New answers tagged

0

The section on modelling 2 level hamiltonians with "fictitious spin" and "fictitious B field" might be useful here https://www.chem.uci.edu/~unicorn/249/Handouts/CTspinhalf.pdf


6

In this picture, both orbital and angular momenta are treated as vector. As the third case illustrates, it is perfectly possible for the sum of two vectors to be smaller in length than the length of both constituents in the sum. This third figure would correspond roughly to the case where the rotation of the Earth about its axis is reversed compared to what ...


2

Invariance of the physical systems under rotation allows the operators $S_z$ and $S_x$ be rotationally equivalent means they can be used one another under a rotation operation. The roots of the rotational invariance can be shown by Noether's theorem. On the other hand if you write the spin operators in matrix form and using a rotation matrix you will see ...


1

Qualitatively, the reason is simple. You distinguish between $S_z$ and $S_x$ because of axis orientation, but in case you rotate your axis, $S_z$ becomes your $S_x$ and vice-versa. Thus these two are equivalent, and the eigenvalues must be the same.


0

Let $\Pi_{+z}=\vert+\rangle\langle +\vert$ and $\Pi_{-z} =\vert -\rangle\langle -\vert$. It seems for any state $\vert\psi\rangle$ you want to compute \begin{align} \Pi_{-z}\Pi_{+z}\vert\psi\rangle \end{align} but this contains one piece of the form $\langle -\vert +\rangle=0$ so if $\Pi_{-z}$ immediately follows $\Pi_{+z}$ the result is $0$ so $\vert -\...


1

Firstly, am I correct in saying that we can imagine the angular momentum (|L|^2 and Lz values) at every point in this angular momentum eigenstate wavefunction as being identical (and equal to the respective |𝐿|2 and 𝐿𝑧 eigenvalues)? Or instead are we meant to interpret the net 'sum' of the angular momentua at each point over the whole wavefunction as ...


0

We're are dealing with magnetic fields which are vector quantities so you have to consider also the direction and not only the module. The excitation field, $B_1$, must be orthogonal to the direction of the $\mathbf{B_1} × \mathbf{μ} \ne 0$, to generate transitions of the nuclear spin state. To convince you that a small magnetic field is able to excite the ...


0

If you consider B0 acting along the z axis then we know the there will be precession about this axis due to the component of the magnetic moment in the x y plane. This precession occurs at the larmor frequency. If you apply a B1 field that acts orthogonal to B0 and rotates at the same rate w about the z axis, then the component of the magnetic moment along ...


1

I am answering the title: Is there an unentangled electron pair in the universe? Please keep in mind that entanglement really means that "there exists a quantum wavefunction that describes the entangled particles". Then one uses the conservation laws to find a way to check the statement experimentally . i.e. it is a quantum mechanical model. ...


2

Your concern is unnecessary. Suppose the universe is divided into two separate regions A and B, and every particle in region A is entangled with a corresponding particle in region B, but not entangled with other particles in region A. No experiment done solely on the particles in region A can reveal that the particles are entangled with particles in the ...


0

You can absolutely represent different kinds of particles if you choose different $\alpha$ or $\beta$ in your first-order wave equation. As a trivial example, consider using $8 \times 8$ block diagonal matrices with $\alpha$ (or $\beta$) in the top-left and bottom-right blocks. That represents a pair of spin $1/2$ particles, which could have spin $0$ or $1$. ...


0

You can write massive spin-0 and spin-1 wave equations using the Dirac equation with the standard Dirac matrices, but with the fields occupying different subspaces of the Clifford algebra than the Dirac field does. The trick is essentially the same as that used to turn the second-order harmonic oscillator equation $\phi'' = -k^2\phi$ into the first-order ...


1

The Lagrangian for a massive vector field (without sources) has the form $$ L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{1}{2}M^{2}A_{\mu}A^{\mu}$$ and is not gauge invariant due to the mass term. The eq.s of motion are $$\partial _{\mu}F^{\mu\nu} + M^{2}A^{\nu} = 0$$ and by deriving a second time $$\partial_{\mu} A^{\mu} = 0$$ that is not a gauge-fixing. ...


