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# 物理代写|量子力学代写Quantum mechanics代考|Scattering amplitude, S-Matrix, T-matrix

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## 物理代写|量子力学代写Quantum mechanics代考|Scattering amplitude, S-Matrix, T-matrix

In general, the observation point $\mathbf{r}$ is located at a distance much greater than the scattering center. So it is appropriate to take the limiting value of the above integral equation for $|\mathbf{r}| \rightarrow \infty$. The first term on the right hand side will not be affected by the limit, since it represents an oscillating plane wave. The integral in the second term has a term $e^{i k\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^{-1}$. Assuming a reasonably short range potential, the $\mathbf{r}^{\prime}$ integration will get its main contribution for small values of $\mathbf{r}^{\prime}$ close to the center. Therefore the denominator of this term will be approximated by $\left|\mathbf{r}-\mathbf{r}^{\prime}\right| \sim|\mathbf{r}|$ for large values $|\mathbf{r}|$. However, in the exponential we perform the following expansion
$$\left|\mathbf{r}-\mathbf{r}^{\prime}\right|=\sqrt{r^2-2 \mathbf{r} \cdot \mathbf{r}^{\prime}+r^{\prime 2}} \simeq r-\hat{\mathbf{r}} \cdot \mathbf{r}^{\prime}+O\left(r^{\prime 2} / r\right)$$
so that for large $r$ we have
$$\Psi_{\mathbf{k}}(\mathbf{r})=\frac{e^{+i \mathbf{k} \cdot \mathbf{r}}}{(2 \pi)^{3 / 2}}-\frac{e^{i k r}}{4 \pi r} \int d^3 \mathbf{r}^{\prime} e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}^{\prime}} v\left(r^{\prime}\right) \Psi_{\mathbf{k}}\left(\mathbf{r}^{\prime}\right) .$$
We define
$$\mathbf{k}^{\prime}=k \hat{\mathbf{r}}$$
and interpret this vector as the momentum of the outgoing particle. This is justified since $\mathbf{r}$ is the location of a detector where the scattered particle is detected, and therefore $\hat{\mathbf{r}}$ is the direction of the scattered particle. Furthermore, since the particle is a free particle after the scattering takes place, and since energy is conserved, its momentum must have absolute value $\hbar k$ so that the energy is $E=\hbar^2 k^2 / 2 m$. Therefore, we have now an expression for the wavefunction that depends on the incoming and outgoing momenta $\left(\mathbf{k}, \mathbf{k}^{\prime}\right)$
\begin{aligned} & \Psi_{\mathbf{k}}(\mathbf{r}) \underset{r \rightarrow \infty}{\rightarrow} \frac{e^{+i \mathbf{k} \cdot \mathbf{r}}}{(2 \pi)^{3 / 2}}-\frac{e^{i k r}}{4 \pi r} \int d^3 \mathbf{r}^{\prime} e^{-i \mathbf{k}^{\prime} \cdot \mathbf{r}^{\prime}} v\left(r^{\prime}\right) \Psi_{\mathbf{k}}\left(\mathbf{r}^{\prime}\right) \ & =\frac{1}{(2 \pi)^{3 / 2}}\left[e^{+i \mathbf{k} \cdot \mathbf{r}}+\frac{e^{i k r}}{r} f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)\right] \end{aligned}
where
\begin{aligned} f\left(\mathbf{k}, \mathbf{k}^{\prime}\right) & =-\frac{(2 \pi)^3}{4 \pi} \int d^3 \mathbf{r}^{\prime} \frac{e^{-i \mathbf{k}^{\prime} \cdot \mathbf{r}^{\prime}}}{(2 \pi)^{3 / 2}} v\left(r^{\prime}\right) \Psi_k\left(\mathbf{r}^{\prime}\right) \ & =-2 \pi^2\left\langle\mathbf{k}^{\prime}|v(r)| \mathbf{k}\right\rangle^{+} \end{aligned}

## 物理代写|量子力学代写Quantum mechanics代考|Differential and total cross section

The differential cross section is defined as follows
$$d \sigma=\frac{\text { number of particles scattered into solid angle } d \Omega \text { per unit time }}{\text { number of incident particles per unit time per unit area }}$$
Thus an experimentalist simply counts particles in the incident beam and in the detector that captures the scattered particles, and reports the ratio described above. The mathematical expression for what he sees is computed as follows. The denominator is just the probability current of the incident particles $\frac{d N_{i n}}{d t(d A)}=\left|\hat{\mathbf{k}} \cdot \mathbf{J}{i n}\right|$, where the current is to be computed from the incident wavefunction far away from the interaction region. The numerator can be expressed as the area of an infinitesimal detector times the number of radially scattered particles per unit area per unit time. This amounts to the probablity current of the outgoing particles $\frac{d N{\text {out }}}{d t(d A)}=\left|\hat{\mathbf{r}} \cdot \mathbf{J}{\text {out }}\right|$ times the area element $r^2 d \Omega$ that represents an infinitesimal detector located at some solid angle. The current is to be computed from the outgoing wavefunction far away from the interaction region. Therefore one should compare the experimentalist measurements to the expression $$d \sigma=\frac{\left|\hat{\mathbf{r}} \cdot \mathbf{J}{\text {out }}\right| r^2 d \Omega}{\left|\hat{\mathbf{k}} \cdot \mathbf{J}_{\text {in }}\right|}$$

