# Sample Space for Free Particle in the general case

1. Oct 4, 2011

### IronHamster

I am a beginner to quantum mechanics and am trying to make sense of Schrodinger's Equation. I am attempting to find probabilities in the case of a free particle in the general case.

It is my understanding that the solution to Schrodinger's Equation in the general case of a free particle is as follows:

$$\psi(X,T) = e^{i/\hslash ( px - Et)}$$

The modulus square of this is 1, which means the probability density function is uniform.

Two questions:
1. Over what values of x is this pdf defined? Can we eliminate all values of x > ct?
2. Am I correct to interpret x as the distance from the (known) starting position of the particle at t = 0?

Thanks.

2. Oct 5, 2011

### Matterwave

Notice that that wave-function is not normalizable. The integral of the modulus square of that wave-function over all space is infinite. A free particle wave function cannot actually be what you gave, but must be a wave-packet.

3. Oct 5, 2011

### IronHamster

So are you saying that the solution I mentioned does not describe a free particle wave? I'm not sure how that could be, I have read from multiple sources that it is.

Is there a different approach that needs to be taken to achieve a normalizable function?

4. Oct 5, 2011

### Matterwave

A real free particle cannot be represented by that function because that function is not normalizable. A real free particle is represented by a wave packet. That function is a function of a particle with exactly 1 momentum (p), but really a particle is represented by a wave packet which has a range of momenta.

You can say that your equation is only a "partial" solution. It hasn't been fixed up yet.

Still, that function is useful for many applications. For example, if we are doing a scattering problem off of a finite square barrier, we tend to just use that function. The central results you obtain by using that function (the transmission and reflection coefficients) is surprisingly good to the results you would get if you made wave packets; however, doing a scattering problem with wave packets is a nightmare.

5. Oct 5, 2011

### IronHamster

Oh ok that makes sense. Thanks!

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