# Are electrons particles?

1. Dec 9, 2008

### emperrotta

I am new to Quantum Mechanics. In my University Physics class, we discussed the concepts of matter waves and the electron cloud. In describing an electron as a probability wave, is that a way of saying that an electron in an atom is a particle that does not move in any clear path such as the model we are taught early in our education of an electron having a circular orbit around the nucleus, rather, the electron is a particle with a random path relative to a circular path that has certain probabilities of bringing the electron to certain locations at any given time? Additionally, the shape of this probability wave is a function of a electron's quantum state within the atom.

Furthermore, I am trying to put my head around orbital angular momentum and spin angular momentum for electrons. Now, I understand that angular momentum in these two terms does not have the same meaning as it does in Newtonian Mechanics. However, let's say, two electrons are in shell n=4. One electron has a value of l=3 for its orbital quantum number and the other electron has a value of l=1. By the equation:

L =$$\sqrt{l(l+1)}$$hbar
L=orbital angular momentum

the electron with l=3 should have a higher orbital angular momentum than the electron with l=1. I know neither electron is moving in a circular path, but does the electron with a higher momentum move faster in its path since it has more momentum.

Thanks for help, anyone.

2. Dec 9, 2008

### malawi_glenn

Since you already ruled out that particles in QM-world don't follow a classical path, you must reprase your question to something else.

You can ask if an electron in state L = 3 has bigger expectation value of momentum or kinetic energy than an electron in state L = 1.

It is quite meaningless to talk about velocities for bound particles.

3. Dec 10, 2008

### emperrotta

I am not sure what you mean by "has bigger expectation value." I think the biggest thing I am struggling with is whether or not an electron has some sort of orbit, just not the clean circular orbit we learned about when we were younger. Isn't an electron a particle with mass, albeit a very small mass, moving in some direction? After all, when we an electron beam is created, are we not giving electrons momentum and kinetic energy? Don't they move with some velocity? Do the concepts of momentum and kinetic energy change when we are referring to an electron bound to a nucleus?

Or is the best we can do is detect the electron at various points around the nucleus, and not think of it as moving, but just thinking of it as existing in certain places at certain times.

4. Dec 10, 2008

### Nick89

I'm certainly no expert, but I think the latter is the way you need to look at things. I think a bound electron in for example a hydrogen atom simply does not have a definitive position / path. As far as QM can tell us, we can only expect it to be in certain positions: the chance of finding the electron (when you do a measurement for its position) is much higher in one particular spot than in another. This is the probability cloud: there is a 'cloud' around the nucleus and the 'cloud density' tells you the probability of finding the electron there.
The expectation value is simply the position where the electron is expected to be most of the time (you could see this as the most probably position). Still, it is only a probability and we won't know certain until we measure it. Before we measure it, for all we know the electron could be on the moon (although that chance is extremely small).

EDIT
Here is a very nice graphical approach to electron 'orbits'.

5. Dec 10, 2008

Excellent Question :)

6. Dec 10, 2008

### Nick89

An electron moving at a certain speed in an electron beam is not the same as an electron bound to a nucleus.
If you know what the shrodinger equation is, you need to solve that to find the wavefunction of the electron.

In the case of a (nearly) free electron (such as in the electron beam) there is no potential (ignoring gravity and the effects of other electrons) in the shrodinger equation.
In the case of a bound nucleus, the shrodinger equation contains a potential term which changes the solution of the equation (the wavefunction) significantly.

Basically, the potential determines the solution to the shrodinger equation and thus the wavefunction. Two different potentials often give radically different wavefunctions.

Whether the concepts of momentum and kinetic energy change for a different potential, I don't know actually, that's a good question... (I think I should really know that by now hehe...)

7. Dec 10, 2008

### Naty1

That's a good start.

Think of the electron as distributed in a cloud...it's exact character is "cloudy, foggy, shrouded in randomness"...read the following...

Look up "expectation value" in Wikipedia....

Then try "electron cloud" which describes the Bohr orbital model and maybe "double slit experiment" ...and you'll be on your way....

8. Dec 10, 2008

### emperrotta

Thank you everyone so far. I think I am getting there. So the way I understand it, a bound electron's position can never be fully pinpointed. The best we can do is use Schrodinger's Equation to form a wave function for the electron's probability of being in certain places at a given time based on the electron's quantum state.

Now if I go back to my original post mentioning orbital angular momentum. Let's say a hydrogen atom is placed in a magnetic field. Is this where the values for the electron's orbital angular momentum and spin angular momentum and the magnetic dipole moments coupled with each come into play? Do the values for these two momentums effect the force placed on the atom. Are the values for these two momentums as calculated by the equations

L = $$\sqrt{l(l+1)}$$hbar
S = $$\sqrt{s(s+1)}$$hbar

statistically likely values? Is that what was meant by expectation value?

Last edited: Dec 10, 2008
9. Dec 11, 2008

### Nick89

I don't think so.

If you have a hydrogen atom you may or may not know that the wavefunction is a function that has a radial part (often called R) and an angle dependent part (often called Y).
These's R and Y have some subscripts which are the quantum numbers (n, l, m) but they are not important for this example.

The (time-independent) wavefunction is thus:
$$\psi = R(r) \, Y(\theta, \phi)$$

If you want to find the expectation value for example for the radius (r) that means you are looking for the predicted mean value, or loosely speaking the radius in which you expect to find the electron when you do a measurement.

You can calculate it by integrating over the square of the wavefuntion and multiplying by the radius:
$$<r> = \iiint r | \psi |^2 dV$$
(Because we are working in spherical coordinates, $$dV = r^2 sin\theta \, dr \, d\theta \, d\phi$$)