Recall that the probability current for any wavefunction $\Psi$ is given by
$$\mathbf{J}=-i \frac{\hbar}{2 m}\left(\Psi^* \nabla \Psi-\Psi \nabla \Psi^\right)=\operatorname{Re}\left(\frac{1}{m} \Psi^(-i \hbar \nabla) \Psi\right)$$
We can thus compute $\mathbf{J}{i n}$ and $\mathbf{J}{\text {out }}$ by using the asymptotic form of the wavefunction given in Eq.(14.2)
$$\Psi_{i n}(\mathbf{r})=e^{+i \mathbf{k} \cdot \mathbf{r}}, \quad \Psi_{\text {out }}(\mathbf{r})=\frac{e^{i k r}}{r} f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)$$
We do not need to worry about the overall normalization since we are interested only in the ratio. Hence for the incident current we have $\left|\hat{\mathbf{k}} \cdot \mathbf{J}{i n}\right|=\frac{\hbar k}{m}$, and for the scattered current we have $$\left|\hat{\mathbf{r}} \cdot \mathbf{J}{\text {out }}\right|=\left|\operatorname{Re}\left(\frac{-i \hbar}{m} \Psi^* \frac{\partial}{\partial r} \Psi\right)\right| \underset{r \rightarrow \infty}{=} \frac{\hbar k}{m r^2}\left|f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)\right|^2$$
Therefore the differential cross section is
$$d \sigma=\left|f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)\right|^2 d \Omega_{k^{\prime}}$$

## 物理代写|量子力学代写Quantum mechanics代考|Scattering amplitude, S-Matrix, T-matrix

$$\left|\mathbf{r}-\mathbf{r}^{\prime}\right|=\sqrt{r^2-2 \mathbf{r} \cdot \mathbf{r}^{\prime}+r^{\prime 2}} \simeq r-\hat{\mathbf{r}} \cdot \mathbf{r}^{\prime}+O\left(r^{\prime 2} / r\right)$$

$$\Psi_{\mathbf{k}}(\mathbf{r})=\frac{e^{+i \mathbf{k} \cdot \mathbf{r}}}{(2 \pi)^{3 / 2}}-\frac{e^{i k r}}{4 \pi r} \int d^3 \mathbf{r}^{\prime} e^{-i k \hat{\mathbf{r}} \cdot \mathbf{r}^{\prime}} v\left(r^{\prime}\right) \Psi_{\mathbf{k}}\left(\mathbf{r}^{\prime}\right)$$

$$\mathbf{k}^{\prime}=k \hat{\mathbf{r}}$$

$$\Psi_{\mathbf{k}}(\mathbf{r}) \underset{r \rightarrow \infty}{\rightarrow} \frac{e^{+i \mathbf{k} \cdot \mathbf{r}}}{(2 \pi)^{3 / 2}}-\frac{e^{i k r}}{4 \pi r} \int d^3 \mathbf{r}^{\prime} e^{-i \mathbf{k}^{\prime} \cdot \mathbf{r}^{\prime}} v\left(r^{\prime}\right) \Psi_{\mathbf{k}}\left(\mathbf{r}^{\prime}\right) \quad=\frac{1}{(2 \pi)^{3 / 2}}\left[e^{+i \mathbf{k} \cdot \mathbf{r}}+\frac{e^{i k r}}{r} f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)\right]$$

$$f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)=-\frac{(2 \pi)^3}{4 \pi} \int d^3 \mathbf{r}^{\prime} \frac{e^{-i \mathbf{k}^{\prime} \cdot \mathbf{r}^{\prime}}}{(2 \pi)^{3 / 2}} v\left(r^{\prime}\right) \Psi_k\left(\mathbf{r}^{\prime}\right) \quad=-2 \pi^2\left\langle\mathbf{k}^{\prime}|v(r)| \mathbf{k}\right\rangle^{+}$$

## 物理代写|量子力学代写Quantum mechanics代考|Differential and total cross section

$$d \sigma=\frac{\text { number of particles scattered into solid angle } d \Omega \text { per unit time }}{\text { number of incident particles per unit time per unit area }}$$

$$d \sigma=\frac{\left|\hat{\mathbf{r}} \cdot \mathbf{J o u t}{\text {out }}\right| r^2 d \Omega}{\left|\hat{\mathbf{k}} \cdot \mathbf{J}{\text {in }}\right|}$$

$\backslash$ |mathbf ${\mathrm{J}}=-i \backslash$ frac ${\backslash$ hbar $}{2 \mathrm{~m}} \backslash$ left( $\backslash$ Psi $\wedge^* \mid$ nabla $\backslash$ Psi- $\backslash$ Psi $\mid$ nabla $\backslash$ Psi^ $\backslash$ right $)=$ loperatorname ${$ Re $} \backslash$ left(

$$\Psi_{i n}(\mathbf{r})=e^{+i \mathbf{k} \cdot \mathbf{r}}, \quad \Psi_{\text {out }}(\mathbf{r})=\frac{e^{i k r}}{r} f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)$$

$$\mid \hat{\mathbf{r}} \cdot \mathbf{J} \text { out }\left.|=| \operatorname{Re}\left(\frac{-i \hbar}{m} \Psi^* \frac{\partial}{\partial r} \Psi\right)\left|\underset{r \rightarrow \infty}{=} \frac{\hbar k}{m r^2}\right| f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)\right|^2$$

$$d \sigma=\left|f\left(\mathbf{k}, \mathbf{k}^{\prime}\right)\right|^2 d \Omega_{k^{\prime}}$$

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