This Quantum World/Implications and applications/Observables and operators< This Quantum World | Implications and applications
Observables and operatorsEdit
Remember the mean values
As noted already, if we define the operators
- ("multiply with ") and
then we can write
By the same token,
Which observable is associated with the differential operator ? If and are constant (as the partial derivative with respect to requires), then is constant, and
Given that and this works out at or
Since, classically, orbital angular momentum is given by so that it seems obvious that we should consider as the operator associated with the component of the atom's angular momentum.
Yet we need to be wary of basing quantum-mechanical definitions on classical ones. Here are the quantum-mechanical definitions:
Consider the wave function of a closed system with degrees of freedom. Suppose that the probability distribution (which is short for ) is invariant under translations in time: waiting for any amount of time makes no difference to it:
Then the time dependence of is confined to a phase factor
Further suppose that the time coordinate and the space coordinates are homogeneous — equal intervals are physically equivalent. Since is closed, the phase factor cannot then depend on and its phase can at most linearly depend on waiting for should have the same effect as twice waiting for In other words, multiplying the wave function by should have same effect as multiplying it twice by :
So the existence of a constant ("conserved") quantity or (in conventional units) is implied for a closed system, and this is what we mean by the energy of the system.
Now suppose that is invariant under translations in the direction of one of the spatial coordinates say :
Then the dependence of on is confined to a phase factor
And suppose again that the time coordinates and are homogeneous. Since is closed, the phase factor cannot then depend on or and its phase can at most linearly depend on : translating by should have the same effect as twice translating it by In other words, multiplying the wave function by should have same effect as multiplying it twice by :
So the existence of a constant ("conserved") quantity or (in conventional units) is implied for a closed system, and this is what we mean by the j-component of the system's momentum.
You get the picture. Moreover, the spatial coordiates might as well be the spherical coordinates If is invariant under rotations about the axis, and if the longitudinal coordinate is homogeneous, then
In this case we call the conserved quantity the component of the system's angular momentum.
Now suppose that is an observable, that is the corresponding operator, and that satisfies
We say that is an eigenfunction or eigenstate of the operator and that it has the eigenvalue Let's calculate the mean and the standard deviation of for We obviously have that
since For a system associated with is dispersion-free. Hence the probability of finding that the value of lies in an interval containing is 1. But we have that
So, indeed, is the operator associated with the component of the atom's angular momentum.
Observe that the eigenfunctions of any of these operators are associated with systems for which the corresponding observable is "sharp": the standard deviation measuring its fuzziness vanishes.
For obvious reasons we also have
If we define the commutator then saying that the operators and commute is the same as saying that their commutator vanishes. Later we will prove that two observables are compatible (can be simultaneously measured) if and only if their operators commute.
Exercise: Show that
One similarly finds that and The upshot: different components of a system's angular momentum are incompatible.
Exercise: Using the above commutators, show that the operator commutes with and