Operators on infinite-dimensional Hilbert space

[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 $L$?

Thanks in advance!

 

[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 $\lambda$.

 

[samalkhaiat]

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 $L$?

Thanks in advance!

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 .

 

[rubi]

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 .

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