# What does not normalizable ential

## Main Question or Discussion Point

What does "not normalizable" ential

the free particle's wave function is not normalizable...what does that mean??
I understand there are mathematical tricks to help, but i still don't understand why it is not normalizable? does that cast the whole statistical interpretation into doubt when it comes to free particles?

Last edited:

Related Quantum Physics News on Phys.org
jtbell
Mentor

When we talk about the "free particle wave function" we usually mean this (in the case of one-dimensional motion):

$$\Psi(x,t) = A e^{i(k x - \omega t)}$$

By "not normalizable" we mean that there is no value of A that makes the following integral true:

$$\int^{+\infty}_{-\infty}{\Psi^* \Psi dx} = 1$$

which makes the total probability of finding the particle somewhere equal to 1.

It simply means that the wave function given above is not actually a valid wave function for a free particle, strictly speaking. Physically, it means that it is not possible for a particle to have a completely definite, exact value of momentum $p = \hbar k$.

To get a valid wave function for a free particle, that is localized in space and is normalizable, you have to superpose wave functions that span a continuous range of momentum values:

$$\Psi(x,t) = \int^{+\infty}_{-\infty} {A(k) e^{i(k x - \omega t) } dk}$$

where a function A(k) gives the amplitude of the wave that has momentum $p = \hbar k$. This is called a wave packet. One common example (because it's relatively easy to analyze) is the Gaussian wave packet:

http://musr.physics.ubc.ca/~jess/hr/skept/GWP/

Last edited by a moderator:
Matterwave
Gold Member

The last integral in jtbell's post should have a "dk" at the end.

Fredrik
Staff Emeritus
Gold Member

I just want to add that all members of a Hilbert space like $L^2(\mathbb R^3)$ have finite norm. So if v is an arbitrary non-zero vector in the Hilbert space, then $v/\|v\|$ is a unit vector in the direction of v, i.e. it's a "normalized" version of v.

So the "not normalizable" wavefunctions aren't really wavefunctions. They also don't correspond to physically realizable states.

Thanks a lot jtbell, matterwave and fredrik.

jtbell
Mentor

The last integral in jtbell's post should have a "dk" at the end.
Oops! I fixed it. Thanks!