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A Transition dipole moment - polarized absorption

  1. Mar 29, 2016 #1
    Hi everyone,

    I am interested how is polarized light absorbed by a molecule or an atom. Unfortunately, I come to a problem in the derivation where a complex vector in a real space appears. This is something I never seen before and I do not know how to interpret it. Therefore I would like to ask you for help about this issue.

    From the harmonic perturbation theory and the dipole approximation we obtain the transition rate between two states, [itex] \vert i> [/itex] - initial and [itex] \vert j> [/itex] - final, and this rate is governed by the following matrix element:

    [tex] |<j| \frac{e}{m} \vec{e}\cdot \vec{p}|i>|^2 [/tex]

    where e is the electron charge, m is the electron mass, [itex] \vec e [/itex] is light polarization vector (it is an unit vector, having information only about direction) and [itex] \vec p [/itex] is the electron momentum operator in the vector form, for more details see Woodgate's book.

    Now, if one assumes that [itex] \vec e = (\cos\alpha,\cos\beta,\cos\gamma) [/itex], where [itex] \alpha [/itex], [itex] \beta [/itex] and [itex] \gamma [/itex] are the angles of the [itex] \vec e [/itex] to the [itex] x [/itex], [itex] y [/itex] and [itex] z [/itex] axes of the coordinate frame and [itex] \vec p = (\hat{p}_x, \hat{p}_y, \hat{p}_z) [/itex], we then obtain the following expression:

    [tex] <j|\vec{e}\cdot \vec{p}|i> = \frac{i\Delta E}{\hbar}[\cos\alpha <j|ex|i>+\cos\beta <j|ey|i>+\cos\gamma<j|ez|i>][/tex]

    which is equal to:

    [tex] <j|\vec{e}\cdot \vec{p}|i> = \cos\alpha D_x+\cos\beta D_y+\cos\gamma D_z= \vec{e}\cdot \vec{D}[/tex]


    Where [itex]\vec{D}[/itex] is the vector proportional to the dipole moment vector.

    Now this is the point where my problems start. Namely, it is obvious that [itex] \vec{D}[/itex] is a complex vector, and I am not sure if the next expression even has a physical meaning:

    [tex]\vec{e}\cdot \vec{D} = |D|\cos\delta[/tex]

    Where [itex] \delta [/itex] is the "angle" between the polarization angle [itex]\vec{e}[/itex] (exactly known direction in the real space) and the dipole moment complex vector [itex]\vec{D}[/itex].

    This is the result that I obtain from the experiment as well, but unfortunately I am not understanding it well enough. I red some mathematical literature on the topic about angles between complex vectors, but I could not understand it very well. Therefore I wish if anyone could help me about understanding what would be the direction of the dipole moment complex vector?


    On the other hand, if we for simplicity assume that the light travels along the [itex] z [/itex] direction, and proceed with the complex value of the [itex]\vec{D}[/itex] we have:

    [tex] |<j| \frac{e}{m} \vec{e}\cdot \vec{p}|i>|^2 =[/tex]
    [tex] =(\cos\phi D_x + \sin\phi D_y)(\cos\phi D^*_x + \sin\phi D^*_y) =[/tex]
    [tex] =\cos^2 \phi |D_x|^2+ \sin^2 \phi |D_y|^2 + (D_x D^*_y+D^*_x D_y)\cos\phi \sin\phi =[/tex]
    [tex] =\cos^2 \phi |D_x|^2+ \sin^2 \phi |D_y|^2 + (D_x D^*_y+D^*_x D_y)\cos\phi \sin\phi =[/tex]
    [tex] =\cos^2 \phi |D_x|^2+ \sin^2 \phi |D_y|^2 + \sin2\phi \cos(\alpha_1-\beta_1) |D_x||D_y|[/tex]

    where [itex] D_x=|D_x| e^{i\alpha_1} [/itex], [itex] D_y=|D_y| e^{i\beta_1} [/itex] and [itex] D_x=|D_x| e^{i\gamma_1} [/itex]. By comparing this result to my measurements there is a problem with the [itex] \sin2\phi [/itex] term, but even more explicitly it is in contradiction to the previously obtained result that has only [itex] \cos\delta[/itex] dependence.

    I would really appreciate if anyone could help me about this problem or even give any kind of comment.

