|q⟩ Bad Qubits

intermediate · Physics · Angular Momentum Theory

Matrix Representations

Abstract operators become concrete once we fix a basis. In the {,m}\{|\ell, m\rangle\} basis, the angular-momentum operators are finite matrices of size (2+1)×(2+1)(2\ell+1)\times(2\ell+1). Writing them out makes the algebra tangible and connects directly to the gates of a quantum computer.

L_z is always diagonal

We label states by L^z\hat{L}_z, so L^z\hat{L}_z is automatically diagonal with the eigenvalues m\hbar m on the diagonal. For spin-12\tfrac12 (basis m=+12,12m = +\tfrac12, -\tfrac12),

Lz=2(1001)=2σz.L_z = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \frac{\hbar}{2}\sigma_z.

For spin-1 (basis m=+1,0,1m = +1, 0, -1),

Lz=(100000001).L_z = \hbar\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}.

Off-diagonal pieces from the ladder operators

The components L^x\hat{L}_x and L^y\hat{L}_y are not diagonal; we build them from the ladder operators whose matrix elements we already know,

,mL^±,m=(+1)m(m±1)  δm,m±1.\langle \ell, m' |\hat{L}_\pm| \ell, m\rangle = \hbar\sqrt{\ell(\ell+1) - m(m\pm 1)}\;\delta_{m',\,m\pm 1}.

For spin-1, L^+\hat{L}_+ has the two entries 2\sqrt{2}\,\hbar above the diagonal:

L+=2(010001000),L=L+.L_+ = \hbar\sqrt{2}\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}, \qquad L_- = L_+^\dagger.

Then L^x=12(L^++L^)\hat{L}_x = \tfrac12(\hat{L}_+ + \hat{L}_-) and L^y=12i(L^+L^)\hat{L}_y = \tfrac{1}{2i}(\hat{L}_+ - \hat{L}_-) give the explicit 3×33\times3 matrices.

Spin-1/2 and the Pauli matrices

For spin-12\tfrac12 the construction reproduces the Pauli matrices exactly,

Lx=2σx,Ly=2σy,Lz=2σz.L_x = \frac{\hbar}{2}\sigma_x, \quad L_y = \frac{\hbar}{2}\sigma_y, \quad L_z = \frac{\hbar}{2}\sigma_z.

A useful check

One can verify the matrices satisfy Lx2+Ly2+Lz2=2(+1)1L_x^2 + L_y^2 + L_z^2 = \hbar^2\ell(\ell+1)\,\mathbb{1}. For spin-1 this gives 222\hbar^2 times the 3×33\times3 identity — a quick way to confirm you assembled the matrices correctly.

Try it

Construct the spin-1 raising operator L+L_+ as a 3×33\times3 matrix in the basis ordered m=+1,0,1m = +1, 0, -1, and return the matrix element 0L+1\langle 0|L_+|{-1}\rangle (with =1\hbar = 1). It should equal 2\sqrt{2}.

Run your code to see the quantum state.

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