Physical difference between various wave functions

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SUMMARY

The discussion confirms that the wave functions \(\Psi(x,0) = 5e^{-ax^2}\), \(\Psi(x,0) = \frac{1+i}{\sqrt{3}}e^{-ax^2}\), and \(\Psi(x,0) = e^{i\pi/7}e^{-ax^2}\) differ only in their constant factors, which do not affect their physical significance when considering the probability density \(|\Psi(x,0)|^2\). All three wave functions can be normalized to yield identical functions, indicating that their physical implications are equivalent. The decay rates of these wave functions as \(x\) moves away from zero depend on the values of \(a\) in the exponents, which are influenced by the constant factors.

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jror
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Homework Statement


Is there a physical difference between the following wave functions? If yes, why? If no, why not?

\Psi(x,0) =5e^{-ax^2}
\Psi(x,0) =\frac{1+i}{\sqrt{3}}e^{-ax^2}
\Psi(x,0) =e^{i\pi/7}e^{-ax^2}

Homework Equations


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The Attempt at a Solution


They only differ in the constant factors, and since we are speaking about physical significance we are interested in |\Psi(x,0)|^2, which contain no imaginary parts. If I assume that these are realizable states they would have to be normalizable, in which case the a's in the exponents would have set values depending on the constant factor in front? If that is the case then yes these wave functions differ in how quickly they decay as we move away from x=0. I am just not completely convinced by my own argument, and if this is what is meant in the exercise. Any input would be very welcome.
 
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jror said:
If I assume that these are realizable states they would have to be normalizable, in which case the a's in the exponents would have set values depending on the constant factor in front?
'Normalisable' only means that they can be converted to a ket of norm (magnitude) 1 by multiplying by a constant. It doesn't mean they already have a norm of 1. That would be 'normalised' not 'normalisable'.

If we normalise those three wave functions we will end up with three identical functions.
 

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