In Weinberg's Lectures on Quantum Mechanics (pg 31), he said that the commutator relation $$[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$$ is true for any vector $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$, where $\textbf{L}$ is the angular momentum operator given by $\textbf{L}=-i\hbar\textbf{x} \times \nabla$.

An example for vector $\textbf{v}$ is the angular momentum $\textbf{L}$ itself: $$[L_i,L_j] = i\hbar \sum_k \epsilon_{ijk} L_k.$$ Other examples include $\textbf{v}=\textbf{x}$ and $\textbf{v}=\nabla: $ $$[L_i,x_j] = i\hbar \sum_k \epsilon_{ijk} x_k,$$ $$[L_i,\frac{\partial}{\partial x_j}] = i\hbar \sum_k \epsilon_{ijk} \frac{\partial}{\partial x_k}.$$

How can it be shown that the commutator relation $[L_i, v_j]=i\hbar\sum_k \epsilon_{ijk}v_k$ is indeed true for any vector $\textbf{v}$ constructed from $\textbf{x}$ and/or $\nabla$?

Edit: I am looking for an answer that does not simply say that this is the definition of a vector operator. In fact, I think that Weinberg refers to $\textbf{v}$ as a vector, not a vector operator.

  • $\begingroup$ I think you have it the wrong way around: this is the definition of a vector operator in quantum mechanics. $\endgroup$
    – Philip
    Commented Jun 24, 2020 at 5:42
  • $\begingroup$ Yes Phillip is right, most authors use equation 1 to define vector operators. $\endgroup$
    – user224659
    Commented Jun 24, 2020 at 5:42
  • $\begingroup$ I think the key question here is what you mean by "constructed" $\endgroup$
    – user224659
    Commented Jun 24, 2020 at 5:49
  • $\begingroup$ @Philip Perhaps it's like this: equation 1 is true for any vector operator constructed from $\vec{x}$ and $\nabla$. All vector operators in QM are constructed from $\vec{x}$ and $\nabla$. Hence we define vector operators in QM as operators that satisfy equation 1. But that brings us back to the orignial question: why is equation 1 is true for any vector operator constructed from $\vec{x}$ and $\nabla$.? $\endgroup$
    – TaeNyFan
    Commented Jun 24, 2020 at 7:02
  • $\begingroup$ Well, total angular momentum (with spin) is a vector operator that isn't constructed solely from $\vec{x}$ and $\vec{\nabla}$... Though I admit that I wouldn't be personally completely convinced by this counter-example in your position. It seems to me that to answer this question we need another definition of a vector operator. How would you want to define a vector operator $\vec{v}$, independent of this above "definition"? To my mind, this, or rather the more general but equivalent $$U^\dagger(R) V_i U(R) = R_{ij}V_j$$ is the only way, but I'm no expert. $\endgroup$
    – Philip
    Commented Jun 24, 2020 at 7:22

4 Answers 4


Note: I suspect we might have a slightly different reading of the line from Weinberg's book:

It can be shown that [the commutation relation you specify] is true of any vector $\mathbf{v}$ that is constructed from $\mathbf{x}$ or $\mathbf{\nabla}$.

I feel that the stress is on the word "vector", not on the words "$\mathbf{x}$ or $\mathbf{\nabla}$". Furthermore, even though the quantity $\mathbf{v}$ is referred to as a vector and not a vector operator, I think it has to be an operator for this to make sense quantum mechanically. I do not know how to derive a commutation such as this without taking the quantities in question to be operators! I think that this early on in the book, Weinberg didn't want to speak of Spin and total angular momentum, and so he keeps it simple by only mentioning external degrees of freedom, especially since his actual goal is to solve the central potential problem. He is much clearer on the subject when he speaks of Rotations and Spin in Chapter 4 (around pg. 100 of your link).

But let's see if I can be convincing. (Apologies if it's too long!)

What is a vector?

