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Coherent States.

  1. Oct 31, 2013 #1
    Hi guys,

    Just a quick question, is the following statement true (it seems to be implied in the article i'm looking at);

    Ʃ(|α|2)n = 1

    (The sum over n=0 to infinity)

    Thanks to anyone who takes a look.
     
  2. jcsd
  3. Oct 31, 2013 #2

    fzero

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    You don't post the reference, so I will refer to generalities. Typically the coherent state for the harmonic oscillator is written as

    $$ | \alpha \rangle = c \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}} | n\rangle.$$

    The norm of this state is

    $$\langle \alpha | \alpha \rangle = |c|^2 \sum_{n=0}^\infty \frac{|\alpha|^{2n}}{n!} = |c|^2 e^{|\alpha|^2}, $$

    so we can normalize the state by choosing ##c = \exp ( - |\alpha|^2/2 )##. If we had instead written

    $$ | \tilde{\alpha} \rangle = \sum_{n=0}^\infty \tilde{\alpha}^n | n\rangle,$$

    then the normalization condition is indeed

    $$\sum_{n=0}^\infty |\tilde{\alpha}|^{2n} =1.$$
     
  4. Oct 31, 2013 #3
    Great! That's what i was hoping, but for some reason i was confusing myself. Will post the workings i was referencing later. Thanks for taking a look!
     
  5. Nov 1, 2013 #4

    naima

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    We have to say that in the relation of hazzattack alpha is not the complex eigenvalue of the annihilation operator..
     
    Last edited: Nov 1, 2013
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