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Let's assume we have a wave function described by a wave equation and it is a function of space and time $\psi : \mathbb{R}^4 \rightarrow \mathbb{C}$.

This function needs to be normalized, so if I understood the bra-ket notation well:

$$\langle \psi | \psi \rangle = {\iint}_{-\infty}^{\infty} \psi^*(x,t) \psi(x,t)dx dt = 1$$

(asterisk means conjugate.)

But I see some problems here:

  • Wavefunctions are time reversible this means half of it at the past, so that would mean there is 50% chance that the particle is never found at all.
  • In order to get finite value for a full domain integral, the value of the function must approach zero as we go towards the infinite. That would mean that finding the particle diminish over time. That's interesting because if there are nothing that would measure it, then it's gone for good.

Am I missing something or is this an integral over space only and not time?

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    $\begingroup$ At any moment in time the probability of finding the particle is unity. So there should not be any integral over time in your equation. $\endgroup$ – Mark Mitchison Jan 21 '15 at 9:46
  • $\begingroup$ @MarkMitchison: That's the right answer. I've hesitated to make the same comment because OP actually mentions this in his first point. We can always go with "as far as we can tell, it just works", but I'd like to know if a theory, where at some time there is a chance of not being able to find the particle at, would have any use. I have no crucial argument against a theory which tells you "in 5 seconds, even if you look everywhere at once, the chance you find a particle at all is only 80%". $\endgroup$ – Nikolaj-K Jan 21 '15 at 9:50
  • $\begingroup$ @NikolajK Sure, you could imagine a particle in a square well with a finite probability of tunneling out over time. If you want to describe this with a wavefunction for only the positions inside the square well, then you can introduce an imaginary term in the Hamiltonian. This reduces the norm of the wavefunction over time, which tells you that the probability of the particle staying in the well reduces over time. $\endgroup$ – Mark Mitchison Jan 21 '15 at 9:54
  • $\begingroup$ @MarkMitchison: Right, do you know off hand if the complex energy models are placed, somehow, inside the theory which respects the "axioms of quantum mechanics" (especially unitary of time evolution), or are those cases considered a deviation of it (a tool to model certain effects without digging into a more detailed version of QM where unitary holds again.) $\endgroup$ – Nikolaj-K Jan 21 '15 at 9:58
  • $\begingroup$ @NikolajK No, complex energy models are definitely not unitary, since they do not preserve the norm. $\endgroup$ – Mark Mitchison Jan 21 '15 at 9:59
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Where did you get that formula? The correct normalization does not involve the time integral.

Denoting with $| \psi(t) \rangle$ the state at time $t$, the normalization condition reads $$ \tag{A} \langle \psi(t) | \psi(t) \rangle = \int d^3 x | \psi(x,t) |^2 = 1, \forall t. $$ What this is telling you is that at any time $t$ the particle must be somewhere.

Moreover, the unitarity of the time-evolution tells you that at any time $t$ the space integral in (A) is 1, so if you also integrate with respect to time you get a trivially infinite result. Vice versa, if the integral you report if finite, the evolution cannot be unitary.

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  • $\begingroup$ Probably because I don't see $\psi(t)$ in the brakets, just $\psi$ or other letters which made me think that the elements of the Hilbert space contains whole space-times not just the space part of them at a given moment. It might be obvious to the writer by not to me. No one pointed this out anywhere in texts I found online. I also needed to play the same with the Laplacian as well, to find out it's also space only... $\endgroup$ – Calmarius Jan 21 '15 at 11:17
  • $\begingroup$ @Calmarius it wasn't my intention to be rude! I just asked because I can conceive a situation where the evolution is non-unitary and things like that (see e.g. Phoenix87's answer), though I'm not expert in those matters. So if you found that equation in some particular text maybe they were meaning something else. But I see this isn't the case. $\endgroup$ – glS Jan 21 '15 at 11:34
  • $\begingroup$ I didn't find this formula anywhere. Typically the mathematical Hilbert-spaces and the braket notation is introduced using function that have only 1 variable and I thought this generalizes seamlessly to 4 dimensional functions as well. So I thought I need to integrate over time as well. $\endgroup$ – Calmarius Jan 21 '15 at 11:54
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To reply to the title, which doesn't specify the domain of integration, the answer is yes unless the whole configuration space is considered.

For concreteness, let $\Gamma$ be the configuration space of your system, and let $D$ be any (Borel) subset of $\Gamma$. Given that a particle is described by a time-dependent wave-function $\psi:\mathbb R\to L^2(\Gamma)$, the probability of finding it in the domain $D$ at the time $t$ is $$\text{Pr}_\psi(t,D) = \int_D|\psi(t,x)|^2\text d\Omega(x),$$ where $\Omega$ is a regular probability measure on $\Gamma$ (usually the Lebesgue measure when $\Gamma = \mathbb R^n$). This probability can diminish over time, but since the total probability of finding the particle in $\Gamma$ is 1, this means that the probability of finding the particle is increasing in the complement $\Gamma\smallsetminus D$. A physical interpretation of this behaviour is that the particle is, on average, drifting away from the region $D$ towards other regions of the configuration space.

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  • $\begingroup$ I need to learn a lot to fully understand this answer. So far I need to look up "configuration space", "Borel subset", "regular probability measure", "Lebesgue measure". So I need to ask even if it's implied: does your answer imply that the particle must always be found somewhere in the configuration space all the time? So it cannot simply go missing? $\endgroup$ – Calmarius Jan 21 '15 at 14:29
  • $\begingroup$ Yes, there must be a region where the probability of finding your particle is non-zero. It can't go missing because the Schrödinger equation implies a conservation law for $|\psi(t,x)|^2$. $\endgroup$ – Phoenix87 Jan 21 '15 at 14:31

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