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

  • #1
JackeTheDog132
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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.
 
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  • #2
:welcome:

Can you express your question more mathematically? I'm not sure what you are asking.
 
  • #3
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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  • #4
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.
 

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

What is the Time-Independent Schrödinger Equation (TISE)?

The Time-Independent Schrödinger Equation (TISE) is a fundamental equation in quantum mechanics that describes how the quantum state of a physical system changes in space. It is given by the equation \( \hat{H} \psi = E \psi \), where \( \hat{H} \) is the Hamiltonian operator, \( \psi \) is the wavefunction, and \( E \) is the energy eigenvalue.

Why should TISE solutions be combinations of eigenstates?

Solutions to the TISE should be combinations of eigenstates because eigenstates of the Hamiltonian operator form a complete basis set for the Hilbert space of the system. Any physical state of the system can be expressed as a linear combination of these eigenstates, ensuring that the system's wavefunction satisfies the superposition principle of quantum mechanics.

What does it mean if a solution to the TISE is not a combination of eigenstates?

If a solution to the TISE is not a combination of eigenstates, it implies that the solution does not properly describe the quantum state of the system according to the principles of quantum mechanics. Such a solution would not satisfy the completeness and orthogonality properties required for a valid quantum state.

Can there be exceptions where TISE solutions are not combinations of eigenstates?

In standard quantum mechanics, there are no exceptions to this rule. Any physically meaningful solution to the TISE must be expressible as a combination of eigenstates. If a solution appears not to be a combination of eigenstates, it may indicate a misunderstanding of the system's Hamiltonian or boundary conditions.

How can we verify if a TISE solution is a valid combination of eigenstates?

To verify if a TISE solution is a valid combination of eigenstates, one can decompose the solution into a sum of eigenfunctions of the Hamiltonian. This can be done using techniques such as Fourier series or spectral decomposition. If the solution can be expressed in this form, it confirms that it is a valid combination of eigenstates.

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