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Function multiplied by its complex conjugate

  1. May 27, 2013 #1
    Hi Guys,

    I have two questions which kind of relate. The first relates to the complex conjugate of a function. Specifically, When a function is multiplied by its complex conjugate, what does that mean physically?

    For instance, I am reading a book on electromagnetic wave scattering, and often they write the

    [itex]I = E E^{\ast}[/itex]

    Meaning the intensity is equal to the electromagnetic wave multiplied by its complex conjugate. Can someone just say in real simple terms what the em wave multiplied by its complex conjugate means?

    Secondly, which leads on from the first question, I have a forumla that says

    [itex]E(q,0) E^{\ast} (q,t) [/itex]

    Which is the electric field (a function of scattering vector q and time) multiplied by its complex conjugate. What is confusing me is what does the 'zero' mean in the brackets of the first E?

    Thanks for all your help
  2. jcsd
  3. May 27, 2013 #2


    User Avatar

    Hello Steve Drake,

    What a function multiplied by its complex conjugate means will really depend on the physical meaning of the function you started with. In Quantum Mechanics, for example, it may mean the probability density of finding a particle in a given state if the function under consideration is the wave function of that particle; in Eletromagnetism, you can use the complex notation to simplify your calculations and get, for example, the intensity of an eletromagnetic wave by multiplying the complex field by its conjugate.

    Focusing on your problem now, there is one important fact that usually is not openly stated in the literature: the fact that because Maxwell's Equations are all linear, you can solve them for complex functions as well - and, therefore, the real physical solution will be just the real part of the complex solution. Mathematically, we have the Maxwell's Equations:

    [itex]\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_{o}}[/itex]

    [itex]\nabla\cdot\mathbf{B} = 0[/itex]

    [itex]\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}[/itex]

    [itex]\nabla\times\mathbf{B} = \mu_{o} \big ( \mathbf{J} + \epsilon_{o}\frac{\partial\mathbf{E}}{\partial t} \big )[/itex]

    Now, if we define the Complex Fields [itex]\mathbf{\tilde{E}}[/itex] and [itex]\mathbf{\tilde{B}}[/itex] as

    [itex]\mathbf{\tilde{E}} = \mathbf{E}_{r} + i\mathbf{E}_{i}[/itex]

    [itex]\mathbf{\tilde{B}} = \mathbf{B}_{r} + i\mathbf{B}_{i}[/itex]

    Where [itex]\mathbf{E}_{r}[/itex], [itex]\mathbf{E}_{i}[/itex], [itex]\mathbf{B}_{r}[/itex] and [itex]\mathbf{B}_{i}[/itex] are fields that satisfy the Maxwell's Equations written above. Now, because Maxwell's Equations are linear, it is easy to see that the fields [itex]\mathbf{\tilde{E}}[/itex] and [itex]\mathbf{\tilde{B}}[/itex] will also satisfy those equations:

    \begin{array}{l l}
    \nabla\cdot\mathbf{E}_{r} = \frac{\rho_{r}}{\epsilon_{o}} & \quad \nabla\cdot\mathbf{E}_{i} = \frac{\rho_{i}}{\epsilon_{o}} \\
    \nabla\cdot\mathbf{B}_{r} = 0 & \quad \nabla\cdot\mathbf{B}_{i} = 0 \\
    \nabla\times\mathbf{E}_{r} = -\frac{\partial\mathbf{B}_{r}}{\partial t} & \quad \nabla\times\mathbf{E}_{i} = -\frac{\partial\mathbf{B}_{i}}{\partial t} \\
    \nabla\times\mathbf{B}_{r} = \mu_{o} \big ( \mathbf{J}_{r} + \epsilon_{o}\frac{\partial\mathbf{E}_{r}}{\partial t} \big ) & \quad \nabla\times\mathbf{B}_{i} = \mu_{o} \big ( \mathbf{J}_{i} + \epsilon_{o}\frac{\partial\mathbf{E}_{i}}{\partial t} \big )

    Multiplying the set of equations on the right by the complex unity [itex]i[/itex] and summing the equations line by line, we arrive at:

    \begin{array}{l l}
    \nabla\cdot\mathbf{\tilde{E}} = \frac{\tilde{\rho}}{\epsilon_{o}} \\
    \nabla\cdot\mathbf{\tilde{B}} = 0 \\
    \nabla\times\mathbf{\tilde{E}} = -\frac{\partial\mathbf{\tilde{B}}}{\partial t} \\
    \nabla\times\mathbf{\tilde{B}} = \mu_{o} \big ( \mathbf{\tilde{J}} + \epsilon_{o}\frac{\partial\mathbf{\tilde{E}}}{\partial t} \big )

    Where we have defined [itex]\tilde{\rho} = \rho_{r} + i\rho_{i}[/itex] and [itex]\mathbf{\tilde{J}} = \mathbf{J}_{r} + i\mathbf{J}_{i}[/itex] as the complex charge density and complex current density, respectively.

