# A question about wave function

1. Jul 8, 2011

### coki2000

Hello physicists,

My question is why a wave function has to be 0 at infinitiy. The author who I was studying him quantum book explains my question by saying that it is necessary to make wave function normalizable. But in my opinion there can be an arbitrary particle which can be found all around the universe therefore at infinity. Please illuminate me thank you all

2. Jul 8, 2011

### Avijeet

There are two kinds of states for wave functions: bound states and scattering states. For bound states the energy of the particle is less than the potential energy at infinity, such a particle cannot be found at infinity and the wave function goes to zero at infinity. On the other hand for scattering states the energy of the particle is more than the potential energy at infinity. In this case the particle can be found anywhere and the wave function does not go to zero at infinity. The wave function for a free particle is such an example. The wave function in this case is not normalizable.

3. Jul 8, 2011

### coki2000

Thank you Avijeet,
But for a partical, the probability to stand anywhere in space must be 1, mustn't it?

4. Jul 8, 2011

### chogg

You should think of it as a physically reasonable assumption you choose to make. If the wavefunction were not zero at infinity, wouldn't you find that surprising? :) We don't tend to encounter quantum delocalization over very large distances. Hence, this assumption.

Also: consider the normalized wavefunction (or its modulus-squared, the probability density). There is only so much probability to go around: it all has to add to 1. If your probability does not go to zero at infinity, the particle will be so "spread out" that you will never detect it in your laboratory -- so why are you analyzing that particle, instead of one that's a little closer to home?

5. Jul 8, 2011

### Avijeet

For a free particle, the amplitude of the wave function at any point is a constant. The wave function is not normalizable, so the probability interpretation of the wave function is not applicable. All you can say is that the particle is equally probable at all the points.