    Best!
     
  2. jcsd
  3. Mar 29, 2016 #2

    DrDu

    User Avatar
    Science Advisor

    Your expression is still completely general if you assume that ##\delta## is complex, too.
    Namely, ##D=D_r+i D_i## where both ##D_r## and ##D_i## are real vectors. Then ##e \cdot D=e\cdot D_r +i e\cdot |D_i| =|D_r| \cos \delta_r+i D_i \cos \delta_i = |D| (|D_r|/|D| \cos \delta_r+i |D_i|/|D|\cos \delta_i)## so that ##\cos \delta=D_r|/|D| \cos \delta_r+i |D_i|/|D|\cos \delta_i##, where ##|D|^2=|D_r|^2+|D_i|^2##.
     
    Last edited: Mar 29, 2016
  4. Mar 29, 2016 #3
    Thank you for your reply, I really appreciate it. In that case both expressions are equivalent:

    [tex] \vec{D}=\vec{D}_r+i \vec{D}_i [/tex]

    with [itex] \vec{D}_r [/itex] and [itex] \vec{D}_i [/itex] being defined as:

    [tex] \vec{D}=( |D_x|e^{i\alpha_1},|D_y|e^{i\beta_1} ,|D_z|e^{i\gamma_1})= [/tex]

    [tex] ( |D_x|\cos\alpha_1,|D_y|\cos\beta_1 ,|D_z|\cos\gamma_1)+i( |D_x|\sin\alpha_1,|D_y|\sin\beta_1 ,|D_z|\sin\gamma_1) [/tex]

    then

    [tex] \vec{e}\cdot\vec{D}= \vec{e} \cdot \vec{D}_r +i \vec{e} \cdot \vec{D}_i [/tex]

    and the absorption is directly proportional to the

    [tex] |\vec{e}\cdot\vec{D}|^2= |\vec{e} \cdot \vec{D}_r|^2 + |\vec{e} \cdot \vec{D}_i|^2[/tex]

    by using the previous definitions of the angles,

    [tex] \vec{e} \cdot \vec{D}_r = |D_x|\cos\alpha_1 \cos\alpha + |D_y|\cos\beta_1 \cos\beta+ |D_z|\cos\gamma_1 \cos\gamma[/tex]

    and

    [tex] \vec{e} \cdot \vec{D}_i = |D_x|\sin\alpha_1 \cos\alpha + |D_y|\sin\beta_1 \cos\beta+ |D_z|\sin\gamma_1 \cos\gamma[/tex]

    and after taking squares of these expressions we get:

    [tex] |\vec{e}\cdot\vec{D}|^2= |D_x|^2\cos^2\alpha + |D_y|^2\cos^2\beta + |D_z|^2\cos^2\gamma
    + 2 |D_x||D_y| \cos\alpha \cos\beta \cos(\alpha_1-\beta_1)
    + 2 |D_x||D_z| \cos\alpha \cos\gamma\cos(\alpha_1-\gamma_1)
    + 2 |D_y||D_z| \cos\beta\cos\gamma\cos(\beta_1-\gamma_1) [/tex]

    which in the case of [itex] \gamma = 90^{\circ} [/itex] is exactly the same as the last expression from my previous post.

    Then in general if a molecule or an atom is absorbing light along an arbitrary direction , the last expression in this post is the expression that I should use?

    Please correct me if I am wrong.
     
  5. Mar 29, 2016 #4

    DrDu

    User Avatar
    Science Advisor

    Sounds good.
     
  6. Mar 29, 2016 #5
    Thanks!
     
  7. Jun 28, 2016 #6
    Hi again,

    To me it seems that there is a problem with the previous derivation.

