New answers tagged

1

You can write the Hamiltonian for the Heisenberg model as: $H = - J_x \sum_i \sigma_i^x \sigma_{i+1}^x - J_y \sum_i \sigma_i^y \sigma_{i+1}^y - J_z \sum_i \sigma_i^z \sigma_{i+1}^z$ where $J_x, \ J_y$ and $J_z$ are the three exchange terms. In the isotropic Heisenberg model (or the $XXX$ model), all three of these terms would be equal to $J$. In the ...


-1

The explanation is very simple. Based on the Stern- Gerlach experiment spin of 1/2 simply means that if you fire electrons through his apparatus... 1/2 of the electrons will spin up and the other 1/2 will spin down


0

To explain simply, without getting into the details of rotation symmetry groups, etc: When one says $s=1/2$, or $m_s=1/2$ or $m_s=-1/2$, we are specifying a quantum number which describes how the eigenvalues of spin operators behave. If we specify an eigenket of the spin angular momentum operators $|s,m_s\rangle = |1/2, \pm 1/2\rangle$, with operators $\hat{...


0

As an addendum to the other answers: When we say a particle is "spin 0" (e.g. Higgs boson), "spin 1/2" (e.g. electron), "spin 1" (e.g. photon), etc., we are talking about their different values for this operator. In general a "spin $s$" particle is described by eigenstates of $\hat{S}^2$ with eigenvalue $\hbar^2 s (s + 1)$.


4

It measures the (square of the) magnitude of the spin angular momentum, ignoring the spin’s direction.


1

If we have a rotating body, it's classical rotational energy is $$ E_{rot} = \frac{L^2}{2\theta} $$ Therefore, if we include a rotational particle in quantum mechanics it seams "natural" to use the same term and replace $L^2$ by it's operator -- the mathematical argument involves some group theory and so called generators [see e.g. the book of Sakurai for a ...


1

In page 41, we are only told that measuring $|r\rangle$ gives us $|u\rangle$ and $|d\rangle$ with equal probabilities. Here we are using the measurement of $\sigma_z$ of an electron in $|r\rangle$ state to determine it in terms of $|u\rangle$ and $|d\rangle$. We are only looking at probabilities here. Not probability amplitudes. Thus when forming the state ...


0

After the apparatus measured the spin along the +x axis, the particle has the state $|r\gt$. Applying the operator $ \sigma_z$ doesn't modify that state. It is not a measurement (yet). So: $$\sigma_{z}|r\gt = \frac{1}{\sqrt{2}}|u\gt-\frac{1}{\sqrt{2}}|d\gt$$ is just the acting of $$ \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1\\ \end{pmatrix} $$ ...


1

As the title of the paper suggests, this is a spontaneous symmetry breaking. That is, it is not at the level of the Hamiltonian (which remains symmetric) but rather at the level of the state of the system. Just like in the 2d Ising model, or in many magnets - the Hamiltonian describing the system is symmetric, and therefore there are two degenerate ground ...


1

There shouldn’t be a 2. ${\textbf S_{1z}}^2|\chi\rangle={\textbf S_{1z}}^2\left(\frac1{\sqrt2}|\uparrow\downarrow\rangle+ \frac1{\sqrt2} |\downarrow\uparrow\rangle \right)$ $= \frac1{\sqrt2}\left({\textbf S_{1z}}^2 |\uparrow\downarrow\rangle+ {\textbf S_{1z}}^2 |\downarrow\uparrow\rangle \right)$ $= \frac1{\sqrt2}\left(\left(\frac\hbar{2}\right)^2 |\...


1

Yes it does, paradoxically. It goes like this. To reach explicitly gauge independent conservation laws the well known GI Lagrangian is adopted. Unfortunately, the Noether currents that come out are not GI. The energy-momentum tensor is asymmetric and not GI. The angular moment consists of an orbital term of the form ${\bf r} \times {\bf P}$, involving the em ...


0

In nuclear magnetic resonance physics we use Bloch vectors to represent the spin state. The spin state $\vec e_z = |up_z\rangle$ is represented by the Bloch vector $(0,0,1)$. Now, if we rotate this state by $180^{\circ}$ w.r.t. the $z$ axis, the state does not change. Hence, a rotation by $180^{\circ}$ does not always change the state. However, if we ...


5

Both $L$ and $S$ are quantum angular momenta. Angular momentum is different in quantum mechanics (QM). Almost everything is different in QM. In QM, angular momentum is complete if you give two numbers: the "angular momentum" and its "third component" $M$. So you give the pairs $|l,\ m_l\rangle$ and $|s, \ m_s\rangle$ So, any angular momentum is ...


0

Answer to your question has been hinted in the comment.The spin-1/2 operator for the itinerant (or conduction) electron, $\mathbf{s}$ has expectation $\langle \uparrow | s_{z} | \uparrow\rangle = 1/2$ and $\langle \uparrow | s_{x} | \uparrow\rangle = \langle \uparrow | s_{y} | \uparrow\rangle = 0$, where $| \uparrow\rangle$ is the eigenstate of the $z$-...


