Why does nuclear energy level with higher j(total angular momentum quantum number) always lie below the smaller j? Here i don't understand the reason for the energy level with j=l+1/2 lies below the j=l-1/2 .From the shell model i see this and it is also written in my book that  energy level with higher  j lies below energy level with lower j .But why?
 A: This is probably best answered by talking about the $j$ splittings in atomic systems, then moving on to consider nuclei.  With atomic electrons, systems with higher $j$ generally have higher energies.  This is part of the fine structure, which is due to relativistic effects and is thus explained by the Dirac equation.  However, a qualitative explanation of how the relativistic effects modify the energy spectrum was already known before the Dirac hydrogen atom was solved in 1928 by Darwin and Gordon.
To make a first stab at understanding the relativistic effects in a single-electron atom, you can imagine going to the electron's rest frame.  In this frame, the positively-charge proton is revolving around the electron.  The motion of the proton produces a magnetic field at the electron's position, which the electron spin interacts with.  As a crude approximation, you could just imagine the proton going around the electron in a circular Bohr orbit and calculate the magnetic field due to the (time-averaged) proton current at the center of the circle.  If you are a bit more careful, you find the following formula for the interaction of the electron with the revolving proton's magnetic field,
$$U_{RS}=\frac{g}{2m^{2}c^{2}}\left(\vec{L}\cdot\vec{S}\right)\frac{1}{r}\frac{dV}{dr},$$
where $g\approx2$ is the gyromagnetic ratio of the electron, $\vec{L}$ and $\vec{S}$ are the orbital and spin angular momenta, and $V$ is the central potential.  This kind of spin-orbit interaction, proportional to $\vec{L}\cdot\vec{S}$ is known as a Russel-Saunders coupling.  According to $U_{RS}$, states with higher $j$ lie higher in energy, because $\vec{J}^{2}=\vec{L}^{2}+\vec{S}^{2}+2\vec{L}\cdot\vec{S}$.
Before the development of the Dirac theory in 1928, it was known, however, that the interaction $U_{RS}$ gave incorrect splittings between different $j$ states.  The measured splitting were half of what the formula predicted.  The reason for this was explained by Thomas in 1927.  The problem is that the frame in which the electron is stationary is not inertial.  The electron is constantly being accelerated around its orbit.  Another way of describing this is that the electron is constantly being boosted from one instantaneous rest frame to another, and the boost direction is always changing.  The product of two Lorentz boosts in different directions is not just another boost; it is actually a combination of a boost and a rotation.  As a result of the continuous boosting, the electron's rest frame is actually rotating, and this frame rotation adds another term to the energy,
$$U'=U_{RS}+\vec{S}\cdot\vec{\omega}_{T}.$$
If you work out the Thomas frequency $\vec{\omega}_{T}$, based just on the kinematics of the electrons' orbit, you find
$$\vec{\omega}_{T}=\frac{\gamma^{2}}{\gamma^{2}+1}\frac{\vec{a}\times\vec{v}}{c^{2}}.$$
Here, $\gamma$ is the Lorentz factor for the electron's motion, $\vec{a}$ is the orbital acceleration (related to $dV/dr$), and $\vec{v}$ the orbital velocity. In the nonrelativistic limit, this reduces the overall coefficient of the Russel-Saunders term by one half, in agreement with experiment.
Now, with all this preparation done, we can turn to the nuclear question.
With nuclear levels, the forces that hold the nucleons together are mediated primarily by pseudoscalar pions.  Since the force carriers are spinless, there is no magnetic interaction associated with the residual strong force in this approximation.  However, the Thomas precession, being a purely kinematic term related to the orbital movement of the constituent particles, still exists.  Therefore, the predominant spin-orbit term is entirely due to the presence of the Thomas precession, $U'=\vec{S}\cdot\vec{\omega}_{T}$, where
$$\vec{\omega}_{T}=-\frac{1}{2m^{2}c^{2}}\left(\vec{L}\cdot\vec{S}\right)\frac{1}{r}\frac{dV}{dr}.$$
Since the coefficient of $U'$ is negative in the nuclear case, states with higher $j$ are pushed to lower energies, producing what are known as "inverted doublets."  Things are bit more complicated in real nuclear models, since real nuclei are not generally well described by single spin-$\frac{1}{2}$ fermions moving in a potentials.  However, the qualitative reasoning here is correct.
This is all explained in further detail in section 11.8 of Classical Electrodynamics (third edition) by Jackson.  It just so happens that I am teaching this topic in my course tomorrow, so it was fresh in my mind.
