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If a particle has a perfectly defined position wavefunction, what is its p wavefunc?

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MadMike1986
#1
Aug7-11, 09:04 PM
P: 23
1. The problem statement, all variables and given/known data

The position and momentum wavefunctions are fourier transform pairs. If a particle has a perfectly defined position wavefunction Psi(x) = delta(x - x0), then what is its momentum wavefunction? Is this function normalizable?

2. Relevant equations

Fourier transform relation (Can't figure out how to use the latex inputs to write this out)

3. The attempt at a solution

I think I have to take the fourier transform of a dirac delta "function"... I believe this is just a single frequency sine or cosine wave, but I only think this based on intuition from what the fourier transform means.

As far as if this function is normalizable, I'm not really sure what is meant by that. Does that mean you take the integral over all space (or in this case all momentums) of the wavefuntion squared and set it equal to 1?


Any help/advice on this would be much appreciated. Its been a while since I've studied quantum.

Thank you.
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nnnm4
#2
Aug7-11, 11:37 PM
P: 113
You're basically right. The fourier transform of the dirac delta will give a single momentum wave (an exponential with an imaginary argument), and the question as to whether it's normalizable is equivalent to asking whether it is square-integrable.
MadMike1986
#3
Aug8-11, 08:45 PM
P: 23
Quote Quote by nnnm4 View Post
You're basically right. The fourier transform of the dirac delta will give a single momentum wave (an exponential with an imaginary argument), and the question as to whether it's normalizable is equivalent to asking whether it is square-integrable.
Thanks for the help.

The integral of (sin(x))^2 dx from negative infinity to infinity does not converge, therefore it is not square integrable and not normalizable.

My intuition is telling me that this means if you know the position exactly, then you cant know what the momentum is at all. Is this correct?

Dick
#4
Aug8-11, 09:14 PM
Sci Advisor
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Thanks
P: 25,235
If a particle has a perfectly defined position wavefunction, what is its p wavefunc?

Quote Quote by MadMike1986 View Post
Thanks for the help.

The integral of (sin(x))^2 dx from negative infinity to infinity does not converge, therefore it is not square integrable and not normalizable.

My intuition is telling me that this means if you know the position exactly, then you cant know what the momentum is at all. Is this correct?
Your intuition is correct. The details you are showing are all wrong. The momentum wavefunction is a function of the momentum 'p'. Not the position 'x'. Why don't you actually try to find the momentum wavefunction corresponding to the position wavefunction delta(x-x0) and see if it confirms your intuition?
nnnm4
#5
Aug9-11, 01:09 AM
P: 113
Dick is of course correct, and sin^2(p) isn't even the relevant function here. The fourier integral is trivial and you should do it out.


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