# Mistake in “Quantum Mechanics” by Auletta, Fortunato and Parisi?

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

$$$$\hat{\mathbf{l}}^2|l, m_l\rangle=l(l+1)|l, m_l\rangle \tag{6.31}$$$$

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.

• The paragraph is very misleading: there is no such thing as the eigenvalue of $\mathbf{l}$, since its components cannot be simultaneously diagonalized. – Javier Feb 2 at 19:02
• I interpreted it as meaning the magnitude, because otherwise as you say it makes no sense – UtilityMaximiser Feb 2 at 19:18
• They are absolutely right: The are looking at the eigenvalues of $\hat l _x ^2 + \hat l _y ^2 +\hat l _z ^2$ and you are misreading the text if you suspect they are confusing $l$ with eigenvalues of a vector operator. Often, in very advanced q-group texts, one writes expressions such as $| \hat{\vec l} |=\sqrt{l(l+1)}$, but don't worry about such here. – Cosmas Zachos Feb 2 at 19:46
• Cosmas, what do you think "eigenvalue of $\mathbf{l}$" means if not the magnitude? – UtilityMaximiser Feb 3 at 2:01
• l is a vector; when its components commute with each other, its eigenvalue is the vector of the eigenvalues of each component; this is simply not a remote possibility here, given the aggressive noncommutativity of the components, so the authors warn you to not go there, which you may well appear anxious to do. – Cosmas Zachos Feb 3 at 15:06

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

• And what do you interpret "eigenvalue of $\mathbf{l}$" as meaning? – UtilityMaximiser Feb 3 at 2:04
• the vector operator $\hat{\mathbf{l}}$ is really a collection of three conventional operators, one corresponding to each coordinate axis, as such it has no corresponding vector eigenvalue unless the component operators commute and you already know that these component operators do not commute. – hyportnex Feb 3 at 14:30
• @UtilityMaximiser Nothing. There’s no such thing as an “eigenvalue of $\hat{\mathbf l}$”. There is such a thing as an eigenvalue of $\sqrt{\widehat{\mathbf l^2}}$, but it does not appear in your quote (nor is it particularly important). – Alex Shpilkin Feb 3 at 14:57
• Okay so if you agree there is no such thing then you agree the textbook contains a mistake. – UtilityMaximiser Feb 3 at 17:02
• Yes, at best it is misleading, at worst it is wrong; well, nobody is perfect... – hyportnex Feb 3 at 17:07

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 $$\hat{\textbf{l}}_x\psi=m\psi$$ then $$\hat{\textbf{l}}_x^2\psi=m^2\psi$$.

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

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 $$\hat{\textbf{l}}\cdot \hat n=\hat{\textbf{l}}_x \sin\theta\cos\phi + \hat{\textbf{l}}_y \sin\theta\sin\phi +\hat{\textbf{l}}_z \cos\theta$$ when the quantization axis is chosen to lie in the $$\hat n$$ direction.

• Well, no, it’s not an abuse of notation if you write $\mathbf v^2 := v_x^2 + v_y^2 + v_z^2$ for the squared length of a c-vector $\mathbf v$, which is rather common. Sometimes even the scalar product of $\mathbf u$ and $\mathbf v$ is written $\mathbf u\mathbf v$ (e.g. in Landau & Lifschitz), but this is less often used. – Alex Shpilkin Feb 3 at 13:38
• @AlexShpilkin as pointed out $\hat{\mathbf{l}}$ doesn’t make sense as an operator, so it’s square isn’t defined by acting twice with the operator. – ZeroTheHero Feb 3 at 14:10
• Yeah, sorry. The abuse might be to write $\hat{\mathbf l}^2$ instead of $\widehat{\mathbf l^2}$ (the quantization of the classical $\mathbf l^2$). “The square of $\mathbf l$” might make sense, but it’s a bit tricky to define, and “an eigenvalue of $\mathbf l$” does not make sense at all, of course. I’ll write this down in an answer in a moment. – Alex Shpilkin Feb 3 at 14:16

