1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix representations of angular momentum operators

  1. Sep 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Write down the 3×3 matrices that represent the operators [itex]\hat{L}_x[/itex], [itex]\hat{L}_y[/itex], and [itex]\hat{L}_z[/itex] of angular momentum for a value of [itex]\ell=1[/itex] in a basis which has [itex]\hat{L}_z[/itex] diagonal.

    3. The attempt at a solution

    Okay, so my basis states [itex]\left\{\left|\ell,m\right\rangle\right\}[/itex] are [itex]\left|1,-1\right\rangle[/itex], [itex]\left|1,0\right\rangle[/itex], and [itex]\left|1,1\right\rangle[/itex]. [itex]\hat{L}_z\left|\ell, m\right\rangle=\hbar m\left|\ell,m\right\rangle[/itex], so the matrix representation of [itex]\hat{L}_z[/itex] is easy: [tex]\hat{L}_z \doteq \left( \begin{array}{ccc} -\hbar & & \\ & 0 & \\ & & \hbar \end{array} \right).[/tex] But I don't know what to do in order to determine [itex]\hat{L}_x[/itex] and [itex]\hat{L}_y[/itex].

    2. Relevant equations

    The commutation relations [itex]\left[ \hat{L}_x, \hat{L}_y \right] = i\hbar \hat{L}_z[/itex], etc., could maybe be useful but I'm not sure how.
     
  2. jcsd
  3. Sep 1, 2013 #2
    Yes you should also use the fact that the matrices should be linearly independent (they form a so(3) basis).

    The commutation relations give you 3 equations, plus if needed you can use the fact that the sum of the squares must be prorportional to the identity.

    You should be able to find them by solving the system of equations.
     
  4. Sep 1, 2013 #3

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You could express ##\hat{L}_x## and ##\hat{L}_y## in terms of the raising and lowering operators. The matrices for the latter are easy to write down.
     
  5. Sep 1, 2013 #4
    Thank you so much for your suggestion. Here's my attempt at following it. I know that$$\begin{align*}
    \hat L_+\left|1,-1\right\rangle &= \hbar\sqrt{2} \left|1,0\right\rangle \\
    \hat L_+\left|1,0\right\rangle &= \hbar\sqrt{2} \left|1,1\right\rangle \\
    \hat L_+\left|1,1\right\rangle &= 0
    \end{align*}$$ and $$\begin{align*}
    \hat L_-\left|1,-1\right\rangle &= 0 \\
    \hat L_-\left|1,0\right\rangle &= \hbar\sqrt{2} \left|1,-1\right\rangle \\
    \hat L_-\left|1,1\right\rangle &= \hbar\sqrt{2} \left|1,0\right\rangle.
    \end{align*}$$
    From these statements we can work out that $$\hat L_+\doteq \left(\begin{array}{ccc}
    0 & 0 & 0 \\
    \hbar\sqrt2 & 0 & 0 \\
    0 & \hbar\sqrt2& 0
    \end{array}\right)$$ and $$\hat L_-\doteq \left(\begin{array}{ccc}
    0 & \hbar\sqrt2 & 0 \\
    0 & 0 & \hbar\sqrt2 \\
    0 & 0 & 0
    \end{array}\right).$$ I know the raising and lowering operators are defined as $$\begin{align*}
    \hat L_+ &= \hat L_x + i\hat L_y \\
    \hat L_-&= \hat L_x - i\hat L_y.
    \end{align*}$$ So, in reverse, we can say that $$\begin{align*}
    \hat L_x &= \tfrac12 \left(\hat L_+ + \hat L_-\right) \doteq \frac{\hbar}{\sqrt2} \left(\begin{array}{ccc}
    0 & 1 & 0 \\
    1 & 0 & 1 \\
    0 & 1 & 0
    \end{array}\right)\\
    \hat L_y &= \tfrac{1}{2i} \left(\hat L_+ - \hat L_-\right) \doteq \frac{\hbar}{\sqrt2} \left(\begin{array}{ccc}
    0 & i & 0 \\
    -i & 0 & i \\
    0 & -i & 0
    \end{array}\right).
    \end{align*}$$ As a check, these matrices uphold both the commutation relations and the sum-of-the-squares identity kevinferreira mentioned above.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Matrix representations of angular momentum operators
Loading...