# 物理代写|量子力学代写quantum mechanics代考|Connection Between Spin and Statistics

## 物理代写|量子力学代写quantum mechanics代考|Connection Between Spin and Statistics

It follows from some of the more esoteric aspects of relativistic quantum field theory that there is a connection between spin and statistics
Half-integral spin particles obey Fermi statistics, while integer spin particles obey Bose statistics.
Electrons and nucleons have spin-1/2, and they obey Fermi statistics. ${ }^4 \mathrm{He}$ atoms have spin zero and photons have unit helicity (the component of spin along the direction of motion); they obey Bose statistics.

We can keep track of the helicity of the photons with the polarization vectors
$$\vec{e}{\vec{k}, \pm 1}=\mp \frac{1}{\sqrt{2}}\left(\vec{e}{\vec{k} 1} \pm i \vec{e}{\vec{k} 2}\right)$$ For the spin-1/2 particles, we keep track of the spin projection along the $z$-axis with a set of two-component column vectors ${ }^6$ $$\phi{\uparrow}=\left(\begin{array}{l} 1 \ 0 \end{array}\right) \quad ; \phi_{\downarrow}=\left(\begin{array}{l} 0 \ 1 \end{array}\right)$$

## 物理代写|量子力学代写quantum mechanics代考|Stern–Gerlach Experiment

In this experiment a beam of neutral particles with internal angular momentum $\hbar \overrightarrow{\mathcal{S}}$, here assumed to be spin- $1 / 2$ with $\mathcal{S}_z= \pm 1 / 2$, and a magnetic moment $\vec{\mu}=2 \mu \overrightarrow{\mathcal{S}}$ in the direction of the spin, is passed through an inhomogeneous magnetic field (see Fig. 8.1).

The particles feel a force in the $z$-direction of
$$F_z=\mu_z \frac{d B_z}{d z}$$
Instead of seeing a continuous distribution of particles coming out of the detector in the $z$-direction, one observes only two beams, corresponding to $\mathcal{S}_z= \pm 1 / 2$. This illustrates the discrete quantization of the angular momentum. One observes just the eigenvalues of $\mathcal{S}_z$.

In quantum mechanics we understand what is happening by saying the initial internal state of each particle is a linear combination of the two spin states $^1$
$$\left|\psi_{\text {int }}(t)\right\rangle=c_{\uparrow}(t)|\uparrow\rangle+c_{\downarrow}(t)|\downarrow\rangle$$
The probability that we will measure spin up is then $\left|c_{\uparrow}(t)\right|^2$, and the probability that we will measure spin down is $\left|c_{\downarrow}(t)\right|^2$, where
$$\left|c_{\uparrow}(t)\right|^2+\left|c_{\downarrow}(t)\right|^2=1$$

# 量子力学代考

## 物理代写|量子力学代写quantum mechanics代考|Connection Between Spin and Statistics

$$\vec{e} \vec{k}, \pm 1=\mp \frac{1}{\sqrt{2}}(\vec{e} \vec{k} 1 \pm i \vec{e} \vec{k} 2)$$

$$\phi \uparrow=\left(\begin{array}{ll} 1 & 0 \end{array}\right) \quad ; \phi_{\downarrow}=\left(\begin{array}{ll} 0 & 1 \end{array}\right)$$

## 物理代写|量子力学代写quantum mechanics代考|Stern–Gerlach Experiment

$$\left|c_{\uparrow}(t)\right|^2+\left|c_{\downarrow}(t)\right|^2=1$$

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