# What does not normalizable ential

• amrhima
In summary, the free particle's wave function is not normalizable, which casts doubt on the validity of the statistical interpretation when it comes to free particles.

#### amrhima

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?

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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/

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

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.

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

Oops! I fixed it. Thanks!

## What does "not normalizable" mean?

The term "not normalizable" refers to a mathematical function or equation that is unable to be transformed into a normalizable form. This means that the function or equation cannot be properly integrated or summed, making it difficult to analyze or solve.

## Why is normalizability important in science?

Normalizability is important in science because it allows for mathematical functions and equations to be properly analyzed and solved. It also ensures that physical quantities and properties are well-defined and can be compared across different systems or experiments.

## What are some examples of "not normalizable" functions or equations?

Some examples of "not normalizable" functions or equations include the wave function in quantum mechanics, the free energy equation in thermodynamics, and the Fourier transform of a function with infinite support.

## How is normalizability related to the concept of infinity?

Normalizability is closely related to the concept of infinity because a function or equation that is not normalizable often involves infinitely large or small values. This can make it challenging to properly integrate or analyze the function or equation.

## Can a "not normalizable" function or equation still be useful in science?

Yes, a "not normalizable" function or equation can still be useful in science. While it may be difficult to solve or analyze, it can still provide valuable insights and understanding of complex systems or phenomena. In some cases, approximations or alternative methods can be used to work with these types of functions or equations.