Coherent States.

  • Thread starter Hazzattack
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
fzero
Science Advisor
Homework Helper
Gold Member
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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

$$\sum_{n=0}^\infty |\tilde{\alpha}|^{2n} =1.$$
 
  • #3
69
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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!
 
  • #4
naima
Gold Member
938
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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:

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