Normalizing the Momentum Eigenfunctions

In summary, the conversation discusses the momentum eigenfunctions and how to normalize them. It is mentioned that the wave functions are non-normalizable unless C is equal to zero. The conversation then moves on to a formula for finding the integral of a complex exponential, which is shown to equal 2 pi times Dirac's delta function. There is confusion about how this formula works and why it is valid. The conversation ends with a question about the placement of 2 pi on the right hand side of the formula.
  • #1
Domnu
178
0
I know that the momentum eigenfunctions are of the form [tex]\phi = Ce^{ikx}[/tex], but how would we normalize them? We just get

[tex]\int_{-\infty}^{\infty} C^2 dx = 1[/tex]

which means that [tex]C[/tex] is infintesimally small...
 
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  • #2
Indeed, so these wave-functions are non-normalizable. You either choose C = 0 and you have a trivial wave function, or you have a diverging integral.
 
  • #3
Well, see the thing is that there's this formula in my book which I have absolutely no idea as to how they got... here it is... could someone show a proof of this?

[tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = 2\pi \cdot \delta(k' - k)[/tex]

It makes absolutely no sense to me... if we have that [tex]k' \neq k[/tex], then

[tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} dx = \int_{-\infty}^{\infty} \cos [(k'-k) x] dx[/tex]

since sin is an odd function. But even this is a large step to claim... it's like saying that 1 - 1 + 1 - 1 + 1 - 1 + ... = 0. But can we just say that cos is a sideways shift of sin, so we can just say that it evaluates to zero as well? Or do the limits mean that we take a value that approaches infinity and it always equals zero? I think I understand why the k'=k case works, however...delta(0) = infinity, which is also the lhs. Lastly, why does the 2 pi have to be on the RHS?
 
Last edited:

What is normalizing the momentum eigenfunctions?

Normalizing the momentum eigenfunctions means scaling the functions so that their squared magnitudes integrate to 1 over all space. This is important in quantum mechanics because it ensures that the probability of finding a particle in any region of space is 1.

Why is normalizing the momentum eigenfunctions important?

Normalizing the momentum eigenfunctions is important because it allows us to interpret the wave function as a probability distribution. It also ensures that our mathematical calculations are physically meaningful.

How do you normalize momentum eigenfunctions?

To normalize momentum eigenfunctions, we use the normalization condition ∫|Ψ(p)|²dp = 1, where Ψ(p) is the momentum wave function. This involves finding the appropriate scaling factor to multiply the wave function by.

What is the relationship between normalization and uncertainty principle?

The normalization of the momentum eigenfunctions is related to the uncertainty principle because it is a necessary condition for the principle to hold. If the wave function is not normalized, the uncertainty principle cannot be satisfied, and our calculations will not accurately reflect the physical system.

Can the momentum eigenfunctions be normalized to values other than 1?

Yes, the momentum eigenfunctions can be normalized to values other than 1. However, normalizing them to 1 is the most common and physically meaningful choice. Normalizing to other values may have mathematical or physical significance in specific cases, but it is not the standard approach.

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