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4 nonlinear optics problems: susceptibility, polarization

  1. Jan 27, 2013 #1
    [Wasn't sure if each problem needed a separate post. Please feel free to edit if needed.]
    Also \~ and \^ are tilde and hat respectively.

    1a. The problem statement, all variables and given/known data

    Use perturbation theory to derive the 3rd order nonlinear susceptibility[itex] \chi^{(3)}(3w;w,w,w)[/itex] (problem gives potential energy, etc. but I already know what I have to do, I just need help calculating it)


    1b. Relevant equations

    The equation I need to solve is this: [itex]\ddot{\tilde{x}}^{(3)}+2\gamma \dot{\tilde{x}}^{(3)} + w_0^2 \tilde{x}^{(3)} +2a\~x^{(1)}\~x^{(2)}=0[/itex]

    I need to solve it for [itex]\~x^{(3)}[/itex].

    The first-order and second-order solutions were given in lecture. I plugged them into the equation, in the 4th term on the left hand side, getting some horrendous expression.

    We are looking for solutions of the form (*) [itex]\~x^{(3)}(t)=x^{(3)}(3w)e^{-3iwt}[/itex] (is that the right form of the solution?) Since the differential equation has derivatives with respect to time, I guess I need to differentiate (*) twice, but is it only the exponential that depends on time? What about the x, and omega? Are those constants?

    Also, E times its complex conjugate is |E|^2, right?


    1c. The attempt at a solution

    (explained in section 1b)



    --------------------------------------


    2a. The problem statement, all variables and given/known data

    Nonlinear crystal has an EM wave propagating in x direction, linear polarization along [itex]1/\sqrt{2}(\^y+\^z)[/itex] direction, frequency w, intensity 1MW/cm2. The second order non-linear optical susceptibility tensor for second harmonic generation has only one nonzero component, [itex]\chi^{(2)}_{zzz}(2w,w,w)=10pm/V (10^{-11} m/V)[/itex]
    - calculate amplitude and direction of nonlinear polarization at frequency 2w. (use Poynting vector to get E field, be careful with geometry and various factors of 2)
    - calculate amplitude and direction of linear polarization P^(1) at frequency w


    2b. Relevant equations
    k vector is in x direction.
    Poynting vector is [itex]\vec{S}=\vec{E} \times \vec{H}[/itex]. But I thought there was no magnetic field in optical materials at optical frequencies, so when taking the cross product wouldn't everything just go to 0? I wasn't given any magnetic field info. I guess the intensity is what I'd plug in for S.
    I also know the formulas for second-order susceptibility and polarization: [itex]\chi^{(2)}(2w)=-\frac{a(e/m)^2 E^2}{D(2w)D^2(w)},\ \~P^{(2)}=\epsilon_0 \chi^{(2)}\~E^2(t)[/itex]

    2c. The attempt at a solution
    given in section 2b


    --------------------------------------

    3a. The problem statement, all variables and given/known data

    Write all elements of d_il matrix (3x6) for lithium niobate (crystal symmetry 3m)
    (values given: d_33, d_31=d_15, and d_22.)

    3b. Relevant equations
    (from table 1.5.1, Boyd)
    Form of the 2nd order susceptibility tensor. Each element denoted by Cartesian indices
    For 3m crystal class: xzx=yzy, xxz=yyz, zxx=zyy,zzz,yyy=-yxx=-xxy=-xyx (mirror plane perpendicular to x^)

    To convert d_ijk to d_il:
    http://snag.gy/kIrzU.jpg

    3c. The attempt at a solution

    The answer is supposed to be this:
    http://snag.gy/WlbCk.jpg
    But I don't know how to get that from the Cartesian indices given above. maybe someone could please do the first one as an example, (xzx=yzy) and then I could figure out the rest?
    Also, when it says "zxx=zyy,zzz,yyy=-yxx=-xxy=-xyx" is that one big equation?


    --------------------------------------

    4a. The problem statement, all variables and given/known data

    Calculate the d_eff value for second harmonic generation in a nonlinear crystal with only one nonlinear coefficient d_33=10pm/V. Input beam at frequency w incident along x-axis, polarized along [itex]1/\sqrt{2}(\^y+\^z)[/itex] and:
    - (case a) emission at frequency 2w along x-axis and polarized along ^z direction
    - (case b) polarized along ^y direction
    - what is the physical meaning of the value of d_eff in case b?

    4b. Relevant equations

    For SHG, [itex]P(2w)=2\epsilon_0 d_{eff}E(w)^2[/itex], and
    http://snag.gy/ALUI9.jpg

    4c. The attempt at a solution

    We're told that d_il has only one nonzero component (d_33) so the right hand side of the matrix equation reduces to [itex]2\epsilon_0 d_{33}E_z(w)^2[/itex], right? But it seems that P and E are both unknown, so how can we solve for d_eff?
     
  2. jcsd
  3. Jan 30, 2013 #2
    For problem 1, the 3rd order equation becomes [itex]-27w^3x^{(3)}e^{-3iwt}-18\gamma iw^2 x^{(3)} e^{-3iwt}=(2a)(\frac{e}{m})(\frac{2E}{D(w)}e^{-iwt}+\frac{2E}{D(w)}e^{iwt})(-2a)(\frac{e}{m})^2 (\frac{2E^2}{D(0)D(w)D(-w)}+\frac{E^2}{D(2w)D^2(w)}e^{-i2wt}+\frac{E^2}{D(2w)D(w)D(-w)}e^{-i2wt}+\frac{E^2}{D(0)D(w)D(-w)})[/itex].

    What I'm supposed to get from that is: [itex]x^{(3)}(3w;w,w,w)=\frac{-4a^2(\frac{e}{m})^3 E^3}{D(3w)D(w)}[/itex]
    But I'm not sure how to get there. Specifically, how to cancel out all the exponentials and omegas, and create D(3w) and get rid of D(0) and D(w).

    [edit: I should mention that [itex] D(w_j)=w_0^2-w_j-2iw_j \gamma [/itex]

    ----------------

    For problem 2, I got the direction as 1/sqrt(2) x^ by doing the cofactor thing, is it right?
    [itex]\begin{matrix}
    \^x & \^y & \^z\\
    0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
    0 & 0 & 1
    \end{matrix}[/itex]

    Also, we need chi1 but the problem gives chi2 but not chi1. According to Miller's rule, we can get chi1 from chi2 but it seems to be only for sum-frequency generation. Is it valid for second harmonic generation as well? Would the constant be the same? I'll paste the lecture slide excerpt as soon as I can. The image hosting site is acting up.
     
    Last edited: Jan 30, 2013
  4. Jan 30, 2013 #3
    Actually for problem 2, according to the professor the problem provides enough information to
    deduce the exact value of chi1 at a given frequency. Miller's rule shouldn't be needed. I'm still stumped though.
     
  5. Jan 30, 2013 #4
    For problem 4, the equation is this:
    [itex]\chi_{eff}=\^e \cdot \underset{\chi}{\leftrightarrow} \cdot \^e_1 \^e_2 = \^z \cdot 10pm/V \cdot \frac{1}{\sqrt{2}}(\^y +\^z) \cdot \frac{1}{\sqrt{2}}(\^y+\^z)[/itex]
    But how do you take a dot product of directions? Also, any hints on what the physical meaning might be?
     
    Last edited: Jan 30, 2013
  6. Apr 10, 2013 #5
    good thread!helpful to me!thanks!
     
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