# Tag Info

## Hot answers tagged angular-momentum

6

Basically the rings don't fall into Saturn for the same reason the Moon doesn't fall into the earth. The rings are billions of little moons, each in it's own stable, or largely stable orbit. The rings are also likely resupplied with new ring material from Enseladus, Saturn's 2nd closest moon. (ice volcanoes due to strong tidal forces that can shoot ice ...

4

The general question is quite hard to tackle I think, because a rigorous motivation of Hilbert space would end up in the theory of operator algebras (see e.g. this answer) and the OP is probably not interested in these aspects at the moment. As for the example of spin, the Hilbert space in this case is still an $L^2$ space, but the functions are no longer ...

4

Yes. This commutation relation is that of the Lie algebra $\mathfrak{so}(3)$ corresponding to the rotation group in three dimensions. Thus the commutation relation states that the Pauli matrices generate rotations. To understand why this is the commutation relation of $\mathfrak{so}(3)$, one can draw a diagram showing that the commutator of two ...

4

That you can only ever know one of the three components of angular momentum is best seen not through the uncertainty principle, but on the states themselves. Since $[L_i,L_j] = \epsilon_{ijk} L_k$, the three momentum operators are pairwise not simultaneously diagonalizable (since simultaneous diagonalizability implies that the operators commute), meaning ...

3

The book where the derivation is described sufficiently pedagogically is Ballentine's Quantum mechanics - A Modern development, chapter 3. I am going to give a sketch of the 30-page chapter. (Beware, I suppress vector notation) Transformations of the quantum state are expressible as unitary transformations. The first order expansion of a unitary ...

3

For $d=3$ the group theoretic meaning of total angular momentum is that it is the Casimir operator of $SO(3)$. For $SO(d)$ where $d>3$ you have more than one Casimir operator, so it's not clear what you mean by "total angular momentum" In particular the number of Casimir operators is $[d]/2$, where $[d]=d$ or $d-1$ depending whether $d$ is even or odd.

3

Moving bodies have inertia which means that they will continue to move at a constant velocity unless acted upon by an external force (this is Newton's first law of motion). Similarly, rotating bodies have a moment of inertia, meaning that they will continue to rotate unless acted upon by an external force (torque). Therefore, torque is only required to ...

2

The angular momentum operators obey the commutation relations $$[L_i,L_j]=i\sum_k\epsilon_{ijk}L_k$$ A nonzero commutation relation means we can't have a state vector which is an eigenstate of more than one angular momentum operator at the same time. This also leads to an uncertainty principle.

2

Let $$\tag{1} \hat{T}_{ik}~:=~\hat{n}_i \hat{n}_k-\frac{1}{3}\delta_{ik}\hat{\bf 1}.$$ The phrasing of the problem in Ref. 1 is indeed not the clearest, but by comparing with the given solution, it seems that Ref. 1 is performing a partial averaging over the Hilbert space of states with fixed value of the orbital angular momentum quantum number $\ell$ and ...

2

Looking at your video, it appears that the boomerang is not turning fast enough to return. This typically means that it is too heavy. I wrote a couple of answers earlier [here](http://physics.stackexchange.com/a/156122/26969_ and here to explain some of the physics of boomerangs; perhaps you will find the physics there hard to understand, but then these are ...

2

Just to clarify to Robin Ekman's answer, superpositions of the Pauli matrices exponentiate to $SU(2)$, not $SO(3)$, but both these Lie groups have $\mathfrak{so}(3)$ as their Lie algebra - but I am sure you already know this. Also, there is another way to look at the problem that you might find helpful, even though it is a mathematical insight rather than a ...

1

What you have to assume a defined motion, where $\theta(t)$ is known. If the mass was spinning with a constant rate there would be no torque needed. The basic equation of motion is $$T = I_{zz} \ddot{\theta}$$ where $I_{zz}=\int r^2 {\rm d}m$ is the mass moment of inertia about z. That internal torque is the translated into shear stress with $$\tau = ... 1 The wave function only contains all the information about the system im so far as you consider it. Meaning each qualitatively different physical system needs its a modified Hilbert space to fit what can happen with the system. In case you have something like spin on its own in  H_{Spin} and you want to look at a freely moving particle in H_{free} that ... 1 For a full shell, the addition of the expectation values of any angular momentum L_i is zero, and similarly for the spin operators \sigma_i. This is not hard to see - for l(l+1) as the expectation value of L^2 for a s,p,d,f subshell, the basis of that subshell is spanned by states indexed by integers between -l and l, and since that l is also the ... 1 Let us put an hat on operators, which acts non-trivially on kets:$$ [\hat{J}_a,\hat{O}^s_{\ell}] ~=~ \hat{O}^s_m~[J_a^s]_{m\ell}, \tag{4.1}$$The matrix element$$[J_a^s]_{\ell^{\prime}\ell}~\in~\mathbb{C}$$is just a complex number and hence commutes with a ket. Hence$$[\hat{J}_a,\hat{O}^s_{\ell}]|j,m,\alpha\rangle ~\stackrel{(4.1)}{=}~ ...

1

The point is that the spin operator is defined to be (1/2) times SU(2) generator while the orbital angular momentum is defined to be only SU(2)(or SO(3), is the same) generator. The proof is the same, and is " representation independent", in the sense that the structure of identity multiplied by something plus a linear combination of sigma matrices ...

1

The identity you used, $$\exp(i\theta \, \hat s)=\cos(\theta)+i\sin(\theta)\,\hat s, \tag{\ast}$$ is crucially dependent on the operator $\hat s$ being idempotent, and particularly on the fact that $\hat{s}^2=\mathbb1$. This is generally not the case for angular momenta other than spin-1/2. In general, the total angular momentum is a scalar within the ...

1

First, angular momentum isn't measured about an axis. It's measured about a point. Second, well, of course the angular momenta about different points will be different in general. But they will each be conserved -- there's no need for the point to be in the axis of rotation or even in the same galaxy as the rotating object you're interested in. Now, about ...

1

Let me assume that the object has spherical symmetry, however, for solving the present problem it is painted on its surface with different colors. So, imagining a plane section that contains the orbit of the object around the earth. The section of the plane through the object is a circle and we will see different points of the circumference painted in ...

1

"By the uncertainty principle" is the answer. In more detail, let's say we're talking about x and y axes. The first measurement puts the electron into an eigenstate of the spin X observable (the question of how it does this is the quantum measurement problem). Whichever of the two X eigenstates this "collapse" ends up in, it not an eigenstate of the spin Y ...

1

Nothing prevents the electron's spin from being measured along a particular axis, and then subsequently measured along an axis perpendicular to the first. In this situation, however, the spins along the perpendicular axes would not be known simultaneously, so the uncertainty principle would not be violated. As an example, say that we perform our own ...

1

The answer to your question is that to a first approximation the direction of the spaceship will not change, so the upper diagram is the correct one. However the direction of the spaceship will change very slightly due to a phenomenon called the geodetic effect. The easiest way to see this is to replace the spaceship by a gyroscope, and make the gyroscope ...

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