This Quantum World/Implications and applications/Observables and operators

Observables and operatorsEdit

Remember the mean values

\langle x\rangle=\int |\psi|^2\,x\,dx \quad\hbox{and}\quad 
\langle p\rangle=\hbar\langle k\rangle=\int |\overline{\psi}|^2\,\hbar k\,dk.

As noted already, if we define the operators

\hat x=x ("multiply with x") and \hat p=-i\hbar\frac\partial{\partial x},

then we can write

\langle x\rangle=\int \psi^*\,\hat x\,\psi\,dx\quad\hbox{and}\quad 
\langle p\rangle=\int \psi^*\,\hat p\,\psi\,dx.

By the same token,

\langle E\rangle=\int \psi^*\,\hat E\,\psi\,dx \quad\hbox{with}\quad \hat E=i\hbar\frac\partial{\partial t}.

Which observable is associated with the differential operator \partial/\partial\phi? If r and \theta are constant (as the partial derivative with respect to \phi requires), then z is constant, and

{\partial\psi\over\partial\phi}={\partial y\over\partial\phi}\,{\partial\psi\over\partial y}+{\partial x\over\partial\phi}\,{\partial\psi\over\partial x}.

Given that x=r\sin\theta\,\cos\phi and y=r\sin\theta\,\sin\phi, this works out at x{\partial\psi\over\partial y}-y{\partial\psi\over\partial x} or

-i\hbar\frac{\partial}{\partial\phi}=\hat x\hat p_y - \hat y\hat p_x.

Since, classically, orbital angular momentum is given by \mathbf{L}=\mathbf{r}\times\mathbf{p}, so that L_z=x\,p_y-y\,p_x, it seems obvious that we should consider \hat x\hat p_y - \hat y\hat p_x as the operator \hat l_z associated with the z 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 \psi(q_k,t) of a closed system \mathcal{S} with K degrees of freedom. Suppose that the probability distribution |\psi(q_k,t)|^2 (which is short for |\psi(q_1,\dots,q_K,t)|^2) is invariant under translations in time: waiting for any amount of time \tau makes no difference to it:


Then the time dependence of \psi is confined to a phase factor e^{i\alpha(q_k,t)}.

Further suppose that the time coordinate t and the space coordinates q_k are homogeneous — equal intervals are physically equivalent. Since \mathcal{S} is closed, the phase factor e^{i\alpha(q_k,t)} cannot then depend on q_k, and its phase can at most linearly depend on t: waiting for 2\tau should have the same effect as twice waiting for \tau. In other words, multiplying the wave function by e^{i\alpha(2\tau)} should have same effect as multiplying it twice by e^{i\alpha(\tau)}:



\psi(q_k,t)=\psi(q_k)\,e^{-i\omega t}=\psi(q_k)\,e^{-(i/\hbar)E\,t}.

So the existence of a constant ("conserved") quantity \omega or (in conventional units) E is implied for a closed system, and this is what we mean by the energy of the system.

Now suppose that |\psi(q_k,t)|^2 is invariant under translations in the direction of one of the spatial coordinates q_k, say q_j:

|\psi(q_j,q_{k\neq j},t)|^2=|\psi(q_j+\kappa,q_{k\neq j},t)|^2.

Then the dependence of \psi on q_j is confined to a phase factor e^{i\beta(q_k,t)}.

And suppose again that the time coordinates t and q_k are homogeneous. Since \mathcal{S} is closed, the phase factor e^{i\beta(q_k,t)} cannot then depend on q_{k\neq j} or t, and its phase can at most linearly depend on q_j: translating \mathcal{S} by 2\kappa should have the same effect as twice translating it by \kappa. In other words, multiplying the wave function by e^{i\beta(2\kappa)} should have same effect as multiplying it twice by e^{i\beta(\kappa)}:



\psi(q_k,t)=\psi(q_{k\neq j},t)\,e^{i\,k_j\,q_k}=\psi(q_{k\neq j},t)\,e^{(i/\hbar)\,p_j\,q_k}.

So the existence of a constant ("conserved") quantity k_j or (in conventional units) p_j 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 r,\theta,\phi. If |\psi(r,\theta,\phi,t)|^2 is invariant under rotations about the z axis, and if the longitudinal coordinate \phi is homogeneous, then

\psi(r,\theta,\phi,t)=\psi(r,\theta,t)\,e^{i m\phi}=\psi(r,\theta,t)\,e^{(i/\hbar) l_z\phi}.

In this case we call the conserved quantity the z component of the system's angular momentum.

Now suppose that O is an observable, that \hat O is the corresponding operator, and that \psi_{\hat O,v} satisfies

\hat O\,\psi_{\hat O,v}=v\,\psi_{\hat O,v}.

We say that \psi_{\hat O,v} is an eigenfunction or eigenstate of the operator \hat O, and that it has the eigenvalue v. Let's calculate the mean and the standard deviation of O for \psi_{\hat O,v}. We obviously have that

\langle O\rangle=\int\psi^*_{\hat O,v}\hat O\,\psi_{\hat O,v}\,dx=\int\psi^*_{\hat O,v}v\,\psi_{\hat O,v}\,dx=v\int|\psi_{\hat O,v}|^2\,dx=v.


\Delta O=\sqrt{\int \psi^*_{\hat O,v}\,(\hat O-v)\,(\hat O-v)\,\psi_{\hat O,v}\,dx }=0,

since (\hat O-v)\,\psi_{\hat O,v}=0. For a system associated with \psi_{\hat O,v}, O is dispersion-free. Hence the probability of finding that the value of O lies in an interval containing v, is 1. But we have that

\hat E\,\psi(q_k)\,e^{-(i/\hbar)E\,t}=E\,\psi(q_k)\,e^{-(i/\hbar)E\,t}
\hat p_j\,\psi(q_{k\neq j},t)\,e^{(i/\hbar)\,p_j\,q_k}=p_j\,\psi(q_{k\neq j},t)\,e^{(i/\hbar)\,p_j\,q_k}
\hat l_z\,\psi(r,\theta,t)\,e^{(i/\hbar)\,l_z\phi}=l_z\,\psi(r,\theta,t) \,e^{(i/\hbar)\,l_z\phi}.

So, indeed, \hat l_z is the operator associated with the z 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

{\hat l}_x=-i\hbar\left(y{\partial\over\partial z}-z{\partial\over\partial y}\right)\quad
\hbox{and}\quad{\hat l}_y=-i\hbar\left(z{\partial\over\partial x}-x{\partial\over\partial z}\right).

If we define the commutator [{\hat A},{\hat B}]\equiv{\hat A}{\hat B}-{\hat B}{\hat A}, then saying that the operators {\hat A} and {\hat B} 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 [{\hat l}_x,{\hat l}_y]\,=i\hbar{\hat l}_z.

One similarly finds that [{\hat l}_y,{\hat l}_z]=i\hbar{\hat l}_x and [{\hat l}_z,{\hat l}_x]=i\hbar{\hat l}_y. The upshot: different components of a system's angular momentum are incompatible.

Exercise: Using the above commutators, show that the operator \hat{\mathbf{L}^2}\equiv{\hat l}_x^2+{\hat l}_y^2+{\hat l}_z^2 commutes with {\hat l}_x, {\hat l}_y, and {\hat l}_z.

Last modified on 6 August 2009, at 02:33