1.1.4. Single Frequency Matter Perturbation

As a first step, we solve single frequency matter perturbation

δλ(x)=Asin(kx+ϕ).

Using the relation between η and δλ, we solve out η.

η(x)=ωm2xcos2θm2Akcos(kx+ϕ),

where we have chosen η(0)=cos2θm2Akcosϕ.

The problem is to solve the equation of motion

iddx(ψb1ψb2)=sin2θm2δλ(x)(0e2iη(x)e2iη(x)0)(ψb1ψb2).

We also define

(1.10)h=sin2θm2δλ(x)e2iη(x)=sin2θm2Asin(kx+ϕ)ei(ωmxAcos2θmkcos(kx+ϕ)),

so that the equation of motion becomes

iddx(ψb1ψb2)=(0hh0)(ψb1ψb2).

Obviously, the exponential terms is too complicate. On the other hand, this equation of motion reminds us of the Rabi oscillation. So we decide to rewrite the exponential into some plane wave terms using Jacobi-Anger expansion. (Refs & Notes: Patton et al)

Jacobi-Anger Expansion

One of the forms of Jacobi-Anger expansion is

(1.11)eizcos(Φ)=n=inJn(z)einΦ.

We define zk=Akcos2θm, with which we expand the term

eicos2θmAkcos(kx+ϕ)=n=inJn(zk)ein(kx+ϕ)=n=(i)nJn(zk)ein(kx+ϕ),

where I used Jn(zk)=(1)nJn(zk) for integer n.

The expansion is plugged into the Hamiltonian elements,

h=Asin2θmsin(kx+ϕ)2eiωmxn=(i)nJn(zk)ein(kx+ϕ)=Asin2θm4i(ei(kx+ϕ)ei(kx+ϕ))eiωmxn=(i)nJn(zk)ein(kx+ϕ)=Asin2θm4i(n=ei(n+1)inJn(zk)ei((n+1)kωm)xn=ei(n1)(i)nJn(zk)ei((n1)kωm)x)=Asin2θm4n=einϕ((i)n)2nzkJn(zk)ei(nkωm)x,

where I have used

Jn1(zk)+Jn+1(zk)=2nzkJn(zk).

Here comes the approximation. The most important oscillation we need is the one with largest period, which indicates the phase to be almost zero,

(1.12)(n+1)kωm0(n1)kωm0.

The two such conditions for the two summations are

nn=Int(ωmk)1nn+=Int(ωmk)+1.

We define Int(ωmk)=n0,

n=n01n+=n0+1.

The element of Hamiltonian is written as

h=Asin2θm2ein0ϕ(i)n0n0zkJn0(zk)ei(n0kωm)x.

To save keystrokes, we define

(1.13)F=Asin2θmein0ϕ(i)n0n0zkJn0(zk),

which depends on n0 and zk=Akcos2θm. Notice that

|F|2=|ktan2θmn0Jn0(zk)|2.

Thus the 12 element of the Hamiltonian is rewritten as

(1.14)h=12Fei(n0kωm)x.

Solving Using Mathematica

The Mathematica code:

In[1]:= sys = I D[{phi1[x], phi2[x]}, x] == {{0, (g0R + I g0I) Exp[ I (-omegam + n0 k) x]}, {(g0R - I g0I) Exp[-I (-omegam + n0 k) x], 0}}.{phi1[x], phi2[x]}
In[2]:= DSolve[sys, {phi1, phi2}, x]// FullSimplify
Out[3]:= {{phi1 -> Function[{x},
E^(1/2 I (k n0 + I Sqrt[-4 (g0I^2 + g0R^2) - (k n0 - omegam)^2] - omegam) x) C[1]
+ E^(1/2 (Sqrt[-4 (g0I^2 + g0R^2) - (k n0 - omegam)^2] + I (k n0 - omegam)) x) C[2]],
phi2 -> Function[{x}, (1/(2 (g0I - I g0R)))
I E^(-I (k n0 - omegam) x +
 1/2 I (k n0 + I Sqrt[-4 (g0I^2 + g0R^2) - (k n0 - omegam)^2] - omegam) x)
 (k n0 + I Sqrt[-4 (g0I^2 + g0R^2) - (k n0 - omegam)^2] - omegam) C[1]
 + (1/(2 (g0I - I g0R))) E^(1/2 (Sqrt[-4 (g0I^2 + g0R^2) - (k n0 - omegam)^2]
 + I (k n0 - omegam)) x - I (k n0 - omegam) x) (Sqrt[-4 (g0I^2 + g0R^2) - (k n0 - omegam)^2]
 + I (k n0 - omegam)) C[2]]}}

