If we find the expectation value $$\langle L_x\rangle = \langle L_y\rangle = 0$$ and $$\langle L_x^2\rangle = \langle L_y^2\rangle,$$ what is the physical significance that their values are equal?

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    $\begingroup$ Spherical symmetry. $\endgroup$
    – J.G.
    Sep 3, 2021 at 13:00
  • $\begingroup$ @J.G. that was my 1st reaction, but on 2nd thought: cylindrical? $\endgroup$
    – JEB
    Sep 3, 2021 at 13:16
  • $\begingroup$ @JEB only if all higher-order moments are also equal, which is not guaranteed $\endgroup$ Sep 3, 2021 at 13:47
  • $\begingroup$ To clarify, my comment was meant to explain why it happens in practice, but technically the given conditions don't imply the full symmetry unless these conditions are invariant under a rotation. $\endgroup$
    – J.G.
    Sep 3, 2021 at 13:49
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    $\begingroup$ If you write these in terms of raising and lowering operators, you note that all eigenstates of $L_z$ satisfy them. $\endgroup$ Sep 3, 2021 at 14:14

2 Answers 2


Use \begin{align} (\Delta L_z)^2(\Delta L_x)^2 &\ge \frac{1}{2}\vert \langle L_y\rangle\vert\, ,\\ (\Delta L_z)^2(\Delta L_y)^2 &\ge \frac{1}{2}\vert \langle L_x\rangle\vert \end{align} Thus, if a system is in an eigenstate of $L_z$, $\Delta L_z=0$ and immediately $\langle L_y\rangle=\langle L_x\rangle=0$

A system with this kind of symmetry is the angular momentum coherent state, where \begin{align} L_k\mapsto J_k= R(\Omega) J_k R^{-1}(\Omega) \end{align} where $R(\Omega)$ is any rotation. The condition $\Delta J_x=\Delta J_y$ indicates that the state is not squeezed.

The Wigner function of an angular momentum coherent state with $L=8$ and $R(\Omega)=R_y(\pi/3)$ is shown below. As a vector, the $J_z$ operator would go through the origin and the maximum of the distribution. Clearly the rotational symmetry of the distribution is apparent: this follows from $\Delta J_x=\Delta J_y$.

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  • $\begingroup$ This is definitely a sufficient condition - do you know if it is necessary to be an eigenstate of $L_z$? I can't immediately see it. $\endgroup$ Sep 3, 2021 at 17:18
  • $\begingroup$ @QuantumMechanic intuitively it should be necessary, but I’m not wading in the muddy waters to find a proof. $\endgroup$ Sep 3, 2021 at 17:27
  • $\begingroup$ I doubt it... take $(i|L;m\rangle+|L,m+2\rangle)/\sqrt{2}$ in the basis of $L_z$ eigenstates. This has $\langle L_x\rangle=\langle L_y\rangle=0$ and $\langle L_x^2\rangle=\langle L_y^2\rangle$ but is not an eigenstate of $L_z$. $\endgroup$ Sep 3, 2021 at 18:02
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    $\begingroup$ @QuantumMechanic good point. was thinking in terms of Pauli matrices, not in terms of higher dimensional reps… $\endgroup$ Sep 3, 2021 at 19:25

One can always define a basis in which $\langle L_x\rangle=\langle L_y\rangle=0$ by an appropriate rotation of the coordinates. Then, the only nontrivial uncertainty relationship is $$\mathrm{Var}(L_x)\mathrm{Var}(L_y)\geq \frac{1}{4}\left|\langle L_z\rangle\right|^2.$$ Since in this scenario the two variances are equal, we find that $$\langle L_x^2\rangle=\langle L_y^2\rangle\geq\frac{1}{2}\left|\langle L_z\rangle\right|.$$

The next notion often considered is squeezing. In this basis, it is defined as when $$\langle L_x^2\rangle<\frac{1}{2}\left|\langle L_z\rangle\right|<\langle L_x^2\rangle$$ or $$\langle L_y^2\rangle<\frac{1}{2}\left|\langle L_z\rangle\right|<\langle L_x^2\rangle.$$ Neither of these sets of inequalities can hold with the present condition, so there is no squeezing in this basis.

Now, the above definitions are not SU(2)-invariant; a rotation about the $z$ axis leaves all of the $L_z$ moments unchanged and the expectation values of $L_x$ and $L_y$ unchanged while changing all of the other moments. So we can ask whether, for any value of $\theta$, $$\mathrm{Var}(L_x\cos\theta+L_y\sin\theta)<\frac{1}{2}\left|\langle L_z\rangle\right|<\mathrm{Var}(-L_x\sin\theta+L_y\cos\theta)?$$ Expanding the variances, we find the conditions \begin{aligned} \langle L_x^2\rangle\cos^2\theta+\langle L_y^2\rangle\sin^2\theta+\langle L_x L_y+L_y L_x\rangle\cos\theta\sin\theta<\frac{1}{2}\left|\langle L_z\rangle\right|<\langle L_x^2\rangle\sin^2\theta+\langle L_y^2\rangle\cos^2\theta-\langle L_x L_y+L_y L_x\rangle\cos\theta\sin\theta\\ \langle L_x^2\rangle+\frac{1}{2}\langle L_x L_y+L_y L_x\rangle\sin2\theta<\frac{1}{2}\left|\langle L_z\rangle\right|<\langle L_x^2\rangle-\frac{1}{2}\langle L_x L_y+L_y L_x\rangle\sin2\theta. \end{aligned} These correlations between $L_x$ and $L_y$ do not automatically vanish (if you assume that they do, which is a stronger assumption, you will guarantee no squeezing). If, for example, we have $\frac{1}{2}\left|\langle L_z\rangle\right|\approx\langle L_x^2\rangle$ and nonzero correlations between $L_x$ and $L_y$, then squeezing is definitely present for most choices of angles $\theta$. This cannot be directly accomplished, because $\frac{1}{2}\left|\langle L_z\rangle\right|=\langle L_x^2\rangle$ would here imply that those correlations are zero, so we can still wonder whether this squeezing can be accomplished.

