Two Fourier transforms and the calculation of Effective Hamiltonian.

In summary, the conversation discusses two questions about Fourier transforms and an example related to quantum mechanics. The first question involves calculating two Fourier transforms and the second question deals with an example of exchanging transverse photons between two charged particles. The conversation also includes a detailed explanation and calculation process for obtaining the second transform of \frac{\textbf{r}}{4 \pi r^3}.
  • #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
[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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  • #2
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.
 
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  • #3
MathematicalPhysicist said:
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.

thanks
 
  • #4
I think one should also mention, that the Fourier transform of 1/r has to be understood as a distribution.
 
  • #5
DrDu said:
I think one should also mention, that the Fourier transform of 1/r has to be understood as a distribution.

Could you please elaborate on this? Does a distribution require normalization?
 
  • #6
phase

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
Discman said:
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.

My two cents. Mathematics does not call for physical interpretation. It is based on operation and why this operation works traces back to its derivation of orthogonality and completeness. Now back to your question. I don't know what is an overall phase for Fourier analysis. But if you are asking about the exp(ik.r) prefactor, then it might have a physical meaning of planewave. In the perspective of angles as you pursue, it points everywhere onto a unit sphere from the origin point for fixed r if expanding to different k. In terms of complex analysis, each Fourier component does contribute an additional phase to f(r) for different k
 
  • #8
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.
 

1. What is a Fourier transform and why is it important in scientific research?

A Fourier transform is a mathematical operation that decomposes a signal or function into its constituent frequencies. It is important in scientific research because it allows us to analyze and understand complex signals and systems, and is used in a wide range of fields such as physics, engineering, and data analysis.

2. How are two Fourier transforms used in the calculation of an Effective Hamiltonian?

In the calculation of an Effective Hamiltonian, two Fourier transforms are used to transform from the time domain to the energy domain and then back to the time domain. This allows for the separation of the high and low energy components of a system, making it easier to analyze and calculate the effective behavior.

3. What is the significance of an Effective Hamiltonian in quantum mechanics?

An Effective Hamiltonian is a simplified version of the full Hamiltonian that describes the behavior of a system. It is significant in quantum mechanics as it allows for the prediction of the system's behavior without having to consider all of the complex interactions and variables that make up the full Hamiltonian.

4. How is the Effective Hamiltonian used in practical applications?

The Effective Hamiltonian is used in practical applications such as in the study of condensed matter, quantum computing, and materials science. It can also be used to model and predict the behavior of complex systems, making it a valuable tool in various scientific and technological fields.

5. Are there any limitations to using the Effective Hamiltonian method?

While the Effective Hamiltonian method is a useful and widely used approach, it does have some limitations. It may not accurately describe the behavior of systems with strong interactions or at extremely low temperatures, and it relies on certain simplifying assumptions that may not always hold true. Additionally, the calculation of an Effective Hamiltonian can be computationally intensive and time-consuming.

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