The general solution to the equation of motion is

ψb1=C1e12i(n0kωm|F|2+(n0kωm)2)x+C2e12i(n0kωm+|F|2+(n0kωm)2)xψb2=C1Fi(n0kωm|F|2+(n0kωm)2)e12i(n0kωm)x12i|F|2+(n0kωm)2+C2Fi(n0kωm+|F|2+(n0kωm)2)e12i(n0kωm)x+12i|F|2+(n0kωm)2.

For simplicity, we define

(1.15)g=n0kωm,q2=|F|2+g2.

To determine the constants, we need intial condition,

(ψ1(0)ψ2(0))=(10),

which leads to

(ψb1(0)ψb2(0))=(eiη(0)0),

using equation wavefunction in background matter basis.

Plug in the initial condition for the wave function,

C1+C2=eizk2cosϕC12Fi(gq)+C2Fi(q+g)=0.

The constants are solved out

C1=eizk2cosϕq+g2q,C2=eizk2cosϕqg2q.

where F is defined in (1.13) and l and g are defined in (1.15).

The second element of wave function becomes

ψb2(x)=Fqeizk2cosϕei2gxsin(12qx).

The transition probability becomes

P12=|ψb2|2=|F|2q2sin2(q2x),

where q is the oscillation wavenumber. Period of this oscillation is given by T=2πq.

Compare The Result with Kneller et al

Kneller et al have a transition probability

P12=κn2qn2sin2(qnx),

where qn2=kn2+κn2 and 2kn=δk~12+nk.

In my notation, k is the same as their k. After the first step of translation, we have g=2kn.

The definition of κn is given by

in Kneller’s notation and

δVee(x)=CVsin(kx+η).

So we conclude that my |F|2 is related to Kneller’s |κn|2 through

|F|2=4|κn|2.

We also have

q2=|F|2+g2=4qn2,

i.e., qn=q2.

Now we see the method we have used gives exactly the same transition probability as Kneller’s.

To make the numerical calculations easier, we rewrite the result by defining the scaled variables

x^=ωmx,k^=kωm,A^=Aωm,g^=gωm=n0k^1,q^=|F^|2+g^2=|k^tan2θmn0Jn0(zk)|+g^2,

so that n0=Round(1/k^), zk=A^k^cos2θm and

(1.16)P12=|k^tan2θmn0Jn0(zk)|2|k^tan2θmn0Jn0(zk)|2+g^2sin2(q^2x^).

1.1.4.1. Mathematical Analysis of The Result

There are several question to answer before we can understand the picture of the math.

  1. What does each term mean in the Hamiltonian?

  2. What exactly is the unitary transformation we used to rotate the wave function?

  3. What is the physical meaning of Jacobi-Anger expansion in our calculation?

To answer the first question, we need to write down the solution to Schrodinger equation assuming the Hamiltonian has only one term. The results are listed below.

Hamiltonian

Solution to The First Element of Wave Function

ωm2σ3
ψ1eiωmx/2
δλ2cos2θmσ3
ψ1eiAcos2θm2kcos(kx+ϕ)
δλ2cos2θmσ3
ψ1=C1eiAsin2θm2kcos(kx+ϕ)+C2eiAsin2θm2kcos(kx+ϕ)

The unitary transformation used is to move our reference frame to a co-rotating one. ωm2σ3 is indeed causing the wave function to rotate and removing this term using a transformation means we are co-rotating with it. δλ2cos2θmσ3 causes a more complicated rotation however it is still a rotation.

As for Jacobi-Anger expansion, it expands an oscillating matter profile to infinite constant matter potentials. To see it more clearly, we assume that δλ=λc is constant. After the unitary transformation, the effective Hamiltonian is

H=sin2θm2λc(0e2iη(x)e2iη(x)0),

where η(x)=ωm2x+cos2θm2λcx and we have chosen η(0)=0.

