02 Jun 2026
Angular Momentum Eigenvalues and Eigenvectors
Eigenvalues of J squared and Jz, ladder construction, spherical harmonics, and angular momentum eigenstates.
The components of angular momentum do not commute, so a quantum state cannot generally have definite values of $J_x$, $J_y$, and $J_z$ at the same time. The useful pair is $J^2$ and one component, conventionally $J_z$, because these two operators commute.
The angular momentum eigenvalue problem is therefore built around simultaneous eigenstates of $J^2$ and $J_z$.
Simultaneous eigenstates
Let the common eigenstate be written as $\lvert j,m\rangle$. It is defined by
\[\boxed{ J^2\lvert j,m\rangle=\hbar^2j(j+1)\lvert j,m\rangle }\]and
\[\boxed{ J_z\lvert j,m\rangle=\hbar m\lvert j,m\rangle. }\]Here $j$ is the angular momentum quantum number and $m$ is the magnetic quantum number.
The allowed values are
\[j=0,\frac12,1,\frac32,\dots\]and, for fixed $j$,
\[m=-j,-j+1,\dots,j-1,j.\]Thus for a given $j$, the number of allowed $m$ values is
\[2j+1.\]Ladder operation
The operators $J_+$ and $J_-$ act as
\[\boxed{ J_\pm\lvert j,m\rangle = \hbar\sqrt{j(j+1)-m(m\pm1)}\, \lvert j,m\pm1\rangle. }\]The coefficient becomes zero at the top and bottom of the ladder:
\[J_+\lvert j,j\rangle=0, \qquad J_-\lvert j,-j\rangle=0.\]This termination condition forces the allowed $m$ values to be finite and evenly spaced.
The ladder picture also explains why $m$ changes in steps of one. Starting from the highest state and applying $J_-$ repeatedly gives the full set of $m$ values for a fixed $j$.
Orbital angular momentum
For orbital angular momentum, the quantum number is usually written as $l$ instead of $j$:
\[L^2\lvert l,m\rangle=\hbar^2l(l+1)\lvert l,m\rangle,\] \[L_z\lvert l,m\rangle=\hbar m\lvert l,m\rangle.\]For orbital angular momentum,
\[l=0,1,2,3,\dots\]and
\[m=-l,-l+1,\dots,l.\]Only integer $l$ values occur for orbital angular momentum because the spatial wavefunction must be single-valued under a rotation by $2\pi$.
Spherical harmonics
In coordinate representation, the simultaneous eigenfunctions of $L^2$ and $L_z$ are the spherical harmonics:
\[L^2Y_l^m(\theta,\phi)=\hbar^2l(l+1)Y_l^m(\theta,\phi),\] \[L_zY_l^m(\theta,\phi)=\hbar mY_l^m(\theta,\phi).\]Since
\[L_z=-i\hbar\frac{\partial}{\partial\phi},\]the azimuthal dependence is
\[Y_l^m(\theta,\phi)\propto e^{im\phi}.\]Single-valuedness requires
\[e^{im(\phi+2\pi)}=e^{im\phi},\]so $m$ must be an integer for orbital angular momentum.
Normalization and orthogonality
The spherical harmonics are orthonormal:
\[\int_0^{2\pi}\int_0^\pi Y_l^{m*}(\theta,\phi)Y_{l'}^{m'}(\theta,\phi) \sin\theta\,d\theta\,d\phi = \delta_{ll'}\delta_{mm'}.\]They also form a complete basis for angular functions on the sphere.
Example for $j=2$
For $j=2$, the allowed magnetic quantum numbers are
\[m=-2,-1,0,1,2.\]There are
\[2j+1=5\]states. The eigenvalue of $J^2$ is
\[\hbar^2j(j+1)=6\hbar^2.\]The eigenvalues of $J_z$ are
\[-2\hbar,\ -\hbar,\ 0,\ \hbar,\ 2\hbar.\]Ladder coefficient example
For $j=1$ and $m=0$,
\[J_+\lvert 1,0\rangle =\hbar\sqrt{1(1+1)-0(0+1)} \lvert 1,1\rangle.\]Thus
\[\boxed{ J_+\lvert 1,0\rangle=\hbar\sqrt2\,\lvert 1,1\rangle. }\]Similarly,
\[J_-\lvert 1,0\rangle=\hbar\sqrt2\,\lvert 1,-1\rangle.\]Main points
- The state $\lvert j,m\rangle$ is a simultaneous eigenstate of $J^2$ and $J_z$.
- The eigenvalue of $J^2$ is $\hbar^2j(j+1)$.
- The eigenvalue of $J_z$ is $\hbar m$.
- For each $j$, there are $2j+1$ possible $m$ values.
- Orbital angular momentum has integer $l$; spin can have integer or half-integer $j$.
Practice questions
- Derive the coefficient in $J_\pm\lvert j,m\rangle$.
- Explain why $m$ runs from $-j$ to $j$.
- Why is orbital angular momentum restricted to integer $l$?
- Show that $Y_l^m(\theta,\phi)$ is an eigenfunction of $L_z$.
- How many states correspond to $l=3$?
Discussion