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

[JackeTheDog132]

TL;DR Summary

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 coefficent 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.

 

[PeroK]

Can you express your question more mathematically? I'm not sure what you are asking.

 

[Vanadium 50]

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.

 

[PeterDonis]

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.