The 12 element of the Hamiltonian becomes

sin2θm2λce2iη(x)=sin2θm2λce2i(ωm2+cos2θm2λc)x.

The significance of it is to show that a constant matter profile will result in a simple exponential term. However, as we move on to periodic matter profile, we have a Hamiltonian element of the form

h=sin2θm2Asin(kx+ϕ)e2i(ωm2x+Acos2θm2kcos(kx+ϕ)),

as derived in equation (1.10). To compare with the constant matter case, we make a table of relevant terms in Hamiltonian.

Constant Matter Profile δλ=λc

Period Matter Profile δλ=Asin(kx+ϕ)

sin2θm2λceicos2θmλcx
Asin2θm4n=einϕ(in)2n+1zkJn(zk)ei(nkωm)x

The periodic profile comes into the exponential. Jacobi-Anger expansion (equation (1.11)) expands the periodic matter profile into infinite constant matter profiles. By comparing the two cases, we conclude that cos2θmλc corresponds to nk.

The RWA approximation we used to drop fast oscillatory terms in the summation is to find the most relevant constant matter profile per se.

The big question is which constant matter profiles are the most important ones? Mathematically, we require the phase to be almost zero, i.e. equation (1.12) or

n0kωm0,

where n0=Round(ωmk).

What is the meaning of this condition in this new basis? If we define a effective matter density out of the Jacobi-Anger expanded series, we should define it to be

λc=n0kcos2θm.

Then we can rewrite the RWA requirement as

λccos2θmωm=0.

A Reminder of MSW Resonance

The MSW Hamiltonian in flavor basis is

H=ωv2(cos2θvσ3+sin2θvσ1)+λ2σ3+ΔI,

where the MSW resonance happens when all the σ3 terms cancel eath other, i.e.,

ωvcos2θv+λ=0.

1.1.4.2. The Resonances

Questions

There are several questions to be answered.

  1. How good is the RWA approximation? What are the conditions?

  2. What can we use for other calculations?

  3. Multiple matter frequency?

Now we check the Hamiltonian again to see if we could locate some physics. In the newly defined basis and using scaled quantities

H^=(0h^h^0),

where

h^=12B^nei(nk^1)x^,

and

B^n=(i)nk^tan2θmnJn(A^k^cos2θm).

It becomes much more clearer if we plug h^ back into Hamiltonian. What we find is that

H^=n=(012B^nei(nk^1)x^12B^nei(nk^1)x^0).

With some effort, we find that the solution to the second amplitude of the wave function is

ψ2=iB^nW^ei(nk^1)x^|B^n|2sin(12(nk^1W^)x^),

where

W^=(nk^1)2+|B^n|2.

At this stage, it is quite obvious that our system is a composite Rabi oscilation system. For each specific n term we write down the hopping probability from light state to the heavy state,

PLH(n)=|B^n|2|B^n|2+(nk^1)2sin2(q^(n)2x^),

where

Γ(n)=|B^n|,width of resonance (nk^ as parameter)q^(n)=|Γ(n)|2+(nk^1)2,frequency of oscillations

Scaled Quantities

As a reminder, the scaled quantities are defined as

x^=ωmx,h^=h/ωm,B^n=Bn/ωm,k^=k/ωm,A^=A/ωm,q^=q/ωm.

Just a comment. Bn is used here because I actually want to use it for multi-frequency case and I just think Bn is better than F.

Rabi Oscillations

The form of Hamiltonian reminds of us Rabi oscillations, whose Hamiltonian is

(ω0/2αω0eiωxαω0eikxω0/2),

where ω0 is the energy gap between the two energy levels and ω is the frequency of the incoming light. To be more specific, we explain this phenomenon using two level systems.

../../_images/rabi-diagram.png

Fig. 1.13 Rabi oscillation system

The system is prepared in low energy state. When the incoming light frequency matches the energy gap between two states, we have a resonance. Otherwise, we still have the transition from low energy state to high energy state but with a smaller transition amplitude.

../../_images/rabi-oscillations.png

Fig. 1.14 Rabi oscillations for two differen incoming light frequencies. ω/ω0=1 is the resonance condition.

What we really mean by resonance is that the transition amplitude is maximized. Or the phase inside the off-diagonal element of Hamiltonian is minimized.

What’s more, we explore the transition amplitude as a function of differen incoming light frequencies.

