Angular Momentum: Quantization, Components, & Direction

In summary, the total angular momentum is quantized, but there is a different quantum number, l, that you have to calculate to get a "total". Unfortunately, without carrying out separate measurements you cannot know the eigenvalue of L^2.
  • #1
Master J
226
0
Well I've been looking at angular momentum...

I have an idea why a function f cannot be an eigenfunction of 2 different non-commutating operators, but has anyone a nicely precise reason??

If the angular momentum is resolved into its components, and we look at one, say L_z, then:

L_z.f = h.m.f

I am letting h be h-bar, m the quantum number, f the wave function, and L_z is the operator/

Is the total angular momentum quantized in m then also?? How do we find the total angular momentum from just this equation for a COMPONENT?

is it possible to select a direction so that the total angular momentum is L_z?
 
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  • #2
Master J said:
I have an idea why a function f cannot be an eigenfunction of 2 different non-commutating operators, but has anyone a nicely precise reason??
Let's see if I can remember this: say that we have two operators A and B. If there is a simultaneous eigenfunction of both, call it f. Then
[tex]Af = af[/tex]
[tex]Bf = bf[/tex]
I can use those to calculate
[tex]BAf = Baf = aBf = abf[/tex]
and
[tex]ABf = Abf = bAf = baf = abf[/tex]
(I've used the fact that real numbers commute with each other and with all linear operators) Now putting the last two together,
[tex][A,B]f = ABf - BAf = abf - abf = 0[/tex]
So if there is a simultaneous eigenfunction f, the operators commute, at least when applied to the function f. Thus if they don't commute (for any function), there can't be any simultaneous eigenstates.
Master J said:
If the angular momentum is resolved into its components, and we look at one, say L_z, then:

L_z.f = h.m.f

I am letting h be h-bar, m the quantum number, f the wave function, and L_z is the operator/

Is the total angular momentum quantized in m then also?? How do we find the total angular momentum from just this equation for a COMPONENT?
Total angular momentum is quantized, but it has a different quantum number, [itex]l[/itex]. Well... what we actually calculate is the total angular momentum squared, [itex]L^2 = L_x^2 + L_y^2 + L_z^2[/itex]. (Since angular momentum is a vector, we need to square it to get a "total".) You can't find the total angular momentum just by knowing one component.

Master J said:
is it possible to select a direction so that the total angular momentum is L_z?
No, because of the uncertainty principle. When you have two noncommuting operators, like [itex]L_z[/itex] and [itex]L_y[/itex], you can't know both of them precisely because there are no simultaneous eigenstates. Any eigenstate of [itex]L_z[/itex] is a mixture of several eigenstates of [itex]L_y[/itex], so the total angular momentum could correspondingly have several different values.

If you have access to David Griffiths' book on introductory quantum mechanics, take a look at chapter 4 where he offers a nice diagram regarding exactly that point.
 
  • #3
Diazona explained this beautifully but I just want to add this:

The total angular momentum is a vector that has three non-commuting components so it's usually impossible to measure it with a single measurement. But in spin-half systems the eigenvalue of L^2 is always equal to the average value you obtain by different measurements:

[tex]
L^2 = L_x^2 + L_y^2 + L_z^2 = 3 \hbar^2 / 4
[/tex]

because you either get hbar/2 or -hbar/2 the squares of which are always h^2/4. And remember that the eigenvalues of L^2 are given by l * (l+1) , l is 1/2 here.

so without carrying out separate measurements, you can know the eigenvalue. Of course this is a very special property and immediately fails for spin-one systems for instance. Because you could get all kinds of results ( 0, hbar, 3 hbar, 2 hbar from different measurements of L^2) and they may not always be 2hbar -- which is the eigenvalue of L^2 for l=1. Edit: The fact taht L^2 is a combination of non-commuting angular momentum components doesn't mean that the eigenvalues of L^2 are not measurable -- they may be inferred from other experiments, such as the energy spectra of molecules - which yield direct information about the number l, hence the eigenvalue of L^2.

I hope I didn't make it worse for you.
 
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Related to Angular Momentum: Quantization, Components, & Direction

1. What is angular momentum and why is it important?

Angular momentum is a physical quantity that describes the rotational motion of an object. It is important because it helps us understand the behavior of rotating objects and is a fundamental concept in fields such as physics, engineering, and astronomy.

2. What is quantization of angular momentum?

Quantization of angular momentum is the concept that the total angular momentum of a system can only exist in discrete, specific values. This is due to the fact that angular momentum is quantized in units of Planck's constant, just like energy is quantized in quantum mechanics.

3. How is angular momentum calculated?

Angular momentum is calculated as the cross product of an object's moment of inertia (a measure of its resistance to rotation) and its angular velocity (a measure of how fast it is rotating). The direction of the angular momentum is perpendicular to both the moment of inertia vector and the angular velocity vector.

4. What are the components of angular momentum?

The three components of angular momentum are the x, y, and z components, which correspond to the three axes of rotation. These components can be calculated by taking the cross product of the moment of inertia and angular velocity vectors along each axis.

5. How does the direction of angular momentum affect the motion of an object?

The direction of angular momentum affects the motion of an object in that it determines the axis of rotation and the orientation of the object as it rotates. If the direction of angular momentum changes, the axis of rotation and orientation of the object will also change, resulting in a change in its motion.

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