TISE solutions should be combinations-of-eigenstates, why this is not?

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SUMMARY

The discussion centers on the Time-Independent Schrödinger Equation (TISE) and its properties regarding linear combinations of solutions. It is established that while linear combinations of solutions are valid for certain cases, such as when dealing with degenerate eigenvalues, this does not hold for free particles represented by the wave function exp[-ikx]. The participants clarify that the TISE encompasses multiple equations corresponding to different eigenvalues, which restricts the formation of new solutions from linear combinations unless degeneracy is present.

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JackeTheDog132
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TL;DR
Why a wave pocket is not a solution of the TISE?
I would really appreciate some help with a question I have aboute the TISE (Sch. tipe indipendent equation). This is a linear equation and linear combination of the solution should be solution too. The problem is that for the free particle, which solution can be written like exp[-ikx], a linear combination using gaussian coefficient is not anymore a solution (we should get a wave pocket this way). Of course taking a combination considering the temporal dipendence give a solution to the TDSE. My question is why that does not appen in the TISE case.
 
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Can you express your question more mathematically? I'm not sure what you are asking.
 
PeroK said:
Can you express your question more mathematically? I'm not sure what you are asking.
I agree. It also sounds like you are trying to model a traveling wave with a time-independent model, which of course will not work.
 
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JackeTheDog132 said:
This is a linear equation and linear combination of the solution should be solution too.
Careful. What you are calling the "TISE" is not a single equation. It is many different equations, one for each different eigenvalue. So the only case where you can form a linear combination of solutions to get another solution is degeneracy, i.e., there are multiple eigenvectors with the same eigenvalue.
 

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