Triangle wavefunction - position uncertainty and probability amplitude

  • Thread starter Dixanadu
  • Start date
  • #1
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Homework Statement


Hey guys. Basically I have a wavefunction that looks like this
http://imageshack.com/a/img843/3691/22r3.jpg [Broken]

I have to find:
(a) The normalization constant N, of course by normalizing it
(b) Find <x> and <x^2> and use this to find Δx.

Homework Equations


I'm just gona insert images, as i'm writing it all with my tablet:
http://imageshack.com/a/img33/5705/czbs.jpg [Broken]


The Attempt at a Solution


Okay, so I've found the normalization constant, N, but I've got 2 different answers depending on the way in which I solve the problem, and I want to know which is correct (and why, if possible). Please view the image below:
http://imageshack.com/a/img11/7232/0hmb.jpg [Broken]

So...which is correct? :S

I need to know the explicit expression for the probability density to proceed further.

Thanks a lot guys!
 
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Answers and Replies

  • #2
Päällikkö
Homework Helper
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The square of a function that is defined in parts, is still a function defined in parts, eg.
From
[itex]f(x) = \left\{\begin{array}(g(x); x<0 \\ h(x); x\geq0\end{array}\right.[/itex]
It follows that
[itex](f(x))^2 = \left\{\begin{array}((g(x))^2 &; x<0 \\ (h(x))^2&; x\geq0\end{array}\right.[/itex],
that is to say
[itex](f(x))^2 \neq (g(x))^2 + (h(x))^2[/itex]

This is a notational problem: your [itex]|\Psi|^2[/itex] is incorrect, although you do manage to integrate it correctly in parts. Your first approach is correct as it is the integral of the square that you want to have as 1 (probability of the particle being anywhere is 1).
 
  • #3
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Great, thanks a lot dude! But my next question is this - how do i go about finding <x> ? i mean I need a single expression for psi mod squared, right?
 
  • #4
Päällikkö
Homework Helper
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I'm not exactly sure what you mean by a single expression. Could you elaborate?

There should be no problem integrating over functions that are piece-wise defined.
For example if
f(x) = 1 for x < 0 and f(x) = 2 for x > 0, the integral [itex]\int_{-1}^1f(x)dx = \int_{-1}^0 1 dx + \int_0^1 2 dx = 1 + 2 = 3[/itex].

Start by writing out the function [itex]\Psi^* x \Psi[/itex]. It will be a piece-wise defined function. Then proceed to integrate it, and you will get <x> by definition.
 

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