Physically valid wave funtions

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In summary, the conversation discusses the validity of the function sinx/(x^2) as a wave function in the context of Born's conditions. The function is found to be discontinuous at x=0 and not square-integrable, making it not a valid wave function in the Lebesgue square integrable approach. However, in the Rigged Hilbert space approach, the function can be considered a valid wave function as it can be extended to a linear functional defined on physically realizable states.
  • #1
kini.Amith
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Is the function sinx/(x^2) a physically valid wave function? since Born's conditions state that the function should be continuous and i think this function is discontinuous at x=0.
 
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  • #2
It's not a representant of a pure quantum state, because its square is not integrable, i.e., the integral
[tex]\int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2[/tex]
doesn't exist due to the singularity at [itex]x=0[/itex].

That it is not continuous or even analytic is not that important, because you deal with functions in the sense of the Hilbert space of square integrable functions, and it doesn't matter, if they are singular on a set of Lebesgue measure 0.
 
  • #3
kini.Amith said:
Is the function sinx/(x^2) a physically valid wave function? since Born's conditions state that the function should be continuous and i think this function is discontinuous at x=0.

I don't think this is "a physically valid wave function", not because it's discontinuous, but because it's not square-integrable.
 
  • #4
SO is continuity of the wave function not a necessary condition for it to represent a physical system?
 
  • #5
There are a couple of ways of looking at wave functions.

You have the Lebesgue square integrable view - valid wave functions are those that are square integrable in the Lebesque sense - they also form a Hilbert space. This is Von Neumann's approach you will find in his classic Mathematical Foundations of QM text. The wave function you gave is not valid in that approach.

Then there is the Rigged Hilbert space approach. The physically realisable states are considered to be a subset of square integrable functions so as to have nice mathematical properties such as being continuously differentiable, or even simply finite dimensional. Of course your function is not a physically realisable state in that approach either. However for mathematical convenience that space is extended to be the linear functionals defined on the physically realizable states. Note there is slightly more to it in that it must be such that you can define a norm on the resultant space - but these are fine points you can consult tomes on it eg:
http://physics.lamar.edu/rafa/webdis.pdf

In that sense it is a valid wavefunction.

Thanks
Bill
 
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FAQ: Physically valid wave funtions

1. What is a physically valid wave function?

A physically valid wave function is a mathematical description of a quantum system that satisfies certain criteria, such as being continuous, single-valued, and square-integrable. It also must accurately represent the probability of finding a particle in a particular location.

2. How do we determine if a wave function is physically valid?

To determine if a wave function is physically valid, we use mathematical techniques such as normalization and integration to check if it satisfies the necessary criteria. We also compare it to known solutions for similar systems to ensure its accuracy.

3. Why is it important for a wave function to be physically valid?

A physically valid wave function is essential for accurately describing the behavior of quantum systems. Without it, we cannot make reliable predictions about the behavior of particles and their interactions.

4. Can a wave function be both physically valid and not physically realistic?

Yes, it is possible for a wave function to meet the necessary criteria for being physically valid but still not accurately represent a real physical system. This could be due to simplifications or assumptions made in the mathematical model.

5. What happens if a wave function is not physically valid?

If a wave function is not physically valid, it means that it does not accurately represent a quantum system. This can lead to incorrect predictions and a lack of understanding of the behavior of particles at the quantum level.

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