# Uncertainty Relation between Lx and Ly

*FaerieLight*
There is an uncertainty relation between the x component and the y component of the angular momentum L of a particle, because [Lx, Ly] = i$\hbar$Lz which is not 0.

But what happens when Lz does equal 0? Would we in principle be able to measure both the x and y components of the angular momentum with no uncertainty? What kind of state would this describe?

I've searched my textbooks and the web for an answer to this question, but I haven't found any, so maybe this is actually something really obvious which I'm not seeing! Thanks for your help!

Staff Emeritus
How do you know Lz = 0? You need to measure it, and now you have two other uncertainty relations to contend with. This only works if Lx = Ly = Lz = 0.

*FaerieLight*
Oh, I see! Thanks!

Let's define

$$\Delta_A = \sqrt{\langle A^2\rangle - \langle A\rangle^2}$$

Then one can derive

$$\Delta_A\,\Delta_B \ge \frac{1}{2} \left| \langle [A,B]\rangle \right|$$

Now you can use this relation by setting A=Lx, B=Ly. But then the expectation value on the r.h.s. means that you have to chose a certain state for which it has to be evaluated. Chosing an eigenstate |m> you get

$$\Delta_{L_x}\,\Delta_{L_y} \ge \frac{1}{2} \left| \langle m| {L_z}|m\rangle \right| = \frac{m}{2}$$

You may want to have a look at http://en.wikipedia.org/wiki/Uncertainty_principle#Robertson-Schr.C3.B6dinger_uncertainty_relations

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Zarquon
How do you know Lz = 0? You need to measure it, and now you have two other uncertainty relations to contend with. This only works if Lx = Ly = Lz = 0.

But those other relations involve (ΔL_x)*(ΔL_z) and (ΔL_y)*(ΔL_z)? What if ΔL_z = 0?

Gold Member
2022 Award
The general Heisenberg-Robertson uncertainty relation is
$$\Delta A \Delta B \geq \frac{1}{2} \left |\langle \psi|[\hat{A},\hat{B}] \psi \rangle \right|$$
here $A$ and $B$ are the observables and $\hat{A}$ and $\hat{B}$ their reprsenting self-adjoint operator. $\Delta A$ and $\Delta B$ are the standard deviations of the observables and $|\psi \rangle$ represents the (pure) state of the system. The uncertainty relation also holds for mixed states, of course, but for the principle it's enough to consider pure states.

For $A=L_x$ and $B=L_y$ you have $[\hat{L}_x,\hat{L_y}]=\mathrm{i} \hbar \hat{L}_z$. This gives
$$\Delta L_x \Delta L_y \geq \frac{\hbar}{2} \left |\langle \psi|\hat{L}_z| \psi \rangle \right|.$$

Zarquon
The general Heisenberg-Robertson uncertainty relation is
$$\Delta A \Delta B \geq \frac{1}{2} \left |\langle \psi|[\hat{A},\hat{B}] \psi \rangle \right|$$
here $A$ and $B$ are the observables and $\hat{A}$ and $\hat{B}$ their reprsenting self-adjoint operator. $\Delta A$ and $\Delta B$ are the standard deviations of the observables and $|\psi \rangle$ represents the (pure) state of the system. The uncertainty relation also holds for mixed states, of course, but for the principle it's enough to consider pure states.

For $A=L_x$ and $B=L_y$ you have $[\hat{L}_x,\hat{L_y}]=\mathrm{i} \hbar \hat{L}_z$. This gives
$$\Delta L_x \Delta L_y \geq \frac{\hbar}{2} \left |\langle \psi|\hat{L}_z| \psi \rangle \right|.$$

Right. But I still have a problem with this: If either of the components is exactly zero, say L_z = 0;
then we get the relations

(ΔLx)(ΔLy) >= 0
(ΔLy)(ΔLz) >= h\2*E(Lx)

and so on. My problem is the second inequality, when ΔLz = 0.

Staff Emeritus
How do you know Lz=0 without measuring it?

why is 0 = 0 a problem?

Gold Member
2022 Award
Of course, if the state is such that $\langle L_z \rangle=0$, then the uncertainty relation becomes trivial, but it's still a true relation. There is no problem!

Zarquon
why is 0 = 0 a problem?

Right, good point.. Thanks!

Of course, if the state is such that $\langle L_z \rangle=0$, then the uncertainty relation becomes trivial, but it's still a true relation. There is no problem!

OK, but what is the minimum requirement on the uncertainties of Lx and Ly in this case (I mean if we measure Lz = 0, L^2 != 0). Seems like there should be one?

Gold Member
Consider the first eigenfunction of the hydrogen Hamiltonian. It is spherically symmetric, so all three angular momentum operators have the same eigenvalue ##0##.

Zarquon
Yes, but how about a simultaneous eigenfunction for the operators L^2 and Lz, such that Lz = 0 and L^2 != 0. In this case Ly and Lx must be uncertain(?), so I'm wondering if there's some relation that sets a limit for how precisely these can be determined.

Staff Emeritus