Where Does the Energy Go in a Quantum System with 3 Possible Measurements?

[romsofia]

Start with a quantum system with some observable, and 3 possible measurements. We can then take some random prepared state as $\left| \psi \right> = 3\left| a_1 \right> - 2 \left| a_2 \right> + 2i \left| a_3 \right>$

Now, am I right in thinking that each of these possible states has some energy associated to them? I would think that because our state has some energy associated with it.

When the outcome is known, what happens to the energy associated with the rest of the states (assuming that we can associate energy with states in superposition)? Does the energy just become noise in our system?

 

[Mentz114]

If one makes an energy measurment the result will be an eigenvalue of $\hat{H}$ the Hamiltonian operator. The fact that the wave function is this or that superposition of some other observable is irrelevant.

If you measure your observable the result is obviously an $a_i$ but the state of the system after this depends on the the physical aspects of the measurement. So I guess the answer is 'it depends'.

(Your $\left| \psi \right>$ needs some adjustment to normalise it.)

 

[romsofia]

Yes, I know the posulates say that, but I'm asking before a measurement takes place should all three outcomes above have an energy associated to them?

To make it clear, I know that operators acting on kets is how we get physical observables, but my question relates I guess to the initial state, before operators are introduced. Am I forcing physics onto the math in this case?

 

[Mentz114]

Yes, I know the posulates say that, but I'm asking before a measurement takes place should all three outcomes above have an energy associated to them?

To make it clear, I know that operators acting on kets is how we get physical observables, but my question relates I guess to the initial state, before operators are introduced. Am I forcing physics onto the math in this case?

I don't understand the question. It's too general. You need to give a Hamiltonian before it can be answered, I think.

 

[PeterDonis]

before a measurement takes place should all three outcomes above have an energy associated to them?

If you don't measure the energy, you can't say the system has an energy.

In fact, that's a good general rule for any quantum observable: if you don't measure the observable, you can't say the system has a value for that observable.

 

[PeterDonis]

Moderator's note: I have changed the level of this thread to "I" since the OP question does not justify an "A".

 

[romsofia]

If you don't measure the energy, you can't say the system has an energy.

In fact, that's a good general rule for any quantum observable: if you don't measure the observable, you can't say the system has a value for that observable.

I guess I'll be more specific, and will start by stop saying energy when I mean information.

What happens to the information given by the other states that were NOT measured? By quantum mechanics itself, information cannot be destroyed, but it seems in the Stern-Gerlach experiments, information is destroyed? How can this be? Does the information from the other possible measurements just become noise? In other words, how does the 5th postulate handle the inherent information of each state if we have no information about it? Or am I trying to force meaning out of math?

The 5 postulate (taken from a textbook) is: After a measurement of A that yields the result $a_n$, the quantum system is in a new state that is the normalized projection of the original system ket onto the ket (or kets) corresponding to the result of the measurement $\left |\psi' \right> =\frac{P_n \left| \psi \right>}{\sqrt{| \left< \psi |P_n| \psi \right>}}$

If you guys need me to make an example, I can write one up involving the Stein-Gerlach experiment..

 

[PeterDonis]

What happens to the information given by the other states that were NOT measured?

That depends on which interpretation of QM you adopt. If you avoid interpretation and just look at the math of QM, there is no answer to this question.

By quantum mechanics itself, information cannot be destroyed

No, by unitary evolution, information cannot be destroyed. But unitary evolution might not be all there is--that depends on which interpretation of QM you adopt. Roughly speaking, "collapse" interpretations say that unitary evolution is not all there is--there are also wave function collapses when measurements occur, which are not unitary and do not necessarily preserve information; while "no collapse" interpretations like the many-worlds interpretation say that unitary evolution is all there is, and information is never destroyed. But there are a lot of nuances.

 

[Demystifier]

If you don't measure the energy, you can't say the system has an energy.

In fact, that's a good general rule for any quantum observable: if you don't measure the observable, you can't say the system has a value for that observable.

Yes, that's probably the best answer from the orthodox operational point of view. But it's interesting to see how other interpretations answer the same question. For instance in Bohmian interpretation, the quantum potential (associated with the superposition of wave functions with different energies) has an explicit time-dependence. Consequently, in the absence of measurement, the energy of the particle is well defined but not conserved.