# Question on wave function of free particle

1. Feb 27, 2015

### Shan K

Hi,
I was studying the solution of Schrodinger equation with no potential and found that the wave function is just a single plane wave eikx for movement of the particle in positive x direction.
But when the phase velocity of a single wave is calculated it turns out to be less than the particle's velocity and from that argument is made that the de broglie wave, which is associated with a particle, is not a single wave but superposition of waves.
So my question is that how one single wave (which we have found by solving TISE for no potential) can describe a particle, when from the calculation of phase velocity and group velocity we know that a particle can be described by a group of waves?
Any kind of help will be highly appreciated.Thanks in advance

2. Feb 27, 2015

### Staff: Mentor

Why do you want to analyse Schroedinger's equation in terms of a theory that was consigned to the dustbin when QM was fully developed?

You are caught up in one of the myths of QM - the wave particle duality - its neither wave or particle:
http://arxiv.org/pdf/quant-ph/0609163.pdf

One reason is taking the wave idea seriously you end up with the kind of issues you raise.

Thanks
Bill

Last edited: Feb 27, 2015
3. Feb 27, 2015

### Shan K

But in the book by Resnick and Eisberg they have come to the wave function by considering the de broglie wave.In one hand they have said that a de Broglie wave cannot be a single wave because the phase and the particle velocity does not match and on the other hand they have said that the de Broglie wave is given by this wave function. That is what raising the difficulty for me to conceptualize .

4. Feb 27, 2015

### Staff: Mentor

Then don't. The De-Broglie hypotheses is wrong as can be seen by what happens when you jump to a frame where the particle is at rest - what's its wavelength then?

The correct derivation of Schroedinger's equation is based on the invariance of probabilities between inertial frames - see chapter 3 Ballentine.

Its often done in beginning texts - they try to justify Schroedinger's equation by hand-wavey arguments something along the lines of this:

Its done so they don't have to say its pulled out of thin air - which is basically what Schroedinger did:
http://arxiv.org/pdf/1204.0653.pdf

The trouble is the correct derivation is complex - you probably know the math to follow it - but its a bit involved with each step slow going.

Thanks
Bill

Last edited: Feb 27, 2015
5. Feb 27, 2015

### Staff: Mentor

The correct description of a free particle in SchrÃ¶dinger wave mechanics is a wave packet constructed by superposing waves $Ae^{i(kx-\omega t)}$ with various values of $k = p/\hbar$ and various amplitudes: $$\Psi(x,t) = \int_{-\infty}^{+\infty} {A(k) e^{i(kx-\omega t)} dk}$$ If the packet contains a very narrow range of momenta, i.e. the function A(k) is zero or almost zero almost everywhere, and very sharply peaked at some central value $k_0$, then the packet has a large spread in position. In the central region of the packet, the single plane wave $Ce^{i(k_0 x-\omega t)}$ is a good approximation to the actual wave function. For some purposes we can represent the particle to a good approximation by using this plane wave.