Suppose $\boldsymbol R,\boldsymbol P$ are the common position and momentum operators, and $\boldsymbol L=\boldsymbol R\times \boldsymbol P$ is the orbital angular momentum . $\boldsymbol K$ is a linear combination of $\boldsymbol R,\boldsymbol P$, and Jordan product of any of their cross product, e.g. $\frac12(\boldsymbol R\times (\boldsymbol R\times \boldsymbol P)-(\boldsymbol R\times \boldsymbol P)\times \boldsymbol R)$, whose coefficients are scalar functions of $ R^2,P^2, \boldsymbol R\cdot \boldsymbol P$, independent from coordinates, i.e. terms like $\boldsymbol R\cdot \boldsymbol{\hat x}$ are not involved.

Let $\boldsymbol J$ be the collection of three angular momentum operators, then a vector operator $\boldsymbol V$ satisfies

$$ [J_i,V_j]=\text i\hbar\epsilon_{ijk}V_k $$

Question: Can we generally proof that $\boldsymbol K$ satisfies the defnition of vector operators, or is there a counter example? In other words, is $$ [L_i,K_j]=\text i\hbar\epsilon_{ijk}K_k $$ true for all $\boldsymbol K$s? One example is the Runge-Lenz operator of a Hydrogen atom

$$ \boldsymbol B=\frac{\boldsymbol P\times\boldsymbol L-\boldsymbol L\times \boldsymbol P}{2me^2}-\frac{\boldsymbol R}R $$

which is confirmed to be a vector operator.

Note: The key is to justify the fact that an operator directly constructed by a classic vector is indeed a vector operator, so simply referring to the definition can't be the proof.


1 Answer 1


What defines a tensor operator is the commutation relation of its components with the angular momenta. Thus, if $[L_a,K_b]=i\hbar\epsilon_{abc}K_c$ you have a vector operator by definition.

More generally you want something like \begin{align} [L_\pm,T^{\lambda}_\mu]&=\sqrt{(\lambda\mp\mu)(\lambda\pm \mu+1)}T^{\lambda}_{\mu\pm 1}\\ [L_0,T^{\lambda}_\mu]&=\mu T^{\lambda}_\mu\, , \end{align} which tends to be easier to work with and goes beyond vectors ($\lambda$ is not restricted to $1$). It is also useful to use spherical components where, for instance, $$ T^1_1=-\frac{V_x+iV_y}{\sqrt{2}} $$

It is a good bet that, if you start with a classical vector with components $x^a p_y^b$ with $a$ or $b=0$ or $1$ (or such types of products), the corresponding observable will also be a vector operator. If you have higher powers in $x$ or $p$ (i.e. both $a$ and $b\ge 2$), there may be ordering issues and you have to be quite careful. (You want to be aware that a product of hermitian operators remains hermitian only if the operators commute.)

Polynomials in $x,y,z$ alone, or in $p_x,p_y,p_z$ alone, can be decomposed into tensors by expressing them in terms of spherical harmonics. Thus, \begin{align} x+iy&= r \sin\theta e^{i\phi} = c r Y^1_{-1}(\theta,\phi)\, ,\\ (x+iy)(x-iy)&= a r^2 + b r^2 Y^2_{0}(\theta,\phi)\, ,\\ \end{align} for some constants $a,b,c$ and are thus linear combinations of tensor operators, with the latter containing a scalar and the component of a tensor with $\lambda=2$.

In the case of more sophisticated products like the ones you have, you need to use the definition of composite tensors to properly combine the individual vectors using Clebsch-Gordan technology. Thus, again using spherical coordinates: $$ Z^{j}_m =\sum_{qq'} T^{\lambda}_q W^{\lambda'}_{q'} C_{\lambda,q;\lambda'q'}^{jm} $$ so it then becomes a matter of checking if the complicated expressions are properly combined using CGs (again note the use of spherical components).

A good source for this is the text by

Baym, Gordon. Lectures on quantum mechanics. CRC Press, 2018.

  • $\begingroup$ The CG method seems to be quite tedious. As for a specific operator, why should I adopt it instead of just working out the commutators? $\endgroup$
    – Po1ynomial
    Commented Dec 20, 2022 at 12:30
  • $\begingroup$ By the way, I have heard that the CGs are for coupling angular momentums, but here's only the orbital one. So is it really necessary to introduce that? (I am not really familiar with the topic) $\endgroup$
    – Po1ynomial
    Commented Dec 20, 2022 at 12:33
  • 1
    $\begingroup$ @Po1ynomia The CG method can be tedious but only because you are starting in a basis which is not spherical. Otherwise it is probably the most efficient way to construct complicated tensors without having to check the commutation relations all over again. I highly recommend you have a look at Baym as composite tensors are extremely useful in general. $\endgroup$ Commented Dec 20, 2022 at 15:19

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