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?

  • 2
    $\begingroup$ Momentum eigenstates are non-normalizable in the position state space. $\endgroup$
    – resgh
    Apr 7, 2014 at 5:56
  • $\begingroup$ @namehere, what does it mean "momentum eigenstates are non-normalizable" $\endgroup$
    – user37343
    Apr 7, 2014 at 22:26

3 Answers 3


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.

  • $\begingroup$ I think your explanation is sane. We impose maximal probability to be unity.Thence make adjustments that follow from there. Here we use mod-squared as per restrictions from the 2-norm. I think this is the essence. $\endgroup$
    – user37343
    Apr 7, 2014 at 22:29

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.

  • $\begingroup$ The actual way we deal with plane waves is by constructing wave packets. Basically, we assume $$e^{i(kx-\frac{\hbar k^2}{2m}t)}$$ are valid solutions (for every $k$) and then go on saying: well, then the general solution is a linear combination of these. $$\Psi \propto \int_{-\infty}^{\infty}{dk\phi(k)e^{i(kx-\frac{\hbar k^2}{2m}t)}}$$ If we choose the $\phi(k)$ right, this function is normalizable. Of course in reality the plane wave is an idealization like the frictionless surface, so unusual behaviour is not entirely unexpected. $\endgroup$
    – Wouter
    Apr 9, 2014 at 8:33

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.


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