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In a finite-dimensional Hilbert space, the eigenvalues of the adjoint $A^*$ of an operator $A$ are the complex conjugates of the eigenvalues of $A$.

But in infinite dimensions this need no longer be the case. For example, an annihilation operator $a$ has all complex numbers as eigenvalues (and coherent states as normalizable eigenstates), while the creation operator $a^*$, its adjoint, has no eigenvalues. Here $a$ and $a^*$ act on a Hilbert space with a fixed countable orthonormal basis written as $|n\rangle$ ($n=0,1,2,,\ldots$) as $$a|n\rangle=\sqrt{n}|n-1\rangle,~~~a^*|n\rangle=\sqrt{n+1}|n+1\rangle.$$ (The terminology is as in https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator.)

What is the general relation between the eigenvalues of an operator $A$ densely defined on a Hilbert space and those of its adjoint? What is their relation with the spectrum defined as the set of $x$ such that $xI-A$ has no bounded inverse?

As the comments so far showed, the spectrum of both $a$ and $a^*$ is the complex plane. For self-adjoint operators, for each point in the interior of the continuous spectrum, there are apparently unnormalizable eigenstates in the dual of a dense nuclear subspace. Does this generalize to general closed operators? What would these generalized eigenstates be in the case of $a^∗$?

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    $\begingroup$ For a closed densely defined linear operator $A$ on a Hilbert space, the spectrum of $A^*$ is in fact always the complex conjugate of the spectrum of $A$, just as in the finite-dimensional case (see for instance Theorem III.6.22 in "T. Kato: Perturbation theory for linear operators" (1980)). To better understand your example, a reference or a precise definition of $a$ would probably be helpful (since the notions "annihilation operator" and "creation operator" seem to be used in various contexts in quantum physics). $\endgroup$
    – Jochen Glueck
    Commented Dec 29, 2019 at 14:32
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    $\begingroup$ I suspect that you are confusing the spectrum with the eigenspectrum. Indeed, eigenvalues of an operator need not be eigenvalues of the adjoint. $\endgroup$
    – Michael Renardy
    Commented Dec 29, 2019 at 15:51
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    $\begingroup$ @ArnoldNeumaier: see here: en.wikipedia.org/wiki/… $\endgroup$
    – András Bátkai
    Commented Dec 29, 2019 at 18:05
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    $\begingroup$ @ArnoldNeumaier There must be something in the air, because I was just mentioning the fact that that isn't true elsewhere. Consider the operator $T|n\rangle = \frac{1}{n+1}|n\rangle$. It is self-adjoint, its eigenvalues are $\frac{1}{n+1}$ for each $n = 0,1,2,\ldots$, but additionally it is not invertible, so $0$ is a spectral value (and part of the continuous spectrum). $\endgroup$
    – Robert Furber
    Commented Dec 30, 2019 at 20:49
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    $\begingroup$ @ArnoldNeumaier But there is no corresponding generalized eigenvector (when $T$ is considered as an operator on the nuclear space $s$ of sequences that rapidly converge to $0$) $\endgroup$
    – Robert Furber
    Commented Dec 30, 2019 at 20:49

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