What does it mean when a degeneracy is lifted? I would like to ask what is the meaning of degeneracy been lifted? For example when the Schrodinger equation is subjected to magnetic field, there is a  $m\ell$ degeneracy is lifted while $\ell$ remains fixed.
 A: To add to the correct answer: A degeneracy means there is a symmetry in your system. There may be a hidden symmetry, leading to what is called an "accidental degeneracy", or an explicit symmetry that leads to an "essential  degeneracy" (aka: "systematic-", "geometric-", "normal-").
In the non-relativistic hydrogen atom, the energy does not depend on angular momentum quantum numbers $l$ and $m$, so the orbital (fixed principle quantum number, $n$) has a degree of degeneracy:
$$ d =2\sum_{l=0}^{n-1}2l+1 = 2n^2$$
where the factor of 2 accounts for the spin-degeneracy.
The $m$ quantum number is a great example of an essential degeneracy. We are taught (in physics) that at fixed $l$ (for example, the P shell, $l=1$), that each available $m$ orbital is filled one-by-one (ignoring the spin degeneracy):  $|1,-1\rangle, |1,0\rangle, |1,1\rangle$, or equivalently, using orbital wave function $Y_1^{-1}(\theta_z, \phi_z)$, $Y_1^{0}(\theta_z, \phi_z)$, $Y_1^{1}(\theta_z, \phi_z)$, where the subscript $z$ indicated the angles are relative to the arbitrary $z$-axis.
Sometimes we forget that the $z$-axis is arbitrary. Suppose we put an electron in the state:
$$ \psi = R_2(r)(Y_1^{-1}(\theta_z, \phi_z) +Y_1^{1}(\theta_z, \phi_z))/\sqrt 2 $$
where $R_2(r)$ is $n=2$ radial wave function.
Is it in a mixed state, since it mixes $m$ values? Well, no: $m=+1$ and $m=-1$ are essentially degenerate. We just made the $z$-axis up, and its definition has no physical consequences.
Had we decided to define spherical harmonics relative to $x$-axis, you'd find that:
$$ \psi = R_2(r)Y_1^0(\theta_x, \phi_x) = R_2(r)|L=1,L_x=0\rangle$$
which is a pure state. The $m$ values must be degenerate because of the spherical symmetry of our problem. We artificially break that symmetry (on paper), by defining a $z$-axis. Any (normalized) linear combination of the $2l+1$ $|n,l,m\rangle$ for fixed $n$, $l$, $m\in(-l,\ldots,l)$ eigenstates will be a pure state for some axis. Cleary, if energy depended on $m$, that would not be possible because the time evolution the states, $e^{iE_{n,l,m}/\hbar}$, would change relative phases, and the state would not be stationary. It really is an essential degeneracy.
Now when you turn on a constant magnetic field (see: Zeeman Effect).
$$\vec B(\vec r) = B\hat z$$
you break the spherical symmetry of the hamiltonian. The $z$-axis is no-longer arbitrary, $E_{n,l,m}$ becomes a function of $m$, and the degeneracy is said to be lifted.
Note that the spherical symmetry of $\hat H$ is the reason $\vec L$ is conserved so that $l, m$ are "good" quantum numbers.
The Start Effect (external electric field) breaks rotational symmetry and lift degeneracy.
The non-relativistic hydrogen atom also has an accidental degeneracy. For fixed principal quantum $n$, the energy does not depend on the angular momentum quantum number, $l$. This may be called an accidental degeneracy, but it is not an accident. There is a hidden symmetry in:
$$ \hat H=\frac{\hat p^2}{2m}-\frac k r $$
that leads to the conservation of the Laplace-Runge-Lenz vector:
$$\vec A\equiv \vec p \times \vec L -mk\hat r$$
(In classical mechanics, this leads to orbital energy being independent of eccentricity).
This degeneracy is lifted by the fine structure imposed by the Dirac equation (in which spin is implicit.). E.g., spin-orbit coupling, $l,m,s$ are no-longer good quantum numbers. Instead one uses total angular momentum $j,m_j$:

Note that rotational symmetry preserves the $m_j$ degeneracy.
The $l$-degeneracy can also broken by an external electric field (Stark Effect), where one can see the phenomenon of "avoided-level crossings" and also exact degeneracies which are due to underlying symmetries in the Coulomb interaction (see figures, from https://en.wikipedia.org/wiki/Stark_effect):

A: When there is a degeneracy in a quantum number $m$, it means that changing $m$ between its eigenvalues will not change the energy of the state.
If your system has a magnetic dipole moment attached to the $m$ quantum number (as it has in the case of e.g. an atom), then it will be energetically favorable for the system to align this moment in the same direction as an external magnetic field. Thus, if $m = -1/2$ is the state which is aligned with the magnetic field, then the energy of this state will be less than the $m = +1/2$ state. Now the energies of the two systems are no longer the same, and so there is no degeneracy. The degeneracy is said to have been lifted.