    Let us for simplicity consider a 2D case.
    [tex]B_{01}=\frac{\pi e^2}{\epsilon\hbar^2\omega^2_{01} m^2}|\langle\Psi_{S_1}|\vec{e}\cdot\vec{p}|\Psi_{S_0}\rangle|^2
    [/tex]

    and define
    [tex]
    \begin{split}
    D_r & =\langle\Psi_{S_1}|\vec{e}\cdot\vec{p}|\Psi_{S_0}\rangle \\
    & =\frac{i\Delta E}{\hbar}(\cos\phi\langle\Psi_{S_1}|x|\Psi_{S_0}\rangle+\sin\phi\langle\Psi_{S_1}|y|\Psi_{S_0}\rangle)\\
    & \propto\cos\phi D_x+\sin\phi D_y\\
    & =\vec{e}\cdot\vec{D}\\
    &=|\vec{D}|\cos\delta
    \end{split}
    [/tex]

    Since being bound states, ## |\Psi_{S_1}\rangle ## and ## |\Psi_{S_0}\rangle ## are real functions normalized with a multiplicative constant, which could be a complex number. Therefore, ## D_x ## and ## D_y ## must be real values multiplied by the same complex number, which is later taken care of with the modulus squared.

    Namely,
    ##
    \begin{split}
    \Psi_{S_0}=e^{i\alpha}|A|\psi_{S_0} \\
    \Psi_{S_1}=e^{i\beta}|B|\psi_{S_1}
    \end{split}
    ##

    Here ## \psi_{S_0} ## and ## \psi_{S_1} ## are real functions and ## e^{i\alpha}|A| ## and ## e^{i\beta}|B| ## complex normalization constants.

    ##
    \begin{split}
    D_x=e^{i(\alpha-\beta)}|A||B|\langle\psi_{S_1}|x|\psi_{S_0}\rangle\\
    D_y=e^{i(\alpha-\beta)}|A||B|\langle\psi_{S_1}|y|\psi_{S_0}\rangle
    \end{split}
    ##

    ##
    D_r=e^{i(\alpha-\beta)}|A||B|(\cos\phi\langle\psi_{S_1}|x|\psi_{S_0}\rangle+\sin\phi\langle\psi_{S_1}|y|\psi_{S_0}\rangle)
    ##

    ##
    \begin{split}
    |D_r|^2&=|A|^2|B|^2|\cos\phi\langle\psi_{S_1}|x|\psi_{S_0}\rangle+\sin\phi\langle\psi_{S_1}|y|\psi_{S_0}\rangle|^2\\
    &=|A|^2|B|^2|\cos\phi D_x+\sin\phi D_y|^2
    \end{split}
    ##

    and further,

    ##
    \begin{split}
    |D_r|^2&=|A|^2|B|^2|R|^2|\cos(\phi-\theta)|^2\\
    &=|A|^2|B|^2R^2\cos^2(\phi-\theta)
    \end{split}
    ##

    Where ## R^2=D_x^2+D_y^2## , and ## tg(\theta)=\frac{Dy}{Dx}##. Where depending on values of ## D_x## and ##D_y## we have ##\theta\in [0,2\pi]##. Although ##D_x## and ##D_y## are known up to a complex multiplicative constant, their ratio is well defined.

    What makes me suspicious of this result comes from a consideration of a symmetric molecule, where wavefunctions are either odd or even. In this case there is absolutely no argument that light absorption should be more pronounced along the angle ##\theta## than ##-\theta##. And in order to have a symmetric absorption, either ##D_x## or ##D_y## need to be zero.

    I assume that this might be some bad math from my side, but so far I did not see it.

    Later I was searching through the literature which deals with this subject, and found in paper ( Don L. Peterson, William T. Simpson, Polarized Electronic Absorption Spectrum of Amides with Assignments of Transitions, J. Am. Chem. Soc. 79 (1957) 2375-2382) the following argument:

    "Crystal spectra must be understood as involving absorption of energy out of two independent beams along the principal directions, or, equivalently, as requiring that the light be represented as a statistical ensemble having parts polarized along the two principal directions. The weights of the two streams of photons, oppositely polarized, are given by the cosine squared law. It is believed that this phenomenon is an example of a disturbance due to the possibility of there having been a “measurement” (absorption of a photon by a crystal oscillator) thus leading to the reduction of the wave function of the light."

    Results from this paper are very well fitted with this argument and therefore it seems to be experimentally validated. There are some newer results which use the same idea that a photon is being absorbed by one or the other absorption axis.

    Although this model would fit my results very well, I am a bit concerned about this argument. Namely, I am not an expert about the collapsion of the wavefunction to the basis functions, and hence I would like to ask some of you here who know much more about it to share it with me and everyone else interested in this topic.
     
    Last edited: Jun 28, 2016
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