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Two Fourier transforms and the calculation of Effective Hamiltonian.

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Robert_G
#1
May16-14, 07:52 AM
P: 36
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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MathematicalPhysicist
#2
May16-14, 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.
Robert_G
#3
May17-14, 02:17 AM
P: 36
Quote Quote by MathematicalPhysicist View Post
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

DrDu
#4
May18-14, 05:48 AM
Sci Advisor
P: 3,596
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.
bsmile
#5
May18-14, 09:45 AM
P: 27
Quote Quote by DrDu View Post
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?
Discman
#6
May18-14, 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.
bsmile
#7
May18-14, 12:03 PM
P: 27
Quote Quote by Discman View Post
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
Robert_G
#8
May20-14, 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.


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