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So I want to calculate the matrix of $\hat{L}_x=i\hbar(sin\phi\partial_\theta+cot\theta cos\phi\partial_\phi)$ and $\hat{L}_y=i\hbar(sin\phi\partial_\theta+cot\theta cos\phi\partial_\phi)$ with the sperical harmonics if $l=1$. So I need to calculate

$$ \hat{L}_x=\left(\begin{eqnarray} <Y_1^1|L_x|Y_1^1> & <Y_1^1|L_x|Y_1^0> & <Y_1^1|L_x|Y_1^{-1}>\\ <Y_1^0|L_x|Y_1^1> & <Y_1^0|L_x|Y_1^0> & <Y_1^0|L_x|Y_1^{-1}> \\ <Y_1^{-1}|L_x|Y_1^1> & <Y_1^{-1}|L_x|Y_1^0> & <Y_1^{-1}|L_x|Y_1^{-1}> \end{eqnarray}\right) $$ and $$ \hat{L}_y=\left(\begin{eqnarray} <Y_1^1|L_y|Y_1^1> & <Y_1^1|L_y|Y_1^0> & <Y_1^1|L_y|Y_1^{-1}>\\ <Y_1^0|L_y|Y_1^1> & <Y_1^0|L_y|Y_1^0> & <Y_1^0|L_y|Y_1^{-1}> \\ <Y_1^{-1}|L_y|Y_1^1> & <Y_1^{-1}|L_y|Y_1^0> & <Y_1^{-1}|L_y|Y_1^{-1}> \end{eqnarray}\right) $$

but I also found the identity

$$ [\hat{L}_x,\hat{L}_y]=i\hbar\hat{L}_z $$ but that implies that at least some of the entries of the $\hat{L}_x$ and $\hat{L}_y$ matrix have to be complex. Since for example take row 1 column 1. There $$ \hat{L}_{z,1,1}=1\Rightarrow i\hbar\hat{L}_{z,1,1}=i\hbar $$ And with the help of matrixcalc.org (no paid advertising) the commutator there is $$ [\hat{L}_x,\hat{L}_y]_{1,1}=<Y_1^0|L_y|Y_1^1><Y_1^1|L_y|Y_1^0>+<Y_1^{-1}|L_y|Y_1^1><Y_1^1|L_y|Y_1^{-1}>-<Y_1^1|L_y|Y_1^0><Y_1^0|L_y|Y_1^1>-<Y_1^1|L_y|Y_1^{-1}><Y_1^{-1}|L_y|Y_1^1\overset{!}{=}i\hbar $$ This implies that something on the left side has to be complex. But the expected value of an operator has to be always real. So is actually the expected value of an operator not always real ? I hope someone can help me with this

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    $\begingroup$ Are these operators hermitian or not? $\endgroup$
    – Gec
    Commented Jan 16, 2022 at 10:52

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Never mind real or complex, it should be intuitively clear that the eigenvalues must be the same by symmetry since there’s nothing special about the choice $\hat x$, $\hat y$ or $\hat z$ axes as quantization axis.

As a result if you believe that the eigenvalues of $L_z$ are real, then the eigenvalues of $L_x$ and $L_y$ will also be real, and as argued above identical to the eigenvalues of $L_z$.

The expressions that you wrote for the matrix elements uses the convenient choice of spherical harmonics as basis states, and this choices makes $L_z$ simpler as an operator because the $\hat z$ axis is distinguished in the construction of polar coordinates: the polar angle $\theta$ is by convention measured from $\hat z$. The spherical harmonics are complex functions so yes in general the matrices for $L_x$ and $L_y$ will be complex, but that doesn’t immediately imply anything about the eigenvalues.

Mathematically, the operators $L_x$ etc are all hermitian (i.e. $L_x^\dagger=L_x$) and one can show mathematically that eigenvalues of hermitian matrices are necessarily real.

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  • $\begingroup$ I think i got it. So the reason that here the matrices of $L_y$ and $L_x$ can have complex entries is that $Y_l^m$ are not eigenstates of $L_y$ and $L_x$. Only for their eigenstates ($\Psi$) applies $$ \hat{L}_{x or y}|\Psi>=q|\Psi>\Rightarrow <\Psi|\hat{L}_{x or y}|\Psi>=q $$ where $l$ is a real number. But because $Y_l^m\neq\Psi$ this equation doesnt hold for $Y_l^m$ and so $<Y_l^m|\hat{L}_{x or y}|Y_l^m>$ can also be complex $\endgroup$
    – peter mafai
    Commented Jan 16, 2022 at 17:17
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The expectation values of $\hat{L}_{k}$ for $k \in \{x, y, z\}$ are indeed always real, since they are hermitian operators.

