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Expectation value <x>

  • #1

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[4]{2}\sqrt{l^3}}{\sqrt{\pi}} ## or ## A = \sqrt[4]{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 :-)
 

Answers and Replies

  • #2
Orodruin
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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##?
 

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