A question on measurment in QM

[wangyi]

Hi,
I don't know whether the conserved quantities in QM are conserved in every conditions. For example, a state can be a non-eigenstate of energy, and when we measure the energy of this state, it falls into an eigenstate of energy. The former state does not have a definite energy, but the latter has. And, another measurment may get another different energy. Does it mean energy is not exactly conserved or it means we give the system some energy when we are measuring them?

Thank you very much and happy new year!

 

[fdarkangel]

being a freshman on the subject, here's a freshman reply:

the energy conservation can be violated by \Delta E \Delta t \geq \frac{\hbar}{2}

 

[abszero]

And, another measurment may get another different energy.

Actually, after the measurement as long as the system is undisturbed until you measure the energy again, you will keep getting the same energy values. This is true for any observable.

 

[linhtm]

Hi,
I don't know whether the conserved quantities in QM are conserved in every conditions

I think until today we know that the conserved quantities in QM are conserved in every conditions.
But in otherwise it may be not ,but it is in the future

 

[Galileo]

As abszero said, a measurement on a system will disturb it in a fundamental manner. The wavefunction will 'collapse' into an eigenstate of the observable measured so an immediate measurement on the same quantity will yield the same result.

This is not the same as conservation of the physical quantity ofcourse. You could wait a while after the first measurement and let the state evolve according to the Schrödinger equation, then measure it again.
Energy eigenstates are called stationary states for the simple reason that the wavefunction only picks up a phase difference over time with no physical consequences, so it remains effectively in the same state. Energy is thus conserved.

So in general, a quantity will be conserved if its eigenstates are also energy eigenstates and that means the observable should commute with the hamiltonian.

 

[George Jones]

Actually, after the measurement as long as the system is undisturbed until you measure the energy again, you will keep getting the same energy values. This is true for any observable.

This is true for any observable A (that doesn't have explicit time dependence) that commutes with the Hamiltonian. After a measurement of A, the state of the system is an eigenstate of A, but if A does not commute with the Hamiltonian, then, in general, as time progresses, the time evolution operator will evolve the system's state into a state that is not an eigenstate of A, even when no further measurements are performed. If A commutes with the Hamiltonian, then after a measurement of A, the time evolution operator always produces an eigenstate of A associated with the measured value of A.

This motivates the quantum mechanical definition that A is conserved if it commutes with the Hamiltonian.

Regards,
George

Edit: I see that while I was composing (I'm a very slow thinker and an even slower typist), Galileo already made the points that I wanted to make.