# The Free particle

1. Apr 14, 2012

### sciboudy

could any one explain the free particle in Quantum mechanics?
when the potential is zero

2. Apr 14, 2012

### Staff: Mentor

This subject occupies an entire chapter (or even more) in most quantum mechanics textbooks. I suggest that you ask specific questions.

3. Apr 17, 2012

### sciboudy

okay i'm using grifth

in the page 45 he transform the forlmla of ψ as function of x "or he intgrate to X "

but the solution was no normalize so he transform it to κ and intfrate to κ dκ how?

and what k represent ?

4. Apr 17, 2012

### Psi^2

5. Apr 17, 2012

### sciboudy

okay why we transform to it ??

6. Apr 17, 2012

### Psi^2

Read it, it’s in Griffiths book. You cannot normalize wave function of a free particle unless you do transform to integral over the continuous variable, which is k.

7. Apr 17, 2012

### sciboudy

why he choose to ingrate to K ? specially ?? how did he transform it to K ?

8. Apr 22, 2012

### sciboudy

i want understand how is we find it's equation and describe it's change in the wave packet

9. Apr 22, 2012

### SergioPL

The wave function can be represented many ways. One of these ways is over the spatial domain, but it can be also represented over the wave vector domain also known as momentum representation. Both representations are related via Fourier transformation.
Wave vector and momenta are related: p = k * h. In the wave vector domain you can get the probability that the particle have certain momenta via Copenhagen interpretation P(k)=|ψ(k)|^2. Any way this is a matter explained on any quantum mechanics textbook.

Best regards,
Sergio

10. Apr 22, 2012

### sciboudy

thank you very much you made it easy :D
i got it
but i want to know how is the probability denisty of finding particle in the wave function increase ?

11. Apr 22, 2012

### sciboudy

and what you mean by Energy is not quantized for a free particle. and how you got this ?

12. Apr 22, 2012

### Psi^2

You simply solve Schrodinger equation for free particle, and look at the result. No boundary conditions, no quantization ;)

13. Apr 22, 2012

### Matterwave

The basic idea is that if you have a free particle, it cannot exist in a energy-momentum eigenstate because those eigenstates are un-normalizable. Additionally, experimentally speaking, knowing the momentum of this particle perfectly well means not knowing the position of this particle at all, which means you can't really measure this particle since you don't know where to measure.

Therefore, the particle must exist in some linear superposition of eigenstates. Because the eigenstates are a complete set of states, one can always do this, just like for any other problem.

The difference in this situation from the situation in e.g. the particle in a box, is that the energy-momentum eigenstates are a continuum of states rather than discrete states. Therefore, a linear superposition is an integral and not a sum over states. Specifically, this integral is a Fourier integral because the eigenstates are the Fourier factor e^ikx.

14. Apr 22, 2012

### the_pulp

Hi there, related to this topic, a free particle spin is something that can take a discrete range of values, as it happens with electromagnetic o colour charge. However the other important observable, impulse, can take a continum range of values. This seems suspicious to me since nature seems to be formed by a finite (inmense, but finite) number of mathematical objects. Isnt out there any model or research where it is supposed that impulse is also a discrete observable? perhaps lattice quantum mechanics or something like that?

Thanks!

15. Apr 23, 2012

### zhangyang

I think it does not exist.

16. Apr 23, 2012

### Psi^2

I agree.

17. Apr 23, 2012

### sciboudy

why you agree

18. Apr 23, 2012

### Psi^2

Because it is approximation. In realty, there is no such a thing as a free particle.

19. Apr 23, 2012

### USeptim

Hello! That's a very interesting matter, but I don't know of any physical model that works on discrete time, space, momentum and energy. As is well known, these variables cannot be perfectly measured by the indeterminacy principia, but they move on the continuum of real numbers.