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# 物理代写|量子力学代写Quantum mechanics代考|Two level system

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## 物理代写|量子力学代写Quantum mechanics代考|Two level system

Consider a physical system that, under certain circumstances and for certain phenomena, can be approximated by two quantum levels. It turns out that this kind of approximation is indeed valid for several important phenomena in Nature, including spin or nuclear magnetic resonance, masers, and the $\mathrm{K}^0-\overline{\mathrm{K}}^0$ elementary particle system.

In the stable configuration the system is described by the two level Hamiltonian $H_0$ given by
$$H_0=\left(\begin{array}{cc} E_1 & 0 \ 0 & E_2 \end{array}\right)=E_0 \times 1+\hbar \omega_{12} \frac{\sigma_3}{2}$$
where we parametrized $E_0=\frac{1}{2}\left(E_1+E_2\right)$ and $\omega_{12}=\frac{1}{\hbar}\left(E_1-E_2\right)$ and introduced the Pauli matrix $\sigma_3$. Now consider disturbing the system by a time dependent perturbation of the form
$$H^{\prime}(t)=\left(\begin{array}{cc} 0 & \hbar \gamma e^{-i \omega t} \ \hbar \gamma^* e^{i \omega t} & 0 \end{array}\right)$$
In a physical setting this could represent the turning on of an external electromagnetic field. Note that $H_0+H^{\prime}(t)$ is of the form of Eq.(13.34) which we studied above. Explicitly we have
$$H_0+H^{\prime}(t)=\left(\begin{array}{cc} e^{-i \omega t / 2} & 0 \ 0 & e^{i \omega t / 2} \end{array}\right)\left(\begin{array}{cc} E_1 & \hbar \gamma \ \hbar \gamma^* & E_2 \end{array}\right)\left(\begin{array}{cc} e^{i \omega t / 2} & 0 \ 0 & e^{-i \omega t / 2} \end{array}\right)$$
which identifies
\begin{aligned} & H_1=\left(\begin{array}{cc} -\frac{\hbar \omega}{2} & 0 \ 0 & \frac{\hbar \omega}{2} \end{array}\right)=-\hbar \omega \frac{\sigma_3}{2}, \ & H_2=\left(\begin{array}{cc} E_1 & \hbar \gamma \ \hbar \gamma^* & E_2 \end{array}\right)=E_0 \times 1+\hbar \omega_{12} \frac{\sigma_3}{2}+\hbar \gamma_1 \sigma_1+\hbar \gamma_2 \sigma_2 \end{aligned}
where $\gamma=\gamma_1-i \gamma_2$. Therefore we expect to solve this problem exactly by using the methods of the previous section.

The interaction picture Hamiltonian $H_I(t)=e^{i t H_0 / \hbar} H^{\prime}(t) e^{-i t H_0 / \hbar}$ becomes
$$H_I(t)=\left(\begin{array}{cc} 0 & \gamma e^{-i\left(\omega-\omega_{12}\right) t} \ \gamma^* e^{i\left(\omega-\omega_{12}\right) t} & 0 \end{array}\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Spin magnetic resonance

An example of the two-level problem discussed before is given by the spin magnetic resonance of a set of nuclei at rest. In fact, if we apply to such system a constant magnetic field $\vec{B}_0$ in the $z$ direction $\mathbf{B}_0=B_0 \hat{z}$, and a “small” oscillating field in the $x-y$ plane, $B_x(t)=b_0 \cos \omega t$ and $B_y(t)=b_0 \sin \omega t$. We can write the Hamiltonian
$$H(t)=-\boldsymbol{\mu} \cdot \mathbf{B}(t)$$
The magnetic moment is proportinal to the spin operator $\boldsymbol{\mu}=\mu \mathbf{S}$ of the nucleus , then
$$H(t)=-\mu\left(B_0 S_z+b_0 \cos \omega t S_x+b_0 \sin \omega t S_y\right)$$
or
$$H=H_0+H^{\prime}(t)$$
with
$$H_0=-\mu B_0 S_z, \quad H^{\prime}(t)=-\mu\left(b_0 \cos \omega t S_x+b_0 \sin \omega t S_y\right)$$
We can write
$$H^{\prime}(t)=e^{-i \omega t S_z / \hbar}\left(-\mu b_0 S_x\right) e^{i \omega t S_z / \hbar}$$
Therefore, this is of the form $H^{\prime}(t)=e^{i t H_1 / \hbar} H_2 e^{-i t H_1 / \hbar}$ which we can solve exactly, with $H_1=\omega S_z$ and $H_2=-\mu b_0 S_x$.