This is the hardest part since I'm not sure exactly how you want to define a vector operator. But let's say we use a very basic definition of a vector: a vector $\mathbf{V}$ is any combination of 3 components which all transform a particular way under rotations. In other words, under a rotation:

$$V_i' = R_{ij} V_j,$$

where $R_{ij}$ are the components of the $3\times 3$ matrix denoting geometric rotations. For example, for a rotation in the $xy-$plane by an angle $\theta$,

$$ R_z(\theta) = \begin{pmatrix}\cos{\theta}&\sin{\theta}&0\\-\sin{\theta}&\cos{\theta}&0\\0&0&1\end{pmatrix}$$

We will be working with infinitesimal rotations, and so it's a good idea to have a general form for $R$ when the angle is infinitesimally small. If we rotate a vector $\mathbf{v}$ in a plane orthogonal to an axis-vector $\mathbf{\hat{u}}$, then

\begin{equation} R_\mathbf{\hat{u}}(\text{d}\theta) \mathbf{v} \approx \mathbf{v} - \text{d}\theta\,\, \mathbf{\hat{u}}\times \mathbf{v}. \tag{1} \end{equation}

I won't prove it this here in full generality, but it's quite instructive to prove that this is true for $R_z$ by doing a Taylor expansion in $\theta$.

How does a general operator transform under symmetry transformations?

Given the transformation properties of physical states, we can easily derive the transformation properties of the operators that act on them. Let us consider a state $|\psi\rangle$ and its image under a rotation $|\psi'\rangle = \mathcal{R}|\psi\rangle$. In this case, $\mathcal{R}$ is the unitary operator in the Hilbert Space that represents the symmetry operation of rotating a state. (This is not the $3\times 3$ matrix $R$.)

When acting on $|\psi\rangle$ with some operator $\mathcal{O}$ we obtain a new state $|\phi\rangle = \mathcal{O}|\psi\rangle$. How does $|\phi\rangle$ transform under a rotation? Well, that's simple:

\begin{equation*} \begin{aligned} |\phi\rangle \rightarrow \mathcal{R}|\phi\rangle &= \mathcal{R O} |\psi\rangle\\ &= \mathcal{ROR^\dagger} \left( \mathcal{R}|\psi\rangle \right)\\ &= \mathcal{O'} |\psi'\rangle \end{aligned} \end{equation*}

Thus under a rotation,

\begin{equation*} \begin{aligned} |\psi\rangle \rightarrow |\psi'\rangle &= \mathcal{R}|\psi\rangle,\\ \mathcal{O} \rightarrow \mathcal{O'}&= \mathcal{R O R^\dagger}. \end{aligned} \end{equation*}

The operator $\mathcal{R}$ in state-space can be represented as $$\mathcal{R}(\theta) = e^{-i\theta \,\mathbf{\hat{u}\cdot L}/\hbar}.$$ This is true under a physical rotation of the coordinates. (i.e. for spin-less particles. It's not too hard to prove this, but I'm omitting it here. If you'd like me to show it for $\mathcal{R}_z$, I could. Let me know in the comments.) Thus, an infinitesimal rotation by $\text{d}\theta$ is represented by

\begin{equation} \mathcal{R}_\mathbf{\hat{u}}(\text{d}\theta) \approx \mathbb{1} - \frac{i}{\hbar} \text{d}\theta\,\mathbf{\hat{u}\cdot L} \tag{2} \end{equation}

What is a vector operator?

Combining the two ideas above, it should hopefully be clear that a vector operator is one that transforms as

\begin{equation*} \begin{pmatrix}\mathcal{V}_x\\\mathcal{V}_y\\\mathcal{V}_z\end{pmatrix} \rightarrow \begin{pmatrix}\mathcal{V'}_x\\\mathcal{V'}_y\\\mathcal{V'}_z\end{pmatrix} = \mathcal{R} \begin{pmatrix}\mathcal{V}_x\\\mathcal{V}_y\\\mathcal{V}_z\end{pmatrix} \mathcal{R^\dagger} = R \begin{pmatrix}\mathcal{V}_x\\\mathcal{V}_y\\\mathcal{V}_z\end{pmatrix} \end{equation*}

i.e. \begin{equation}\boxed{\mathcal{R} \mathcal{V}_i \mathcal{R^\dagger} = R_{ij} \mathcal{V}_j} \tag{3} \end{equation}

(Notice that the symbols for $\mathcal{R}$ and $R$ are different as they act on vectors in different vector spaces. Weinberg uses $U(R)$ instead of $\mathcal{R}$) In other words, under a rotation, its components transform under a unitary transformation exactly as vectors would in 3D space.

