Skip to main content
Chemistry LibreTexts

Worksheet 5B: Angular Momentum and Commutators (L^2 corrected)

  • Page ID
    120357
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Name: ______________________________

    Section: _____________________________

    Student ID#:__________________________

    Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

    We can express the quantum mechanical angular momentum (in Cartesian coordinates) using operators:

    • \(\hat L_x=\hat y\hat p_z-\hat z\hat p_y=-i\hbar \left(y\dfrac{\partial }{\partial z}-z\dfrac{\partial}{\partial y}\right)\)
    • \(\hat L_y=\hat z\hat p_x-\hat x\hat p_z=-i\hbar \left(z\dfrac{\partial }{\partial x}-x\dfrac{\partial}{\partial z}\right)\)
    • \(\hat L_z=\hat x\hat p_y-\hat y\hat p_x=-i\hbar \left(x\dfrac{\partial }{\partial y}-y\dfrac{\partial}{\partial x}\right)\)
    • \(\hat L^2=\hat L_x^2+\hat L_y^2+\hat L_z^2 = \hat L_x \hat L_x+\hat L_y \hat L_y+\hat L_z \hat L_z =\\ -\hbar^2((y^2+z^2)\dfrac{\partial^2}{\partial x^2} + (x^2+z^2)\dfrac{\partial^2}{\partial y^2} + (y^2+x^2)\dfrac{\partial^2}{\partial z^2}-2 (xy\dfrac{\partial^2}{\partial x \partial y}+xz\dfrac{\partial^2}{\partial x \partial z}+zy\dfrac{\partial^2}{\partial z \partial y})-2(x \dfrac{\partial }{\partial x}+y \dfrac{\partial }{\partial y}+z \dfrac{\partial }{\partial z}))\)

    Q1

    What exactly does each of the the four operators above measure (more specifically, what do their eignevalues represent)?

    Given these definitions, do \(\hat L^2\) and \(\hat L_x\) commute? (at least set up this problem - you can skip to the next question if you get stuck on the math and review outside of class).

    If \([\hat {L}^2,\hat {L}_x]=0\), should \(\hat L^2\) commute with \(\hat {L}_y\) and/or \(\hat L_z\)? Explain why or why not.

    If \([\hat {L}^2,\hat {L}_x]=0\), what does this tell you about the eigenfunctions for these operators and the "measurability" of their corresponding eigenvalues?

    If \([\hat L^2,\hat L_x]=0\) and \([\hat L^2,\hat L_z]=0\), does this tell you anything about whether the individual \(x\), \(y\), and \(z\) components of \(\hat L^2\) commute with each other? Why or why not?

    Q2

    As it turns out (you should confirm this mathematically before the next exam)

    • \([\hat L_x,\hat L_y]=i\hbar\hat L_z\),
    • \([\hat L_y,\hat L_z]=i\hbar\hat L_x\),
    • \([\hat L_z,\hat L_x]=i\hbar\hat L_y\)

    What does this tell you about the eigenfunctions for \(\hat L_x\), \(\hat L_y\), and \(\hat L_z\)?

    How does the fact that \(\hat L_x\), \(\hat L_y\), and \(\hat L_z\), the individual components of \(\hat L^2\), do not commute with each other affect the commutators below?

    • \([\hat L^2,\hat L_x]=0\),
    • \([\hat L^2,\hat L_y]=0\)
    • \([\hat L^2,\hat L_z]=0\)

    If the angular moment operators were converted to spherical coordinates, then

    • \(\hat L_x=i\hbar(\sin{\phi}\dfrac{\partial}{\partial \theta}+\cot{\theta}\cos{\phi}\dfrac{\partial}{\partial \phi})\),
    • \(\hat L_y=-i\hbar(\cos{\phi}\dfrac{\partial}{\partial \theta}-\cot{\theta}\sin{\phi}\dfrac{\partial}{\partial \phi})\), and
    • \(\hat L_z=-i\hbar\dfrac{\partial}{\partial \phi}\),
    • \(\hat L^2=-\hbar^2 \left[ \dfrac{1}{\sin{\theta}}\dfrac{\partial}{\partial \theta} \left(\sin{\theta}\dfrac{\partial}{\partial \theta}\right)+\dfrac{1}{\sin^2{\theta}}\dfrac{\partial^2}{\partial\phi^2}\right]=-\hbar^2 \left(\dfrac{\partial^2}{\partial \theta^2}+cot{\theta}\dfrac{\partial}{\partial \theta}+\dfrac{1}{\sin^2{\theta}}\dfrac{\partial^2}{\partial\phi^2}\right)\)

    Based on the simplicity of the problem, which commutator problem, \([\hat L^2,\hat L_x]=0\), \([\hat L^2,\hat L_y]=0\), or \([\hat L^2,\hat L_z]=0\) looks the easiest? Does switching to spherical coordinates change the resulting commutator relationships discussed above?


    Worksheet 5B: Angular Momentum and Commutators (L^2 corrected) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?