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I am new to quantum physics, help me out please on this question. I am confuse about how to compute the norm of the states describe below:

Given the operators $a_m, a_m^{\dagger}, m=1,2,\dots, M$, satisfying $[a_m,a_m^{\dagger}]=\delta _{nm}$, $[a_m,a_m]=[a_m^{\dagger},a_m^{\dagger}]=0$. If $s_1,\dots, s_M\in \mathbb{N}$ given that for the state $|0\rangle$ such that $a_m|0\rangle=0$ and $\langle 0|0\rangle=1$. So the question here presicely is to compute the norms of the states $|s_1,s_2,\dots,s_M\rangle=\Pi_{m=1}^M({a_m^\dagger})^{s_m}|0\rangle$.

What do you suggest, should I use the definition in the reference https://en.wikipedia.org/wiki/Fock_space whichwe can compute the norm as $$|||s_1,\dots, s_M\rangle||^2=\sum a^*_{j1,\dots, jm}a_{k1,\dots, jk}\langle \psi_{j1}|\psi_{k1}\rangle\dots \langle \psi_{jM}|\psi_{kM}\rangle ?$$ However I found that not convincing as the in this case produces at each braket $\langle 0|a^{s_1}(a^\dagger)^{s_1}|0\rangle=\dots=\langle 0|a^{s_M}(a^\dagger)^{s_M}|0\rangle$. This to me does not make sense because the sum yields the norms equal $M$. Thanks for your response.

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Start with $$ \vert s_k\rangle :=(a_k^\dagger)^{s_k}\vert 0) = \sqrt{s_k!}\vert s_k) \, . $$ where the round ket $\vert s_k)$ is a normalized harmonic oscillator ket. This follows by induction since $a^\dagger_k\vert s_k)=\sqrt{s_k+1}\vert s_k+1)$ for normalized kets.

Suppose for simplicity $M=2$. Then \begin{align} \vert s_1,s_2\rangle=\left(a_1^\dagger\right)^{s_1}\left(a^\dagger_2\right)^{s_2}\vert 0)&= \left[(a_1^\dagger)^{s_1}\vert 0)\right] \otimes \left[(a_2^\dagger)^{s_2}\vert 0) \right]\, ,\\ &= \sqrt{s_1!s_2!}\vert s_1) \otimes \vert s_2)\\ &:=\sqrt{s_1!s_2!}\vert s_1,s_2) \end{align} and so $$ ( 0\vert a_1^{s_1}a_2^{s_2}=\sqrt{s_1!s_2!}( s_1,s_2\vert $$ and thus the norm of $\vert s_1,s_2\rangle$ is then $$ \langle s_1,s_2\vert s_1,s_2\rangle= s_1!s_2! (s_1,s_2\vert s_1,s_2) =s_1!s_2! $$ since the round kets are normalized.


Edit: To see that $\vert s_1\rangle$ as defined by the OP is not normalized, we suppose that $\langle s_1\vert s_1\rangle=1$ but show that $\langle s_1+1\vert s_1+1\rangle\ne 1$. To this end consider $\vert s_1+1 \rangle= a_1^\dagger \vert s_1\rangle$, which holds by induction. Then clearly

$$ \langle s_1+1 \vert = \langle s_1\vert a_1 \, , \qquad \langle s_1+1\vert s_1+1\rangle = \langle s_1\vert a_1a_1^\dagger\vert s_1\rangle = (s_1+1)\langle s_1\vert s_1\rangle $$ so that, if $\vert s_1\rangle$ is normalized, then $\vert s_1+1\rangle$ is not. In fact, the states $\vert s_1\ldots s_M\rangle$ as properly normalized if they are defined as $$ |s_1,s_2,\dots,s_M\rangle=\displaystyle\prod_{m=1}^M \frac{({a_m^\dagger})^{s_m}}{\sqrt{s_m!}}|0\rangle $$ as can be shown by induction.

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  • $\begingroup$ Yes that's great. Thanks. I still wonder how $\langle s_1, s_2,\dots , s_M |s_1, s_2,\dots , s_M \rangle=1$, the states are not noramlaized however normalized here, right? $\endgroup$
    – unknown
    Commented Jan 13, 2018 at 13:13
  • $\begingroup$ This approached was also good, thanks very much. My thought here is that we have to simplify using the vacuum states property $\langle 0 | 0\rangle=1$, but at least I got some hint, just want to be sure to continue. $\endgroup$
    – unknown
    Commented Jan 13, 2018 at 13:33
  • $\begingroup$ @M.U.Faruq I've added some additional material for clarification. $\endgroup$ Commented Jan 13, 2018 at 18:57

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