# What does vector operator for angular momentum measure?

Consider the vector operator for angular momentum $$\hat L=\hat L_x \vec i +\hat L_y \vec j + \hat L_z \vec k$$.

Does this mean that if we want to measure the angular momentum of a particle in state $$\psi$$, we take $$\hat L$$ and let it act on $$\psi$$ to give us three possible eigenvalues of $$\hat L_x$$, $$\hat L_y$$ and $$\hat L_z$$, which will correspond to the $$x,y,z$$ components of the angular momentum?

But since $$\hat L_x$$, $$\hat L_y$$ and $$\hat L_z$$ does not commute, this should not be the meaning of $$\hat L$$ because if we first make a measurement of $$\hat L_x$$, the state of the particle will be changed, and $$\hat L_y$$ should no longer act on the original state $$\psi$$. What then does the vector operator $$\hat L$$ gives us? More precisely, what is the measurement this operator $$\hat L$$ is trying to measure?

The fact that $$\hat L_x$$, $$\hat L_y$$ and $$\hat L_z$$ do not commute means that you cannot have a function which is eigenfunction of the three operators. Your function can only be an eigenfunction of one of the component operators, say $$\hat L_z$$ (it is the typically chosen one). However, the operator $$\hat L^2$$ does commute with each of the component operators. So if we have an eigenfunction of $$\hat L_z$$, it is also an eigenfunction of $$\hat L^2$$. (But not an eigenfunction of $$\hat L_x$$ and $$\hat L_y$$).

Therefore, the answer to your question is that the operator $$\hat L$$ is used to give us the magnitude of the angular momentum, by operating on the eigenfunction with its square: $$\hat L^2$$. The eigenvalue is then $$l(l+1)ħ^2$$, which is the magnitude of the angular momentum squared.

You can only measure operators. A vector operator is a set of three operators, which we package into one piece of notation because they transform under rotations in a nice way. You can't measure it all in one go, any more than you can try to measure $$\hat{x}$$ and $$\hat{p}$$ at the same time. However, you can measure the components of $$\hat{L}$$, which are, as you would expect, the components of the angular momentum.

• Also, aside from transformation under rotations, using this not-fully-measurable vector "operator" also lets us build other operators corresponding to such quantities as square of angular momentum $L^2$ or Laplace-Runge-Lenz vector $\mathbf A=mk\frac{\mathbf{r}}r-\frac12(\mathbf{p}\times\mathbf{L}-\mathbf{L}\times\mathbf{p})$. Dec 21, 2018 at 7:15

I like this question. What you describe is taking:

$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y \hat L_x|l\rangle {\bf \hat y} ... = {\bf \vec L}|l\rangle$$

which would lead to problems. I think an eigenvalue equation for a vector operator should be:

$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y|l\rangle {\bf \hat y} + \hat L_z|l\rangle {\bf \hat z} = L_x|l\rangle {\bf \hat x} + L_y|l\rangle {\bf \hat y} + L_z|l\rangle {\bf \hat z} = {\bf \vec L}|l>$$

where

$${\bf \hat L} = -i\hbar (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}){\bf \hat x} + (permutations)$$

is the operator and

$${\bf \vec L} = L_x {\bf \hat x}+ L_y {\bf \hat y} + L_z {\bf \hat z }$$

is an ordinary vector.

Now with:

$$|l\rangle \equiv F_l(x, y, z)$$

the equation:

$${\bf \hat L}F_l(x, y, z) = {\bf \vec L}F_l(x, y, z)$$

doesn't have any non-trivial solutions: there are no eigenstates of completely known (non zero) angular momentum.

Regarding your question on "meaning": note that:

$$L_z = -i\hbar\frac{\partial}{\partial\phi}$$

and eigenstates of that function look like:

$$F(r, \theta, \phi) = R(r)\Theta(\theta)e^{\pm i m \phi}$$

which means that the function is invariant under rotations of $$2\pi/m$$.