Let's explore further. Using the Schwinger mapping $$L_x=(a^\dagger b+a b^\dagger)/2,\quad L_y=-i(a^\dagger b-a b^\dagger)/2,\quad L_z=(a^\dagger a+b^\dagger b)/2,$$ the assumed conditions are $$\boxed{\langle a^\dagger b\rangle=\langle a b^\dagger\rangle=0}$$ and $$\boxed{\langle a^{\dagger \,2} b^2+a^2 b^{\dagger \,2}\rangle=0}.$$ These imply that $$\langle L_x^2\rangle=\langle L_y^2\rangle=\frac{1}{2}\langle a^\dagger a b^\dagger b\rangle+\frac{1}{4}\langle a^\dagger a +b^\dagger b\rangle$$ and $$\langle L_x L_y+L_y L_x\rangle=\frac{-i}{4}\langle a^{\dagger \,2} b^2- a^2 b^{\dagger \,2}\rangle=\frac{-i}{2}\langle a^{\dagger \,2} b^2\rangle.$$ Clearly any state with $\langle a^\dagger a\rangle=\langle b^\dagger b\rangle$ has $\langle L_z\rangle=0$ and thus cannot be squeezed. Without loss of generality, we can inspect $\langle a^\dagger a\rangle>\langle b^\dagger b\rangle$, for which the squeezing condition becomes \begin{aligned} \frac{1}{2}\langle a^\dagger a b^\dagger b\rangle+\frac{1}{4}\langle a^\dagger a +b^\dagger b\rangle-i\frac{\sin2\theta}{2}\langle a^{\dagger \,2} b^2\rangle<\frac{1}{4}\langle a^\dagger a-b^\dagger b\rangle\\ \langle a^\dagger a b^\dagger b\rangle+\langle b^\dagger b\rangle-i{\sin2\theta}\langle a^{\dagger \,2} b^2\rangle<0. \end{aligned} We know that, since the covariance matrix with elements $\frac{1}{2}\langle L_i L_j+L_j L_i\rangle-\langle L_i \rangle\langle L_j\rangle$ is positive semidefinite, $$\langle a^\dagger a b^\dagger b\rangle+\frac{1}{2}\langle a^\dagger a+ b^\dagger b\rangle>\left|\langle a^{\dagger \,2} b^2\rangle\right|$$ and we also know that, with our assumptions, $$\langle a^\dagger a b^\dagger b\rangle+\frac{1}{2}\langle a^\dagger a+ b^\dagger b\rangle>\langle a^\dagger a b^\dagger b\rangle+\langle b^\dagger b\rangle,$$ so I do not yet see a reason why there should be no states for which \begin{aligned} \langle a^\dagger a b^\dagger b\rangle+\langle b^\dagger b\rangle-i{\sin2\theta}\langle a^{\dagger \,2} b^2\rangle<0\\ \Leftarrow \langle a^\dagger a b^\dagger b\rangle+\langle b^\dagger b\rangle<\left|\langle a^{\dagger \,2} b^2\rangle\right|. \end{aligned} I wonder if this can be accomplished...

Anyway, using the boxed equations, we can specify some states satisfying the assumptions. Obviously the eigenstates of $L_z$, $$||L,m\rangle=|L+m\rangle_a\otimes|L-m\rangle_b,$$ will suffice, but so too will states that are not eigenstates of $L_z$, such as $$\frac{||L,m\rangle+||L,m+2\rangle}{\sqrt{2}}.$$

Going back to the original angular momentum notation, if we use the standard raising and lowering operators $L_\pm=L_x\pm iL_y$ that enact $L_\pm||L,m\rangle\propto ||L,m\pm 1\rangle$, all of the conditions become \begin{aligned} \boxed{\langle L_\pm\rangle=0,\quad \langle L_+^2+L_-^2\rangle=0.} \end{aligned} This makes it easy to find other exemplary states, such as $$||L,m\rangle+||L,m+k\rangle+||L,m+k+l\rangle+\cdots,\qquad k>2, l>2, \cdots.$$

  • $\begingroup$ Since averages of $a^\dagger a b^\dagger b$ and $b^\dagger b$ are real, can your guarantee that the term in $\sin\theta$ in your last inequality will also be real else you are comparing a complex number to $0$ :( $\endgroup$ Sep 3, 2021 at 20:55
  • $\begingroup$ @ZeroTheHero yes of course! The first boxed equation (from one of the assumptions) implies that $\langle a^{\dagger\,2}b^2\rangle$ is purely imaginary $\endgroup$ Sep 3, 2021 at 23:48

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