Wave function not normalizable

Question

Does the solution of the Schrodinger equation always have to be normalizable? By normalizable I mean, given a wavefunction $\psi(x)$

$$\int_{-\infty}^{\infty}|\psi(x)|^2 dx<\infty \qquad \text{or}\qquad \int_{0}^{\infty}|\psi(x)|^2 dx <\infty$$

What would be the physical implications if one (or both) of those integrals diverges. From the viewpoint that the Copenhagen interpretation is one of the most popular and the wave function is interpretated as a probability distribution in this case; will the wavefunction diverging be valid for any other interpretation of quantum mechanics. Does anyone know any wavefunctions which are not normalizable? What if there was a singularity at 0 that made it diverge at all times. For instance

$$\int_{0}^{\infty}|\psi(x)|^2 dx \to\infty\quad \text{but}\quad \int_{a}^{\infty}|\psi(x)|^2 dx <\infty $$

where $a>0.$ Would the second integral count as a valid pdf?

Answer 1

From my limited knowledge of this subject, I would say that a non-normalizable wave-function wouldn't really make any physical sense.

Remember, the wave-function is a function whose value squared evaluated between two points represents the probability that the particle will be found between those two points. So, the restriction that wave functions be normalized is basically just a nod to reality - the particle must be found SOMEWHERE.

Normally, the restriction is: $$ \int_{-\infty}^ {\infty} |\psi(x)|^{2} dx = 1, $$ i.e. the probability of finding the particle if you looked between $-\infty$ and $\infty$ is 1. Having a probability greater than one of finding the particle between these bounds would not make any physical sense.

Having a wave-function described by the equations you posted above would imply that there is an infinite chance of finding the particle anywhere.

Answer 2

The wavefunction must be either normalizable or the limit of a sequence of normalizable functions which in general are known as distributions (generalizations of functions). A well known example of a distribution is the Dirac delta "function," $\delta(x)$. If the spatial wavefunction is $\psi=\delta(x_0)$, then the momentum wavefunction will be of the form $\psi\propto e^{-ipx_0},$ which is not strictly normalizable. The opposite example is $\psi(p)=\delta(k)$ which are infinite plane waves with spatial wavefunctions of the form $\psi\propto e^{ikx}.$

The way we deal with this mathematically is assume such states are normalizable i.e. $$\int |e^{ikx}|^2dx=\int dx\equiv 1,$$ even though this integral is divergent. The physical solution is the fact that 'infinite plane-waves' are never actually physically infinite.

Answer 3

yes, good example will be solution of free particle.Where solution is like a plane wave solution hence such sols do not represent physically accepted states.this is the reason why any problem related to free particle should have a initial wave function which can be normalized other wise we cannot proceed further.