Fock space-Weyl Algebra 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. 
 A: 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.
