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Consider a system where the Hamiltonian is time independent, the wavefunction which is say a linear combination of the eigenfunction of the Hamiltonian (with different eigenvalues) is not the solution of the time independent Schrodinger equation (TISE). This wavefunction is in fact a solution of the time dependent Schrodinger equation. My question is why is it so? The system has a time independent Hamiltonian so it should be described by a TISE, why do we get a wavefunction which is solution of the general Schrodinger Equation?

Edit: I should rephrase my question to make it more comprehensible. The linear combination of the eigenfunctions gives solution to the Schrodinger equation. For a system with time independent Hamiltonian this linear combination should be a solution of the TISE. It is not always true(when eigenvalues are not equal). Why is it so?

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I thing the phrase "time-independent Schrodinger equation", though widely used, can be a cause for confusion because it implies that it is in some sense a different version of the Schrodinger equation, which isn't true. The Schrodinger equation is a first-order differential equation $$i\hbar \psi'(t) = \hat H \psi(t) \qquad (\star)$$ where $\psi(t)$, which is an element of whatever Hilbert space underlies your theory, can be loosely thought of as the state of the system at time $t$. This equation tells you how to evolve your state vector forward in time.

If $\psi(t)$ is an eigenstate of $\hat H$ with eigenvalue $E$, then this equation becomes extremely easy to solve:

$$i\hbar \psi'(t) = \hat H \psi(t) = E\psi(t) \implies \psi(t) = e^{-iEt/\hbar} \psi(0)$$ If $\psi(t)$ is a linear combination of such eigenvectors, then the solution is similarly very simple. If $\psi(t) = \sum_n c_n(t) \phi_n$ with $\hat H \phi_n = E_n \phi_n$, then $$\psi(t) = \sum_n c_n(t) \phi_n =\sum_n e^{-iE_n t/\hbar} c_n(0) \phi_n$$

The fact that $\hat H$ is self-adjoint means that every state can be written as a linear combination of its eigenstates$^\ddagger$. Therefore, in order to understand how to evolve any arbitrary vector forward in time, we need only find the set $\{\phi_n\}$ of eigenvectors of $\hat H$ and then expand the initial vector in terms of them. From there, time evolution is simple - each eigenvector evolves via the corresponding phase factor $e^{-iE_n t/\hbar}$.

Therefore, we turn our attention to finding all of the eigenvectors and eigenvalues of $\hat H$. This requires finding all pairs eigenvector/eigenvalue pairs $\{\phi,E\}$ which are solutions to $\hat H\phi = E\phi$. Once we have all such solutions, we have all the ammunition we need to evolve state vectors forward in time. This equation $\hat H\phi = E\phi$ is what we call the time-independent Schrodinger equation, but I really dislike that name.

As you can see, the TISE really of completely different character than $(\star)$ - the former being the eigenvalue equation for the Hamiltonian operator, and the latter telling us how to evolve a general state forward in time. Solving the TISE is an important and very useful step in solving the "general" Schrodinger equation, but they aren't different versions of the same thing.


$^\ddagger$This assumes that the spectrum of $\hat H$ is discrete, and that $\hat H$ is not explicitly time-dependent. Further complications arise if either of these two assumptions fail, but the spirit of this answer remains the same.

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  • $\begingroup$ Does this mean that the eigen value equation of the time-independent Hamiltonian, not the TISE? $\endgroup$ Commented May 16, 2021 at 17:25
  • $\begingroup$ @Physicsfreak I'm afraid I don't understand your question. Can you rephrase it? $\endgroup$
    – J. Murray
    Commented May 16, 2021 at 17:32
  • $\begingroup$ This equation H^ϕ=Eϕ is what we call the time-independent Schrodinger equation, but I really dislike that name. Can you tell what does this statement mean? $\endgroup$ Commented May 16, 2021 at 17:34
  • $\begingroup$ @Physicsfreak The time-independent Schrodinger equation is just the eigenvalue equation $\hat H\phi = E\phi$. I don't like that name because it makes it sound like it's a different version of the Schrodinger equation $i\hbar \psi'(t) = \hat H\psi(t)$, when in reality those two equations have very little to do with one another, except for the fact that solving the former is an important step in solving the latter. $\endgroup$
    – J. Murray
    Commented May 16, 2021 at 17:36
  • $\begingroup$ thanks for your comment, it cleared the confusion I had all this time around the SE. Is it possible for you to answer some more doubts regarding the SE? $\endgroup$ Commented May 16, 2021 at 17:44

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