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# 物理代写|量子力学代写Quantum mechanics代考|PHYS3001 Addition of angular momentum

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## 物理代写|量子力学代写Quantum mechanics代考|Addition of angular momentum

In many physical applications one needs to consider combining the angular momenta of two or more parts of the system. In classical physics this is done by the adding the vectors that represent the angular momenta of the various parts. For example the total angular momentum of the Earth orbiting around the sun is obtained by adding the spin of the Earth around itself to the orbital angular momentum for rotating around the sun. How is this done in quantum mechanics? We must answer this question in order to understand a host of problems in quantum systems such as atoms, molecules, solids, nuclei, particles composed of quarks, scattering, etc., where the spins of particles must be combined with their orbital angular momenta, and the total angular momentum of the system is obtained by adding all the spins and all the orbital angular momenta.

To understand the process of addition of angular momentum it may be helpful to first consider the addition of ordinary linear momentum. Thus consider two particles (or two parts of a system) that have momenta $\mathbf{p}_1$ and $\mathbf{p}_2$. In classical physics the total momentum of the system is $\mathbf{p}=\mathbf{p}_1+\mathbf{p}_2$. In quantum mechanics each particle is described by a state $\left|\mathbf{k}_1\right\rangle,\left|\mathbf{k}_2\right\rangle$ labelled with the eigenvalues of the commuting operators $\mathbf{p}_1 \rightarrow \hbar \mathbf{k}_1, \mathbf{p}_2 \rightarrow \hbar \mathbf{k}_2$. The total system is described by the direct product state
$$\left|\mathbf{k}_1\right\rangle \otimes\left|\mathbf{k}_2\right\rangle \equiv\left|\mathbf{k}_1, \mathbf{k}_2\right\rangle .$$
When a function of the operators $f\left(\mathbf{p}_1, \mathbf{p}_2\right)$ acts on the direct product state, each operator acts on its corresponding label, leaving the other label untouched. In particular, the total momentum operator $\mathbf{p}=\mathbf{p}_1+\mathbf{p}_2$ may act on this state, and pick up an overall eigenvalue $\mathbf{p} \rightarrow \hbar \mathbf{k}_1+\hbar \mathbf{k}_2=\hbar \mathbf{k}$. Thus, the state $\left|\mathbf{k}_1, \mathbf{k}_2\right\rangle$ is already an eigenstate of the total momentum, and therefore, it is possible to relabel it in terms of total angular momentum, plus other labels corresponding to the eigenvalues of operators that commute with the total momentum operator $\mathbf{p}$ (e.g. relative angular momentum, if this is convenient for the application)
$$\left|\mathbf{k}_1\right\rangle \otimes\left|\mathbf{k}_2\right\rangle \equiv\left|\mathbf{k}_1, \mathbf{k}_2\right\rangle=|\mathbf{k}, \cdots\rangle$$

## 物理代写|量子力学代写Quantum mechanics代考|Total angular momentum

Consider two rotating systems with angular momentum operators $\mathbf{J}$ (1) and $\mathbf{J}$ (2) respectively. Some examples are: the orbital angular momentum of an electron $\mathbf{J}^{(1)}=\mathbf{L}$ and its spin $\mathbf{J}^{(2)}=\mathbf{S}$, the spins of two electrons in a multi-electron atom $\mathbf{J}^{(1)}=\mathbf{S}^{(1)}, \mathbf{J}^{(2)}=\mathbf{S}^{(2)}$, etc. The commutation rules are
\begin{aligned} & {\left[J_i^{(1)}, J_j^{(1)}\right]=i \hbar \varepsilon_{i j k} J_k^{(1)}} \ & {\left[J_i^{(2)}, J_j^{(2)}\right]=i \hbar \varepsilon_{i j k} J_k^{(2)}} \ & {\left[J_i^{(1)}, J_j^{(2)}\right]=0} \end{aligned}

Each system is described by a state $\left|j_1 m_1,\right\rangle,\left|j_2 m_2\right\rangle$, while the combined system has the direct product state
$$\left|j_1 m_1\right\rangle \otimes\left|j_2 m_2\right\rangle \equiv\left|j_1 m_1 j_2 m_2\right\rangle$$
We will study how to express this state in terms of total angular momentum states, where the definition of total angular momentum is consistent with classical mechanics $\mathbf{J}=\mathbf{J}^{(1)}+\mathbf{J}^{(2)}$.

How does the state $\left|j_1 m_1 j_2 m_2\right\rangle$ rotate? This must be obtained from the rotation of each part, thus
\begin{aligned} \left|j_1 m_1\right\rangle^{\prime} \otimes\left|j_2 m_2\right\rangle^{\prime} & \left.=e^{-\frac{i}{\hbar} \mathbf{J}^{(1)} \cdot \omega} \cdot j_1 m_1\right\rangle \otimes e^{-\frac{i}{\hbar} \cdot \mathbf{J}^{(2)} \cdot \boldsymbol{\omega}}\left|j_2 m_2\right\rangle \ & \equiv\left(e^{-\frac{i}{h} \mathbf{J}^{(1)} \cdot \omega} e^{-\frac{i}{\hbar} \mathbf{J}^{(2)} \cdot \omega}\right)\left|j_1 m_1, j_2 m_2\right\rangle \end{aligned}
where in the second line each operator $\mathbf{J}^{(1)}, \mathbf{J}^{(2)}$ is understood to act on the corresponding labels (1 or 2), leaving the others untouched. Note that the same rotation angles $\boldsymbol{\omega}$ must appear in each exponent since the same rotation is applied on the entire system. The exponents may be combined because $\left[\mathbf{J}^{(1)}, \mathbf{J}^{(2)}\right]=0$,
$$\left(e^{-\frac{i}{\hbar} \mathbf{J}^{(1)} \cdot \boldsymbol{\omega}} e^{-\frac{i}{\hbar} \mathbf{J}^{(2)} \cdot \boldsymbol{\omega}}\right)=e^{-\frac{i}{\hbar}\left(\mathbf{J}^{(1)}+\mathbf{J}^{(2)}\right) \cdot \boldsymbol{\omega}}$$

## 物理代写|量子力学代写Quantum mechanics代考|Total angular momentum

$$\left|j_1 m_1\right\rangle \otimes\left|j_2 m_2\right\rangle \equiv\left|j_1 m_1 j_2 m_2\right\rangle$$

$$\left(e^{-\frac{i}{h} \mathbf{J}^{(1)} \cdot \boldsymbol{\omega}} e^{-\frac{i}{h} \mathbf{J}^{(2)} \cdot \boldsymbol{\omega}}\right)=e^{-\frac{i}{\hbar}\left(\mathbf{J}^{(1)}+\mathbf{J}^{(2)}\right) \cdot \boldsymbol{\omega}}$$

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