# Intrinsic and Spectroscopic Electric Quadrupole Moments

1. Oct 26, 2011

### James_1978

Hi...I am trying to calculate the Spectroscopic Electric Quadrupole Moment. I have found an explaination of this in a book by E. Segre called "Nuclei and Particles". However, I am unable to get the correct answer. I hope someone can help me fill in the blanks.

The energy for the quadrupole term is as follows

$$W_{Q} = \frac{1}{2} \psi_{zz} \int \rho(r) \left[\frac{3}{2} z^{2} - \frac{r^{2}}{2}\right] d \tau$$

Here Segre introduces nuclear coordinates

$$eQ = \int \rho_{n} (3 \zeta^{2} - r^{2}) d\tau = 3Q_{\zeta \zeta} - Q_{r r}$$

The nuclear charge distribution is symmetric around $\zeta$. This gives

$$\int \rho_{n} \xi^{2} d \tau = e Q_{\xi \xi} = Q_{\eta \eta}$$

and

$$\int \rho_{n} \xi \eta d \tau = e Q_{\xi \eta} = Q_{\eta \zeta} = Q_{\xi \zeta} = 0$$

The relationship between the x,y,z and \xi, \eta, \zeta give

$$\xi^{2} + \eta^{2} + \zeta^{2} = x^{2} + y^{2} + z^{2} = r^{2}$$

Therefore

$$Q_{r r} = Q_{x x} + Q_{y y} + Q_{z z} = Q_{\xi \xi} + Q_{\eta \eta} + Q_{\zeta \zeta}$$

Moreover

$$z = \xi cos \xi z + \eta cos \eta z + \zeta cos \zeta z$$

With

$$cos^{2} \xi z + cos^{2} \eta z + cos^2 \zeta z = 1$$

Calculating $z^{2}$ and $Q_{z z}$

$$Q_{z z} = Q_{\xi \xi} ( 1 - cos^{2}) + Q_{\zeta \zeta}$$

Then on the next page Segre shows how using the equations above we can get

$$3Q_{z z} - Q_{r r} = Q\left(\frac{3}{2} cos^{2} - \frac{1}{2}\right) = Q P_{2} (cos \theta)$$

I have struggled with this and would appreaciate the help. Also, this is my second post. I am familiar with latex but unsure if it shows up.

Last edited: Oct 26, 2011