Register to reply 
Two Fourier transforms and the calculation of Effective Hamiltonian. 
Share this thread: 
#1
May1614, 07:52 AM

P: 36

Hi, The following contains two questions that I encountered in the books of Claude CohenTannoudji, "AtomPhoton 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 [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_pE_{p''}\hbar\omega}+\frac{1}{E_p'E_{p''}\hbar\omega}]\langle\mathbf{p}'_\alpha, \mathbf{p}'_\beta;0H_{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_pE_{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 \otimes0\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;0H_{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;0H_{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. 


#2
May1614, 08:56 AM

P: 3,243

For your first question, you get the second transform of [tex]\frac{\textbf{r}}{4 \pi r^3}[/tex] by taking [tex]\nabla_r[/tex] on your first relation for [tex]\frac{1}{4\pi r}[/tex].
BTW, the fourier transform has a minus sign in exp(ik\cdot r). As for the first just use the fact that [tex]d^3k = k^2 \sin(\theta) d\theta d\phi dk[/tex] and [tex] k\cdot r = kr cos(\theta)[/tex] the limits of integration are obvious k goes from zero to infinity, theta from 0 to pi and phi from zero to 2pi. Now because [tex]\sin(\theta)d\theta = d(\cos \theta)[/tex] First integrate this integral which is from 1 to 1. You are left with an integrand of the form: [tex] [exp(ikr)exp(ikr)]/(ikr) [/tex] which is an integral of sinc function. I leave it to you to rearrange all the factors. 


#3
May1714, 02:17 AM

P: 36




#4
May1814, 05:48 AM

Sci Advisor
P: 3,626

Two Fourier transforms and the calculation of Effective Hamiltonian.
I think one should also mention, that the Fourier transform of 1/r has to be understood as a distribution.



#5
May1814, 09:45 AM

P: 27




#6
May1814, 11:27 AM

P: 14

As a total layman I stumble straight into the discussion with a basic question.
Can somebody in simple words explain what is the phase in Fourier analysis. I understand it has something to do with angles. But visionary it is incomprehensible for me. 


#7
May1814, 12:03 PM

P: 27




#8
May2014, 08:18 PM

P: 36

For the second question, I think I understand it. The author of the book is not trying to calculate the element of the effective Hamiltonian, but focusing on how to represent the operator of the effective Hamiltonian in the complex numbers.
PS: I am the one who start this post. 


Register to reply 
Related Discussions  
Fourier transforms  Advanced Physics Homework  3  
Fourier transforms  Calculus  2  
From Fourier Series to Fourier Transforms  Calculus  5  
Need help to find application of the Fourier series and Fourier Transforms!  Introductory Physics Homework  8  
Fourier Transforms  Calculus & Beyond Homework  0 