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We have:

$$|\psi\left(t\right)\rangle=e^{-iH_{NL}t}|N\rangle_{a}|0\rangle_{b}$$

Also

$$|\psi\left(t\right)\rangle=\sum_{n=0}^{N}c_{n}\left(t\right)|N_{1}\rangle_{a}|N_{2}\rangle_{b}$$

I would like to know whether

$$c_{n}\left(t\right)=\langle N_{2}|\langle N_{1}|e^{-iH_{NL}t}|N_{1}\rangle_{a}|N_{2}\rangle_{b}$$

is the right way to express.

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    $\begingroup$ I'm guessing something's wrong in the way you've written the second equation. Are you sure it isn't something like $\lvert \psi(t) \rangle = \sum_{N_1,N_2=0}^N c_{N_1,N_2}(t) \lvert N_1 \rangle_a \lvert N_2 \rangle_b$ ? $\endgroup$ May 17 at 5:02
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If we label the expansion coefficients as $c_{N_1,N_2}(t)$ as suggested in the comments, the correct expression will be $$c_{N_1,N_2}(t)= \vphantom{a}_a\langle N_1| \vphantom{a}_b\langle N_2|\psi(t)\rangle= \vphantom{a}_a\langle N_1| \vphantom{a}_b\langle N_2|e^{-i H_{NL}t}|N\rangle_a|0\rangle_b.$$

Responding to the comment: some people write their Hilbert spaces in opposite order when using bras, preferring to write this expression as $$c_{N_1,N_2}(t)= \vphantom{a}_b\langle N_2|\vphantom{a}_a\langle N_1|\psi(t)\rangle= \vphantom{a}_b\langle N_2|\vphantom{a}_a\langle N_1|e^{-i H_{NL}t}|N\rangle_a|0\rangle_b.$$ This does not change the answer, so long as the appropriate subscripts are used to appropriately match the Hilbert spaces.

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    $\begingroup$ Can you explain the reason why N_1 is coming first? Are there any references? $\endgroup$
    – Jasmine
    May 18 at 0:56
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    $\begingroup$ I simply wrote the $a$ Hilbert space on the left of the $b$ Hilbert space; you find the same result if you write them in another order. In all of these expressions, we must use that $\vphantom{a}_a\langle N_1|\otimes \vphantom{a}_b\langle N_b| (A\otimes B) |N_1\rangle_a\otimes|N_2\rangle_b=(\langle N_1|A|N_1\rangle)*(\langle N_2|B|N_2\rangle)$, etc. $\endgroup$ May 18 at 1:12

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