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6

You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron. In quantum mechanics, we describe the states of objects as elements of a Hilbert space $\mathcal{H}$. The crucial thing is that not all elements ...


4

There are a number of ways to represent a 720 degree spin of a particle, but the particle has to be a little more complex then a spinning ball. Imagine this wiki image (from http://en.wikipedia.org/wiki/Spin-1/2) as the inside of a larger ball and you can see an example of 720 degree spin. My favorite representation is the idea of breaking the electron into ...


4

The way you've written them, those are the spin operators in the $\hat{S}_z$ eigenbasis for a spin-1/2 particle. The two $\hat{S}_z$ eigenstates are spin up (written as $|+\rangle$ or $\uparrow$) and spin down ($|-\rangle$ or $\downarrow$), which can be written as $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ in the $\hat{S}_z$ ...


3

First of all you should not think of spin of an electron as a ball spinning on is own axis. At least this comparison should not go beyond your intuitive understanding of an electron spin. The quantity $s=\sqrt{s(s+1)}= \frac{\sqrt3}{2} \hbar$ is an intrinsic property of an electron with spin one-half and has nothing to do with the axis of rotation. I don't ...


3

The diagram is simply a way to remember two variables at the same time, one called the total spin $\hbar\sqrt{3}/2$, the other called the z-component which has two possibilities $\pm\hbar/2$. If you draw a picture of a vector of length $\hbar\sqrt{3}/2$ that is at an angle so that the z-component of the vector is $\pm\hbar/2$ (so it lies on two cones, one ...


3

The propability $P$ that a system in state $|\Psi\rangle$ will eventually transit into state $|\Theta\rangle$ is given by: $$ P = | \langle\Theta | \Psi \rangle |^2 $$ So for example if you have a System in Spin-X-State "up" denoted by $ |\Psi\rangle \hat{=} \frac{1}{\sqrt{2}} (1, 1)^T$, the propability of getting Spin-Z-State "up" (e.g. by measuring along ...


2

These are mainly conventions. Conventionally, the kets $|+\rangle$ and $|-\rangle$ are taken to be eigenkets of the z-spin operator with, respectively, z-spin of $+\hbar/2$ and -$\hbar/2$. S_x and S_y are chosen such that they obey the canonical commutation relations for angular momenta $$ [S_i,S_j]=i\epsilon_{ijk}S_k $$ E.g., $$ [S_x,S_y]=iS_z $$ and so ...


2

You are basically asking how to compute $$e^{i\gamma B \hat{S}_z t/\hbar} |+\rangle.$$ Now, $e^{i\gamma B \hat{S}_z t/\hbar} =\sum_{n=0}^{n=\infty} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!},$ so you can approximate it quite well by $$\sum_{n=0}^{n=N} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!}.$$ And you can figure out how to operate ...


1

since every sample has both protons and electrons, and all have magnetic spin, But the electron spins cancel out usually, because normal matter only has electrons in Pauli pairs. EPR is restricted to radicals in organic chemistry or transition metal complexes, or O2 gas :=)


1

In quantum mechanics an "observable" like the position of an atom, $\hat X$, is represented by a Hermitian operator. An "observable" is some physical quantity you can actually go out and measure, like the position. The eigenvalues of that operator are the possible measurement results. In quantum mechanics a "system" (some physical entity that has ...


1

but I'm stuck to carry the computation forward: what does $e^{i\hat{S}_zt}$ do to an eigenvector like $|+\rangle$? I'm stuck because the exponential contains an operator, and I want to get an expression without operator. Luckily, you only have to act the operator on its own eigenvector (|+> and |->), so you can replace the operator with the ...


1

Notice that $\hat{S_z}= \frac{\hbar}{2}\sigma_3$ and that $\sigma_3^2=\mathbb{1}$, where $\sigma_3$ is the Pauli matrix. I don't know if you know the identity $$\hat{O}^2= \mathbb{1} \implies e^{i \alpha \hat{O}}= \mathbb{1} \cos(\alpha) + i \hat{O} \sin(\alpha) \tag{1} $$ It is fairly easy to derive it via Taylor expansion. I set $\omega = \frac{\gamma ...


1

Because you want the maximal spin particle allowed to be 2 (since there is no higher spin field theory interacting non trivially), thus in a supermultiplet starting with a particle with helicity $0$, say in $d=4$, the maximal number of susy you can apply is $8$, thus $N=8$ in $d=4$ is the maximal supersymmetry admitted, which means $32$ supercharges.


1

Actually I "believe" that this topic is not really "settled"... The expression $$ \frac12 \left(\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z\right) $$ is introduced by P. Dirac in his famous book The Principles of Quantum Mechanics IV ed. chap IX p. 221 where he uses this expression (actually he does not write explicitly the ...


1

Actually I "believe" that this topic is not really "settled"... The expression 1/2 (σx⊗σx +σy⊗σy +σz⊗σz) is introduced by P. Dirac in his famous book "The Principles of Quantum Mechanics" IV ed. chap IX p. 221 where he uses this expression (actually he does not write explicitly the tensor or Kronecker product sign ⊗ but one "undersdands" that it corresponds ...



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