You will recall from last time that we defined the electric quadrupole interaction
be performing a taylor expansion on the electromagnetic four-potential
and identifying the term that transformed as a two index symmetric traceless tensor
under rotations,

where is the electric charge density of the localized object, in this case the nucleus. It is clear that this is a object by inserting , in which case we find .

Just as we did for the magnetic moment interaction and hence defining the
magnetic moment of a nucleus wish
to use eq. (38) to define the quadrupole moment of a nucleus.
Let us choose to evaluate the diagonal matrix element of
between
states defined by
(summing over all the electrically charged objects inside the nucleus)

This sets the conventional definitions of and . This also leads to the generalization to arbitrary multipoles , where denotes the multipolarity,

We still have to define the operator ,
which we do in terms of .
We note that we can use similar
arguments that lead us to the form for the magnetic moment operator.
We realize that acting in the space of states defined by
,
the
matrix element of any operator in
this space must be reproduced by an analogous operator formed from the .
The quadrupole operator is a
two index, symmetric, traceless tensor, and hence the operator given in terms
of the
must also have
these properties. Therefore we find the general form for
must be

where is a yet to be determined constant. It is again convenient to examine the matrix element of between states of

and hence we have solved for the constant

We can then write the operator as

A couple of general comments are worth making at this point. Notice that the charge density is the square modulus of the wavefunction and as such it is not possible for them to have odd-electric moments if our world respects time reversal invariance. This is most easily seen by considering the time reversal properties of an odd number of combined together into a traceless, symmetric tensor. This clearly has the same property under time-reversal as a single , which is odd. Combined with the fact that is even under time-reversal, we find that such odd-electric moments are forbidden by Time-Reversal Invariance which we know is pretty well respected in nature (i.e. we will forget about it!).

Also, we note that a nucleus of spin can have a static moment of highest multipolarity (i.e. ). This means that a object cannot have an electric quadrupole moment but a object such as the deuteron can.

It is worth spending a few minutes to get a physical grip on what a
quadrupole moment is
(I hope I am not boring you).
Imagine we have a ellipsoidal object with semi-major axis of length "a"
and a semi-minor axis of length "b".
Lets us start by computing the charge in terms of the charge density
which we will assume to be
uniform 9 out to the nuclear surface and then falls abruptly to zero),
and the lengths of axes .
You will recall that the trick for performing volume integrals over
ellipsoids is to
rescale the axis by the length of the semi-major or -minor axis,
e.g
and
.
We then have that

Next, lets consider the quadrupole moment of this object where we define the z-axis to be the symmetry axis along the semi-major direction, i.e. the axis of cylindrical symmetry. This is called the body-axis, and I will denote coordinates in this frame with a subscript b, i.e. . It is straightforward to compute the quadrupole moment in this frame,

where we have used the fact that in the coordinates the ellipsoid is a sphere, and set . Further we can show that the quadrupole moment about any axis, we call z, corresponding to a rotation from the body-axis by an angle is given by

where we have again used symmetry properties of the integration.

Our discussion of this object has been entirely classical up to this
point.
However, we are interesetd in
asking about quantum systems, and states defined by
and .
The quantum nature of the nuclei tells us that its angular momentum
and its angular orientation are
conjugate variables, and if we knew that the ellipsoid was aligned
with its semi-major axis along the
quantization axis, then we would not know what its angular momentum
is!
What this means is that there is a minimum angle between the
body-axis and the quantization axis which we can
determine by a hand waving argument!
The maximum
is equal to J, while the magnitude of the angular
momentum vector is
,
leading to a minimum angle between the axes of
.
Therefore, by our previous definitions, the observable quadrupole moment of
this ellipsoidal object, is related to its quadrupole moment as computed
along its body-axis by

Classically, we can have a charge distribution with an quadrupole deformation, that has vanishing rotation, so one would have guessed that in fact a object would have had a quadrupole moment. In fact, it is a quantum mechanical effect that such an object does not. It is the uncertainty principle and the fact that the angular momentum and angle are conjugate variables. One cannot make a rank two tensor from 0 or 2 magnetic substates. We see that in fact a and a nucleus could have a large quadrupole deformation, but we would not be able to observe it in any experiment. However, we might imagine spinning up the nucleus and observing a characteristic type rotator spectrum, and by measuring transition strengths of decays, and also the moment of inertia (from the energy level spacings) we could determine the quadrupole moment that is independent of . This is done in light nuclei and one can conclude that the level does in fact have a quadrupole moment, but it can never be observed. The question to ask is if a tree falls and no one is around does it make a noise, and my answer to this is who cares! If I don't here it I really don't care. Same thing with this quadrupole moment. It is interesting in principle, but who cares if it can't be measured and has no consequences.