Finding fourier transfrom of the following wavefunction

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The wavefunction Psi(x,0) = E^(ik0x) is considered a valid representation for a free particle, as any function can be expressed as a wavepacket. To find the probability density P(x,0), the square of the modulus of the wavefunction is used, yielding a uniform probability distribution of 1/a within the specified range. The challenge arises in calculating P(k,0) through the Fourier transform, where a delta function is expected to appear, indicating a precise momentum k0. The uncertainty principle suggests that if k0 is definite, the particle's position is highly uncertain, leading to implications for the wavefunction's normalizability. Ultimately, the Fourier transform reveals that while position probabilities are uniform, momentum is precisely defined, aligning with the principles of quantum mechanics.
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Let Psi(x,0)=E^(ik0x) when x=(-a/2,a/2) and zero elsewhere.

Can this be a wavefunction of a free particle. I believe it is so because every function of x can be expressed as a wavepacket. Is this correct?

If I want to calculate P(x,0), probability to find the particle between x, x+dx it's just the square of the modulus. But what about P(k,0)? I'm having trouble calculating it's Fourier transform, I think that the delta function must show somewhere but I don't know how.

k seems to be certain k=k0 , right? However, P(x,0)=1/a is the same everywhere.
 
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Feynmanfan said:
Let Psi(x,0)=E^(ik0x) when x=(-a/2,a/2) and zero elsewhere.

Can this be a wavefunction of a free particle. I believe it is so because every function of x can be expressed as a wavepacket. Is this correct?

If I want to calculate P(x,0), probability to find the particle between x, x+dx it's just the square of the modulus. But what about P(k,0)? I'm having trouble calculating it's Fourier transform, I think that the delta function must show somewhere but I don't know how.

k seems to be certain k=k0 , right? However, P(x,0)=1/a is the same everywhere.

If k_0 is one definite value, then the particle has one precise momentum. What does the uncertainty principle say about the position of such a particle?
 
Is the wavefunction normalizable...?It is a generalized eigenfunction of the Hamiltonian...?

Daniel.
 
Yes it is normalizable. But it's not a generalized function of the Hamiltonian, is it? It's a particular case where k=ko.

I'm asked to draw P(x,0) and P(k,0) and find out delta(x) and delta(k) and justify it using Heisenberg's uncertainty principle.

By doing Psi's Fourier transform I get a complicated function and I don't know if that's the way I can justify the following: we know nothing about the position (cause all probabilities are the same) but k is certain.
 
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