1

A vector representing the spin state of a spin-1/2 particle is two-dimensional. Spin is associated with the group $SU(2)$. The Pauli sigma matrices form a basis for a two-dimensional representation of this group. For example, if we let the operator that measures spin along the $z$-axis be proportional to the Pauli matrix $$\sigma_z=\left(\begin{matrix} 1 &...


1

The states given in your table seem to be states of a system of N spins in the $\hat{S}^z_{\text{tot}}$ and $\hat{S}^2_{\text{tot}}$ basis which are also eigenstates of the cyclic permutation operator $\hat{P}_{\text{cycl}}$ with $P_{\text{cycl}}$ being the eigenvalue of the cyclic permutation operator. An Example definition for the cyclic permutation ...


1

Frankly, If you only sat down and wrote down all 16 states involved, and confirmed the multiplets, you'd see them all together, including their symmetries, and dispel your magical misconception on symmetry and antisymmetry. (You'd have 6 states of total M=0, of which two combinations would amount to your singlets!) In any case, distributing $$ (1\oplus 0)\...


0

This is a comment. The Dirac equation as defined, has wavefunctions that can model spin 1/2 particles. Any changes define a different differential equation and it has to be checked what the solutions are good for. For example: In theoretical physics, the Rarita–Schwinger equation is the relativistic field equation of spin-3/2 fermions. It is similar to ...


0

Angular momentum is defined as : In three dimensions, the angular momentum for a point particle is a pseudovector $r × p$, the cross product of the particle's position vector r (relative to some origin) and its momentum vector; So by definition, if momentum can be infinite so can angular momentum. Electrons, though quantum mechanical particles, when free ...


1

Extend your assumption to the entire visible universe. As others have pointed out, there will be at least two universes (a few more I would say, but still a minimal fraction of the total) in which the statistical results are such that quantum mechanics as we know it do not represent the laws of physics, even if they were true. Observers in such universe ...


2

This question is actually easier than many related questions about statistics, because this flavor of MWI lends itself to frequentist interpretations of probability. According to MWI (as you seem to be using it), there exist $2^n$ universes total after the measurement you describe. Of these exactly one universe has all copies of spin pointing up, and exactly ...


2

It doesn't explain it. Neither does any interpretation of quantum mechanics that reduces quantum amplitudes to classical probabilities. Suppose I claim that a coin comes up heads with some classical probability $p\in(0,1)$. What does that mean, operationally? In other words, how can you test it? You can flip the coin $n$ times, and if I'm right you should ...


1

The first problem with this question is that it is somewhat like the Inverse gambler's fallacy. It assumes that in a universe where things have frequently been "wrong" things will continue to go wrong. In any such universe, the chance of something going wrong in the future is the same as the chance of something going wrong in our own "perfect&...


1

The answer here addresses a common misconception about statistics and reality. If you’re asking why we live in a universe where the odds when flipping a coin are 50/50, remember that the universe is of finite size, so there are finitely many coins to be flipped. There’s a good change the actual prevalence of heads vs. tails as far as spin states are ...


4

Clearly $\exp(\hat A)\hat A\exp(-\hat A)=\hat A$ by Baker-Campbell-Hausdorff. Here, set $\hat A=\hat H=-it\mu\vec B\cdot \vec \sigma/\hbar$. Note that \begin{align} \exp(-i t\mu \vec B\cdot \vec \sigma/\hbar)\ne \exp(-it B_x\sigma_x)\exp(-it B_y\sigma_y)\exp(-it B_z\sigma_z) \end{align} or any other kind of factorization suggested by your snippet of Mma ...


0

There is a general formula for the Kronecker multiplication of an arbitrary number of doublets. Applied to your case, it yields $$ \frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}= 2\oplus 1\oplus 1\oplus 1\oplus 0\oplus 0, $$ a total of 16 states. These multiplicities are specified by the Catalan triangle. The general formula for n doublets ...


