# Can’t specify the exact energy of system in a distinc timet

1. Jul 24, 2010

### alimehrani

According to uncertainty principle we can’t specify the exact energy of system in a distinc time
On the other hand we say that for example: at t=0 system is in basic state of energy and according to time revolution we can specify it’s energy at the next time. There are contradiction!

2. Jul 24, 2010

### tom.stoer

There is no uncertainty principle for time and energy.

The uncertainty principle for two observables (hermitean operators) A and B is derived via their commutator [A, B]. But time t is no observable in qm, it is just a parameter.

3. Jul 24, 2010

### alimehrani

ok
but
what is the heisenberg uncertainty perinciple?
you have seen at any modern physics books that heisenberg uncertainty principle is
1.uncertainty between E and t
&
2. uncertainty between x and p
and according to uncertainty between E and t, you can say that:
the lowest state of atom has distinct energy because the atom spend infinite time at this state.

4. Jul 24, 2010

### alxm

No, the Heisenberg uncertainty principle is not between E and t, and only x-p uncertainty is a 'proper' uncertainty relation as these are operators. The so-called 'time-energy' uncertainty principle is an entirely different thing, valid only in some cases. It is not a fundamental relationship.

See, e.g. http://arxiv.org/abs/quant-ph/0609163 and any number of older threads here on the subject.

5. Jul 24, 2010

### tom.stoer

The Heisenberg uncertainty principle is not between E and t as there is no operator for t!

6. Jul 26, 2010

### charlesworth

It is important to understand the difference between what can be called the Robertson inequality, and the Heisenberg inequality. It seems that these two notions have been mixed up in the above posts. The Robertson inequality refers to an inequality involving the expectation value of the commutator of two operators, as well as their individual statistical variances, and is derivable in a very general setting. That is to say, it places a limit on the information extractable over an ensemble of measurements. You do N measurements on iid systems measuring A, and then N measurements on N more iid systems measuring B, and then the Robertson inequality tells you something about the product of the variances.

On the other hand, the Heisenberg inequality refers to ONE experiment on a SINGLE system, and it tells you something a little bit obvious; disturbing a system with a measurement disturbs the system. There is no general derivation of the Heisenberg inequality, if it is to be derived rigorously one needs the specifics of the system.

If you think about it, these two inequalities really have nothing to do with each other.

As to the original poster's question, I have no comment.

7. Jul 26, 2010

### tom.stoer

I have never seen the so-called Robertson inequality. Can you explain a bit more about it? And about its relation with the Heisenberg inequality?

8. Jul 26, 2010

### charlesworth

What we choose to call things is often arbitrary. The "Robertson Inequality" and the "Heisenberg Inequality" were simply names I learned first in one of my undergrad QT courses; when someone mentions the Heisenberg inequality to me, I wait for context to decide what they mean; the two inequalities are often muddled up.

By "Robertson Inequality" I mean the formula that is derived in this section: http://en.wikipedia.org/wiki/Heisenberg_uncertainty_principle#Mathematical_derivations

By "Heisenberg Inequality" I mean inequalities that are derived in a Heisenberg's miscroscope-esque way, ie., an experimenter trying to measure the position and momentum of an electron by shooting a photon at it, and finding out that he has a lower limit on accuracy.

Neither of these implies the other. The former deals with statistical information over an ensemble of different (but iid) systems. The latter deals with a single shot experiment.

9. Jul 26, 2010

### tom.stoer

OK; I see the difference.

Nevertheless I think that not everybody would agree that the Robertson inequality applies ony to an ensemble of identical particles. This is of course true in the ensemble interpretation, but I think that most people today belief that this is property of every individual particle.

That means that an individual particle does not have both definite position and momentum.

10. Jul 27, 2010

### charlesworth

I suppose I am always reluctant to admit that I know anything at all about what nature actually is. But I'm happy so long as people are thinking carefully about the physical conclusions they draw from the math, and not taking anything for granted.

11. Jul 27, 2010