# Particle in superposition of energy eigenstates and conservation of energy.

1. Dec 10, 2012

### amirhdrz

When a particle is in superposition of energy eigenstates and has a probability of being found in either state, what does that say about the energy of the particle and conservation of energy.
What I mean is, since the energy eigenstates have different energy values, where's the rest of the energy of the particle, if it is found to be in the lower energy eigenstate?

Let's say $|\psi> = c_1 |E_1> + c_2 |E_2>$

Take the system to be a free particle in a box where $\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x})$ and

$V(x)=\{ \begin{array}{1 1} 0 \quad |x|<a/2\\ V_0 \quad |x|>a/2 \end{array}$

I guess my questions could be rephrased as, how do you calculate the total energy of a quantum particle?

Last edited: Dec 10, 2012
2. Dec 10, 2012

### Avodyne

The energy is uncertain. If you measure it, you have probability |c1|^2 to get E1, and probability |c2|^2 to get E2. These probabilities do not change with time. This is what it means for energy to be conserved in QM.

3. Dec 10, 2012

### amirhdrz

I was just expecting quantum mechanics to always give me definite answer for the total energy of the system.

4. Dec 11, 2012

### tom.stoer

A conservation law in quantum mechanics is usually formulated in terms of an operator algebra. In the Heisenberg picture which corresponds to the Hamiltonian framework in classical mechanics the time-dependence is shifted from the states to the operators; the state vectors are time-independent, therefore there is no conservation law on the level of the states (trivial). The conservation law (time-independence) of a certain observable Q is formulated in terms of its commutator with the Hamiltonian H, i.e. [H,Q] = 0 which corresponds to the Heisenberg equations of motion for a conserved quantity Q with dQ/dt = 0. Here I always assume that we start with time-indep. Hamiltonian operator in the Schrödinger picture. In that case the time-independence of the Hamiltonian in the Heisenberg piucture is trivial: [H,H] = 0

5. Dec 11, 2012

### Sonderval

I think you should also take into account that your particle has to enter the superposition state in some way. If you start with an excited atom, for example, after a while the atom will be in a superposition of being excited or not. But the full superpositon state has a photon taking up the energy in one case and not in the other, so in any case the total energy is conserved.

Thsi leads to a follow-up question which I cannot answer: Is there any way to prepare a particle in an energy superposition state that does not involve entanglement with another system that is in a complementary superposition?

6. Dec 11, 2012

### JK423

That's a very interesting question, i have thought about it as well but i cannot answer it either.. Although my intuition tells me that it cannot be done.

7. Dec 11, 2012

### Demystifier

Yes there is. See e.g.
L. E. Ballentine, Quantum Mechanics: A Modern Development
Sec. 12.4 Quantum Beats

8. Dec 11, 2012

### JK423

At the cited section the magnetic field is treated classicaly and that's how the superposition is achieved. It's like exciting an atom with a classical electromagnetic field. However if you treat the E/M field quantum mechanically, the atomic states are entangled with the field and total energy is exactly conserved.

9. Dec 11, 2012

### DrDu

The states of the electromagnetic field which behave the most classical are coherent states which are themselves not energy eigenstates and neither contain a definite number of photons.
Exciting an atom with these states leaves it in a superposition of energy eigenstates and does not change the coherent state, so there is also no entanglement created.

10. Dec 11, 2012

### JK423

You're right but i'm not sure about the "does not change the coherent state" statement, it seems strange that an interaction that involves energy-exchange changes one of the subsystems but not the other. A photon should be absorbed from somewhere..

11. Dec 11, 2012

### DrDu

Yes, but the coherent states are eigenstates of the photon anihilation operator. So the removal of a photon does not change the coherent state.

12. Dec 11, 2012

### JK423

Correct, and that's because a coherent state involves an infinite number of photons in superposition. However such a coherent state is unnatural. In reality you can find approximate coherent states that involve a very large number of photons in superposition, but still it's a finite number and a subtraction of a photon makes a (small to negligible) difference.
Coherent states are created by lasers, and the process of their creation exactly preserves the total energy. So we are always 'tracing back' to find the closed system that includes all the subsystem involved.
I think that the correct assertion would be: In a closed system with well defined energy, all its subsystems that interact with each other unitarily are entangled. Each subsystem is in a superposition of eigenstates, but the overall energy is conserved. If you assume the universe to be closed and with well defined total energy (that's a big assertion to make!:P ) and described by QM then energy is always exactly conserved.

Last edited: Dec 11, 2012
13. Dec 11, 2012

### Sonderval

@JK423 &others
Thanks - that's what I always assumed, but I've never found it discussed anywhere.

14. Dec 11, 2012

### DrDu

Nobody doubts that energy is exactly conserved. The question is about the correlations generated.
I always find the assumption that the whole universe is a closed quantum system quite dubious.
The - even approximate - construction of closed systems is de facto impossible and they are much more unnatural than coherent states.
All measurements are ultimately performed with instruments that are classically or arbitrarily closely so. These systems can swallow and delete any kind of correlation. The coherent states show how this becomes possible in the limit of large system size.

15. Dec 12, 2012

### Demystifier

As long as a closed system (e.g., the whole Universe) evolves unitarily, an energy eigen-state must necessarily evolve into an energy-eigenstate. This seems to imply that creation of coherent superpositions of different energies is impossible. But for such a system the state does not change with time, so where does the dependence on time come from?

The catch is that in all interpretations of QM, the system as a whole does NOT evolve unitarily. (For example, the wave function collapse is not unitary.) The only exception is many-world interpretation, but even there the physical time needs to be identified with some clock-configuration variable, which again leads to an effective non-unitary "evolution" with respect to the clock time. As a consequence, creation of coherent energy superpositions is possible, in all interpretations.

http://arxiv.org/abs/1209.5196

16. Dec 12, 2012

### Demystifier

Why? If it is not closed then it interacts with something else, but if there is something else to interact with, then the "whole" universe was not really whole in the first place.. In other words, by DEFINITION, the whole universe must be closed.

17. Dec 12, 2012

### Demystifier

See the link in post #15.

18. Dec 12, 2012

### K^2

Incorrect. Hamiltonian of the closed universe includes all the measurements in said universe. That means if you are in an eigen state of Hamiltonian, the measurement won't change that either. It's only when you are considering sub-systems that distinction is relevant.

The only way you can have dynamics in the universe is if universe is not in an eigen state of the Hamiltonian. And why should it be in an eigen state? There are infinitely more states than there are eigen states.

So we don't need to discuss collapse here. The universe is in some super-position of eigen states, and that's enough.

Note that regardless of what kind of superposition you have, [H, H] = 0. So expectation value of energy does not change in time. Yes, energy of the universe is not an eigen energy. But it's always the same, so there are no issues with conservation laws.

19. Dec 12, 2012

### Sonderval

@Demystifier
Thanks for that, but I'm still mystified ;-)

Consider an isolated H-atom in an excited state (for example 2p).
This is an energy eigenstate and stationary.

However, we know that the H-atom can emit a photon and go to the ground state. The total energy of the system does not change. So can the system evolve from being excited into a superpositon of excited + (ground state +photon)?

If not, is the actual photon emission of an excited atom due to some external disturbance? An interaction with the vacuum state of the photon field?

20. Dec 12, 2012

### K^2

It's not a closed system. In a perfect closed system, an atom in 2p state stays in 2p state. It will never decay. It takes a small perturbation from outside, plus the electromagnetic field to get the decay started. If you place an atom in a perfectly closed box, even if you "nudge" it to decay, it will eventually return into an excited state again. It's only in an open system that decay is irreversible.