Operators with continuous eigenbasis I came across the following relation, regarding commutators :
if $[\hat{a},\hat{b}] = k$, then we can write the following,
$\hat{a} = k\frac{\partial }{\partial \hat{b}}$ and $\hat{b} = -k\frac{\partial }{\partial \hat{a}}$
It was written that the above is true, only if we are in a continuous eigenbasis. However, I don't seem to understand this line. What exactly do they mean by continuous eigenbasis?  Does it mean that the two operators above, have continuous spectra like position or momentum?
Do continuous spectra automatically imply continuous eigenbasis ? I know that for an operator with discrete eigenvalues or spectra, we have a discrete basis, for example, the states of the harmonic oscillator, well, angular momentum etc.
How can we find out if an operator has a continuous or a discrete eigenbasis or spectra ? What about operators with no eigenvalues, for example the creation operator of the harmonic oscillator ? What category does that fall into ? I can't seem to figure it out, for the continuous case atleast. How to show that an operator has a continuous eigenbasis ? Any explanation would be highly appreciated.
 A: That's a lot of questions!

*

*We can argue directly from the commutator: let $|q>$ be an eigenket of $\hat a$, $\hat a|q>=q|q>$. Take the expectation value of the commutator:
$$
<q|\hat a \hat b-\hat b \hat a|q>=<q|k|q>=k<q|q>
$$
where the second step uses the fact that $k$ is a constant. Using the Hermitian property of $\hat a$ we can transform the left hand side:
$$
<q|\hat a \hat b-\hat b \hat a|q>=<q|q\hat b-\hat b q|q>=q(<q|\hat b|q>-<q|\hat b|q>)=0
$$
where the final step assumes that the quantities are finite.
Hence we have $0=k<q|q>$. But this is a contradiction: the right hand side is non-zero, assuming $|q>$ has finite norm. So we must have made an incorrect assumption: either $\hat a$ is not Hermitian, or it has no eigenstates, or the norm $q$ and the expectation value of $\hat  b$ are not finite.

So we have a choice. We could say $a$ is not Hermitian and has no eigenvalues, and some mathematicians would say that. But the physcist normally takes the route of saying that $a$ (and $b$) have eigenvalues and eigenkets but they aren't normalisable.


*Yes, continuous eigenvalues and non-normalisable kets go together. I can't prove that but doubtless someome else can


*The argument doesn't apply to creation and destruction operators because they are definitely not Hermitian.