../../_images/rabi-resonance.png

Fig. 1.15 Resonance of Rabi oscillations.

../../_images/stimulated-probability-apmlitude-vs-k.svg

Fig. 1.16 Probabiity Amplitude as a function of k/ωm within RWA, with parameters A=0.1,θm=π/5,ϕ=0.

../../_images/stimulated-probability-apmlitude-vs-k-non-RWA.svg

Fig. 1.17 Probabiity Amplitude as a function of k/ωm for each term in Jacobi-Anger expansion, with parameters A=0.1,θm=π/5,ϕ=0.

To look at the resonances I define a Mathematica function to calculate the FWHM.

Find FWHM Using Mathematica

The Mathematica code:

fwhm[n_] := First@Differences[k /. {ToRules@Reduce[amplitude[k, 0.1, Pi/5, n] == 0.5 &&k > (1 - 0.5/Exp[n]/n^2)/n && k < (1 + 0.5/Exp[n]/n^2)/n, k]}]
../../_images/stimulated-probability-apmlitude-vs-k-resonance-width.svg

Fig. 1.18 Width of the resonances for A=0.1,θm=π/5,ϕ=0.

How do we understand the resonance? Resonance width of each order of resonance (each n) should be calculated analytically.

Lorentzian Distribution

Three-parameter Lorentzian function is

fx0,σ,A(x)=1πσσ2+(xx0)2,

which has a width 2γ.

To find the exact width is hopeless since we need to inverse Bessel functions. Nonethless, we can assume that the resonance is very narrow so that |F|2 doesn’t change a lot. With the assumption, the FWHM is found be setting the amplitude to half, which is

Γ=|F^n0|=|k^tan2θmJn0(n0A^cos2θm/k^)n0|.

To verify this result, we compare it with the width found numerically from the exact amplitude.

Given this result, and equation (1.14), we infer that the coefficient in front of the phase term of 12 element in Hamiltonian is related to the width, while the the deviation from the exact resonance is given by g^=n0k^1.

Guessing The Width

Given a Hamiltonian 12 element here

h=F2ei(n0k^1)x^=F2eig^x^,

the width of the resonance is

(1.17)Γ=|Fn0|.
../../_images/stimulated-single-frequency-width-approximation-amp-point1.png

Fig. 1.19 Comparison of approximated width and numerical results for perturbation amplitude A^=Aωm=0.1.

../../_images/stimulated-single-frequency-width-approximation-amp-1.png

Fig. 1.20 Comparison of approximated width and numerical results for perturbation amplitude A^=Aωm=1.

A Special Property of Bessel Function

A special relation of Bessel function is that [Ploumistakis2009]

Jn(n\sechα)en(tanhαα)2πntanhα

for large n. As a matter of fact, for all positive α, we always have tanhαα<0.

Using this relation and defining \sechα=Acos2θm, which renders

(1.18)α=2nπi+ln(1±A2cos22θm+1Acos2θm),nIntegers,

where the Mathematica code to solve it is shown below,

In[1]:= Solve[Exp[z] + Exp[-z] == 2/(A Cos[2 Subscript[[Theta], m]]), z] // FullSimplify

we find out an more human readabale analytical expression for the width

Γ=|2k^tan2θmen(tanhαα)n02πn0tanhα|

where α is solved out in (1.18).

For small α, we have expansions for exponentials and hyperbolic functions tanhααα33,

Γ2tan2θmenα3/32παn03/2.

However, it doesn’t really help that much since n is large and no expansion could be done except for significantly small α.

Ploumistakis2009
  1. Ploumistakis, S.D. Moustaizis, I. Tsohantjis, Towards laser based improved experimental schemes for multiphoton pair production from vacuum, Physics Letters A, Volume 373, Issue 32, 3 August 2009, Pages 2897-2900, ISSN 0375-9601, http://dx.doi.org/10.1016/j.physleta.2009.06.015.

(http://www.sciencedirect.com/science/article/pii/S0375960109007166) Keywords: Pair production; Multiphoton processes; High intensity lasers

1.1.4.3. Perturbation Amplitude and Transition Probability

../../_images/pltPertAmpPertWaveNumTransitionAmp.svg

Fig. 1.21 Transition probability amplitude at different perturbation amplitude and perturbation wavenumber.


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