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?
Asked
Active
Viewed 87 times
1 Answers
3
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
- 42,352