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Next: Quadrupole Moment of the Up: Electromagnetic Interactions. Previous: Magnetic Moments

Quadrupole Interactions.

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,

    $\displaystyle H_{\rm int}^{L=2} = -{1\over 6}\ Q^{ij}\ {\partial E^i\over\partial x_j}$  
    $\displaystyle Q^{ij} = \int\ d^3x\ \rho (x)\ ( 3 x^i x^j - x^2 \delta^{ij} )
\ \ \ ,$ (38)

where $\rho(x)$ is the electric charge density of the localized object, in this case the nucleus. It is clear that this is a $L~=~2$ object by inserting $i~=~j~=~z$, in which case we find $Q^{zz}\sim Y_2^0(\Omega)$.

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 $Q^{zz}$ between states defined by $\vert J,M=J\rangle$(summing over all the electrically charged objects inside the nucleus)

    $\displaystyle Q =
\langle J,J\vert Q^{zz}\vert J,J\rangle =
\sum_{k=1}^Z\ \int\ d^3x\ \rho^{(k)} (x)\ ( 3 z^{(k)} z^{(k)} - (r^{(k)})^2 )$  
    $\displaystyle = \sum_{k=1}^Z\ \int\ d^3x\ \rho^{(k)} (x)\ (r^{(k)})^2 \sqrt{16\pi\over 5}
\ Y_2^0(\Omega^{(k)})$  
    $\displaystyle = \sqrt{16\pi\over 5} \int\ d^3x\ \Psi_{J,J}^* \ \sum_{k=1}^Z\ e\ (r^{(k)})^2
\ Y_2^0(\Omega^{(k)})\
    $\displaystyle = \sqrt{16\pi\over 5} \langle Q_2\rangle
\ \ \ .$ (39)

This sets the conventional definitions of $Q$ and $Q_2$. This also leads to the generalization to arbitrary multipoles $Q_\lambda$, where $\lambda$ denotes the multipolarity,
    $\displaystyle Q_\lambda = \sum_{k=1}^Z\ e\ (r^{(k)})^\lambda \ Y_\lambda^0(\Omega^{(k)})
\ \ \ .$ (40)

We still have to define the operator $Q^{ij}$, which we do in terms of $Q$. 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 $\vert J,M\rangle$, the matrix element of any operator in this space must be reproduced by an analogous operator formed from the $J_i$. The quadrupole operator is a two index, symmetric, traceless tensor, and hence the operator given in terms of the $J_i$ must also have these properties. Therefore we find the general form for $Q^{ij}$ must be

    $\displaystyle \langle J,M\vert Q^{ij}\vert J,M\rangle = C\ \langle J, M\vert {3...
... 2 }
\left( J^i J^j+J^j J^i\right)
- \delta^{ij} J^2 \vert J , M\rangle
\ \ \ ,$ (41)

where $C$ is a yet to be determined constant. It is again convenient to examine the matrix element of $Q^{zz}$ between states of $\vert J,M=J\rangle$
    $\displaystyle \langle J,J\vert Q^{zz}\vert J,J\rangle = C\ \langle J, J\vert {3\over 2 }
\left( J^z J^z+J^z J^z\right)
- J^2 \vert J , J\rangle$  
    $\displaystyle = C\ 3 J^2 - J(J+1)$  
    $\displaystyle = C\ J\ (2J-1)
\ \ \ ,$ (42)

and hence we have solved for the constant
    $\displaystyle C = {Q\over J (2J-1)}
\ \ \ .$ (43)

We can then write the operator as
    $\displaystyle Q^{ij} = {Q\over J (2J-1)} \left[ {3\over 2 }\left( J^i J^j+J^j J^i\right)
- \delta^{ij} J^2 \right]
\ \ \ .$ (44)

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 $J_i$ combined together into a traceless, symmetric tensor. This clearly has the same property under time-reversal as a single $J_i$, which is odd. Combined with the fact that $E_j$ 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 $J$ can have a static moment of highest multipolarity $2J$ (i.e. $J\otimes J=\lambda$). This means that a $J=0,{1\over 2}$ object cannot have an electric quadrupole moment but a $J=1$ object such as the deuteron can.

Figure: A non-spherical object with a quadrupole moment.

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 $\rho$ which we will assume to be uniform 9 out to the nuclear surface and then falls abruptly to zero), and the lengths of axes $a,b$. 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 $z^\prime~=~z/a, x^\prime~=~x/b$ and $y^\prime y/b$. We then have that

    $\displaystyle Z = \int\ d^3x\ \rho (x)$  
    $\displaystyle = a b^2 \ \int\ d^3x^\prime\ \rho (x^\prime)$  
    $\displaystyle = a b^2 \rho {4\pi\over 3}
\ \ \ .$ (45)

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. $z_b$. It is straightforward to compute the quadrupole moment $Q^{z_b z_b}$ in this frame,
    $\displaystyle Q^{z_b z_b} = \int\ d^3x_b\ \rho(x_b)\ (3 z_b^2 - r_b^2 )$  
    $\displaystyle = \int\ d^3x_b\ \rho(x_b)\ (2 z_b^2 - x_b^2 - y_b^2 )$  
    $\displaystyle = e a b^2 \int\ d^3x_b\ \rho(x_b)\ (2 a^2 z_b^{\prime 2} -
b^2 x_b^{\prime 2} - b^2 y_b^{\prime 2} )$  
    $\displaystyle = 2 e a b^2 (a^2-b^2) \int\ d^3x_b\ \rho(x_b)\ z_b^{\prime 2}$  
    $\displaystyle =
{2\over 5} Z e (a^2-b^2)
\ \ \ ,$ (46)

where we have used the fact that in the $x^\prime$ coordinates the ellipsoid is a sphere, and set $\langle x^{\prime 2} \rangle = \langle y^{\prime 2} \rangle =
\langle z^{\prime 2} \rangle$. Further we can show that the quadrupole moment about any axis, we call z, corresponding to a rotation from the body-axis $z_b$ by an angle $\theta$ is given by
    $\displaystyle Q^{z z} = \int\ d^3x\ \rho(x)\ (3 z^2 - r^2 )$  
    $\displaystyle = \int\ d^3x_b\ \rho(x_b)\ (2 (z_b\cos\theta + x_b\sin\theta )^2
- (x_b\cos\theta - z_b\sin\theta )^2 -y_b^2)$  
    $\displaystyle = {1\over 2}\left(3\cos^2\theta-1\right) Q^{z_b z_b}
\ \ \ ,$ (47)

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 $J$ and $M_J$. The quantum nature of the nuclei tells us that its angular momentum $J$ 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 $J_Z$ is equal to J, while the magnitude of the angular momentum vector is $\sqrt{J(J+1)}$, leading to a minimum angle between the axes of $\cos\theta_{\rm min} = \sqrt{J\over J+1}$. 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

    $\displaystyle Q = {2J-1\over 2(J+1)} Q_b
\ \ \ .$ (48)

Classically, we can have a charge distribution with an quadrupole deformation, that has vanishing rotation, so one would have guessed that in fact a $J~=~0, {1\over 2}$ 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 $J=0$ and a $J={1\over 2}$ 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 $J(J+1)$ type rotator spectrum, and by measuring transition strengths of $\gamma$ decays, and also the moment of inertia (from the energy level spacings) we could determine the quadrupole moment that is independent of $J$. This is done in light nuclei and one can conclude that the $J~=~{1\over 2}$ 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.

next up previous
Next: Quadrupole Moment of the Up: Electromagnetic Interactions. Previous: Magnetic Moments
Martin Savage