# Operators on infinite-dimensional Hilbert space

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1. Oct 9, 2014

### linbrits

Hello all!

I have the following question with regards to quantum mechanics.

If $H$ is a Hilbert space with a countably-infinite orthonormal basis $\{ \left | n \right \rangle \}_{n \ \in \ \mathbb{N} }$, and two operators $R$ and $L$ on $H$ are defined by their action on the basis elements as follows:

$\begin{eqnarray} R \left | n \right \rangle & = & \left | n + 1\right \rangle ,\\ L \left | n \right \rangle & = & \left\{ \begin{array}{11} \left | n - 1 \right \rangle & \text{for n > 1} \\ 0 & \text{for n = 1}. \end{array} \right. \end{eqnarray}$

What are the the eigenvalues and eigenvectors of $R$ and $L$, if they do exist? Also, what are the hermitian conjugates of $R$ and $L$?

2. Oct 9, 2014

### rubi

$R$ can't have an eigenvector, because if it had an eigenvector $\left|\psi\right>$, then there would be a smallest $n_0$ such that $\left<n_0|\psi\right>\neq 0$ (well ordering of $\mathbb N$), but $\left<n_0|R|\psi\right> = 0$.

For the eigenvectors of $L$, you need to solve $\lambda \sum_n a_n \left|n\right> = L\sum_n a_n \left|n\right> = \sum_n a_n \left|n-1\right>$, so you get $\lambda a_n = a_{n+1}$. For $a_0 := c$, you get $a_n = \lambda^n c$. The sum will converge for $|\lambda|\lt 1$ (geometric series) and you get an eigenvector for each such $\lambda$.

3. Oct 9, 2014

### samalkhaiat

Since the set $\{ | n \rangle \}$ is complete, you can write
$$R = \sum_{ n = 0 } | n + 1 \rangle \langle n | , \ \ \ \ \ (1)$$
$$L = \sum_{ n = 1 } | n - 1 \rangle \langle n | = R^{ \dagger } . \ \ \ \ (2)$$
So, you can set $L = A$ and $R = A^{ \dagger }$. From (1) and (2), you can show $[ A , A^{ \dagger } ] = 1$.
Now, suppose that for some $\alpha \in \mathbb{ C }$, we have
$$A^{ \dagger } | \alpha \rangle = \alpha | \alpha \rangle . \ \ \ \ \ \ (3)$$
Write
$$| \alpha \rangle = c_{ 0 } | 0 \rangle + c_{ 1 } | 1 \rangle + c_{ 2 } | 2 \rangle + \cdots .$$
Substitute this expansion in (3) and equate coefficients, you find that $c_{ n } = 0$ for all $n$. So, $A^{ \dagger }$ cannot have renormalized eigen-states. If you do the same with $A$, you find $c_{ n } = \alpha^{ n } c_{ 0 }$ so that the renormalized eigen-states of $A = L$, for any $\alpha \in \mathbb{ C }$, are given by
$$| \alpha \rangle = \frac{ 1 }{ \sqrt{ \sum_{ n } | \alpha |^{ 2 n } } } \sum_{ n } \alpha^{ n } | n \rangle \sim \frac{ 1 }{ \sqrt{ \sum_{ n } | \alpha |^{ 2 n } } } \sum_{ n } ( \alpha A^{ \dagger } )^{ n } | 0 \rangle .$$

4. Oct 10, 2014

### rubi

The sums don't converge for $|\alpha|\ge 1$, so you don't get eigenstates for these values of $\alpha$.