We all learn that quantum states need to be normalized, as they are associated to probabilities which needs to sum up to one. However, I would like to know whether you have other valid reasons to motivate this normalisation. Do you have examples in which non-normalized states give rise to paradoxical situations, except for probabilities larger than one? One scenario could be the case of nonlinear Schrödinger equations (e.g. Gross-Pitaevskii equation) non-normalized wave functions clearly give rise to different results.

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    $\begingroup$ My understanding of this is that since measurement projects onto a subspace, the magnitude of the vector before projection is irrelevant. The normalisation to 1 is I think just a convention. $\endgroup$
    – jacob1729
    Nov 20 '18 at 17:37

There is no particular reason to normalize a quantum state if you just define, say an expectation value of some observable $A$, as

$$ \langle A \rangle := \frac{\left\langle \psi | A \psi \right\rangle}{\left\langle \psi | \psi \right\rangle}. $$

This is actually often done. It is just a simplification and therefore a sensible convention to set $\langle \psi | \psi \rangle = 1$.

Also, the Schrödinger equation is linear in the variable $\psi$, so if we take as initial value $c \times \psi$, with $c \in \mathbb{C}$, the solution will also just be multiplied by $c$ and nothing physical changes.

To say something about non-linear equations like the Gross-Pitaevskii equation, it is correct that you would have to correct the interaction term in this equation by a factor of $1/c$ to care for the different normalization.


Do you have examples in which non-normalized states give rise to paradoxical situations, except for probabilities larger than one?

I don't, but perhaps the following will help.

If $\psi$ is not normalized but is normalizable, there is actually no issue - all functionals of $\psi$ can be modified to account for the lack of normalization - just normalize in the functional formula.

However, if $\psi$ is not normalizable for some region $\Omega$, it means one cannot use the Born rule

$$ \frac{\int_{x\in \omega} |\psi(x) |^2 dx}{\int_{x\in\Omega} |\psi(x)|^2dx} = \text{probability particle is in }\omega, $$ because the denominator integral does not exist or is infinite, so one either has to adopt some different meaning for $\psi$, or one decides such $\psi$ is not admissible as far as the Born meaning is used. For example, if $\Omega$ is the whole infinite space,

$$ \psi(x) = e^{ipx/\hbar} $$ is not normalizable, so it is not applicable as a Bornian description of the system in that space. However, if we choose $\Omega$ to be a finite volume, say a box, the above psi function becomes normalizable and is admissible. So the normalizability depends just as much on the configuration space region as on the function itself.

On the other hand, if we have something like the Dirac delta distribution $$ \psi(x) = \delta(x-x_0) $$ this is not normalizable in the sense of the Born rule at all. It can be used as initial condition for the Schr. equation, but the resulting solution $G(x,t)$ is not a realistic psi function for the whole infinite space. It has other uses - it is the Green function of the Schr. equation for the whole infinite space.


Consider a classical scattering problem: you project, say, N particles per second at a target and you count the number of particles scattered in a solid angle $d\Omega$: $dN\propto N$ so the ratio $dN/N$ is always smaller than 1. Reported to one particle, it is its probability to be scattered into $d\Omega$.

Now, in QM one observes superpositions of states of scattered particles. It implies linearity of their equation, kind of a wave equation. The wave equation solution must be normalized in order to obtain the value of calculated $dN$ without ambiguity. The rest is similar to the classical case: for one particle you have the corresponding probability, but the equation of motion must admit superposition of amplitudes (solutions).

The third thing is that we always need many particles to have a reliable statistics and certain judgments, so in reality we count the number of particles (per second or the total number, whatever). Thus the probability is only a part of the whole picture, which implies many-many particles involved in order to cover all corners of the wavy picture. Neither in classical nor in quantum physics one-particle "experiment" is sufficient.


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