Triangle wavefunction - position uncertainty and probability amplitude

Click For Summary
The discussion revolves around calculating the normalization constant N for a triangle wavefunction and determining the expected values <x> and <x^2> to find the position uncertainty Δx. The user has encountered two different values for N based on their approach and seeks clarification on which is correct. There is an emphasis on the importance of using the correct probability density expression, particularly for piecewise-defined functions. Participants advise that integrating piecewise functions is valid and suggest starting with the expression for <x> using the wavefunction. The conversation highlights the need for a clear understanding of probability density in quantum mechanics.
Dixanadu
Messages
250
Reaction score
2

Homework Statement


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

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


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

So...which is correct? :S

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

Thanks a lot guys!
 
Last edited by a moderator:
Physics news on Phys.org
The square of a function that is defined in parts, is still a function defined in parts, eg.
From
f(x) = \left\{\begin{array}(g(x); x&lt;0 \\ h(x); x\geq0\end{array}\right.
It follows that
(f(x))^2 = \left\{\begin{array}((g(x))^2 &amp;; x&lt;0 \\ (h(x))^2&amp;; x\geq0\end{array}\right.,
that is to say
(f(x))^2 \neq (g(x))^2 + (h(x))^2

This is a notational problem: your |\Psi|^2 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).
 
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?
 
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 \int_{-1}^1f(x)dx = \int_{-1}^0 1 dx + \int_0^1 2 dx = 1 + 2 = 3.

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

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
4
Views
3K
  • · Replies 9 ·
Replies
9
Views
8K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
1K