# What does the definition the energy is not continuous mean?

1. Nov 22, 2013

### Brucezhou

What does the definition" the energy is not continuous" mean?

Title is the whole question

2. Nov 22, 2013

### atyy

Energy in general is continuous, in the sense that any value of the energy can be measured.

However, for systems such as the electrons in atoms which are stable, only certain energies are permitted. The permitted energies are discrete, not continuous. When an electron transitions between stable states, light is emitted. Only certain frequencies of light are seen, because the transitions are between these discrete energy levels.

Here is an example of the discrete frequencies of light emitted by hydrogen. http://en.wikipedia.org/wiki/Balmer_series

3. Nov 22, 2013

### SteamKing

Staff Emeritus
I guess it means that the energy comes in little packets called 'quanta'.

4. Nov 23, 2013

### tom.stoer

This is a common misconception. In general not energy but action is quantized.

5. Nov 23, 2013

### Staff: Mentor

Yep.

Its not a definition, if it is continuous, or discrete, is dictated by the differential equation that is energy, the Schrodinger equation.

Why the Schrodinger equation? - the deep answer is, believe it or not, symmetry. Yes, this abstract intellectual thing you would think has nothing to do with it, is in fact the deep reason. Curios? Read Chapter 3 Ballentine - QM - A Modern Development.

At a less detailed level check out:
http://www.pnas.org/content/93/25/14256.full

Thanks
Bill

6. Nov 23, 2013

### Naty1

Is there any 'action' that does not involve energy?

In other words, that sounds like a semantic distinction..... a more modern description to be sure.

I am more used to 'quantum of action'.....but couldn't we just as well say "energy is
exchanged in discrete quanta"....photon emissions, black body radiation,etc....
just wondering.....

7. Nov 24, 2013

### tom.stoer

All I am saying is that we have quantum states with non-discrete energy, i.e.

$(H-E)|\psi\rangle = 0$

But the action is discrete in terms of these energy quanta, i.e.

$S_n = nE;\;n = 1,2,\ldots$

8. Nov 24, 2013

### Staff: Mentor

Hmmmm. I thought you meant it in a slightly different sense - namely when the wave function is written using the path integral formalism we need the quantum of action dividing the action so the wavefunction is dimensionless.

But for a free particle the action is not quantisized nor is the wave-function.

Thanks
Bill

9. Nov 24, 2013

### Naty1

Is the OP's question pertaining to relativistic or non relativistic theory or both? How about the above statement?

10. Nov 24, 2013

### Staff: Mentor

Its trivial.

Put the classical free particle action in the path integral formalism - see the following:
http://en.wikipedia.org/wiki/Path_integral_formulation

For a free particle the action can take any value.

The quantitization comes from the fact it must be divided by the quantum of action, Planks constant, so the wavefunction is dimensionless. That's how quantitization enters into it, that's the sense action is quantisized. If Planks constant was zero you get Classical Mechanics - the fact its not is what QM is all about.

It has been discussed on Physics forums before - check out:

Again - action is not quantizied in the sense in all situations it can only take on particular values, it is in the sense a quantum of action, Planks constant, needs to be introduced.

Thanks
Bill

Last edited: Nov 24, 2013
11. Nov 24, 2013

### tom.stoer

I think instead of discussing the PI one should look at Fock space (even if there is no "action operator"); the Fock space of free particles represents directly the fact that there _are_ discrete quanta; of course for non-interacting particles (not confined in a box) energy is not discrete; but adding one quantum of a certain energy always increases the action by an amount of hf in the sense of Planck

I know that this is not the way looking at it in QFT, but it could allow us to make sense of "quantized action" in the canonical approach as well

12. Nov 24, 2013

### atyy

That's an interesting way of thinking about it. But given that the Fock states are only basis states, does changing a state from one Fock state into a superposition also change the action by a quantum?