In Laundau & Lifshitz Quantum Mechanics. Non-relativistic theory in $\S29$ a problem is given:

PROBLEM Average the tensor $n_in_k-\frac13\delta_{ik}$ (where $\mathbf{n}$ is a unit vector along the radius vector of a particle) over a state where the magnitude but not the direction of the vector $\mathbf{l}$ is given (i.e. $l_z$ is indeterminate).

The solution then starts with this (italics mine):

$\def\sc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \sc{SOLUTION.}$ The required mean value is an operator which can be expressed in terms of the operator $\mathbf{\hat l}$ alone. We seek it in the form

$$\overline{n_in_k}-\frac13\delta_{ik}=a[\hat l_i\hat l_k+\hat l_k\hat l_i-\frac23\delta_{ik}l(l+1)];$$

this is the most general symmetrical tensor of rank two with zero trace that can be formed from the components of $\mathbf{\hat l}$. ...

What confuses me is the italicized part: "mean value is an operator". As I understand, mean value in a given state $|\psi\rangle$ of a quantity $\kappa$ is given by


Here $\overline\kappa$ is not an operator, but $\hat\kappa$ is. Do L&L try to abbreviate some clearer phrase by their statement? Or do I understand something wrongly?

  • $\begingroup$ Maybe they mean that this is an equivalent operator that acts in the space of fixed "$\ell$", spanned by states of different "m", from $-\ell$ to $\ell$? $\endgroup$ – hft Feb 24 '15 at 19:56
  • $\begingroup$ @Ruslan - I looked in the book of Landau and Lifschitz. They explain you how to solve the exercise. Just I notice that in the end they give a formula in which appears $\hat {{\vec {\ell}}^2} = l(l+1)$. Of course, an operator is not equal to its eigenvalue. But what they do, given that you were told that the state $|\psi\rangle$is an eigenvalue of $\hat {{\vec {\ell}}^2}$, and given that they obtained that $\hat k$ in terms of $\hat {{\vec {\ell}}^2}$, they indeed do $\langle \psi|\hat k|\psi \rangle$, where $\hat k |\psi\rangle$ is replaced by the eigenvalue of $\hat k$ times$|\psi\rangle$. $\endgroup$ – Sofia Feb 24 '15 at 20:58

Let $$\tag{1} \hat{T}_{ik}~:=~\hat{n}_i \hat{n}_k-\frac{1}{3}\delta_{ik}\hat{\bf 1}.$$

The phrasing of the problem in Ref. 1 is indeed not the clearest, but by comparing with the given solution, it seems that Ref. 1 is performing a partial averaging over the Hilbert space of states with fixed value of the orbital angular momentum quantum number $\ell$ and keeping the magnetic quantum number $m$ as a lone indeterminate. In practice, this means averaging over a radial direction.

In other words, Ref. 1 is considering an irreducible $(2\ell+1)$-dimensional representation $R$ of the operator algebra [and of the Lie group $SO(3)$], with a vector space $V$, spanned by vectors $|\ell m \rangle$, $m\in\{-\ell,\ldots,\ell\}$. Denoting the averaging procedure with an overline, we have

$$\tag{2} \overline{\hat{T}_{ik}}~=~R(\hat{T}_{ik}), \qquad \hat{\ell}_i~:=~R(\hat{L}_i).$$

We would like to calculate the matrix elements

$$\tag{3}\langle \ell m | \overline{\hat{T}_{ik}}|\ell m^{\prime} \rangle ~=~\langle \ell m |R(\hat{T}_{ik})|\ell m^{\prime} \rangle ~=~f_{ik}(\ell,m,m^{\prime}), $$

which are some functions of $i,k,\ell,m,m^{\prime}$. Instead of considering matrix elements, we may consider the operator/matrix $R(\hat{T}_{ik})\in{\rm End}(V)$. It is natural to assume that

$$\tag{4}R(\hat{T}_{ik})~=~\sum_{m,m^{\prime}}|\ell m \rangle\langle \ell m |R(\hat{T}_{ik})|\ell m^{\prime} \rangle\langle \ell m^{\prime} | ~=~\hat{f}_{ik}(\hat{\ell}_1,\hat{\ell}_2,\hat{\ell}_3;\ell).$$

It follows from the tensor structure that $R(\hat{T}_{ik})$ must be of the form

$$\tag{5}R(\hat{T}_{ik})~\propto~\{\hat{\ell}_i,\hat{\ell}_k\}_{+}-\frac{2}{3}\delta_{ik}\ell(\ell+1)\hat{\bf 1}.$$

See Ref. 1 for further details.


  1. L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S29$.
  • $\begingroup$ Hmm, I don't quite understand, over what we are averaging. As it is now, it seems $l$ is still a parameter of $\overline{\hat T_{ik}}$. As I understood the procedure, it's something like switching to a basis where $\hat T_{ik}$ is block-diagonal, picking one of these blocks (irreps) $R_l(\hat T_{ik})$, expressing it in terms of $R_l(\hat L_i)$, and then going back to original basis. Is it right? $\endgroup$ – Ruslan Feb 25 '15 at 16:57
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Feb 25 '15 at 17:29
  • $\begingroup$ Thanks, that's a bit clearer. So, does "averaging over a radial direction" actually mean "averaging over radial quantum numbers", like here $$\langle lm|\overline{\hat T_{ik}}|lm'\rangle=\sum_n\langle nlm|\hat T_{ik}|nlm'\rangle?$$ Or do you actually mean an integral over configuration space direction $r$? (Or maybe it's somehow the same, and I'm confusing something?) $\endgroup$ – Ruslan Feb 26 '15 at 8:46

Apologies, I did not read the question very clearly. I leave my old answer below as it answers directly the title and therefore may help future comers.

This does appear to be slightly sloppy language (though I wouldn't necessarily blame it on L&L if it's not present in the Russian original, but my Spanish copy has an equivalent form).I would read the text as

The required mean value is that of an operator which can be expressed in terms of the operator $\mathbf{\hat l}$ alone.

This is indeed a correct statement, though you need fairly clunky machinery for it. The wavefunction $\psi(\mathbf r)$ can be split into an $r$-dependent part and a wavefunction on the unit sphere. The components $n_i$ are equal to spherical harmonics on the sphere, which means that they are a function of the components of angular momentum. Their product is therefore in the algebra, but since it transforms in a specific way it is reduced to the combinations claimed by L&L.

Their equation

$$\overline{n_in_k}-\frac13\delta_{ik}=a[\hat l_i\hat l_k+\hat l_k\hat l_i-\frac23\delta_{ik}l(l+1)]$$

definitely has operators on both sides, so I think it's fairly safe to put this one to a language (or even translation) quirk.

Any complex number can be seen as an operator.

More specifically, any complex number $z$ acts on states $|\psi\rangle\in\mathcal H$ in a natural way via scalar multiplication: $$|\psi\rangle\mapsto z|\psi\rangle.$$ This operator is the natural embedding of $z$ into the space of linear operators on $\mathcal H$. As Landau and Lifshitz note, this is simply $z$ times1 the identity operator $\mathbf 1$, which sends any $|\psi\rangle$ to itself.

There's really not much more to it. You're just applying this general concept to the specific complex number $\overline\kappa\in\mathbb C$.

1 Where "times", of course, is the scalar multiplication in $\mathcal L(\mathcal H)$ when it is seen as a vector space.

  • $\begingroup$ @EmilioPisanty regretfully, this is not the issue, see my remark. The book indeed, was a bit non-rigorous and equated an operator with its eigenvalue. $\endgroup$ – Sofia Feb 24 '15 at 21:16

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