Putting it all together:

Let's now perform an infinitesimal rotation and see what the equation above becomes. Using Equation (2),

$$\mathcal{R}(\text{d}\theta) \mathcal{V}_i \mathcal{R^\dagger}(\text{d}\theta) = \left( \mathbb{1} - \frac{i}{\hbar} \text{d}\theta\,\mathbf{\hat{u}\cdot L} \right)\, \mathcal{V}_i \left(\mathbb{1} + \frac{i}{\hbar} \text{d}\theta\,\mathbf{\hat{u}\cdot L}\right) = \mathcal{V}_i - \frac{i}{\hbar} \text{d}\theta\,[\mathbf{\hat{u}\cdot L},\mathcal{V}_i]. $$

Similarly, using Equation (1),

$$R_{ij}(\text{d}\theta) \mathcal{V}_j = \mathcal{V}_i - \text{d}\theta\,\epsilon_{ijk}\hat{u}_j \mathcal{V}_k,$$

where I've used the definition $(\mathbf{A}\times\mathbf{B})_i = \epsilon_{ijk}A_j B_k$, and the summation over repeated indices is implied.

Thus, Equation (3) becomes

$$[\mathbf{\hat{u}\cdot L},\mathcal{V}_i] = -i\hbar \epsilon_{ijk}\hat{u}_j \mathcal{V}_k.$$

To get the specific results for $L_i$, the components of $\mathbf{L}$ in the Cartesian axes, we can successively choose $\mathbf{\hat{u}}$ as the unit vectors $\mathbf{\hat{x}},\mathbf{\hat{y}},$ and $\mathbf{\hat{z}}$, so that we get

\begin{equation*} [L_j, \mathcal{V}_i] = -i\hbar \epsilon_{ijk}\mathcal{V}_k \end{equation*}

or, by exchanging the indices $i \leftrightarrow j$ and using the antisymmetry properties of the $\epsilon$ symbol and commutator,

\begin{equation*} \boxed{[L_i, \mathcal{V}_j] = i\hbar \epsilon_{ijk}\mathcal{V}_k} \end{equation*}

You'll notice that I haven't mentioned anything to do with the operators $\mathbf{x}$ and $\mathbf{\nabla}$ here. The point is (as has been indicated in other answers) that not every combination of the above two operators is a vector operator. So how would one know that one particular combination was a vector operator? Why is (for example) $\mathcal{A \times B}$ a vector operator and not $\mathcal{A\cdot B}$? Well, it must transform as a vector, i.e. as Equation (3)!

In other words, this particular proof is much more general than just proving it for all vector operators composed of $\mathbf{x}$ and $\mathbf{\nabla}$, which seems to me to be harder to prove even though it's a weaker statement than the one we've just proved! So I don't see the point in trying to prove it.

Also, while it is true that so far as I know all vector operators are in fact "composed" of $\mathbf{x}$ and $\mathbf{\nabla}$ I see no reason in QM that that be the case. Tomorrow, if a funky observable was found that "vector" observable was found that didn't depend on $\mathbf{x}$ and $\mathbf{\nabla}$, it would still satisfy such equations!

In particular, this argument also works for Spin, which is a vector operator that isn't composed of position and momentum operators, provided me make the simple generalisation to shift $\mathbf{L} \to \mathbf{J}$ to account for the internal degrees of freedom. Weinberg himself discusses all of this in his chapter on Rotations (around pg. 100 of the link you've posted).

Further reading: All the amazing answers to this question.

