Matrix representations of angular momentum operators

  • #1

Homework Statement



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.

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].

Homework 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.
 

Answers and Replies

  • #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.
 
  • #3
vela
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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.
 
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  • #4
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.
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.
 

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