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Operators on infinite-dimensional Hilbert space

  1. Oct 9, 2014 #1
    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!
     
  2. jcsd
  3. Oct 9, 2014 #2

    rubi

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    Science Advisor

    ##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##.
     
  4. Oct 9, 2014 #3

    samalkhaiat

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    Science Advisor

    Since the set [itex]\{ | n \rangle \}[/itex] is complete, you can write
    [tex]R = \sum_{ n = 0 } | n + 1 \rangle \langle n | , \ \ \ \ \ (1)[/tex]
    [tex]L = \sum_{ n = 1 } | n - 1 \rangle \langle n | = R^{ \dagger } . \ \ \ \ (2)[/tex]
    So, you can set [itex]L = A[/itex] and [itex]R = A^{ \dagger }[/itex]. From (1) and (2), you can show [itex][ A , A^{ \dagger } ] = 1[/itex].
    Now, suppose that for some [itex]\alpha \in \mathbb{ C }[/itex], we have
    [tex]A^{ \dagger } | \alpha \rangle = \alpha | \alpha \rangle . \ \ \ \ \ \ (3)[/tex]
    Write
    [tex]| \alpha \rangle = c_{ 0 } | 0 \rangle + c_{ 1 } | 1 \rangle + c_{ 2 } | 2 \rangle + \cdots .[/tex]
    Substitute this expansion in (3) and equate coefficients, you find that [itex]c_{ n } = 0[/itex] for all [itex]n[/itex]. So, [itex]A^{ \dagger }[/itex] cannot have renormalized eigen-states. If you do the same with [itex]A[/itex], you find [itex]c_{ n } = \alpha^{ n } c_{ 0 }[/itex] so that the renormalized eigen-states of [itex]A = L[/itex], for any [itex]\alpha \in \mathbb{ C }[/itex], are given by
    [tex]| \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 .[/tex]
     
  5. Oct 10, 2014 #4

    rubi

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    Science Advisor

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