    Having said that, let's analyse the wave equation satisfied by the Complex Fields in a region where there are no charges and currents. The derivation of the wave equation is analogous to that of the Real Fields, and we can write:

    [itex]\nabla^2\mathbf{\tilde{E}} - \frac{1}{c^2}\frac{\partial^2 \mathbf{\tilde{E}}}{\partial t^2} = \mathbf{0}[/itex]

    [itex]\nabla^2\mathbf{\tilde{B}} - \frac{1}{c^2}\frac{\partial^2 \mathbf{\tilde{B}}}{\partial t^2} = \mathbf{0}[/itex]

    Now, it is clear that a possible solution for the first of these equations is the plane wave expression, given by:

    [itex]\mathbf{\tilde{E}}(\mathbf{r}, t) = \mathbf{\tilde{E}_{o}}\exp{i(\mathbf{k}\cdot\mathbf{r}-\omega t)} \quad[/itex] with [itex]\quad \mathbf{\tilde{E}_{o}} = E_{o}\exp{(i\varphi)} \quad[/itex] and [itex]\quad k = \omega/c[/itex]

    Given our previous discussion, it is nice to see that for the plane wave we have

    [itex]\mathbf{E}_{r}(\mathbf{r}, t) = Re( \mathbf{\tilde{E}}(\mathbf{r}, t) ) = E_{o}\cos{(\mathbf{k}\cdot\mathbf{r}-\omega t + \varphi)} \quad[/itex] and

    [itex]\mathbf{E}_{i}(\mathbf{r}, t) = Im( \mathbf{\tilde{E}}(\mathbf{r}, t) ) = E_{o}\sin{(\mathbf{k}\cdot\mathbf{r}-\omega t + \varphi)}[/itex]

    Notice that here we get a complex exponential instead of a linear combination of sines and cosines. This is the underlying reason why physicists prefer to work with complex numbers instead of real numbers - it is much easier to manipulate exponentials, which have simple multiplication properties; than to manipulate a lengthy combination of sines and cosines with properties that are a bunch of complicated trigonometric identities.

    Knowing about the complex notation now makes every kind of calculation easier. The average of quantities can be calculated in a much straight forward way, as well as the intensity of the wave. Below I write the expressions for the case of the time average of the Poynting Vector and the intesity of the wave:

    [itex]\langle\mathbf{S}\rangle = \frac{Re( \mathbf{\tilde{E}} ) \times Re( \mathbf{\tilde{B}} )}{\mu_{o}} = \cdots = \frac{Re( \mathbf{\tilde{E}} \times \mathbf{\tilde{B}^*} )}{\mu_{o}}[/itex]

    [itex]I \sim E_{o}^2 = E_{o}\exp{i(\mathbf{k}\cdot\mathbf{r}-\omega t + \varphi)}E_{o}\exp{-i(\mathbf{k}\cdot\mathbf{r}-\omega t + \varphi)} = \mathbf{\tilde{E}}\mathbf{\tilde{E}^*}[/itex]

    So, basically, the complex notation makes it easier to write things! There is no new physics associated with it, it is just a more convenient way of writing the expressions so that you don't have carry sines and cosines into every calculation.

    Last edited: May 27, 2013
  4. May 28, 2013 #3
    Dude, huge response. Thanks a lot for that deep insight!
  5. Jul 10, 2015 #4
    Thanks Zag!
  6. Jul 10, 2015 #5
    another way to look at E*E: its the amplitude squared of the real wave E = Eo cos(k⋅r−ωt+φ) which is easily seen if E'(r,t)=E'o expi(k⋅r−ωt) then E'E'* = E0^2

    for you second equation, where is from? I'm guessing your squaring an electric field which was solved using a green's function.
    where ever the equation came from if you have an average like <E*(t) E(t') > (this might represent the scattering amplitude or something), this is a correlation between the value of the electric field at different times (which would be important in a scattering problem)

    To answer your question the correlation: C(t - t') = <E*(t) E(t') > will only depend on how much you waited, and not on the absolute values of the times, so you might as well set t' to zero and get the same answer.
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