- #1

- 69

- 0

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.

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- Thread starter Hazzattack
- Start date

- #1

- 69

- 0

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

Ʃ(|α|

(The sum over n=0 to infinity)

Thanks to anyone who takes a look.

- #2

fzero

Science Advisor

Homework Helper

Gold Member

- 3,119

- 289

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

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

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- 0

- #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..

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