# Conservation of energy after measurement

1. Aug 10, 2011

### Amir Livne

Is conservation of energy, momentum, and other physical properties absolutely true in quantum mechanics, or only on average?

As an example, think of a single particle in free space. Measure its energy, and write down the result. Then look at where it is, and measure the energy again. You'll probably get a different result the second time. I can see that the average of that distribution - at least in this case - is the original energy.

This situation does not seem to mean too much though, since if you measure a particle it's not a closed system, so you don't expect energy to be conserved. Is there a more general statement of conservation of energy?

2. Aug 10, 2011

### kith

Mathematically speaking, the generators of the symmetry operators of your system are conserved quantities (this corresponds to Noether's Theorem of classical physics). Since sharp energies, momenta, etc. are only defined if your system is in an eigenstate of the corresponding observable, the conservation law is only true for these eigenstates.

If you measure other quantities (for example position), your system leaves the eigenstate of the conserved observable. So yes, measurements break conservation, but as you noted, this is due to introducing an environment which interacts with the system.

Last edited: Aug 10, 2011
3. Aug 10, 2011

### Amir Livne

I was actually wondering about the system together with its environment.

Is the total energy of both parts still conserved? Is there any way to even state conservation of energy, momentum, spin, charge and so on? Since the observer interacts with the system, she should be included in the description, so these properties might not even have exact values?

From what I read (on popular level) about QM physics, when measuring we always collapse to the position basis. So is energy/momentum of a system really ever defined?

4. Aug 10, 2011

### kith

Only if the state of the whole is an eigenstate of the complete Hamiltonian (which consists of the system Hamiltonian, the environment Hamiltonian and the interaction Hamiltonian).

Yes, atoms can be easily prepared into energy eigenstates. From measurements, you can then infer in which state your system has been prior to measurement.

5. Aug 10, 2011

### Amir Livne

I'm not sure I understand you correctly. True, after measuring the energy the system collapses to an eigenvector of the Hamiltonian, and this can easily be seen to stay this way. But is energy not conserved in every circumstance? That's certainly the experience from classical physics, so there should be some limit theorem.

I'll try to present another thought-experiment, this time where angular momentum is not conserved. In classical physics it can't happen of course.

Take a bunch of particles (electrons, or nuclei, whatever), all aligned with their spin pointing toward the positive direction of the X axis. Measure their spin along the Z axis. This will give a completely random distribution, half one one and half the other way. Now measure the spin along the X axis. You'll get a random distribution again - losing angular momentum in the process. It seems to me that if you charge a lump of metal with enough such electrons you could actually get it to rotate. Is this situation possible? And if not, and the momentum escapes elsewhere, is it conserved exactly or only on average?

6. Aug 10, 2011

### Drakkith

Staff Emeritus
I don't see the issue here Amir. Measuring a system disturbs it, so any changes would be a result of that correct? Also, in your first example, how can you measure the energy of a lone particle? What energy are you measuring? Without something else to compare it with then how could you say it has X amount of energy?

7. Aug 10, 2011

### Amir Livne

Measuring the energy of a single particle isn't too hard. If it doesn't interact with the surrounding material there are only kinetic and mass energy. If that material is a translucent, it emits Cherenkov radiation in a small amount. The angle of the radiation cone gives you a measurement that allows you to compare the energy at different times.

Measuring position is more difficult, and will generally require putting a lot of energy in and around the particle. That's why I described the spin-experiment, since measuring spins is much simpler technically.

On another note, you say "any changes would be a result of that [measurement] correct?". That is exactly what I'm asking in this thread. Is the change in the energy/momentum of a system exactly equal to the energy/momentum you put into it? Is there some mathematical argument to show why this follows from Schrodinger's equation?

8. Aug 10, 2011

### Drakkith

Staff Emeritus
That's not what I mean. Without another object to compare it against, the particle has no kinetic energy nor can you measure its mass.

Position compared to what? Either the observer or another object is necessary in order to measure energy. Without prior knowledge of the mass of the particle you would not be able to calculate the mass/energy.

All I'm getting at is that a "free particle" cannot be measured without something interacting with it.

Yes. Conservation of energy and mass is the simplest way to look at it. I don't know enough to put it in terms of Schrodingers equations or similar.