Skip to main content
Chemistry LibreTexts

15.9: Problems

  • Page ID
    107059
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
    Problem \(\PageIndex{1}\)

    Given

    \[\mathbf{A}=\begin{pmatrix} 2&3&-1 \\ -5&0&6\\ 0&2&3 \end{pmatrix}\; ;\mathbf{B}=\begin{pmatrix} 2 \\ 1\\0 \end{pmatrix}\; ;\mathbf{C}=\begin{pmatrix} 0&1\\ 2&0\\-1&3 \end{pmatrix} \nonumber \]

    Multiply all possible pairs of matrices.

    Problem \(\PageIndex{2}\)

    The matrix representation of a spin \(1/2\) system was introduced by Pauli in 1926. The Pauli spin matrices are the matrix representation of the angular momentum operator for a single spin \(1/2\) system and are defined as:

    \[\mathbf{\sigma_x}=\begin{pmatrix} 0&1 \\ 1&0 \end{pmatrix}\; ;\mathbf{\sigma_y}=\begin{pmatrix} 0&-i \\ i&0 \end{pmatrix}\; ;\mathbf{\sigma_z}=\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \nonumber \]

    1. Show that \(\mathbf{\sigma_x}\mathbf{\sigma_y}=i\mathbf{\sigma_z}\),\(\mathbf{\sigma_y}\mathbf{\sigma_z}=i\mathbf{\sigma_x}\) and \(\mathbf{\sigma_z}\mathbf{\sigma_x}=i\mathbf{\sigma_y}\)
    2. Calculate the commutator \(\left[\mathbf{\sigma_x},\mathbf{\sigma_y} \right]\).
    3. Show that \(\mathbf{\sigma_x}^2=\mathbf{\sigma_y}^2=\mathbf{\sigma_z}^2=\mathbf{I}\), where \(\mathbf{I}\) is the identity matrix. Hint: as with numbers, the square of a matrix is the matrix multiplied by itself.
    Problem \(\PageIndex{3}\)

    The inversion operator, \(\hat i\) transforms the point \((x,y,z)\) into \((-x,-y,-z)\). Write down the matrix that corresponds to this operator.

    symmetry_i.jpg
    Figure \(\PageIndex{1}\): The inversion operator (CC BY-NC-SA; Marcia Levitus)
    Problem \(\PageIndex{4}\)

    Calculate the inverse of \(\mathbf{A}\) by definition.

    \[\mathbf{A}=\begin{pmatrix} 1&-2 \\ 0&1 \end{pmatrix} \nonumber \]

    Problem \(\PageIndex{5}\)

    Calculate the inverse of \(\mathbf{A}\) by definition.

    \[\mathbf{A}=\begin{pmatrix} \cos \theta&-\sin \theta \\ \sin \theta&\cos \theta \end{pmatrix} \nonumber \]

    Problem \(\PageIndex{6}\)

    Find the eigenvalues and nomalized eigenvectors of

    \[\mathbf{M_1}=\begin{pmatrix} 2&0 \\ 0&-3 \end{pmatrix} \nonumber \]

    \[\mathbf{M_2}=\begin{pmatrix} 1&1+i \\ 1-i&1 \end{pmatrix} \nonumber \]

    Problem \(\PageIndex{7}\)

    Given,

    \[\mathbf{M_3}=\begin{pmatrix} 1&1-i \\ 1+i&1 \end{pmatrix} \nonumber \]

    1. Show that the matrix is Hermitian.
    2. Calculate the eigenvectors and prove they are orthogonal.

    This page titled 15.9: Problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Marcia Levitus via source content that was edited to the style and standards of the LibreTexts platform.

    • Was this article helpful?