Do Wave Functions in Different Positions Form an Orthonormal Set?

[einai]

Quantum question again...

What's the wave function in coordinate space Ψ_x0(x') of a particle (in 1-D) located at a certain position x₀? What about the wave function Φ_x0(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 Ψ_x0(x'), I can Fourier transform it to Φ_x0(p')? But what's Ψ_x0(x')?

Thanks in advance!

 

[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.

 

[einai]

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. 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?

 

[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.)

 

[einai]

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.

 

[arcnets]

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.

 

[arcnets]

How can I make these math symbols show?

 

[einai]

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! :)

 

[arcnets]

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!

 

[einai]

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.