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 $$\hat l_x, \hat l_y, \hat l_z :\mathcal H\to\mathcal H$$ 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 $$\left|\psi\right\rangle\in\mathcal H$$ 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 $$l_z\equiv\lambda$$, and the eigenvector, $$\left|l_z\right\rangle\equiv\left|\psi\right\rangle$$, leading to the funny-looking equation $$\hat l_z\left|l_z\right\rangle = l_z\left|l_z\right\rangle$$. I repeat that for such a vector, $$\hat l_x\left|l_z\right\rangle =\mu\left|l_z\right\rangle$$ 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 $$\mathbf l^2\equiv l_x^2 + l_y^2 + l_z^2$$ (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 by\begin{align*} \mathcal H\hphantom{\rangle} &{}\to\mathcal H\otimes\mathbf R^3\\[.5em] \left|\psi\right\rangle &{}\mapsto\mathbf l\left|\psi\right\rangle\equiv \pmatrix{ l_x\left|\psi\right\rangle \\ l_y\left|\psi\right\rangle \\ l_z\left|\psi\right\rangle }, \end{align*} that is, it returns a block vector composed of the component operators; the space of such vectors is called $$\mathcal H\otimes\mathbf R^3$$, 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 $$\hat{\mathbf A}, \hat{\mathbf B} : \mathcal H\to\mathcal H\otimes\mathbf R^3$$, 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: $$\hat{\mathbf A}\hat{\mathbf B} :=\hat A_x\hat B_x +\hat A_y\hat B_y +\hat A_z\hat B_z$$ (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 $$\widehat{\mathbf A\mathbf B} =\hat{\mathbf A}\hat{\mathbf B}$$ (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.

• (Somehow I never realized before that this was, indeed, $\mathcal H\otimes_{\mathbf R}\mathbf R^3$ and not $\mathcal H\otimes_{\mathbf C}\mathbf C^3$...) – Alex Shpilkin Feb 3 at 15:12
• Totally tangential remark, irrelevant to the question and answer, regarding your "Because quantization is linear and quantizes powers as powers, we also have that ..." Not quite! Not only are the expressions following that not equal (they differ by 3$\hbar^2$/2, accounting, e.g., for the peculiar non-vanishing angular momentum of the lowest Bohr orbit) but the difference might vary with quantization prescription, since the square of the angular momentum is quartic in phase-space variables. – Cosmas Zachos Feb 3 at 16:08
• @CosmasZachos Wait, what? You mean the quantization of $\mathbf l^2$ is (not $0$ but) $3\hbar^2/2$ in the s state? Whoops. Tangential or not, it’s a false statement; thanks! Now if only I understood why =/ (Although I did indeed look at all the nice squares and totally missed that $\hat{\mathbf l}^2$ was in fact quartic and so not safe from prescription-dependent corrections... And neither are powers unless they’re powers of $\alpha\hat x +\beta\hat p$ or something like that dependening on the prescription. Guess I’ll go fix this now.) – Alex Shpilkin Feb 3 at 20:31
• @CosmasZachos Huh. Indeed, Weyl gives $\hat l_z^2 = \widehat{l_z^2} -\hbar^2/2$. Makes me wonder how I’ve never seen that before. Thanks again! – Alex Shpilkin Feb 3 at 21:23
• It's all described an a concise treatise ... book of ours. Mind you, this is in the Weyl prescription. In the Husimi ordering prescription the story is much darker. So, the expectation value of $\hat{ l\cdot l}$ in the s-state of hydrogen is 3$\hbar^2$/2 in Weyl's prescription, even though the expectation value of the conventional $\widehat l \cdot \hat l$ vanishes. The paradox was why the ground state of the Bohr atom does not have l=0.... – Cosmas Zachos Feb 3 at 21:32