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It is well knwon that an eigenstate can be obtained by superposing wavepackets. In other words, if $\Psi({\bf x},t)$ is the solution of the time dependent Schroedinger equation for an initial wavepacket $\Psi({\bf x},0)$, the eigenstate $\psi_n({\bf x})$ corresponding to the eigenenergy $E_n$ is related to $\Psi({\bf x},t)$ by \begin{equation} \int\Psi({\bf x},t)e^{iE_nt/\hbar} dt=2\pi \hbar\sum_m a_m\psi_m({\bf x})\delta(E_m-E_n)\propto\psi_n({\bf x}), \end{equation} where the coefficients $a_m$ are defined by $\psi({\bf x},0)=\sum_m a_m\psi_m$.

Now comes my question, what happens when the spectrum is degenerate (or quasidegenerate)? i.e. when two eigenvalues $E_n$ and $E_{n+1}$ are equal (or very close to each other). Does the same relationship hold?

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First of all, not every Hamiltonian admits a basis of eigenvectors. An operator has to be either compact or with compact resolvent to admit a basis of eigenvectors. With those type of operators, eigenvalues can accumulate (i.e. being distinct, but getting arbitrarily close) only at zero (compact) or infinity (compact resolvent). A discrete eigenvalue can have, however, multiplicity bigger than 1.

So:

1) if the operator is not compact, or with compact resolvent, you cannot write $\psi(x,0)$ as a linear combination of eigenvectors and thus the reasoning you outlined does not hold.

2) if the operator is compact/with compact resolvent and degenerate: the decomposition $\psi(x,0)=\sum_m \alpha_m\psi_m $ is not unique. Suppose $a_{\bar{m}}$ is an eigenvalue with multiplicity $n$. Denote by $A_{\bar{m}}$ the $n$-dimensional subspace of the Hilbert space spanned by the eigenvectors of $a_{\bar{m}}$. Then $\int \psi(x,t)e^{iE_\bar{m}t/\hslash}dt$ will be a vector in $A_{\bar{m}}$.

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