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In Fock space, the action of annihilation operator $a_i$ on state $N$-particle state $|n_1, n_2, n_3...n_p\rangle$ is directly defined as

$$a_i |n_1, n_2 ..n_i..\rangle = \sqrt{n_i}|n_1, n_2 ...n_i-1...\rangle. $$ In standard books they say that action of $a_i$ on this state remove one particle on state i. They also mention that normalization $\sqrt{n_i}$ comes from fact that, $$a_i |n_i\rangle = \sqrt{n_i} |n_i -1\rangle. $$I do not understand how this normalization translate into,

$$a_i|n_1, n_2..n_i.. n_p\rangle = \sqrt{n_i}|n_1, n_2 n_i-1..\rangle $$

Instead can I assume that as it is done in normally in the case of harmonic oscillator (see Ashok Das Lectures in QM)

$$|n_1, n_2...n_i...n_p\rangle = |n_1\rangle \otimes |n_2\rangle \otimes |n_3\rangle ...\otimes |n_p\rangle $$

$a_i$ is operator which acts on state labeled by $i$ is hence $$a_i|n_1, n_2..n_i..n_p> = I_1|n_1> \otimes I_2 |n_2> ..\otimes a_i |n_i> \otimes I_p |n_p>$$ (see R shankar QM)

$$= \sqrt{n_i} |n_1> \otimes |n_2> \otimes |n_i-1>...|n_p>$$ $$= \sqrt{n_i} |n_1, n_2....n_i-1, ...n_p>$$

Please help me understand whether this procedure is OK.

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Your understanding is completely correct -- that is precisely how the normalisation works.

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  • $\begingroup$ Thank you Emilio $\endgroup$
    – Vivek
    Commented Jan 8, 2022 at 15:40

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