In QM its obvious that we need to normalise quantum states since their inner product squared represents a probability. This normalization leads to physical states in QM being represented by 'rays' of vectors in the Hilbert space (i.e. with an arbitrary overall phase)

However in QFT, the inner product of the states no longer has this probability meaning, so why do we need to normalise states (e.g. choosing to normalise $|p\rangle = \sqrt{2 E_p} a_p^\dagger |\Omega\rangle$).

What would break if we did not normalise the states? Also what would be wrong with a state $k | p \rangle$ representing a different physical state to $| p \rangle$?

Is it just so that the inner product $\langle q | p \rangle$ is well-defined, even if it no longer represents probability amplitude?

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    $\begingroup$ "However in QFT, the inner product of the states no longer has this probability meaning," - why do you say this? Of course the inner product is still a probability amplitude, QFT is still quantum mechanics! $\endgroup$
    – ACuriousMind
    Commented Dec 29, 2021 at 15:19
  • 2
    $\begingroup$ I'm obviously confused about something. If $\langle q| p \rangle = (2\pi)^3 2 E_p \delta(p-q)$ then how exactly is this a probability amplitude? My point in the question was that the probability is not as simple as this value squred. $\endgroup$
    – Alex Gower
    Commented Dec 29, 2021 at 15:21
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    $\begingroup$ Well, that's because $\lvert q\rangle$ and $\lvert p\rangle$ aren't really nice properly normalizable states - but neither are they in ordinary QM, you have $\langle x\vert x'\rangle = \delta(x-x')$, there, too! $\endgroup$
    – ACuriousMind
    Commented Dec 29, 2021 at 15:23
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    $\begingroup$ There are plenty of new issues in QFT, but normalization of states is not one of them. $\endgroup$
    – ACuriousMind
    Commented Dec 29, 2021 at 15:42
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    $\begingroup$ QFT is just quantum mechanics but with infinitely many degrees of freedom. States have to be normalized in QFT just like they have to be in QM. What makes you say otherwise? $\endgroup$
    – Prahar
    Commented Dec 29, 2021 at 16:47

1 Answer 1


QM and QFT are probabilistic (hence unitary) theories, and in each theory a given outcome has a specific probability in a given physical state. One well-studied special case of QM is a $1$-particle system; but just because that's not how the states in QFT look, it doesn't mean we don't still have a notion of unitarity, which is all normalization demands. We can then recover probability distributions from suitably computed expectations, in much the same manner as in QM.

For example, the equivalent of $\langle m|\hat{H}|n\rangle$ becomes $\langle p|\hat{H}|q\rangle$; in both cases, complete knowledge of the matrix elements gives the probability distribution of the vacuum's classical Hamiltonian $H$ when observed. (All the footnotes and asterisks about normal ordering, how we make a field theory's Hamiltonian finite over $\Bbb R^3$ and so on are left to the reader.)


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