0

Spin is by definition one special kind of angular momentum. More precisely if $\mathbf{L}$ is the orbital angular momentum and $\mathbf{S}$ is the spin angular momentum the total angular momentum is $\mathbf{J}=\mathbf{L}+\mathbf{S}$ and since $\mathbf{L}$ vanishes when the particle is at rest one may define spin as the residual angular momentum when the ...


1

In a typical collider experiment the momentum vectors are $$ p_1=\begin{pmatrix}E\\0\\0\\p\end{pmatrix}\qquad p_2=\begin{pmatrix}E\\0\\0\\-p\end{pmatrix}\qquad p_3=\begin{pmatrix} E\\ \rho\sin\theta\cos\phi\\ \rho\sin\theta\sin\phi\\ \rho\cos\theta \end{pmatrix} \qquad p_4=\begin{pmatrix} E\\ -\rho\sin\theta\cos\phi\\ -\rho\sin\theta\sin\phi\\ -\rho\cos\...


2

I think there's a misprint in your question so I'll go with what I think is right. Using the coupled basis (which is already permutation symmetric) \begin{align} \vert 11\rangle = \vert +\rangle_1\vert +\rangle_2\, ,\qquad \vert 10\rangle &= \frac{1}{\sqrt{2}}\left(\vert +\rangle_1\vert -\rangle_2 +\vert -\rangle_1\vert +\rangle_2\right)\, ,\quad \vert ...


1

For spin-1 particle, I don't quite understand how the following relationship is derived The numbers inside the kets on the left-hand side represent the eigenvalue of $L_z$. Therefore, we can figure out the right-hand side by how they transform under physical rotations $R_z(\theta)$ about the $z$-axis. The $|0 \rangle$ state corresponds to $\hat{e}_z$, ...


3

No it is not wrong. The intrinsic angular momentum of a photon is spin. But note that photons can also have orbital angular momentum (and so can electrons). Orbital angular momentum is defined as $$ \hat{L} = \hat{\bf x} \times \hat{\bf p} $$ where $\hat{\bf x}$ is the position operator and $\hat{\bf p}$ is the linear momentum operator. Just as in ...


2

Light can have orbital and intrinsic angular momentum. It is the latter that is related to polarisation and spin. The problem is that there is no gauge invariant expression for it. In the standard, gauge theory of light all angular momentum is described as orbital momentum! This paradox is at the root of your question. Light has spin but the present theory ...


0

A 360° rotation of a spin 1/2 wavefunction does indeed produce a '-' sign. You can find more details in the chapter on angular momentum in Sakurai's Modern Quantum Mechanics. Of course, this minus sign does not affect any observables, because we are calculating probabilities or expectation values, where the - signs on the bra and ket cancel each other out. ...


1

They don't have to be postulated and follow from the main postulates of the Quantum Mechanics. Whenever you have a quantum state (i.e. wave function) different observers that are related to each other by rotations can observe the state and compute amplitudes with values independent of observer. The physical object is independent of the observer too, but its ...


2

The commutators for spin taken as postulates because spin itself is not a classical thing. Nevertheless, there is a crude analogy between the spin and the proper rotation of objects in classical motion. And since the rotation in the classical case is described by the angular momentum, it is reasonable to simply assume that this can be similar to describe the ...


0

The thermal expectation value of $m_2$ is $ \left\langle m_2 \right\rangle = \frac{Tr[(I_{2 \times 2} \otimes \hat{\sigma}) \cdot \exp( - \beta H_{eff})]}{Tr[\exp( - \beta H_{eff})]}$ in the basis you use, and it will fix the dimension issue. Usually we don't need to diagonalize the matrix, some function like MatrixExp in mathematica can tell you the ...


3

A key point which the other answers miss: Your equation (*) giving the time evolution is incorrect! For a time dependent Hamiltonian, you have to integrate the Schroedinger equation, which is more complicated than just giving an exponential of $iHt$. Thus, your final result is incorrect.


1

The initial state is an eigenstate of $S_x$, so it is not an eigenstate of the Hamiltonian (which is proportional to $S_z$); therefore, the spin will precess around the direction of the field (i.e. the $z$ axis) half of the time in a clockwise fashion (when $\cos \omega t>0$) and the other half in a counterclockwise fashion. If the field was constant, ...


1

"If it started on an eigenstate it should remain in it forever, are my calculations wrong?" - The state you started with is not an eigenstate of the Hamiltonian. This is also clear from the result you got after evaluation of the time evolved state. If your magnetic field had been in the $ \hat{i} $ direction, then the initial state would be an eigenstate and ...


-1

Disclaimer: I don't know any of the proper functional analysis to make these arguments rigorous. This is going to be very "physicists attempting math" so follow at your own risk. In fact, this is an argument I've sort of made up myself so there might be some glaring issue with it and I would be happy to be corrected if that is the case. I follow the ...


0

Start with the time-dependent Schrodinger equation $$ i\hbar\frac{d}{dt}\vert\psi(t)\rangle=H\vert \psi(t)\rangle\, , \tag{1} $$ and assume $U(t)$ so that $$ \vert \psi(t)\rangle =U(t)\vert\psi(0)\rangle \tag{2} $$ Insert (2) in (1) to get \begin{align} i\hbar \frac{d}{dt}U(t) \vert\psi(0)\rangle&=H U(t)\vert\psi(0)\rangle\, ,\\ \frac{dU}{dt}&=-\...


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