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 $xI-A$ is not invertible?

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 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 $a^∗$?

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 $a^∗$?

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 $xI-A$ is not invertible?

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 continuous spectrum, there are 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 $a^∗$?

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 $xI-A$ is not invertible?

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 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 $a^∗$?

In a finite-dimensional Hilbert space, the spectrum of the adjoint $A^*$ of an operator $A$ is the complex conjugate of the spectrum of $A$.

But in infinite dimensions this need no longer be the case. For example, an annihilation operator $a$ has the complex plane as eigenspectrum (and coherent states as eigenstates), while the creation operator $a^*$, its adjoint, has empty eigenspectrum. 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.)

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 continuous spectrum, there are 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 $a^∗$?

What is the general relation between the eigenspectrum of an operator $A$ densely defined on a Hilbert space and that of its adjoint? And its relation with the spectrum defined as the set of $x$ such that $xI-A$ is invertible?

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 $xI-A$ is not invertible?

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 continuous spectrum, there are 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 $a^∗$?