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Suppose we have solved for the energy eigenstates of some Hamiltonian operator $\hat{H}$.

We call the energy eigenstates $\psi_n (x)$, where:

  • $n=1$: $\psi_1 (x)$ is the ground states

  • $n=2$: $\psi_2 (x)$ is the first excited states

It says that the general wavefunction can be expanded in such eigenstates via:

$$\psi(x)=\sum_{n=1}^\infty a_n\psi_n(x)$$

I don't what it means by general wavefunction in this case? Does it mean that all wavefunction that solve the particular Hamiltonian or does it mean eigenfunctions of all (different) Hamiltonians?

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  • $\begingroup$ You don't solve a Hamiltonian. The Hamiltonian acts/"operates" on a wavefunction and maps it to a new wavefunction. This mapping is particularly easy to evaluate for eigenfunctions of the operator since the mapping reduces to a simple multiplication with the corresponding eigenvalue in that case. This means that once you decompose your wavefunction on which the operator acts into a sum of eigenfunctions, you know how to evaluate that operators action on that particular wavefunction. $\endgroup$
    – Hans Wurst
    Commented Apr 1, 2021 at 8:24

2 Answers 2

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Any function that satisfies the boundary conditions for the problem - even one that doesn’t solve the time-independent Schrodinger equation - can be written as a sum of functions that do solve the TISE.

The solutions $\psi_n(x)$ are like “basis vectors” in the same way that $\hat x$, $\hat y$ and $\hat z$ are basis vectors in 3d space:

  1. any vector in 3D can be written as a sum $a_x\hat x+a_y\hat y+a_z\hat z$; any function can be written as $\sum_n a_n\psi_n(x)$,
  2. the vectors $\hat x$ etc have unit length; the functions $\psi_n(x)$ have unit length when the length is computed as $\int dx \psi^*(x)\psi(x)$.
  3. Unlike the vectors $\hat x$ etc, which are valid for any 3d space, the set of basis functions $\{\psi_n(x)\}$ are specific to a given Hamiltonian. The basis set for the harmonic oscillator is not the same set as for the infinite well or the hydrogen atom.

One difference is that there are in some cases such as harmonic oscillator or hydrogen atom or infinite well, infinitely many basis functions $\psi_n(x)$ rather than just $3$ unit vectors.

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    $\begingroup$ Not any function. E.g. $x\mapsto e^x$ can't. $\endgroup$
    – Ruslan
    Commented Apr 1, 2021 at 14:03
  • $\begingroup$ @Ruslan good point. Edited my answer to reflect your comment. $\endgroup$ Commented Apr 1, 2021 at 14:49
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In general, If $$H|n\rangle =n|n\rangle $$

Then we can expand $|\psi\rangle$ on this basis

$$|\psi\rangle =\sum_n|n\rangle \langle n|\psi\rangle$$


Note that in general, I mean you can do the above procedure for all systems. It doesn't mean that you only have to solve the eigenvalue problem for one system and you can expand the state vector of any system.

So the procedure is valid for all systems but things like eigenvectors or wavefunction or hamiltonian would be different for different systems.

I don't what it means by general wavefunction in this case?

Any general wavefunction of the system can be expanded this way.

Does it mean that all wavefunction solves the particular Hamiltonian or does it mean eigenfunctions of all (different) Hamiltonians?

It should be clear from above.

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