Index
Chapter 6 · Item 6.5
Matrix representation
From fixed-l matrix elements to the x-basis example
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Guided reading

This item builds the matrix representation in a fixed-\(l\) subspace and then develops the \(l=1\) change of basis from the \(z\)-basis to the \(x\)-basis.

How the matrix is built

For a fixed angular momentum \(l\), the allowed \(m\) values define a \((2l+1)\)-dimensional subspace. The basis dyads \(|l,m'\rangle\langle l,m|\) carry the matrix entries.

\[\hat L_z=\sum_{m',m}\langle l,m'|\hat L_z|l,m\rangle |l,m'\rangle\langle l,m|\]
\[\hat L^2=\sum_{m',m}\langle l,m'|\hat L^2|l,m\rangle |l,m'\rangle\langle l,m|\]
\[\hat L_\pm=\sum_{m',m}\langle l,m'|\hat L_\pm|l,m\rangle |l,m'\rangle\langle l,m|\]

The matrix elements are fixed by the eigenvalue equations and by the ladder action:

\[\langle l,m'|\hat L_z|l,m\rangle=m\hbar\delta_{m',m}\]
\[\langle l,m'|\hat L^2|l,m\rangle=l(l+1)\hbar^2\delta_{m',m}\]
\[\langle l,m'|\hat L_\pm|l,m\rangle=\sqrt{l(l+1)-m(m\pm1)}\,\hbar\delta_{m',m\pm1}\]
The \(l=1\) basis and matrices
\[|1,+1\rangle=\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad |1,0\rangle=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad |1,-1\rangle=\begin{pmatrix}0\\0\\1\end{pmatrix}\]
\[\hat L_z=\hbar|1,+1\rangle\langle1,+1|+0|1,0\rangle\langle1,0|-\hbar|1,-1\rangle\langle1,-1|\]
\[\hat L_z=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}\]
\[\hat L_+=\sqrt2\hbar\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\qquad \hat L_-=\sqrt2\hbar\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}\]
\[\hat L_x={\hbar\over\sqrt2}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix},\qquad \hat L_y={i\hbar\over\sqrt2}\begin{pmatrix}0&-1&0\\1&0&-1\\0&1&0\end{pmatrix}\]
Eigenvectors of \(\hat L_x\)

The \(x\)-basis is obtained from the eigenvalue equation written in the \(z\)-basis:

\[\hat L_x|l,m\rangle_x=\lambda |l,m\rangle_x\]

For \(l=1\), the determinant gives

\[-\lambda^3+2\lambda\hbar^2=0,\qquad \lambda=+\hbar,0,-\hbar.\]

The normalized eigenvectors are

\[|+1\rangle_x={1\over2}|+1\rangle+{\sqrt2\over2}|0\rangle+{1\over2}|-1\rangle\]
\[|0\rangle_x={\sqrt2\over2}|+1\rangle-{\sqrt2\over2}|-1\rangle\]
\[|-1\rangle_x={1\over2}|+1\rangle-{\sqrt2\over2}|0\rangle+{1\over2}|-1\rangle\]
The matrix \(\hat U\)

The three equations above are summarized as \(|m\rangle_x=\hat U|m\rangle\), with

\[\hat U={1\over2}\begin{pmatrix}1&\sqrt2&1\\ \sqrt2&0&-\sqrt2\\ 1&-\sqrt2&1\end{pmatrix}\]
\[|m\rangle=\hat U^{-1}|m\rangle_x,\qquad \hat U^{-1}=\hat U^\dagger\]

For this \(l=1\) example, the matrix is unitary and diagonalizes \(\hat L_x\) in the \(x\)-basis:

\[(\hat L_x)_x=\hat U(\hat L_x)_z\hat U^\dagger=\hbar\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}\]
Physical reading

The \(z\)-axis was chosen arbitrarily as reference. If \(x\) or \(y\) is chosen instead, the measured component has the same eigenvalues \(m\hbar\), but the basis vectors are changed.

The change-of-basis matrix expresses the same angular momentum operator in the basis adapted to the measured component.

Calculator link

The companion calculator builds these matrices for \(l=\frac12,1,\frac32,2,\frac52\) and shows the chosen \(x\), \(y\), or \(z\) basis decomposed in the \(\hat L_z\) basis.

Open the angular-momentum matrix calculator.

Exercise-ready boundary

This page supports guided exercises on: constructing angular-momentum matrices in the fixed-\(l\) basis and using \(\hat U\) to diagonalize \(\hat L_x\).

  • Use from this page: the definitions, physical setup, highlighted equations and conceptual links needed for a compact calculation or explanation.
  • Long-form material: complete proofs, long demonstrations, extended examples and the full exercise set remain in the textbook.
  • Boundary: do not introduce results, notation or physical claims outside the sequence presented here and in Chapter 6.
Practice anchors

Use these anchors only within the material introduced on this page.

  • Focus: constructing angular-momentum matrices in the fixed-\(l\) basis and using \(\hat U\) to diagonalize \(\hat L_x\).
  • Conceptual check: identify which basis makes the measured component diagonal.
  • Key equation 1:
    • \[\langle l,m'|\hat L_\pm|l,m\rangle=\sqrt{l(l+1)-m(m\pm1)}\,\hbar\delta_{m',m\pm1}\]
  • Key equation 2:
    • \[(\hat L_x)_x=\hat U(\hat L_x)_z\hat U^\dagger\]
  • Typical task: build one matrix, find one basis vector, or verify the diagonalized form.
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Source note: Original auxiliary summary for this book-app, based on Chapter 6 of Mario Reis, Quantum Mechanics, Elsevier, 2026. Book text and figures are copyright © 2026 Elsevier Inc.; consult the original book for the complete presentation, proofs, examples and exercises.