- #1
Robert_G
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Hi, The following contains two questions that I encountered in the books of Claude Cohen-Tannoudji, "Atom-Photon Interactions" and "Atoms and Photons: Introduction to Quantum Electrodynamics". The first one is about how to calculate two Fourier transforms, and the second one is a example of which I have been confused about for a very long time. Since I am teaching myself the quantum mechanics, so the question are maybe easy for some of you.
1st.
The transform of
\frac{1}{4\pi r}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{1}{k^2}
\frac{\textbf{r}}{4 \pi r^3}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{-i\textbf{k}}{k^2}
e.g. the first one is ...
\frac{1}{4\pi r}=\frac{1}{(2\pi)^{3/2}}\int d^3 k \frac{1}{k^2}\exp(i\textbf{k}\cdot \textbf{r})
For years I just assumed that those two are correct, now I really want to know why.2nd
The example here is about the exchange the transverse photons between two charged particles. A pair of particles moves from state \textbf{p}_\alpha, \textbf{p}_\beta to the state \textbf{p}'_\alpha, \textbf{p}'_\beta by exchanging a transverse photon \mathbf{k}\mathbf{\epsilon}, here \alpha \beta indicate the two atoms, and \textbf{k} and \mathbf{\epsilon} are the wave vector and the polarization respectively. so the system goes from |\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle to |\textbf{p}''_\alpha, \mathbf{p}''_\beta;\textbf{k}\mathbf{\epsilon}\rangle and then ends at the state |\textbf{p}'_\alpha, \textbf{p}'_\beta;0\rangle.
The effective Hamiltonian is
\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta|\delta V| \textbf{p}_\alpha, \textbf{p}_\beta\rangle=\sum_{\textbf{k}\mathbf{\epsilon}}\sum_{\textbf{p}''_\alpha \textbf{p}''_\beta}\frac{1}{2}[\frac{1}{E_p-E_{p''}-\hbar\omega}+\frac{1}{E_p'-E_{p''}-\hbar\omega}]\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|H_{I1}|\mathbf{p}''_\alpha,\mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}\rangle \langle\mathbf{p}''_\alpha, \mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}|H_{I1}|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle (1)
Where
H_{I1}=-\sum_\alpha \frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)
\mathbf{A(\mathbf{r}_\alpha)}=\sum_j \sqrt{\frac{\hbar}{2\epsilon}\omega_j L^3}(\hat{a}\mathbf{\epsilon}_j e^{i \mathbf{k}_j\cdot\mathbf{r}_\alpha}+\hat{a}^{\dagger}\mathbf{\epsilon}_j e^{-i \mathbf{k}_j\cdot\mathbf{r}_\alpha})
According to the book, E_p-E_{p''} and E_{p'}-E_{p''} is much smaller than \hbar\omega, and the summation over \textbf{p}''_\alpha and \textbf{p}''_\alpha introduces a closure relation, the above equation is
\delta V=-\sum_{\mathbf{k}\mathbf{\epsilon}}\frac{1}{2\epsilon_0 L^3 \omega^2}\frac{q_\alpha q_\beta}{m_\alpha m_\beta}(\mathbf{\epsilon} \cdot \textbf{p}_\beta)(\mathbf{\epsilon} \cdot \textbf{p}_\beta)e^{i \mathbf{k} \cdot (\mathbf{r}_\alpha-\mathbf{r}_\beta)}+(\alpha\leftrightarrow\beta) (2)
Questions
(1) the state |\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle should be consider as |\textbf{p}_\alpha\rangle \otimes|\textbf{p}_\beta\rangle \otimes|0\rangle, right?
(2) I do not know how to get (2) from (1). The following is how I proceed with the calculation: Let's disregard all the constants, and calculate only the Dirac bracket: Considering the closure relation, we have
\sum_{\mathbf{p}''_\alpha \mathbf{p}''_\beta}\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|H_{I1}|\mathbf{p}''_\alpha,\mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}\rangle \langle\mathbf{p}''_\alpha, \mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}|H_{I1}|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle=\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|H_{I1}| \mathbf{k}\mathbf{\epsilon}\rangle \langle \mathbf{k}\mathbf{\epsilon}|H_{I1}|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle
and then
=\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|[\frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)+\frac{q_\beta}{m_\beta}\mathbf{p}_\beta \cdot \mathbf{A}(\mathbf{r}_\beta )]| \mathbf{k}\mathbf{\epsilon}\rangle \langle \mathbf{k}\mathbf{\epsilon}|[\frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)+\frac{q_\beta}{m_\beta}\mathbf{p}_\beta \cdot \mathbf{A}(\mathbf{r}_\beta )]|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle (3)
Now let's focus on the second Dirac braket:
\langle \mathbf{k}\mathbf{\epsilon}|[\frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)+\frac{q_\beta}{m_\beta}\mathbf{p}_\beta \cdot \mathbf{A}(\mathbf{r}_\beta )]|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle
The second term of operator \textbf{A}_\alpha can transform |0\rangle into |\textbf{k}\mathbf{\epsilon}\rangle, while the first term containing \hat{a} is zero.
But, Here is the problem, \mathbf{p}_\alpha can not transform |\textbf{p}_\alpha \rangle into |\textbf{p}'_\alpha \rangle. It should be some number times |\textbf{p}_\alpha \rangle, because |\textbf{p}_\alpha \rangle is an eigenvector of operator \mathbf{p}_\alpha , So without further calculation, the total result of Eq. (3) is zero. because |\textbf{p}_\alpha \rangle and |\textbf{p}'_\alpha \rangle are orthogonal to each other.
