zero mean value operator

[luxxio]

it is possible to build a non trivial operator which the mean value is always zero?

[xepma]

That would mean it's mean value is always a constant. In other words, it's outcome doesn't depend on the state of the system. That means the operator can be written as a constant times the identity operator.

So you automatically end up with the trivial operator.

[luxxio]

That would mean it's mean value is always a constant.
no. a constant will not return a zero mean value.

[xepma]

I was referring to the more general case: what does it mean to have an operator O which always has a mean value equal to some constant C?

That means that <O> = C, irrespective of the state. Therefore, the operator can be represented by a the constant C times the identity opeator 1, so O = C*1. This leas to:

<O> = C<1> = C, which is what we desire.

You're asking for the special case when C=0. This automatically leads to the trivial operator 0.

[luxxio]

I was referring to the more general case: what does it mean to have an operator O which always has a mean value equal to some constant C?

That means that <O> = C, irrespective of the state. Therefore, the operator can be represented by a the constant C times the identity opeator 1, so O = C*1. This leas to:

<O> = C<1> = C, which is what we desire.

You're asking for the special case when C=0. This automatically leads to the trivial operator 0.
this is not true. a little example:
the operator [itex]\left[-i\partial_t-\nabla^2 + cost]|\psi>=0[\itex].
so it's mean value is zero

[RedX]

your question is ambiguous in that you don't specify what you mean by "always". "Always" as in for any time, or as in any state?

if the former then I think xepma has it right. certainly right if the operator is Hermitian, as you just go into the diagonal basis, and each diagonal element has to be the same value or the expectation value of the eigenstates are different from each other. if you rotate the basis then your operator is still diagonal with the same value- this is to be expected from a totally degenerate eigenspace.

if the latter, then if an operator P commutes with the Hamiltonian then the expectation value ought to be unchanging with time. whatever that expectation value is, create a new operator as P-<P> which will have expectation value zero for all time.

[CPL.Luke]

I'm pretty sure that in general the only operator that could always be guarenteed to commute with any hamiltonian would be the hamiltonian itself or a constant, thus the only operators which have a constant mean value in time are the hamiltonian and some constant operator.

[RedX]

I'm pretty sure that in general the only operator that could always be guarenteed to commute with any hamiltonian would be the hamiltonian itself or a constant, thus the only operators which have a constant mean value in time are the hamiltonian and some constant operator.

I think that's true. You could build a function of the Hamiltonian I guess, and that would commute with the Hamiltonian. So a linear combination of Constant, H, H^2... H^n etc.