The states of $H_0$ are angular momentum states $|j m\rangle$ for a fixed $j$, and has eigenvalues $E_m=-\mu B_0 m$. The transitions will be between different rotation states and can be computed for any $j$ in terms of the D-functions for rotation matrices. The general case is left as an exercise for the student. Here we discuss the case of $\operatorname{spin} j=1 / 2$. For that case the spin operator is represented by Pauli matrices $\mathbf{S}=\hbar \boldsymbol{\sigma} / 2$. Therefore the Hamiltonian takes the $2 \times 2$ matrix form
$$H=H_0+H^{\prime}(t)=\left(\begin{array}{cc} -\frac{1}{2} \hbar \mu B_0 & 0 \ 0 & \frac{1}{2} \hbar \mu B_0 \end{array}\right)+\left(\begin{array}{cc} 0 & \hbar \gamma e^{-i \omega t} \ \hbar \gamma^* e^{+i \omega t} & 0 \end{array}\right)$$
with
$$|\gamma|=\frac{1}{2} \mu b_0$$
This is precisely the problem we studied in the previous section. By making the measurments of the resonance frequency $\omega$ and of the width of the curve $q$ indicated in the previous section we learn
$$\mu B_0=\omega, \quad \frac{1}{2} \mu b_0=\frac{q}{4}$$
Since an experimentalist has control of the external fields $B_0, b_0$ and can vary them, these measurements allow us to extract the value of the magnetic moment.
The setup can also be used in reverse to make instruments for which $\mu$ is known, and use it to measure electromagnetic properties of systems.

## 物理代写|量子力学代写Quantum mechanics代考|Two level system

$$H^{\prime}(t)=\left(\begin{array}{lll} 0 & \hbar \gamma e^{-i \omega t} \hbar \gamma^* e^{i \omega t} & 0 \end{array}\right)$$

$$H_0+H^{\prime}(t)=\left(\begin{array}{llll} e^{-i \omega t / 2} & 0 & 0 & e^{i \omega t / 2} \end{array}\right)\left(\begin{array}{lllll} E_1 & \hbar \gamma \hbar \gamma^* & E_2 \end{array}\right)\left(\begin{array}{llll} e^{i \omega t / 2} & 0 & 0 & e^{-i \omega t / 2} \end{array}\right)$$

## 物理代写|量子力学代写Quantum mechanics代考|Spin magnetic resonance

$$H(t)=-\boldsymbol{\mu} \cdot \mathbf{B}(t)$$

$$H(t)=-\mu\left(B_0 S_z+b_0 \cos \omega t S_x+b_0 \sin \omega t S_y\right)$$

$$H=H_0+H^{\prime}(t)$$

$$H_0=-\mu B_0 S_z, \quad H^{\prime}(t)=-\mu\left(b_0 \cos \omega t S_x+b_0 \sin \omega t S_y\right)$$

$$H^{\prime}(t)=e^{-i \omega t S_z / \hbar}\left(-\mu b_0 S_x\right) e^{i \omega t S_z / \hbar}$$

$$H=H_0+H^{\prime}(t)=\left(\begin{array}{llll} -\frac{1}{2} \hbar \mu B_0 & 0 & 0 & \frac{1}{2} \hbar \mu B_0 \end{array}\right)+\left(\begin{array}{lll} 0 & \hbar \gamma e^{-i \omega t} \hbar \gamma^* e^{+i \omega t} & 0 \end{array}\right)$$

$$|\gamma|=\frac{1}{2} \mu b_0$$

$$\mu B_0=\omega, \quad \frac{1}{2} \mu b_0=\frac{q}{4}$$

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