0

So the related question is not so much help to you because you have a much more fundamental misunderstanding. Once we clear that up, any further questions should be answered by the above. There is no such $-1$ or $-1/2$ system. For this we have to be clear on the difference between two different angular momentum quantum numbers, $\ell$ and $m$. They come ...


1

This equation is an equation for representations of the Lie algebra of rotations. It is a fact proven in every quantum mechanics book that the irreducible representations of the angular momentum algebra $[J_i,J_j]=\epsilon_{ijk}J_k$ are spin $j$ systems where $j\in \mathbb{N}/2=\{0,1/2,1,3/2,2,\dots\}$. We thus call this representation by the integer $j$. ...


3

You are essentially trying to change basis from eigenvectors of $\hat S_z$ to eigenvectors of $\hat S_x$, for example, for an arbitrary fixed S. (The other two cases are similar, and you may trivially carry them out yourself, as suggested here.) Feynman, vIII does your S = 1/2 case. The short answer is that your quantities are Wigner rotation d matrices for ...


0

Greetings fellow layperson, I would say that my own speculation is yes. I think that it absolutely does. The reason I think so is that the best reason for every particle not being a black hole is naked singularities, and in order for this to be the case, the chargeless, unspinning higgs boson needs to be superextremal, and the only property that it has ...


0

In classical theory, a charge oscillating in one dimension introduces transverse components into its electric field. These move away as part of a linearly polarized electromagnetic wave. If the charge moves in a circle (or ellipse) the electric fields in the wave rotate and can impart angular momentum to a receiving charge.


1

The wave field is not just a scalar field. The field value is "spinny". For a spin-0 particle the field has scalar values, but for a spin-1 particle the field has vector values. For a spin-½ particle the field has spinor values.


1

You want to substitute $$ S_m^- = (2S)^{1/2} a_m^\dagger $$ (on the A sublattice) as well as $$ \tilde S_m^- = (2S)^{1/2} a_m^\dagger $$ (on the B sublattice). Then you will obtain the desired results. Note that since $S$ ($\tilde S$) appears only on the A (B) sublattice, there is a clear meaning what $a_m$ is for each lattice site. Also note that in the ...


-1

You can ask the same question classically. An electromagnetic wave is fluctuations of the electric force and magnetic force, which point in particular directions. If the directions follow the right-hand-rule, the wave has positive spin. If these directions follow the left-hand-rule, the wave has negative spin.


0

Hopefully your homework deadline has expired, so I'm not doing your homework for you. Recall that "direction" is a unit vector, so, in spherical coordinates, $\hat n= (\sin \theta \cos\phi, \sin\theta\sin\phi,\cos\theta)^T$. Consequently, $$ \hat n\cdot \vec \sigma= \begin{pmatrix} \cos\theta & e^{-i\phi}~ \sin\theta \\ e^{i\phi}~ \sin\theta &...


1

Regarding the apparatus (b), you are correct that the "classically" expected result would be a roughly even distribution of measurements in a spectrum between $S_z=+\hbar/2$ and $S_z=-\hbar/2$. The surprising result is that the the particles exiting the $SGz$ apparatus have exactly two distinct beams, one with $S_z=+\hbar/2$ and one with $S_z=-\...


3

Let us start with a general remark. Why there are typically at most only first-order derivatives in the Lagrangian (density) is discussed in e.g. this Phys.SE post. This implies that the Euler-Lagrange EL equations are at most of second order, cf. e.g. this Phys.SE post. Now let us return to OP's question. OP is interested in the case where the Lagrangian (...


2

Let me discuss just one aspect of your question. I don't understand the statement about "first-order nature of the Dirac equation". Note that the Dirac equation is a system of four first-order partial differential equations (PDEs) for four components of the Dirac spinor. However, it is well-known that any system of PDEs can be rewritten as a system ...


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