For a wave-function to be an eigenstate of $${\bf \hat L}$$ it would have to invariant under some non-trivial rotation about not just the $$z$$-axis, but the $$x$$ and $$y$$ axes also--which would mean it would have to be invariant under some rotation about any axis.

The only thing I can think of that can do that is a sphere, and this is, of course, the zero angular momentum eigenstate.

Basically you are asking does a state exist which is a simultaneous eigenstate of the three components. The answer is no. You can construct a state that is an eigenfunction of $$L^2$$ and any one component, usually the z-component is chosen. This is due to the commutation relations. If a state exists whose all three components are known then a rotation about any three axis will not change the state. But this would mean that the state would be spherically symmetric which is only possible if $$L = 0$$.

This is right. You cannot have an eigenstate and eigenvalue of this vectorial angular momentum operator $$\hat{\mathbf{L}}$$. In fact, one doesn't even need the commutation to see that this cannot work. The reasons are purely mathematical. Just consider that given an operator $$\hat{O}$$, its eigenvalue equation is

$$\hat{O} |\psi\rangle = O |\psi\rangle$$

where $$O$$ is the eigenvalue, which is generally a complex number, but should be a real number for operators representing physical parameters. Now consider what this should look like for vectorial operator $$\hat{\mathbf{L}}$$. We should, following the same logic, have it associated with an eigenvalue that is itself a vector, representing a specific angular momentum vector $$\mathbf{L}$$, giving

$$\hat{\mathbf{L}} |\psi\rangle = \mathbf{L} |\psi\rangle$$

But look at the right hand side. We have a vector multiplying a quantum state. That's not mathematically possible. You can only multiply a quantum state by a complex scalar, because that's the only other operation than addition and taking the inner product that is defined on the Hilbert space. So by that token alone, $$\hat{\mathbf{L}}$$ cannot have "eigenvalues" in this sense. And this makes sense: the output of applying your $$\hat{\mathbf{L}}$$ to a quantum state isn't even another quantum state, but instead the "vector of quantum states" (a "meta-vector", perhaps?)

$$\hat{\mathbf{L}} |\psi\rangle = (\hat{L}_x |\psi\rangle) \mathbf{i} + (\hat{L}_y |\psi\rangle) \mathbf{j} + (\hat{L}_z |\psi\rangle) \mathbf{k}$$

Thus even worse, technically your proposed "operator" isn't really a proper operator at all, but a map between two rather different vector spaces, and no such maps can have eigenvalues, only self-maps of the same vector space.

• I thought that the equation $$\hat{\mathbf{r}} |\psi\rangle = \mathbf{r} |\psi\rangle$$ is meaningful Oct 7, 2020 at 21:37

Does this mean that if we want to measure the angular momentum of a particle in state $$ψ$$, we take $$\hat L$$ and let it act on $$ψ$$ to give us three possible eigenvalues of $$\hat L_x$$, $$\hat L_y$$ and $$\hat L_z$$, which will correspond to the x,y,z components of the angular momentum?

As you know $$L_x$$, $$L_y$$ and $$L_z$$ are incompatible observables, but note that $$L^2$$ does commute with the three components of the angular momentum, so:

$$[L^2, Lx] = [L^2, Ly] =[L^2, Lz]=0$$

Which more compactly is:

$$[L^2, L] = 0$$

Based on it, it is feasible to find simultaneous eigenstates of $$L^2$$ and one of the three components of the angular momentum. Using as an example $$L_z$$ we would get:

$$L^2 f = \lambda f$$

$$L_z f = \lambda' f$$

What is the measurement this operator $$\hat L$$ is trying to measure?

Because of compatibility reasons, $$L^2$$ and the components of the angular momentum are worked out separately.

Actually, to know $$\hat L$$ you should know simultaneously the three components of the angular momentum which is not the case. The uncertainty principle denies it.