Normalising superposition of momentum eigenfunctions

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SUMMARY

The discussion centers on the normalization of the wave function Ψ(x,0)=A(iexp(ikx)+2exp(−ikx), where A is a constant. It is established that plane waves cannot be normalized in the traditional sense, as they do not satisfy the normalization condition ∫ψ⋆ψ=1. However, the concept of "box normalization" is introduced, suggesting that a wave packet formed from a superposition of plane waves can be normalized under certain conditions. The conversation highlights the confusion surrounding the normalization of momentum eigenfunctions and the implications of the probability coefficients Cn².

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  • Quantum Mechanics fundamentals
  • Understanding of wave functions and their properties
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  • Familiarity with momentum eigenfunctions and their representations
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alec_grunn
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Hi all, I asked for help with one part of this question here. But after thinking about another part of the question, I realized I didn't understand it as well as I'd thought.

Homework Statement



Ψ(x,0)=A(iexp(ikx)+2exp(−ikx)) is a wave function. A is a constant.

Can Ψ be normalised?

Homework Equations



Code:
 ##
{\langle}p{\rangle}
= \Big( \sum_{n=1} \hbar C[SUB]n[/SUB] ^2) ##

Where Cn 2 is the probability that the associated momentum will be observed.

The Attempt at a Solution



My initial thought was, plane waves can't be normalised, since that would violate the normalisation condition.
But the equation above (from textbook 'foundations of modern physics'), implies one of two options in my mind. Either:
1) The wavefunction can be normalised by ## A= \frac{1}{\sqrt{5}} ##
2) The allowed momentum values are not ## p= ± \hbar k##, but ## p = ± \frac{\hbar k}{5A2}##

Both of these seem to have their own problems. The first because I've read in other places that plane waves can not be normalised (unless you have some a Fourier series which gives you a finite integral). And the second because the momentum should not vary due to its coefficient.

Cheers,
Alec
 
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A plane wave cannot be normalized in the sense that ##\int_\infty\psi^\star\psi = 1## but you can "box normalize" a plane wave, and you can also normalize a wavepacket consisting of a superposition of plane waves under some situations.

In what way do you feel the textbook implies one of those two conditions?
It is unclear why the following comments are a problem - can you articulate the issue and how they apply to the examples?
Often, going through a QM issue in careful detail leads to a better understanding.
 
Hi Simon, thanks for the response. I've just realized the second option I gave "The allowed momentum values are not p=±ℏk, but ## p = ± \frac{\hbar k}{5A2}##" is not implied by anything above - please forget about it.

First off, I forgot to mention it's not in a potential well. So this is for all real values of x.
So my question is, doesn't this equation ⟨p⟩=∑ℏkCn^2 imply ∑Cn^2 = 1, since the textbook also says that each Cn2 is the probability of observing the corresponding momentum value? And therefore we can normalise any linear combination of momentum eigenfunctions, even Ψ=Aexp(-ikx). But that makes me uneasy because (a) it clearly doesn't meet the normalisation condition, and (b) I've read elsewhere that plane waves can't be normalised, for instance here.

Maybe it's got something to do with my definition of 'normalisation'. I just read this post which basically says normalisation is strictly defined as making your wavefunction satisfy ∫ψ⋆ψ=1 over all x values. If so, then the wavefunction I was working with can't be normalised. If it just means something like 'fixing the constant outside the brackets', then I can normalise this wavefunction.

Also, apologies for the poor formatting - I'm still coming to terms with using Latex.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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