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Why do we omit solutions that do not converge at $\pm\infty$ from the physical Hilbert space, what is the argument for us being allowed to do so?

Thomas Fritsch
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gyzgyz123
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    Who says that we are removing these solutions (solutions to what?), and in what context? Perhaps also have a look at https://physics.stackexchange.com/q/331976/50583 and its linked questions. – ACuriousMind Jul 28 '19 at 16:18
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    I second the questions of the previous comment, but it's worth noting that in most situations where you're dealing with functions and you want to make them converge at $\pm \infty$, the ones that do not converge aren't even in the Hilbert space, as this space is usually $L^2$ or some extension of it to tempered distributions – user2723984 Jul 28 '19 at 16:21
  • @ACuriousMind Shankar when solving problems in his book Principles of quantum mechanics. I can give plenty of examples from there, the most recent one that kinda made me ask why( he probably did explain it somewhere but I have since forgotten where and can't find it) is on page 192. https://imgur.com/a/nB3grCe – gyzgyz123 Jul 28 '19 at 16:22

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We are only interested in solutions $\psi(x)$ which are normalizable, i.e. $$ \int_{-\infty}^{+\infty} |\psi(x)|^2 \text{d}x = 1.$$

If $\psi(-\infty)$ or $\psi(+\infty)$ diverge, then this normalization would not be possible.

Edit:

Actually the above is nearly true (for practical physical purposes), but not completely true (in strict mathematical sense). There are some pathological counter-examples, where $\psi(x)$ is normalizable, but $\psi(-\infty)$ and $\psi(+\infty)$ are divergent. See question "Normalizable wavefunction that does not vanish at infinity" and its answers.

Thomas Fritsch
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