# Tag Info

## Hot answers tagged angular-momentum

3

The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by: $$D_t = d_t + \omega(t)\times\tag{1}$$ where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body. Since there there are no torques ...

2

I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity $\boldsymbol q$, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by ... 2 As the other answers have said, the true reason for this result is that the angular momentum of a quantum system cannot really be thought of as a classical vector with a magnitude and a direction, and well-defined values on all three components simultaneously. What Beiser is trying to do is to go against this direction as far as possible, and see how ... 2 The answer from WetSavannaAnimal is much more comprehensive (and better) than my own will be. But just in case this helps you, I will have a go. No offence intended, I have a feeling you might be getting the spin axis and orbital axis of the electron mixed up. Just a quick analogy with the earth, it's tilted axis precesses, that means it describes a ... 1 Torque is the angular force (the moment) - i.e. \tau = r \times F, not r \times p, which is angular momentum. When you are dealing with a rotating or rotational system (like a disc on an axis), then when the torque and angular momentum are always aligned along that axis, it is much easier to use the magnitude of the torque and the magnitude of the ... 1 Larmor precession is the steady rotation of the direction of a magnetic moment of a particle about a magnetic field I'm pretty sure the word is simply meant to be an analogy with the precession of a spinning rigid body's angular momentum when the body is acted on by a torque that is not aligned with the angular momentum. The reason for the analogy is that ... 1 Both negative sign and positive sign are correct. When you make an infinitesimal rotation with angle d\phi about the z-axis, then both two following representations for transformed coordinates are true: \left\{ \begin{array}{ll} x'=x-d\phi y \\ y'=y+d\phi x \end{array} \right. $$and$$ \left\{ \begin{array}{ll} x'=x+d\phi y \\ y'=y-d\phi x \end{array} ...

1

I don't know your textbook, but I might have understood the point. Considering a $\frac{1}{2}$-spin particle (for example) you should know that a suitable Hilbert space to represent the system is $\mathbb{C}^2$. Moreover Pauli matrices $\sigma_i$ for $i = x,y,z$ toghether with the identity matrix form a base for the vector space of linear operators ...

1

If you have an observable, that is like a symmetric matrix. So its eigenvectors span the whole space so you can form a basis for the whole space. So the span of the eigenvectors really is the whole space. What about eigenspaces? You did have to use all the eigenvectors to get the whole space, not just ones that share the same eigenvalue. If you only use ...

1

Never seen that before, so I just tried it. Cool. I believe that the membrane between the yolk and the white is elastic, so when you first, gently, give the egg a little angular momentum, you are only spinning the white. As the yolk catches up the effective moment of inertia drops, and conservation of momentum therefor implies a higher angular velocity.

1

Are you sure that's what the book is asking you to find? $\hbar/2$ is the eigenvalue of the $S_{x}$ operator corresponding to spin up, but it is not part of the state vector. If the question is really asking you to express the $\mid S_{x};+\rangle$ ket in the $S_{z}$ basis, then you're nearly correct, just a minor sign error: \mid S_{x};+\rangle = ...

1

I think you could work it like this: $X_+ ={1 \over \sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix}) =a (\begin{matrix} 1 \\ 0 \end{matrix} ) +b(\begin{matrix} 0 \\ 1 \end{matrix} )$.where $X_+$ is the eigenvector on the positive axon of $S_x$ Solve and find a,b and there you are. Note also that you can write a general spinor as $(\begin{matrix} cos\theta ... 1 I think the statement is nonsensical. In quantum mechanics the vector$\mathbf L$is not a world-vector. Its components are not numbers, they are operators. However, the expectation value of$\mathbf L$in a given state,$\langle \mathbf L \rangle\$ is a world-vector and it can be aligned with any given axis. (The magnetic field is a red herring.) It doesn't ...

1

Honestly, it's because the ideas of "parallel" and "antiparallel" and "perpendicular" don't quite apply to quantum angular momentum. The angular momentum of a quantum particle is not really a vector in the sense you're probably used to thinking about it. What I mean is, it's not really something that points in a direction and has a magnitude. It's more ...

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