- #1
Niles
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Hi
When the dielectric function of a system is time-invariant, solutions of Maxwell's Equations are separable and they are usually written as (I only write the E-field)
<br /> E(r, t) = E(r) \exp(-i\omega t)<br />
Now, in my book they write an optical field as
<br /> E(t) = E_0\exp(-i\omega t) + E_0^*\exp(+i\omega t)<br />
Taking the real part of the two expressions, the time-dependence will be the same to a multiplicative factor, so all OK there. But why is it that I am allowed to neglegt the spatial part in the second way of writing the field? Is it simply because the spatial part is not a part of my Hamiltonian for the system?
Any help is appreciated.Niles.
When the dielectric function of a system is time-invariant, solutions of Maxwell's Equations are separable and they are usually written as (I only write the E-field)
<br /> E(r, t) = E(r) \exp(-i\omega t)<br />
Now, in my book they write an optical field as
<br /> E(t) = E_0\exp(-i\omega t) + E_0^*\exp(+i\omega t)<br />
Taking the real part of the two expressions, the time-dependence will be the same to a multiplicative factor, so all OK there. But why is it that I am allowed to neglegt the spatial part in the second way of writing the field? Is it simply because the spatial part is not a part of my Hamiltonian for the system?
Any help is appreciated.Niles.