Let's say we start with no particles: $\mid0\rangle$. We have $\vec{p}\vert0\rangle = 0$, $H\vert0\rangle = 0$, where we are ignoring $\infty$ vacuum energy. Also, $a(\vec{k})\vert0\rangle = 0$ for all $\vec{k}$. The Hilbert space now splits into sectors of $r$ particle states, $r = 0$, $1$, $2, \dots$. A general $r$ particle state has the form$$\vert r\rangle = {1\over{\sqrt{r! \int {{d^3 \vec{k}_1}\over{(2\pi)^3}} \dots \int {{d^3 \vec{k}_r}\over{(2\pi)^3}} \left| F(\vec{k}_1, \dots, \vec{k}_r)\right|^2}}}$$$$\cdot \int {{d^3 \vec{k}_1}\over{(2\pi)^3}} \dots \int {{d^3 \vec{k}_r}\over{(2\pi)^3}} F(\vec{k}_1, \dots, \vec{k}_r)\, a^\dagger (\vec{k}_1) \dots a^\dagger(\vec{k}_r) \,\vert0\rangle.$$My question is, must $\vert r\rangle$ necessarily satisfy $\langle r\, \vert \,r \rangle = 1$?
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$\begingroup$ If the question is: "must one include the normalisation factor?" the answer is no, one does not necessarily have to. $\endgroup$– gentedCommented Oct 5, 2015 at 21:43
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1$\begingroup$ No, if you define the expectation values on a vector $|r\rangle$ as $\operatorname{EXP}[O] := \frac{\langle r|O|r\rangle}{\langle r|r\rangle}$. $\endgroup$– Phoenix87Commented Oct 6, 2015 at 1:36
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The short answer is "no, the normalization condition is not always necessary." I think this really depends on how do you use that normalization condition. You can always make a state normalized to 1 as you have correctly stated. But that condition can be arbitrarily chosen for different purposes. For example, a coherent state, $|\alpha\rangle$, satisfies $\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$ and $⟨\alpha|\alpha⟩=e^{|\alpha|^2-\alpha^2}$, which is referred to overcomplete-basis state. This definition is useful when considering a laser field with uncertain photon numbers, and $\alpha$ is measure of photon statistical properties.
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