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# 物理代写|量子力学代写Quantum mechanics代考|PHYS402 Free-Particle States

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## 物理代写|量子力学代写Quantum mechanics代考|Free-Particle States

For a free particle the Hamiltonian is just the kinetic-energy operator, which obviously commutes with the momentum operator. We note, however, that the free-particle Hamiltonian also commutes with $\mathbf{L}^2$ and $L_z$. Thus it is possible to consider a simultaneous eigenket of $H_0, \mathbf{L}^2$, and $L_z$. Ignoring spin, such a state is denoted by $|E, l, m\rangle$, often called a spherical wave state.

More generally, the most general free-particle state can be regarded as a superposition of $|E, l, m\rangle$ with various $E, l$, and $m$ in much the same way as the most general free-particle state can be regarded as a superposition of $|\mathbf{k}\rangle$ with different $\mathbf{k}$, different in both magnitude and direction. Put in another way, a free-particle state can be analyzed using either the plane wave basis ${|\mathbf{k}\rangle}$ or the spherical wave basis ${|E, l, m\rangle}$.

We now derive the transformation function $\langle\mathbf{k} \mid E, l, m\rangle$ that connects the plane wave basis with the spherical wave basis. We can also regard this quantity as the momentumspace wave function for the spherical wave characterized by $E, l$, and $m$. We adopt the normalization convention for the spherical wave eigenket as follows:
$$\left\langle E^{\prime}, l^{\prime}, m^{\prime} \mid E, l, m\right\rangle=\delta_{l l^{\prime}} \delta_{m m^{\prime}} \delta\left(E-E^{\prime}\right) .$$
In analogy with the position-space wave function, we may guess the angular dependence to be
$$\langle\mathbf{k} \mid E, l, m\rangle=g_{I E}(k) Y_l^m(\hat{\mathbf{k}}),$$
where the function $g_{I E}(k)$ will be considered later. To prove this rigorously, we proceed as follows. First, consider the momentum eigenket $|k \hat{\mathbf{z}}\rangle$, that is, a plane wave state whose propagation direction is along the positive $z$-axis. An important property of this state is that it has no orbital angular-momentum component in the $z$-direction:
$$L_z|k \hat{\mathbf{z}}\rangle=\left(x p_y-y p_x\right)\left|k_x=0, k_y=0, k_z=k\right\rangle=0 .$$

## 物理代写|量子力学代写Quantum mechanics代考|Partial-Wave Expansion

Let us now come back to the case $V \neq 0$. We assume that the potential is spherically symmetric, that is, invariant under rotations in three dimensions. It then follows that the transition operator $T$, which is given by (6.71), commutes with $\mathbf{L}^2$ and $\mathbf{L}$. In other words, $T$ is a scalar operator.

It is now useful to use the spherical wave basis because the Wigner-Eckart theorem [see (3.481)], applied to a scalar operator, immediately gives
$$\left\langle E^{\prime}, l^{\prime}, m^{\prime}|T| E, l, m\right\rangle=T_l(E) \delta_{l l^{\prime}} \delta_{m m^{\prime}} .$$
In other words, $T$ is diagonal both in $l$ and in $m$; furthermore, the (nonvanishing) diagonal element depends on $E$ and $l$ but not on $m$. This leads to an enormous simplification, as we will see shortly.
Let us now look at the scattering amplitude (6.58):
$$f\left(\mathbf{k}^{\prime}, \mathbf{k}\right)=-\frac{1}{4 \pi} \frac{2 m}{\hbar^2} L^3\left\langle\mathbf{k}^{\prime}|T| \mathbf{k}\right\rangle$$
\begin{aligned} \longrightarrow &-\frac{1}{4 \pi} \frac{2 m}{\hbar^2}(2 \pi)^3 \sum_l \sum_m \sum_{l^{\prime}} \sum_{m^{\prime}} \int d E \int d E^{\prime}\left\langle\mathbf{k}^{\prime} \mid E^{\prime} l^{\prime} m^{\prime}\right\rangle \ & \times\left\langle E^{\prime} l^{\prime} m^{\prime}|T| E l m\right\rangle\langle E l m \mid \mathbf{k}\rangle \ =&-\left.\frac{1}{4 \pi} \frac{2 m}{\hbar^2}(2 \pi)^3 \frac{\hbar^2}{m k} \sum_l \sum_m T_l(E)\right|{E=\hbar^2 k^2 / 2 m} Y_l^m\left(\hat{\mathbf{k}}^{\prime}\right) Y_l^{m^}(\hat{\mathbf{k}}) \end{aligned} $$=-\left.\frac{4 \pi^2}{k} \sum_l \sum_m T_l(E)\right|{E=\hbar^2 k^2 / 2 m} Y_l^m\left(\hat{\mathbf{k}}^{\prime}\right) Y_l^{m^}(\hat{\mathbf{k}}) .$$

## 物理代写|量子力学代写Quantum mechanics代考|Free-Particle States

$$\left\langle E^{\prime}, l^{\prime}, m^{\prime} \mid E, l, m\right\rangle=\delta_{u \prime} \delta_{m m^{\prime}} \delta\left(E-E^{\prime}\right) .$$

$$\langle\mathbf{k} \mid E, l, m\rangle=g_{I E}(k) Y_l^m(\hat{\mathbf{k}})$$

$$L_z|k \hat{\mathbf{z}}\rangle=\left(x p_y-y p_x\right)\left|k_x=0, k_y=0, k_z=k\right\rangle=0 .$$

## 物理代写|量子力学代写Quantum mechanics代考|Partial-Wave Expansion

$$\left\langle E^{\prime}, l^{\prime}, m^{\prime}|T| E, l, m\right\rangle=T_l(E) \delta_{l l^{\prime}} \delta_{m m m^{\prime}}$$

$$f\left(\mathbf{k}^{\prime}, \mathbf{k}\right)=-\frac{1}{4 \pi} \frac{2 m}{\hbar^2} L^3\left\langle\mathbf{k}^{\prime}|T| \mathbf{k}\right\rangle$$

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