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dextercioby said:You question is not quite inteligible. You need to specify who the symbols stand for. Who's H, Q ?
Outrageous said:For commutator, HQ-QH = 0 .
But for this case as shown below, complex ψQHψ - HcomplexψQψ= 0?
If the operator Q is in term of (∂/∂t) and (∂/∂x) ,then the HQ-QH may not be zero.
Is there any restriction for Q operator?
dextercioby said:, H=f(Q) is a trivial solution of [Q,H]=0.
dextercioby said:Yes, if H=H(Q) such as e.g. H = Q^2, then [Q^2,Q]=0 on all vectors on which the commutator makes sense.
The Ehrenfest theorem is a fundamental equation in quantum mechanics that relates the time derivative of an expectation value to the expectation value of the time derivative of an operator. It is important because it allows us to make connections between classical mechanics and quantum mechanics, and to understand how the classical laws of motion arise from the quantum description of particles.
The Ehrenfest theorem is derived by applying the Heisenberg equation of motion to an operator, using the commutation relations between operators, and then taking the expectation value of the resulting equation.
The operator Q in the Ehrenfest theorem represents the classical observable or physical quantity that is being measured. It can be position, momentum, energy, or any other measurable quantity.
Yes, there is a condition for the operator Q in the Ehrenfest theorem. It must be a Hermitian operator, meaning that it is equal to its own adjoint. This ensures that the expectation value of Q is a real number, which is necessary for the classical interpretation of the theorem.
Yes, the Ehrenfest theorem can be generalized to include time-dependent operators. This is known as the extended Ehrenfest theorem and it takes into account the time dependence of the operator Q, allowing us to describe the time evolution of systems more accurately.