Determine eigeinvalue and eigenvector of two operators R and L Question: Let H be a Hilbert space with countable-infinite orthonormal basis ${|n>}_{n \in N}$. The two operators R and L on H are defined by their action on the basis elements 
\begin{align}
R|n\rangle &= |n+1\rangle \\
L|n\rangle &=\begin{cases}|n-1\rangle & {\rm for}\,n>1 \\ 0&{\rm for} n=1\end{cases}  \end{align}
Determine the eigenvalue and eigenvectors of $R$ and $L$ and find their Hermitian conjugate.
Attempt: $R$ looks like the creation operator and $L$ looks like the annihilation operator. So I did: 
\begin{align}aR|n\rangle&=a|n+1\rangle\\  
 &=\sqrt{n+1}|n\rangle \end{align}
Now I have a simultaneous eigenstate for the annihilation and $R$ operator with eigenvalue $\sqrt{n+1}$. I did the same with the $L$ and creation operator and found an eigenvalue of $\sqrt{n}$ The hermitian conjugates are then just $\langle n|R^\dagger = \langle n+1|$
I do not think my steps are correct, can someone help me in the right direction please.
 A: I assume that $n =1,2,\ldots$ and I indicate by $\psi_n$ the unit vector $|n\rangle$.
A generic vector in the Hilbert space can therefore be written as
$$\psi = \sum_{n=1}^{+\infty} c_n \psi_n$$
where $\sum_n |c_n|^2 < +\infty$.
The action of $R$ and $L$ on that vector respectively is:
$$R \psi = \sum_{n=1}^{+\infty} c_n \psi_{n+1}$$ 
and 
$$L \psi = \sum_{n=2}^{+\infty} c_n \psi_{n-1}\:.$$ 
Thus, assuming $\phi = \sum_{m=1}^{+\infty} b_m \psi_m$,
$$\langle \phi | R \psi \rangle = \sum_{m=1}^{+\infty} \sum_{n=1}^{+\infty} b^*_m c_n \langle \psi_m | R \psi_n \rangle = \sum_{m=1}^{+\infty} \sum_{n=1}^{+\infty} b^*_m c_n \langle \psi_m |  \psi_{n+1} \rangle $$ $$=
\sum_{m=1}^{+\infty} \sum_{n=1}^{+\infty} b^*_m c_n\delta_{m,n+1}
=  \sum_{n=1}^{+\infty} b^*_{n+1} c_n\:.\tag{1}$$
Similarly,
$$\langle L\phi |  \psi \rangle = \sum_{m=1}^{+\infty} \sum_{n=1}^{+\infty} b^*_m c_n \langle L \psi_{m} | R \psi_n \rangle = \sum_{m=2}^{+\infty} \sum_{n=1}^{+\infty} b^*_m c_n \langle \psi_{m-1} |  \psi_{n+1} \rangle $$ $$ =
\sum_{m=2}^{+\infty} \sum_{n=1}^{+\infty} b^*_m c_n\delta_{m-1,n}
=  \sum_{n=1}^{+\infty} b^*_{n+1} c_n\:.\tag{2}$$
We have found that 
$$\langle \phi | R \psi \rangle = \langle L\phi |  \psi \rangle$$
which means $R^\dagger=L$. Taking the complex conjugate of (1) and (2), we also get
$$\langle R\psi |  \phi \rangle = \langle \psi | L \phi \rangle$$
which means $L^\dagger=R$. (I deliberately omitted to discuss several issues about domains of the involved operators and convergence of series, but everything goes right taking all mathematical subtleties into account...)
Regarding eigenvectors, $R\psi = \lambda \psi$ can be rewritten as
$$R\psi = \sum_{n=1}^{+\infty} c_n R\psi_n =  \sum_{n=1}^{+\infty} c_n \psi_{n+1} = \lambda \psi =   \sum_{n=1}^{+\infty} \lambda c_n \psi_{n}$$
that is 
$$ \sum_{n=1}^{+\infty} c_n \psi_{n+1} = \sum_{n=1}^{+\infty} \lambda c_n \psi_{n}$$
that is, in turn,
$$ \sum_{n=2}^{+\infty} c_{n-1} \psi_{n} =  \lambda c_1 \psi_1 +\sum_{n=2}^{+\infty} \lambda c_n \psi_{n}\:,$$
which implies $\lambda c_1 =0$ and $\lambda c_n = c_{n-1}$ for $n>2$.
These equations only admit the solution $c_n=0$, the proof being immediate, both for $\lambda =0$ and $\lambda \neq 0$. No eigenvectors exist for $R$.
With the same approach we see the $L\psi = \lambda \psi$ is equivalent to $c_{n+1}= \lambda c_n$, so that $c_n = \lambda^{n-1} c_1$. Hence a candidate eigevector with eigenvalue $\lambda \neq 0$ is 
$$\psi_\lambda = \frac{1}{\lambda}\sum_{n=1}^{+\infty} \lambda^n \psi_n\:.$$
It is an eigenvector provided the series converges in the Hilbert space, i.e.
$$\sum_{n=1}^{+\infty} |\lambda|^{2n} <+\infty\:.$$
It in fact happens only for $|\lambda|^2 <1$. So, there is an eigenvector of $L$ with eigenvalue $\lambda$ for every complex $\lambda$ with $0 <|\lambda|<1$