  • $\begingroup$ I do not think all of this is relevant and second guessing Weinberg is really unwise. However, you did correctly point out that one does not need to prove Weinbergs statement since L is generator of rotations, eijk is group structure tensor so if one had a group theory class this is what follows. The question was, however, posed in constructive way, please prove this to me. and that can be done, if needed will do below. $\endgroup$
    – JohannR
    Commented Jul 5, 2020 at 20:50
  • $\begingroup$ Dear @JohannR, thank you for your feedback. As I pointed out, I was not sure how one would show that any operator composed of $\mathbf{x}$ and $\mathbf{\nabla}$ was a vector operator without using Equation (3). As for the relevance of this, I felt that it was necessary to motivate the idea of a vector operator. If there is another way, I'd be very happy to know, please let me know when you post your answer. $\endgroup$
    – Philip
    Commented Jul 5, 2020 at 20:56
  • $\begingroup$ See below, and I hope you agree and show your approaval! $\endgroup$
    – JohannR
    Commented Jul 5, 2020 at 21:09
  • $\begingroup$ @JohannR I would like to point out that I have nowhere used the fact that "$\epsilon_{ijk}$ is the structure of the rotation group", I have merely used the definition of the cross-product. $\endgroup$
    – Philip
    Commented Jul 5, 2020 at 21:25
  • $\begingroup$ which is the only way to define antisymmetric product on group manifold...... $\endgroup$
    – JohannR
    Commented Jul 5, 2020 at 21:38

It follows from the fact that $\hat L_i$ are generators of the rotations.

Generally, the vector operator $\hat V_i$ is not just "a list" of operators. It has to transform like a vector under rotations $\hat V_i' = R_{ij} \hat V_j$ (summation implied).

This identity can be proven by considering an observable in a rotated frame of reference. The frame rotation can be carried out through a transformation $R$ acting on $\hat V$ or a unitary operator $\hat{U}(R)$: \begin{equation} \left._R\right. \langle\alpha|\hat{V}_i|\beta\rangle_R = \langle\alpha|\hat U^\dagger \hat V_i \hat U|\beta\rangle = R_{ij}\langle\alpha| \hat V_j |\beta\rangle \end{equation}

Thus, we have: $\hat U^\dagger \hat V_i \hat U = R_{ij} \hat V_j$. By considering infinitesimal rotation $\hat U = 1 - \frac{i \epsilon \bf{J}\cdot \bf{n}}{\hbar}$ and corresponding $R = \left(\begin{matrix}1& -\epsilon & 0\\ \epsilon & 1 &0 \\ 0 & 0 & 1\end{matrix}\right)$ we find at first order in $\epsilon$: \begin{equation} [V_i, J_j] = i \hbar \epsilon_{ijk} V_k \end{equation}

A nice proof of this identity can be found here: https://www.oulu.fi/tf/kvmIII/english/2004/09_tensop.pdf.

Similar question: Angular and linear momentum operators' commutation

Another useful reference: https://www.wikiwand.com/en/Tensor_operator#/Vector_operators


Theorem: Let $\mathbf{A},\mathbf{B}$ be vector operators. then $\mathbf{A}\times\mathbf{B}$ is a vector operator. But e.g. $\mathbf{A}\mathbf{B}$ is a scalar. That is $[L_j,\mathbf{A}\mathbf{B}]=0$

The proof is done by straight forward algebra, using the definition of a vector operator $[L_j,A_i]=i\hbar\epsilon_{jik}A_k$.

But I think your confusion isn't related to the algebra, but stems from the term "constructed". This is very imprecise language. As the theorem shows, not all combinations or even products of two vector operators are vector operators.

