# Uncertainty Relation between Lx and Ly

## Main Question or Discussion Point

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!

## Answers and Replies

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Staff Emeritus
2019 Award
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.

Oh, I see! Thanks!

tom.stoer
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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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?

vanhees71
Gold Member
2019 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
$$\Delta L_x \Delta L_y \geq \frac{\hbar}{2} \left |\langle \psi|\hat{L}_z| \psi \rangle \right|.$$

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
2019 Award
How do you know Lz=0 without measuring it?

tom.stoer
why is 0 = 0 a problem?

vanhees71
Gold Member
2019 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!

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?

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

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.