# Quantum mechanics of a particle

1. Mar 4, 2005

### sarabellum02

I have a couple things I don't understand:

1. Why is it that the more you conifne a particle, the higher its energy is?
2. Why is it that the more nodes there are in the wavefunction the higher the energy is?
3. What causes the energy of a particle to be quantized?

thanks!

2. Mar 4, 2005

### dextercioby

1.What do you mean by "conifne a particle"...?If you mean "confine a particle",i have to ask you what do you mean by this...?
2.What are nodes of a wave function...?I've never heard that a function would have nodes...
3.Principles of QM...Essentially the second.

Daniel.

3. Mar 4, 2005

### marlon

Are you referring to nodal planes of orbitals???then your answer is the magnetic quantumnumber...(at least to some extent)

Mother Nature

marlon

4. Mar 4, 2005

### masudr

It's not true that the more you confine something the more energy it has -- what is true is that the more you confine something, the variance in it's momentum will increase.

As for the node thing; I think sarabellum means that in a "particle in a box", the number of nodes refers to the number of points where the wavefunction $\psi(x)=0$. There is no reason as to why that is -- it's just how the solutions to Schrodinger's equation works out.

And for the third point, as marlon says, that's just how nature is. If energy wasn't quantised, one could ask "Why is energy continuous?" and so on.

Masud.

5. Mar 4, 2005

### marlon

This is not entirely correct. Just look at quarks...

Besides, what exactly do you mean by this and how does it apply to QM?

marlon

6. Mar 4, 2005

### Edgardo

sarabellum probably means, if you make the well smaller, the energies become higher.
Click on this link http://www.quantum-physics.polytechnique.fr/en/pages/p0203.html, and change the width of the well, by dragging the left bottom corner of the well.

sarabellum probably means for example the harmonic oscillator eigenstates.
The higher the energy, the more zeros (sarabellum calls them nodes) the wavefunction has http://encyclopedia.laborlawtalk.com/Quantum_harmonic_oscillator [Broken]

I think that's just the result of QM calculations.

Hmm....as marlon said, it's nature.

Last edited by a moderator: May 1, 2017
7. Mar 4, 2005

### masudr

It's the celebrated Uncertainty Principle.

8. Mar 4, 2005

### BlackBaron

I think I know what you mean, bur truth is, I'm not sure how exactly to give you a good answer.

As some mentioned before, nodes are the points where the wavefuncion equals 0 (like the nodes of a vibrating string).
The fact that the more nodes there are, the higher is the energy actually happens with any kind of wave. Again, think in a vibrating string, the stronger you make it vibrate (i.e. the more energy you give to it), the bigger is the number of nodes.
The explanation on why that is so, is pretty much like this: the higher the number of nodes per unit distance, the more oscilations there are, wich means the bigger is the wavenumber, and thus the higher is the energy (remember that in the case of the wavefunction, the momentum of the particle it's proportional to the wavenumber).
In fact, in more than one QM problem (I'm not sure if there's a general rule, maybe not), the number of nodes is actually closely related to the quantum number you use for the energy.

More or less, this comes as a result of SchrÃ¶dinger equation, energy is quantized whenever there are bound states (i.e. the potential confine the wavefunction to be in a certain place), when that happens, you have to force the wavefunction to be zero at certain point in space (like infinite, or the extremun of a infinite square well) and those boundary conditions can only be met by certain values of the wavenumber (and thus, the energy).
There are good discussions on this point in some standard Quantum Mechanics textbooks, like Eisberg & Resnick and Cohen-Tanoudji.

9. Mar 5, 2005

### sarabellum02

$$\frac{-\hbar^2}{2m} \frac{d^2 \Psi}{dx^2}$$