Jtdylktuy

On page 200 of Auletta, Fortunato and Parisi's textbook on Quantum Mechanics they write:

begin{equation}
hat{mathbf{l}}^2|l, m_lrangle=l(l+1)|l, m_lrangle tag{6.31}
end{equation}

This is a peculiarity of quantum mechanics, in that the eigenvalue of the square of $hat{mathbf{l}}$ is not the square of the eigenvalue of $hat{mathbf{l}}$. As we see from Eq. (6.27), it is a direct consequence of the fact that the angular momentum's components (in particular $hat{l}_x$ and $hat{l}_y$) do not commute with each other. Indeed, if they did commute, the last term in brackets in Eq. (6.28) would vanish and the eigenvalue of $hat{mathbf{l}}$ would be equal to $l^2$.

Here $hat{mathbf{l}}$ is the operator corresponding to the magnitude of angular momentum and $l$ its maximal eigenvalue of $hat{l}_z$. I have read through the section carefully and they seem to be confusing the quantum number $l$ with the eigenvalue of $hat{mathbf{l}}$, which is $sqrt{l(l+1)}$.

The statement that the eigenvalues of an operator $A^2$ are not the square of the eigenvalues of $A$ seems like complete nonsense to me. Clearly if $$A psi = lambda psi$$ Then $$A^2 psi = lambda^2 psi$$ from the simple fact that operators commute with scalar multiplication.

$hat{mathbf{l}}$ is a 3-component vector operator and as such its "square" is not the same as it being applied twice.

Writing $hat{textbf{l}}^2$ is a bit of an abuse of notation but conventional since $hat{textbf{l}}^2$ is actually the sum of squares of operators:
$$
hat{textbf{l}}^2:=hat{textbf{l}}_x^2+hat{textbf{l}}_y^2+hat{textbf{l}}_z^2, .
$$
Of course if $hat{textbf{l}}_xpsi=mpsi$ then
$hat{textbf{l}}_x^2psi=m^2psi$.

The notation is meant to reflect the fact that
$vec Lcdotvec L=L^2=L_x^2+L_y^2+L_z^2$ is the length squared of the angular momentum vector
$vec L=hat x L_x+hat yL_y+hat zhat L_z$.

The "operator" $hat{textbf{l}}$ doesn't make a lot of of sense, although
$hat{textbf{l}}cdot hat n$ is often used to denote the operator
$$
hat{textbf{l}}cdot hat n=hat{textbf{l}}_x sinthetacosphi
+ hat{textbf{l}}_y sinthetasinphi
+hat{textbf{l}}_z costheta
$$
when the quantization axis is chosen to lie in the $hat n$ direction.

The mistake in the textbook is the claim that there is such a thing as an eigenvalue of $mathbf l$. There isn’t.

What do we have at hand?

• Three operators $hat l_x, hat l_y, hat l_z :mathcal Htomathcal H$ that take vectors in the Hilbert space $mathcal H$ of the system and return vectors in the same space. These are the quantum counterparts (“quantizations”) of the classical observables $l_x$, $l_y$, $l_z$, but they do not commute and therefore do not have a common eigenbasis.




• For a vector $left|psirightrangleinmathcal H$ in the eigenbasis of a single one of those operators (say $hat l_z$), the corresponding eigenvalue $lambda$. By a common abuse of notation, this eigenvalue is also denoted $l_zequivlambda$, and the eigenvector, $left|l_zrightrangleequivleft|psirightrangle$, leading to the funny-looking equation $hat l_zleft|l_zrightrangle = l_zleft|l_zrightrangle$. I repeat that for such a vector, $hat l_xleft|l_zrightrangle =muleft|l_zrightrangle$ does not hold for any scalar $mu$, and similarly for $hat l_y$.




• For the whole state space $mathcal H$ (or its subspace, see below), the largest among the possible eigenvalues of $hat l_z$. By yet another abuse of notation, this is usually denoted $l$.




• The operator $widehat{mathbf l^2}$ (note the positioning of the hat), i.e. the quantum counterpart of the classical observable $mathbf l^2equiv l_x^2 + l_y^2 + l_z^2$ (note that the $l_z$ here is not the same thing as in the previous two bullet points). Because quantization is linear and quantizes powers as powers (up to higher-order corrections in $hbar$), we also have that $widehat{mathbf l^2} = widehat{(l_x^2 + l_y^2 + l_z^2)}$ is equal to $hat l_x^2 +hat l_y^2 +hat l_z^2 + O(hbar^2),$.

What would, then, be the correct formulation of the statement of the book? As follows: the eigenvalues of $widehat{mathbf l^2}$ are of form $l(l+1)$ for integral or half-integral $l$. Also, if we take an eigenspace of $widehat{mathbf l^2}$ corresponding to an eigenvalue $l(l+1)$ (it will in general be more than one-dimensional), then $hat l_z$ would map it into itself, so we can discuss the eigenvalues of $hat l_z$ restricted to that subspace. The maximum such eigenvalue will turn out to be $l$.

This is surprising, because in the classical case (or, equivalently but more confusingly, in an alternative world where $hat l_x$, $hat l_y$, and $hat l_z$ commuted) the maximum value of the observable $l_z$ among states with a fixed value of $mathbf l^2$ (call it $lambda$) would be $l = sqrt{lambda}$ (that’s a different $l$!), i.e. $lambda = l^2$. Not so in the quantum case.

That was the required part. However, if you want to recover your $hat{mathbf l}^2$, read on...

• So far there is no operator $hat{mathbf l}$ that would allow us to define the operator $hat{mathbf l}^2 := bigl(hat{mathbf l}bigr)^2$. Can such an operator be defined? Yes, but it is a bit tricky: it acts bybegin{align*}
  mathcal Hhphantom{rangle} &{}tomathcal Hotimesmathbf R^3\[.5em]
  left|psirightrangle &{}mapstomathbf lleft|psirightrangleequiv
  pmatrix{
  l_xleft|psirightrangle \
  l_yleft|psirightrangle \
  l_zleft|psirightrangle
  },
  end{align*}
  that is, it returns a block vector composed of the component operators; the space of such vectors is called $mathcal Hotimesmathbf R^3$, the “tensor product” of $mathcal H$ and $mathbf R^3$ (you’ll encounter it in the discussion of composite system, although this particular use of it is unrelated to them). Because it is not an operator from a vector space to itself, it does not even make sense to speak of its eigenvectors or eigenvalues.



• 
  

  We still haven’t defined anything worthy of the name $hat{mathbf l}^2$, let alone shown it were equal to $widehat{mathbf l^2}$. It can’t be the usual square of an operator, because that definition would require $hat{mathbf l}$ to map $mathcal H$ to itself, and it does not.

  
  
  

  The idea is that if we have two operators $hat{mathbf A}, hat{mathbf B} : mathcal Htomathcal Hotimesmathbf R^3$, we can compose their “$mathcal H$ parts” while taking the scalar product of their “$mathbf R^3$ parts”. This is, of course, fancy language for what you would’ve written anyway: $hat{mathbf A}hat{mathbf B} :=hat A_xhat B_x +hat A_yhat B_y +hat A_zhat B_z$ (note that it does map $mathcal H$ to itself). This definition, indeed, somethimes used in both physics and mathematics. In the particular case of quantum observables, we also have $widehat{mathbf Amathbf B} =hat{mathbf A}hat{mathbf B}$ (up to higher-order corrections in $hbar$) provided that the corresponding $hat A_i$ and $hat B_i$ commute (in particular, if they are equal to each other, $hat{mathbf A} =hat{mathbf B} =hat{mathbf l}$), but not otherwise.