I am having trouble differentiating between the two.

If we have a Hilbert space $H$, which has an ortho-normal basis ${|n\rangle}$. Then is it right to say that ${|n\rangle}$ are the eigenvectors of the Hilbert space, belonging to some eigenvalues?

Do we find the eigenvalues by having an hermitian operator A act on the eigenvectors: $A|n\rangle = b |n\rangle$.

But if we have $|\Psi\rangle =\Sigma_n c_n|n\rangle$ as a linear superposition of the eigenvectors of the Hilbert space, then we can do the following:

$A|\Psi\rangle=k |\Psi\rangle $ where: $k$ is the eigenvalue of the hermitian operator, represented by a matrix and $|\Psi\rangle$ is the eigenvector of the eigenvalue $k$.

Does the hermitian operator $A$ and the Hilbert space $H$, each have their own eigenvalues and vectors?


1 Answer 1


Eigenvectors and eigenvalues are properties of operators, not of Hilbert spaces. Hence, answering the first part of your question, if $| n \rangle$ is an orthonormal basis of $\mathcal{H}$ (the Hilbert space), it is not correct to say they are the eigenvalues of the Hilbert space. We simply say they are an orthonormal basis.

Suppose now you are given an operator $A \colon \mathcal{H} \to \mathcal{H}$ on the Hilbert space. Loosely speaking, a matrix. A vector $| \Psi \rangle \in \mathcal{H}$ is said to be an eigenvector of $A$ associated to the eigenvalue $a$ when $A | \Psi \rangle = a | \Psi \rangle$. Notice that I need the operator to define what I mean by eigenvector: eigenvectors are the vectors in which $A$ acts as if it was simply a number, this number being called an eigenvalue.

Now if $A$ is hermitian, there are two cool results:

  1. the eigenvalues of $A$ are all real;
  2. the eigenvectors of $A$ provide an orthonormal basis of the Hilbert space.

Hence, if you are given a hermitian operator on the Hilbert space, you can use it to obtain a basis. We usually pick the Hamiltonian, for example, because then each state in the basis has a simple time-evolution in terms of its energy (more specifically, in terms of the eigenvalue of the Hamiltonian to which it is associated). By the very definition of what a basis is, we can then write any vector in the Hilbert space in terms of eigenvectors of the Hamiltonian, i.e., in terms of states of definite energy. This provides a nice way to write down the time evolution of the state, since the Hamiltonian eigenstates have simple time-evolution rules.

We could pick other operators, if we so desired. Instead of the Hamiltonian, you could pick some other hermitian operator to obtain a basis, but it would probably be less convenient to work with.

Finally, it is worth mentioning that sometimes a few linearly independent states might be associated to the same eigenvalue. We call this degeneracy. In this case, we often need a few more operators (such as angular momentum squared and angular momentum in the $\vec{z}$ direction) to properly label all the states in the basis in a unique way. This is what happens when we solve for the hydrogen atom and need three labels to define uniquely which state in the basis is which, because a few of them have the same energy (as a consequence, we distinguish them by using their angular momentum properties).

  • $\begingroup$ First of all thank you for the clarification, now I understand mostly everything. One question regarding the hydrogen atom and its wave equation. In wikipedia the wave equation presented, which is characterised by the quantum numbers n.l.m.s usually is the wave equation of the electron of the H-atom. That is different then the wave equation of the whole atom, including the proton,translation movement of the whole atom etc? Is this the case? $\endgroup$
    – imbAF
    Commented Aug 10, 2021 at 18:52
  • $\begingroup$ @imbAF you're welcome! The wave equation with everything (electron + proton moving around) is certainly far more complicated, but I meant the wave equation you saw, for the electron alone. For $n > 1$, there are multiple states with the same energy (or the same $n$, if you prefer), so we tell them apart with the numbers $l$ and $m$, which are associated to the orbital angular momentum of the electron around the proton (which is assumed to be at rest, as an approximation). If you want to take into account the electron's spin, you should also throw in $m_s$, for spin adds more degeneracy $\endgroup$ Commented Aug 10, 2021 at 20:25
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    $\begingroup$ Thinking a bit in terms of the planetary model of an atom, think that in some shells there can be more than one electron, so you'd need to distinguish them apart by using their angular momentum. Beware that this is just a picture that shouldn't be taken too seriously, for I'm ignoring some nuances about indistinguishable particles and other stuff for the purposes of trying to be a bit more pedagogical. The key point is that given some energy value, there might be more than one state with having it as an eigenvalue $\endgroup$ Commented Aug 10, 2021 at 20:29

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