Wavefunction of a particle

  • Thread starter einai
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  • #1
einai
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Quantum question again....

What's the wave function in coordinate space Ψx0(x') of a particle (in 1-D) located at a certain position x0? What about the wave function Φx0(p') in momentum space? Now, consider the totality of these wave functions for different values of x0. Do they form an orthonormal set?

The only thing I know is that if I know Ψx0(x'), I can Fourier transform it to Φx0(p')? But what's Ψx0(x')?

Thanks in advance!
 

Answers and Replies

  • #2
HallsofIvy
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The wave function of a particle AT a specific x0 is the Dirac delta function δ(x-x0). Of course, the wave function in momentum space is the fourier transform: the constant function. If you know the position exactly, then you have no information at all about the momentum.
 
  • #3
einai
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Originally posted by HallsofIvy
The wave function of a particle AT a specific x0 is the Dirac delta function δ(x-x0). Of course, the wave function in momentum space is the fourier transform: the constant function. If you know the position exactly, then you have no information at all about the momentum.

Thank you! That makes a lot of sense. :smile: However, I'm not sure whether I understand this part of the question -

Now, consider the totality of these wave functions for different values of x0. Do they form an orthonormal set?

Does it mean whether all the delta functions at different x0 are orthogonal or not?
 
  • #4
HallsofIvy
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What does "orthogonal" mean for these functions?

(The delta function is not a true function. It is a "generalized function" or "distribution". But, the same concepts apply.)
 
  • #5
einai
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Originally posted by HallsofIvy
What does "orthogonal" mean for these functions?

(The delta function is not a true function. It is a "generalized function" or "distribution". But, the same concepts apply.)

I think it means if they're the same function, the product should be integrated to one, otherwise it's zero?

Hm.....I multiplied 2 wavefunction and integrate them. It gave me another delta function.
 
  • #6
arcnets
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einai,
I think the trick is that you use substitution.
You got
&int &delta (x-x0) &delta (x-x1) dx.
Now, let y = x-x0.
Then, dx = dy, and
&delta (x-x0) = &delta (y), and
&delta (x-x1) = &delta (y+x0-x1).
So,
&int &delta (x-x0) &delta (x-x1) dx
=
&int &delta (y) &delta(y+x0-x1) dy
=
&delta (x0-x1).
You probably already got that, and it's the orthogonality relation that you want.
 
Last edited:
  • #7
arcnets
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How can I make these math symbols show?
 
  • #8
einai
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Originally posted by arcnets
How can I make these math symbols show?


You need to put ; after &delta :D.

And thanks for answering my question. I did get the same thing, but I wasn't sure whether that implies orthonormality. Now I know, since I got the solution from the prof. It does imply orthonormality, and I got it right! :)
 
  • #9
arcnets
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You need to put ; after &delta :D.

Just testing...
∫ δ (x-x0) δ (x-x1) dx = ?
Now, let y = x-x0.
Then, dx = dy, and
δ (x-x0) = δ (y), and
δ (x-x1) = δ (y+x0-x1).
So, ∫ δ (x-x0) δ (x-x1) dx
= ∫ δ (y) δ (y+x0-x1) dy
= δ (x0-x1).

Fine! Learned something!
 
Last edited:
  • #10
einai
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Originally posted by arcnets
Just testing...
∫ δ (x-x0) δ (x-x1) dx = ?
Now, let y = x-x0.
Then, dx = dy, and
δ (x-x0) = δ (y), and
δ (x-x1) = δ (y+x0-x1).
So, ∫ δ (x-x0) δ (x-x1) dx
= ∫ δ (y) δ (y+x0-x1) dy
= δ (x0-x1).

Fine! Learned something!

Looks much better :)!

I didn't use substitution. I just treated
∫δ (x-x0) δ (x-x1) dx
as ∫f(x) δ(x-x1) dx

so when I integrate it, it gives f(x) -> f(x1) = δ (x0-x1).

I'm not sure whether this is a correct method, although it does give me the answer :D.
 

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