Of course, I am wrong, but I don't know where is the mistake. Please tell me, I am so close to kill myself.
1st.
The transform of
\frac{1}{4\pi r}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{1}{k^2}
\frac{\textbf{r}}{4 \pi r^3}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{-i\textbf{k}}{k^2}
e.g. the first one is ...
\frac{1}{4\pi r}=\frac{1}{(2\pi)^{3/2}}\int d^3 k \frac{1}{k^2}\exp(i\textbf{k}\cdot \textbf{r})
For years I just assumed that those two are correct, now I really want to know why.2nd
The example here is about the exchange the transverse photons between two charged particles. A pair of particles moves from state \textbf{p}_\alpha, \textbf{p}_\beta to the state \textbf{p}'_\alpha, \textbf{p}'_\beta by exchanging a transverse photon \mathbf{k}\mathbf{\epsilon}, here \alpha \beta indicate the two atoms, and \textbf{k} and \mathbf{\epsilon} are the wave vector and the polarization respectively. so the system goes from |\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle to |\textbf{p}''_\alpha, \mathbf{p}''_\beta;\textbf{k}\mathbf{\epsilon}\rangle and then ends at the state |\textbf{p}'_\alpha, \textbf{p}'_\beta;0\rangle.
The effective Hamiltonian is
\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta|\delta V| \textbf{p}_\alpha, \textbf{p}_\beta\rangle=\sum_{\textbf{k}\mathbf{\epsilon}}\sum_{\textbf{p}''_\alpha \textbf{p}''_\beta}\frac{1}{2}[\frac{1}{E_p-E_{p''}-\hbar\omega}+\frac{1}{E_p'-E_{p''}-\hbar\omega}]\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|H_{I1}|\mathbf{p}''_\alpha,\mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}\rangle \langle\mathbf{p}''_\alpha, \mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}|H_{I1}|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle (1)
Where
H_{I1}=-\sum_\alpha \frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)
\mathbf{A(\mathbf{r}_\alpha)}=\sum_j \sqrt{\frac{\hbar}{2\epsilon}\omega_j L^3}(\hat{a}\mathbf{\epsilon}_j e^{i \mathbf{k}_j\cdot\mathbf{r}_\alpha}+\hat{a}^{\dagger}\mathbf{\epsilon}_j e^{-i \mathbf{k}_j\cdot\mathbf{r}_\alpha})
According to the book, E_p-E_{p''} and E_{p'}-E_{p''} is much smaller than \hbar\omega, and the summation over \textbf{p}''_\alpha and \textbf{p}''_\alpha introduces a closure relation, the above equation is
\delta V=-\sum_{\mathbf{k}\mathbf{\epsilon}}\frac{1}{2\epsilon_0 L^3 \omega^2}\frac{q_\alpha q_\beta}{m_\alpha m_\beta}(\mathbf{\epsilon} \cdot \textbf{p}_\beta)(\mathbf{\epsilon} \cdot \textbf{p}_\beta)e^{i \mathbf{k} \cdot (\mathbf{r}_\alpha-\mathbf{r}_\beta)}+(\alpha\leftrightarrow\beta) (2)
Questions
(1) the state |\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle should be consider as |\textbf{p}_\alpha\rangle \otimes|\textbf{p}_\beta\rangle \otimes|0\rangle, right?
(2) I do not know how to get (2) from (1). The following is how I proceed with the calculation: Let's disregard all the constants, and calculate only the Dirac bracket: Considering the closure relation, we have
\sum_{\mathbf{p}''_\alpha \mathbf{p}''_\beta}\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|H_{I1}|\mathbf{p}''_\alpha,\mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}\rangle \langle\mathbf{p}''_\alpha, \mathbf{p}''_\beta; \mathbf{k}\mathbf{\epsilon}|H_{I1}|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle=\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|H_{I1}| \mathbf{k}\mathbf{\epsilon}\rangle \langle \mathbf{k}\mathbf{\epsilon}|H_{I1}|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle
and then
=\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0|[\frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)+\frac{q_\beta}{m_\beta}\mathbf{p}_\beta \cdot \mathbf{A}(\mathbf{r}_\beta )]| \mathbf{k}\mathbf{\epsilon}\rangle \langle \mathbf{k}\mathbf{\epsilon}|[\frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)+\frac{q_\beta}{m_\beta}\mathbf{p}_\beta \cdot \mathbf{A}(\mathbf{r}_\beta )]|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle (3)
Now let's focus on the second Dirac braket:
\langle \mathbf{k}\mathbf{\epsilon}|[\frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)+\frac{q_\beta}{m_\beta}\mathbf{p}_\beta \cdot \mathbf{A}(\mathbf{r}_\beta )]|\mathbf{p}_\alpha,\mathbf{p}_\beta; 0\rangle
The second term of operator \textbf{A}_\alpha can transform |0\rangle into |\textbf{k}\mathbf{\epsilon}\rangle, while the first term containing \hat{a} is zero.
But, Here is the problem, \mathbf{p}_\alpha can not transform |\textbf{p}_\alpha \rangle into |\textbf{p}'_\alpha \rangle. It should be some number times |\textbf{p}_\alpha \rangle, because |\textbf{p}_\alpha \rangle is an eigenvector of operator \mathbf{p}_\alpha , So without further calculation, the total result of Eq. (3) is zero. because |\textbf{p}_\alpha \rangle and |\textbf{p}'_\alpha \rangle are orthogonal to each other.
Of course, I am wrong, but I don't know where is the mistake. Please tell me, I am so close to kill myself.
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