The commutator of $\hat{L}_{j}$ and $\hat{L}_{k}$ for $j \neq k$ on the other hand is not hermitian, but anti hermitian: \begin{align} [\hat{L}_{j}, \hat{L}_{k}]^{\dagger} &= (\hat{L}_{j}\hat{L}_{k})^{\dagger} - (\hat{L}_{k}\hat{L}_{j})^{\dagger}\\ &= \hat{L}_{k}^{\dagger}\hat{L}_{j}^{\dagger} - \hat{L}_{j}^{\dagger}\hat{L}_{k}^{\dagger} \\ &= \hat{L}_{k}\hat{L}_{j} - \hat{L}_{j}\hat{L}_{k} \\ &= [\hat{L}_{k},\hat{L}_{j}] \\ &= -[\hat{L}_{j},\hat{L}_{k}] \,. \end{align} Another way of seeing that directly is by using the angular momentum algebra: \begin{align} [\hat{L}_{j}, \hat{L}_{k}] = i \hbar \sum_{l = 1}^{3} \epsilon_{jkl} \hat{L}_{l} \end{align} If you take the hermitian conjugate of this equation you get an extra minus on the right hand side due to the implicit complex conjugation of the imaginary unit, while $\hat{L}_{l}$ is mapped to itself.

The eigenvalues of anti hermitian operators must be purely imaginary. Therefore, your result is perfectly valid.

I would suggest to try calculating the matrix elements of $\hat{L}_{x}$ and $\hat{L}_{y}$ by first evaluating all matrix elements of the Ladder operators \begin{align} \hat{L}_{\pm} &= \hat{L}_{x} \pm i \hat{L}_y \end{align} which fulfill the relations [Source: https://en.wikipedia.org/wiki/Ladder_operator#Angular_momentum]: \begin{align} \hat{L}_{\pm}Y_{l,m} &= \hbar \sqrt{l(l+1) - m(m \pm 1)} Y_{l,m \pm 1}\,. \end{align} I hope this was helpful to you!

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  • $\begingroup$ So as a result for the ladder operators i get $$ L_+=\left(\begin{eqnarray} 0 & \sqrt2 & 0\\ 0 & 0 & \sqrt2\\ 0 & 0 & 0 \end{eqnarray}\right)\hbar $$ $$ L_-=\left(\begin{eqnarray} 0 & 0 & 0\\ \sqrt2 & 0 & 0\\ 0 & \sqrt2 & 0 \end{eqnarray}\right)\hbar $$ Then with $L_\pm=L_x\pm iL_y$ i get $$ L_y=\left(\begin{eqnarray} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{eqnarray}\right) $$ and $$ L_x=\left(\begin{eqnarray} 0 & \sqrt2 & 0\\ \sqrt2 & 0 & \sqrt2\\ 0 & \sqrt2 & 0 \end{eqnarray}\right)\hbar $$ is that right ? $\endgroup$
    – peter mafai
    Commented Jan 16, 2022 at 11:45
  • $\begingroup$ But then $[L_x,Ly]=L_xL_y-L_yL_x=0\neq i\hbar L_z$ since $$ L_z=\left(\begin{eqnarray} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{eqnarray}\right)\hbar $$ $\endgroup$
    – peter mafai
    Commented Jan 16, 2022 at 11:49
  • $\begingroup$ I am a little confused that you get the zero operator for \begin{align} \hat{L}_y = -\frac{i}{2}(\hat{L}_+ - \hat{L}_-) \end{align} using your results you should get \begin{align} \hat{L}_y = -\frac{i}{2} \hbar \begin{pmatrix} 0 & \sqrt{2} & 0 \\ -\sqrt{2} & 0 & \sqrt{2} \\ 0 & -\sqrt{2} & 0 \end{pmatrix} \end{align} Moreover, I think you should reverse the order of rows in your matrices, since your first matrix element should e.g. look like $\langle Y_{1, -1} \vert \hat{L}_j \vert Y_{1,-1} \rangle$. But that may be just a choice of convention as long as you do it consistently. $\endgroup$
    – Bobsn
    Commented Jan 16, 2022 at 12:01
  • $\begingroup$ And $\hat{L}_x = \frac{1}{2}(\hat{L}_{+} + \hat{L}_{-})$. So you forgot a factor of $\frac{1}{2}$ in your calculation, I think. $\endgroup$
    – Bobsn
    Commented Jan 16, 2022 at 12:03
  • $\begingroup$ ahh. okay. i took the easy route. i said "because $L_\pm=L_x+iL_y$ is real it means that $L_y=0$". Okay but does that mean then that $L_y$ is an anti hermitian operator? And if yes: Is it not measureable ? or what does it physically mean that an operator is anti hermitian? $\endgroup$
    – peter mafai
    Commented Jan 16, 2022 at 12:33

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