# Time evolution in quantum mechanics of states not contained in the Hilbert space

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.

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.

• Great answer!!! – DanielC Jun 20 at 17:13
• Yep great answer as always, thank you @J.Murray – Charlie Jun 20 at 22:41
• Perhaps clarify that the state does not "re-enter" the space $\mathcal L^2(\Bbb R)$ as OP suggested. – nanoman Jun 21 at 2:34

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.