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## Main Question or Discussion Point

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.

The transform of

[itex]\frac{1}{4\pi r}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{1}{k^2}[/itex]

[itex]\frac{\textbf{r}}{4 \pi r^3}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{-i\textbf{k}}{k^2}[/itex]

e.g. the first one is ...

[itex]\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})[/itex]

For years I just assumed that those two are correct, now I really want to know why.

The example here is about the exchange the transverse photons between two charged particles. A pair of particles moves from state [itex]\textbf{p}_\alpha[/itex], [itex]\textbf{p}_\beta[/itex] to the state [itex]\textbf{p}'_\alpha[/itex], [itex]\textbf{p}'_\beta[/itex] by exchanging a transverse photon [itex]\mathbf{k}\mathbf{\epsilon}[/itex], here [itex]\alpha[/itex] [itex]\beta[/itex] indicate the two atoms, and [itex]\textbf{k}[/itex] and [itex]\mathbf{\epsilon}[/itex] are the wave vector and the polarization respectively. so the system goes from [itex]|\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle[/itex] to [itex]|\textbf{p}''_\alpha, \mathbf{p}''_\beta;\textbf{k}\mathbf{\epsilon}\rangle[/itex] and then ends at the state [itex]|\textbf{p}'_\alpha, \textbf{p}'_\beta;0\rangle[/itex].

The effective Hamiltonian is

[itex]\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta|\delta V| \textbf{p}_\alpha, \textbf{p}_\beta\rangle[/itex]=[itex]\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[/itex] (1)

Where

[itex]H_{I1}=-\sum_\alpha \frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)[/itex]

[itex]\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})[/itex]

According to the book, [itex]E_p-E_{p''}[/itex] and [itex]E_{p'}-E_{p''}[/itex] is much smaller than [itex]\hbar\omega[/itex], and the summation over [itex]\textbf{p}''_\alpha[/itex] and [itex]\textbf{p}''_\alpha[/itex] introduces a closure relation, the above equation is

[itex]\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)[/itex] (2)

(1) the state [itex]|\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle[/itex] should be consider as [itex]|\textbf{p}_\alpha\rangle \otimes|\textbf{p}_\beta\rangle \otimes|0\rangle[/itex], 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

[itex]\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[/itex]

and then

[itex]=\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[/itex] (3)

Now let's focus on the second Dirac braket:

[itex]\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[/itex]

The second term of operator [itex]\textbf{A}_\alpha[/itex] can transform [itex]|0\rangle [/itex] into [itex]|\textbf{k}\mathbf{\epsilon}\rangle [/itex], while the first term containing [itex]\hat{a}[/itex] is zero.

But, Here is the problem, [itex]\mathbf{p}_\alpha[/itex] can not transform [itex]|\textbf{p}_\alpha \rangle[/itex] into [itex]|\textbf{p}'_\alpha \rangle[/itex]. It should be some number times [itex] |\textbf{p}_\alpha \rangle[/itex], because [itex]|\textbf{p}_\alpha \rangle[/itex] is an eigenvector of operator [itex]\mathbf{p}_\alpha[/itex] , So without further calculation, the total result of Eq. (3) is zero. because [itex]|\textbf{p}_\alpha \rangle[/itex] and [itex]|\textbf{p}'_\alpha \rangle[/itex] 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

[itex]\frac{1}{4\pi r}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{1}{k^2}[/itex]

[itex]\frac{\textbf{r}}{4 \pi r^3}\leftrightarrow\frac{1}{(2\pi)^{3/2}}\frac{-i\textbf{k}}{k^2}[/itex]

e.g. the first one is ...

[itex]\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})[/itex]

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 [itex]\textbf{p}_\alpha[/itex], [itex]\textbf{p}_\beta[/itex] to the state [itex]\textbf{p}'_\alpha[/itex], [itex]\textbf{p}'_\beta[/itex] by exchanging a transverse photon [itex]\mathbf{k}\mathbf{\epsilon}[/itex], here [itex]\alpha[/itex] [itex]\beta[/itex] indicate the two atoms, and [itex]\textbf{k}[/itex] and [itex]\mathbf{\epsilon}[/itex] are the wave vector and the polarization respectively. so the system goes from [itex]|\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle[/itex] to [itex]|\textbf{p}''_\alpha, \mathbf{p}''_\beta;\textbf{k}\mathbf{\epsilon}\rangle[/itex] and then ends at the state [itex]|\textbf{p}'_\alpha, \textbf{p}'_\beta;0\rangle[/itex].

The effective Hamiltonian is

[itex]\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta|\delta V| \textbf{p}_\alpha, \textbf{p}_\beta\rangle[/itex]=[itex]\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[/itex] (1)

Where

[itex]H_{I1}=-\sum_\alpha \frac{q_\alpha}{m_\alpha}\mathbf{p}_\alpha \cdot \mathbf{A}(\mathbf{r}_\alpha)[/itex]

[itex]\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})[/itex]

According to the book, [itex]E_p-E_{p''}[/itex] and [itex]E_{p'}-E_{p''}[/itex] is much smaller than [itex]\hbar\omega[/itex], and the summation over [itex]\textbf{p}''_\alpha[/itex] and [itex]\textbf{p}''_\alpha[/itex] introduces a closure relation, the above equation is

[itex]\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)[/itex] (2)

**Questions**(1) the state [itex]|\textbf{p}_\alpha, \textbf{p}_\beta;0\rangle[/itex] should be consider as [itex]|\textbf{p}_\alpha\rangle \otimes|\textbf{p}_\beta\rangle \otimes|0\rangle[/itex], 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

[itex]\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[/itex]

and then

[itex]=\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[/itex] (3)

Now let's focus on the second Dirac braket:

[itex]\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[/itex]

The second term of operator [itex]\textbf{A}_\alpha[/itex] can transform [itex]|0\rangle [/itex] into [itex]|\textbf{k}\mathbf{\epsilon}\rangle [/itex], while the first term containing [itex]\hat{a}[/itex] is zero.

But, Here is the problem, [itex]\mathbf{p}_\alpha[/itex] can not transform [itex]|\textbf{p}_\alpha \rangle[/itex] into [itex]|\textbf{p}'_\alpha \rangle[/itex]. It should be some number times [itex] |\textbf{p}_\alpha \rangle[/itex], because [itex]|\textbf{p}_\alpha \rangle[/itex] is an eigenvector of operator [itex]\mathbf{p}_\alpha[/itex] , So without further calculation, the total result of Eq. (3) is zero. because [itex]|\textbf{p}_\alpha \rangle[/itex] and [itex]|\textbf{p}'_\alpha \rangle[/itex] 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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