  • 1
    $\begingroup$ The author also specified that the operator $\textbf{v}$ is a vector, so that already leaves out the combination of dot products. $\endgroup$
    – TaeNyFan
    Commented Jun 24, 2020 at 6:56
  • $\begingroup$ Also please see my comment on the question and tell me what you think! $\endgroup$
    – TaeNyFan
    Commented Jun 24, 2020 at 7:01
  • $\begingroup$ @TaeNyFan It's not clear to me what kind of operations you are interested in. The sum and the cross product of vector operators are once again vector operators. If you want to extend the result to other operations you need to state about which operations you are talking about. I can think of a million funny ways to "construct" a operator by two operators. Many of these will not produce vector operators. $\endgroup$
    – user224659
    Commented Jun 24, 2020 at 7:43
  • $\begingroup$ Furthermore you should be careful to differentiate between a group of 3 operators, ordered in a coloumn vector and a vector operator. $\endgroup$
    – user224659
    Commented Jun 24, 2020 at 7:43
  • $\begingroup$ What I mean is that $\textbf{v}$ is any combination of $\textbf{x}$ and $\nabla$, with the criteria that the combination is a vector operator. Hence, sums and cross with $\textbf{x}$ and $\nabla$ are valid, while dot products and 'division' are not. One of my queries is also what are all the possible combinations of $\textbf{x}$ and $\nabla$ that give a vector operator $\textbf{v}$. $\endgroup$
    – TaeNyFan
    Commented Jun 24, 2020 at 7:46

The question asks for a constructive procedure: any vector $\mathbf{v}$ can be written in terms of two scalar functions multiplied by the two allowed vectors $\mathbf{x}$, and the gradient, respectively. These scalar functions depend on $\mathbf{x\cdot\nabla}$, $\mathbf{x}^2$ and $\nabla^2$. For simplicity let us forget about poles and cuts, so use for each (analytical) function a power series in these 3 variables. Remembering rules of commutation relation for $\mathbf{x}$ with powers of $\mathbf{p}$ and $\mathbf{p}$ with powers of $\mathbf{x}$ (sign change) and applying them to these power series one indeed finds the result presented in Weinberg.

NOTE: this is how one shows constructively that $\mathbf{L}$ is the generator of rotations, and $\epsilon_{ijk}$ is the structure of the rotation group, all valid answers above assumed this, thus these were not answers to the question posed.

  • 1
    $\begingroup$ It's not clear to me how this answers the question beyond effectively stating that the result is correct. Moreover, are there not 3 functions (rather than 2): $\textbf{x}\cdot \nabla$, $\textbf{x}^2$ and $\nabla^2$? $\endgroup$ Commented Jul 5, 2020 at 21:24
  • $\begingroup$ Furthermore, I am not certain I understand what you mean by "any vector $\mathbf{v}$ can be written in terms of two scalar functions multiplied by the two allowed vectors $\mathbf{x}$, and the gradient." Do you mean that $$\mathbf{v} = f(\mathbf{x},\nabla) \mathbf{x}+ g(\mathbf{x},\nabla) \mathbf{\nabla},$$ where $f$ and $g$ are scalar functions depending only on the quantities you mention? In which case, how would one represent $\mathbf{L}$, for example, using this? And even if that were true, I am not certain how one would go about actually proving it, which was what the OP asked anyway. $\endgroup$
    – Philip
    Commented Jul 5, 2020 at 21:30
  • $\begingroup$ v=f(x^2,∇^2, x.∇)x+g(x^2,∇^2, x.∇)∇, expand both f,g in powers x^2n,∇^2k, x.∇^m and the rest is just keeping track of signs, powers, remembering rules of commutations. $\endgroup$
    – JohannR
    Commented Jul 5, 2020 at 21:37
  • $\begingroup$ Ok, and what about representing a vector like $\mathbf{L}$ in this case? What would your functions $f$ and $g$ be? In fact, since classically $\mathbf{L}$ is "orthogonal" to the plane produced by $\mathbf{x}$ and $\mathbf{p}$, I doubt it could be written as vector sum of $\mathbf{x}$ and $\mathbf{p}$... $\endgroup$
    – Philip
    Commented Jul 5, 2020 at 21:43
  • $\begingroup$ You need to construct L such that it is a group generator, this is for rotation group a unique construction - I believe presented explicitly in red cover Sakurai QM book for example (that is original edition and that part is written by Sakurai so is correct, Sakurai died writing the book, alas. $\endgroup$
    – JohannR
    Commented Jul 5, 2020 at 22:03

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