# This Quantum World/Implications and applications/Observables and operators

## Observables and operatorsEdit

Remember the mean values

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

As noted already, if we define the operators

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

then we can write

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

By the same token,

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

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

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

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

${\displaystyle -i\hbar {\frac {\partial }{\partial \phi }}={\hat {x}}{\hat {p}}_{y}-{\hat {y}}{\hat {p}}_{x}.}$

Since, classically, orbital angular momentum is given by ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} ,}$  so that ${\displaystyle L_{z}=x\,p_{y}-y\,p_{x},}$  it seems obvious that we should consider ${\displaystyle {\hat {x}}{\hat {p}}_{y}-{\hat {y}}{\hat {p}}_{x}}$  as the operator ${\displaystyle {\hat {l}}_{z}}$  associated with the ${\displaystyle z}$  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 ${\displaystyle \psi (q_{k},t)}$  of a closed system ${\displaystyle {\mathcal {S}}}$  with ${\displaystyle K}$  degrees of freedom. Suppose that the probability distribution ${\displaystyle |\psi (q_{k},t)|^{2}}$  (which is short for ${\displaystyle |\psi (q_{1},\dots ,q_{K},t)|^{2}}$ ) is invariant under translations in time: waiting for any amount of time ${\displaystyle \tau }$  makes no difference to it:

${\displaystyle |\psi (q_{k},t)|^{2}=|\psi (q_{k},t+\tau )|^{2}.}$

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

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

${\displaystyle e^{i\alpha (2\tau )}=[e^{i\alpha (\tau )}]^{2}=e^{i2\alpha (\tau )}.}$

Thus

${\displaystyle \psi (q_{k},t)=\psi (q_{k})\,e^{-i\omega t}=\psi (q_{k})\,e^{-(i/\hbar )E\,t}.}$

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

Now suppose that ${\displaystyle |\psi (q_{k},t)|^{2}}$  is invariant under translations in the direction of one of the spatial coordinates ${\displaystyle q_{k},}$  say ${\displaystyle q_{j}}$ :

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

Then the dependence of ${\displaystyle \psi }$  on ${\displaystyle q_{j}}$  is confined to a phase factor ${\displaystyle e^{i\beta (q_{k},t)}.}$

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

${\displaystyle e^{i\beta (2\kappa )}=[e^{i\beta (\kappa )}]^{2}=e^{i2\beta (\kappa )}.}$

Thus

${\displaystyle \psi (q_{k},t)=\psi (q_{k\neq j},t)\,e^{i\,k_{j}\,q_{k}}=\psi (q_{k\neq j},t)\,e^{(i/\hbar )\,p_{j}\,q_{k}}.}$

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

${\displaystyle \psi (r,\theta ,\phi ,t)=\psi (r,\theta ,t)\,e^{im\phi }=\psi (r,\theta ,t)\,e^{(i/\hbar )l_{z}\phi }.}$

In this case we call the conserved quantity the ${\displaystyle z}$  component of the system's angular momentum.

Now suppose that ${\displaystyle O}$  is an observable, that ${\displaystyle {\hat {O}}}$  is the corresponding operator, and that ${\displaystyle \psi _{{\hat {O}},v}}$  satisfies

${\displaystyle {\hat {O}}\,\psi _{{\hat {O}},v}=v\,\psi _{{\hat {O}},v}.}$

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

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

Hence

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

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

${\displaystyle {\hat {E}}\,\psi (q_{k})\,e^{-(i/\hbar )E\,t}=E\,\psi (q_{k})\,e^{-(i/\hbar )E\,t}}$
${\displaystyle {\hat {p}}_{j}\,\psi (q_{k\neq j},t)\,e^{(i/\hbar )\,p_{j}\,q_{k}}=p_{j}\,\psi (q_{k\neq j},t)\,e^{(i/\hbar )\,p_{j}\,q_{k}}}$
${\displaystyle {\hat {l}}_{z}\,\psi (r,\theta ,t)\,e^{(i/\hbar )\,l_{z}\phi }=l_{z}\,\psi (r,\theta ,t)\,e^{(i/\hbar )\,l_{z}\phi }.}$

So, indeed, ${\displaystyle {\hat {l}}_{z}}$  is the operator associated with the ${\displaystyle z}$  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

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

If we define the commutator ${\displaystyle [{\hat {A}},{\hat {B}}]\equiv {\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}},}$  then saying that the operators ${\displaystyle {\hat {A}}}$  and ${\displaystyle {\hat {B}}}$  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 ${\displaystyle [{\hat {l}}_{x},{\hat {l}}_{y}]\,=i\hbar {\hat {l}}_{z}.}$

One similarly finds that ${\displaystyle [{\hat {l}}_{y},{\hat {l}}_{z}]=i\hbar {\hat {l}}_{x}}$  and ${\displaystyle [{\hat {l}}_{z},{\hat {l}}_{x}]=i\hbar {\hat {l}}_{y}.}$  The upshot: different components of a system's angular momentum are incompatible.

Exercise: Using the above commutators, show that the operator ${\displaystyle {\hat {\mathbf {L} ^{2}}}\equiv {\hat {l}}_{x}^{2}+{\hat {l}}_{y}^{2}+{\hat {l}}_{z}^{2}}$  commutes with ${\displaystyle {\hat {l}}_{x},}$  ${\displaystyle {\hat {l}}_{y},}$  and ${\displaystyle {\hat {l}}_{z}.}$