Tag Info

New answers tagged

3

Your matrix is wrong. It should be det {$\begin{pmatrix} -\lambda & \hbar/2 (1-i) \\ \hbar/2 (1+i) & -\lambda \end{pmatrix}$} = 0 Now when you compute the determinant, it should be okay.


0

Indeed as was commented before usually in physics these derivatives are 'derived' by postulating how the object transforms. Obviously there are more rigorous definitions, as is usually the case. In this particular case you could try to find the structure which gives you these 'covariant derivatives' in the first place. In books they are often referred to ...


0

The eigenvalues (or measurable values) of each spin operator are ±1, corresponding to the only two mutually exclusive outcomes that could occur after a measurement of an electron's spin has been made. The mutual exclusivity is an experimental fact, but follows from the definition of the eigenstates above.In classical mechanics an electron has a magnetic ...


1

The principle behind STT is conservation of total angular momentum of the system. The equation that describes spin dynamics is called the LLG equation, named after Landau, Lifshitz and Gilbert. I will try to illustrate its physical meaning by using a general example. Consider an s-d model, where the sp-band electrons are itinerant and thus contribute to ...


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

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 = ...


3

Does the invariant helicity property contribute to the the concept of a photon and an "anti-photon" being the same entity? Not really, and I think you're getting mixed up between helicity and chirality here. Take a look at this deep-water wave image by Kraaieniest. See how the red-dot test particles move in a helical-like fashion? They can't move "the other ...


0

A photon should be discussed as a wave, instead of a particle (as it ceases to exist in rest state). As such two waves of the same wavelength and frequency may or may not be "negative" to each other, depending on their phase difference. When their phase difference is an odd multiple of pi, they will invariably cancel out each other upon interference.


0

Perhaps this isn't quite the answer you're seeking, but you may be interested in the phenomena of thermoremanent and isothermal remanent magnetization. Basically, if you perform a deep quench on a spin glass (i.e., freeze the spins) in a uniform external magnetic field and then, after some time, switch off the magnetic field, (or alternatively, quench ...


0

I guess I will take my own answer as the possible solution: taking the Hamiltonian in the interaction picture and applying the rotating wave approximation. The transformed Hamiltonian includes a term with the Rabi vector in the form: ${\boldsymbol \omega}_R(t)\cdot {\bf{S}}$. The only problem is that I also obtain an additional time dependent term of the ...


1

With such a two-electron spin wave function both electrons cannot be in the 1s state due to the electron parity (the wave function should be antisymmetric relative to exchange of two particles). When both electrons are in the orbital 1s, the only possible spin function reads $$\frac{1}{\sqrt{2}} \left( |\uparrow \downarrow >-| \downarrow \uparrow> ...


1

The ground state nuclear spin quantum number and parity, $J^{\pi}$ for all even-even nuclei is $0^+$. The isospin can vary, but for the ground state of even-even will probably be either 0 or 1. The isospin quantum number, $I$, is limited to the range of $$\frac{|Z-N|}{2}\le I \le \frac{Z+N}{2}.$$ The $J$ for odd-mass-number nuclei will be a half-integer ...


0

The quantum numbers are "parameters" that characterize the states of a particle or an atom. If I understand correctly what is your question, it is not always possible to measure directly all the quantum numbers, but sometimes you have to use some tricks to calculate them (for example special experiments or calculations based on the properties of the quantum ...


0

Your statement "Maxwell's equations imply that magnetic fields are due to changes in electric fields." is not complete. A corrected statement is that Maxwell's equations imply that magnetic fields are due to changes in electric fields AND due to currents (which can be stationary): $$ \nabla\times\mathbf{H} = \mathbf{J}+\partial\mathbf{D}/\partial t $$ As ...


1

If you imagine a beam entering a Stern-Gerlach device, at first the beam is behind the device and then the beam could be deflected left or deflected right after it goes through the device. A Stern-Gerlach device sends a spin up $\vert+\rangle$ to the (say) left and a spin down $\vert-\rangle$ to the right. So it sends $\vert+\rangle\vert behind\rangle$ to ...


1

$\newcommand{\ket}[1]{\left| #1 \right>}$$\newcommand{\bk}[2]{\left< #1 | #2 \right>}$Notice that the eigenvectors of the operator $S_z$ spans the whole space, which means that you can write any state as a superposition, (if you prefer as a linear combination) of these states. The situation is akin to the basis vectors of usual 3d Euclidean space. ...


0

A two component spinor can be geometrically interpreted as representing a point on the Riemann sphere, defined by the ratio of its two complex components, and its stereographic projection onto the xy-plane. Similarly, a four component spinor can be interpreted, by a more complicated ratio defined by its four complex components, as a point on the Riemann ...


2

The spin of a quantum field is related to the representation of the Lorentz group they transform under: scalar fields transform under the trivial representation, spinors transform under the spinorial representation, gauge bosons under the vectorial representation, gravitons (if they exist) under the second-rank tensorial representation... If you restrict to ...



Top 50 recent answers are included