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Eigenstates of, for example, $\hat p$, are not elements of the standard quantum mechanical Hilbert space, i.e. $\psi(x)=e^{ipx}\notin\mathcal L^2(\Bbb R)$. This prompts the question of - given that after measurement the state of the system becomes one of these seemingly problematic states - how the time evolution can be defined such that we are able to "re-enter" the space $\mathcal L^2(\Bbb R)$ in such a way that the time-evolution is a continuous operation.

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Generalized eigenfunctions are most naturally formalized as tempered distributions - linear maps from $\mathcal S\subset L^2(\mathbb R)$ to $\mathbb C$, where $\mathcal S$ is the Schwartz space of rapidly decreasing functions. For example, we can define the distribution

$$\mathcal F_k: \varphi \mapsto \frac{1}{\sqrt{2\pi\hbar}}\int \mathrm dx \ e^{-ikx} \varphi(x)$$

This looks exactly like the inner product $\langle f_k,\varphi\rangle$ with $f_k(x) = e^{ikx}/\sqrt{2\pi\hbar}$, except for the fact that $f_k\notin L^2(\mathbb R)$, as you say. However, this will provide a guiding intuition.

If an operator $\hat A$ is defined on the Schwartz space $\mathcal S$, we can extend its action$^\ddagger$ to a tempered distribution $D$ via $$(\hat A D)[\varphi] = D[\hat A^\dagger\varphi]$$

This definition is motivated by the fact that if $D = \langle \psi,\cdot \rangle$ for some $\psi\in L^2(\mathbb R)$, then we should have $\hat A D = \langle \hat A \psi,\cdot \rangle = \langle \psi, \hat A^\dagger \cdot \rangle$.

This extension allows us to define a notion of a generalized eigenvector. Note that for $\hat P := -i\hbar \frac{d}{dx}$, $$(\hat P \mathcal F_k)[\varphi] =\frac{1}{\sqrt{2\pi}} \int \mathrm dx \ e^{-ikx} \big(-i\hbar \varphi'(x)\big) = \frac{\hbar k}{\sqrt{2\pi}}\int\mathrm dx\ e^{-ikx}\varphi(x) = \hbar k \mathcal F_k[\varphi]$$

Therefore, $\mathcal F_k$ is a generalized eigenvector of $\hat P$ with eigenvalue $\hbar k$.

In developing this technology, we have also answered your question. If $\hat U_t = e^{-it\hat H/\hbar}$ is the time evolution operator, then the time evolution of $\mathcal F_k$ is given by $\hat U_t \mathcal F_k$. In the case of a free particle, this yields

$$\mathcal F_k(t) [\varphi] = \frac{1}{\sqrt{2\pi}}\int\mathrm dx\ e^{-ikx} e^{i\frac{\hbar k^2}{2m}t} \varphi(x)$$

which leads us to say somewhat less formally that the time evolution of $e^{ikx}$ yields $e^{ikx}e^{-i \frac{\hbar k^2}{2m} t}$.


$^\ddagger$Strictly speaking we should also specify that $\mathrm{range}(\hat A^\dagger)\subseteq \mathcal S$ as well.

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    $\begingroup$ Great answer!!! $\endgroup$ – DanielC Jun 20 at 17:13
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    $\begingroup$ Yep great answer as always, thank you @J.Murray $\endgroup$ – Charlie Jun 20 at 22:41
  • $\begingroup$ Perhaps clarify that the state does not "re-enter" the space $\mathcal L^2(\Bbb R)$ as OP suggested. $\endgroup$ – nanoman Jun 21 at 2:34
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The issue doesn't come up in practice, because perfect momentum eigenstates are idealizations that don't occur in the real world.

In order to measure a particle's momentum with infinite precision and end up with a perfect plane wave, your measurement apparatus would need to be infinitely spatially large. Any real-world measurement apparatus comes with a range of experimental uncertainty, so the post-measurement state will be some kind of wave packet (in $\mathcal{L}^2(\mathbb{R}^3)$) narrowly but not perfectly centered around some average momentum.

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