectation value I'm not sure why it's "obviously" $x_0$. I tried setting it up mathematically:

$<x> = |A|^2 \int \frac{x}{(x-x_0)^4+l^4} dx$

But I can't see how to get there (or really anywhere). My teacher suggests the same "substitution" as before (setting $x_0 = 0$) to get:

$<x> = |A|^2 \int \frac{x}{(x-x_0)^4+l^4} dx = |A|^2 \int \frac{x}{x^4+l^4} dx + x_0 |A|^2 \int \frac{1}{x^4+l^4} dx$

But I'm not following how like, what is the last part? And if I were to try and substitute like before, the variable would disappear, so I don't get his point.

I hop I TeXed everything correct. Welp pwease :-)

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Orodruin
Staff Emeritus
The wave function is symmetric around $x_0$, you cannot choose this to be zero if it is not. However, you can make the substitution $t = x-x_0$ and evaluate the integral in $t$ instead, which amounts to what you did.
For the expectation value, you can do the same substitution. The second term comes from the fact that not only is $t = x-x_0$, leaving you with the denominator which is symmetric around $t=0$, but also $x = t + x_0$, leaving you with a numerator that is not.
As for the "obviousness" of the expectation value being $x_0$: What is true for the probability density $|\psi(x)|^2$ when you study its symmetry properties around $x = x_0$?