Uncertainty Relation between Lx and Ly

  • #1

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[itex]\hbar[/itex]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

  • #2
Vanadium 50
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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.
 
  • #3
Oh, I see! Thanks!
 
  • #4
tom.stoer
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Let's define

[tex]\Delta_A = \sqrt{\langle A^2\rangle - \langle A\rangle^2}[/tex]

Then one can derive

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

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

[tex]\Delta_{L_x}\,\Delta_{L_y} \ge \frac{1}{2} \left| \langle m| {L_z}|m\rangle \right| = \frac{m}{2}[/tex]

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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  • #5
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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?
 
  • #6
vanhees71
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The general Heisenberg-Robertson uncertainty relation is
[tex]\Delta A \Delta B \geq \frac{1}{2} \left |\langle \psi|[\hat{A},\hat{B}] \psi \rangle \right|[/tex]
here [itex]A[/itex] and [itex]B[/itex] are the observables and [itex]\hat{A}[/itex] and [itex]\hat{B}[/itex] their reprsenting self-adjoint operator. [itex]\Delta A[/itex] and [itex]\Delta B[/itex] are the standard deviations of the observables and [itex]|\psi \rangle[/itex] 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 [itex]A=L_x[/itex] and [itex]B=L_y[/itex] you have [itex][\hat{L}_x,\hat{L_y}]=\mathrm{i} \hbar \hat{L}_z[/itex]. This gives
[tex]\Delta L_x \Delta L_y \geq \frac{\hbar}{2} \left |\langle \psi|\hat{L}_z| \psi \rangle \right|.[/tex]
 
  • #7
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The general Heisenberg-Robertson uncertainty relation is
[tex]\Delta A \Delta B \geq \frac{1}{2} \left |\langle \psi|[\hat{A},\hat{B}] \psi \rangle \right|[/tex]
here [itex]A[/itex] and [itex]B[/itex] are the observables and [itex]\hat{A}[/itex] and [itex]\hat{B}[/itex] their reprsenting self-adjoint operator. [itex]\Delta A[/itex] and [itex]\Delta B[/itex] are the standard deviations of the observables and [itex]|\psi \rangle[/itex] 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 [itex]A=L_x[/itex] and [itex]B=L_y[/itex] you have [itex][\hat{L}_x,\hat{L_y}]=\mathrm{i} \hbar \hat{L}_z[/itex]. This gives
[tex]\Delta L_x \Delta L_y \geq \frac{\hbar}{2} \left |\langle \psi|\hat{L}_z| \psi \rangle \right|.[/tex]
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.
 
  • #8
Vanadium 50
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How do you know Lz=0 without measuring it?
 
  • #9
tom.stoer
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why is 0 = 0 a problem?
 
  • #10
vanhees71
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Of course, if the state is such that [itex]\langle L_z \rangle=0[/itex], then the uncertainty relation becomes trivial, but it's still a true relation. There is no problem!
 
  • #11
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why is 0 = 0 a problem?
Right, good point.. Thanks!

Of course, if the state is such that [itex]\langle L_z \rangle=0[/itex], 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?
 
  • #12
Jano L.
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Consider the first eigenfunction of the hydrogen Hamiltonian. It is spherically symmetric, so all three angular momentum operators have the same eigenvalue ##0##.
 
  • #13
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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.
 
  • #14
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. 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.
[Lx, Lz] = iLy
 
  • #15
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But is there an explicit requirement on ΔLx and ΔLy? I still don't get it.
 

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