Derive the form of the fields for TEM waves in a waveguide In my book it says that for TEM waves in a waveguide, with: $$\textbf E = \textbf E_0(x,y)e^{i(kz-\omega t)}$$ and $$\textbf H = \textbf H_0(x,y)e^{i(kz-\omega t)},$$ where $z$ is the direction of the waveguide, that's the longitudinal direction, $$\nabla^2_{\perp}\textbf E=0 $$ and $$\nabla^2_{\perp}\textbf H=0 ,$$ should hold. $\perp$ denotes the transverse direction (x,y).
My question: How do I derive this? 
 A: For the dielectric parts of the waveguide, make an ansatz of the form:
$$\mathbf{E}(x,\,y,\,z,\,t) = -E_z(z,\,t)\,\boldsymbol{\nabla} \phi(x,\,y)$$
$$\mathbf{H}(x,\,y,\,z,\,t) = -H_z(z,\,t)\,\boldsymbol{\nabla} \psi(x,\,y)$$
into the Faraday and Ampère laws. Note that the two Gauss laws then force $\nabla^2\phi=\nabla^2\psi=0$. You should now have:
$$\partial_z H_z\;\hat{\mathbf{z}}\times \boldsymbol{\nabla}\psi =\epsilon\,\partial_t E_z\;\boldsymbol{\nabla}\phi$$
$$\partial_z E_z\;\hat{\mathbf{z}}\times \boldsymbol{\nabla}\phi =-\mu_0\,\partial_t H_z\;\boldsymbol{\nabla}\psi$$
Form $\hat{\mathbf{z}}\times$ of both sides of either equation; note that $\hat{\mathbf{z}}\times(\hat{\mathbf{z}}\times\mathbf{A})$ for a transverse $\mathbf{A}$ is simply $-\mathbf{A}$. You now have a way to eliminate $\boldsymbol{\nabla}\phi$ and  $\boldsymbol{\nabla}\psi$ from both equations and, at the same time, derive the one dimensional wave equation for both $E_z$ and $H_z$.
You can now substitude fields of the form $a\,f^+(z-c\,t)+b\,f^-(z+c\,t)$ back into your equations and prove that solutions of Maxwell's equations with the form assumed in the ansatz exist if:
$$E_z(z, t) = f_+(z - c\,t) + f_-(z + c\,t)$$
$$H_z(z, t) = \sqrt{\frac{\epsilon_0}{\mu_0}}\left(f_+(z - c\,t) - f_-(z + c\,t)\right)$$
$$\hat{\mathbf{z}}\times \boldsymbol{\nabla}\phi = \boldsymbol{\nabla}\psi$$
$$\nabla^2\phi=\nabla^2\psi=0$$
where $f^+$ and $f^-$ are arbitrary $C^2$ functions.
A very neat and elegant way to express the above conditions is to represent the transverse fields as complex number fields. Then the above conditions are all summarized by requiring that the fields should be of the form:
$$\mathbf{E}(\zeta,\,z,\,t)= \left(\frac{\mathrm{d} \Psi}{\mathrm{d} \zeta}\right)^* \left(f_+(z - c\,t) + f_-(z + c\,t)\right)$$
$$\mathbf{H}(\zeta,\,z,\,t) = -i \sqrt{\frac{\epsilon_0}{\mu_0}} \left(\frac{\mathrm{d} \Psi}{\mathrm{d} \zeta}\right)^*\left(f_+(z - c\,t) - f_-(z + c\,t)\right)$$
where $\zeta = x + i\, y$ and $\Psi$ is any holomorphic function of the complex transverse co-ordinate $\zeta$.
A caveat to heed is that, because the transverse variation in the fields, namely $-\boldsymbol{\nabla}\phi$ and  $-\boldsymbol{\nabla}\psi$ have their electrostatic / magnetostatic form, a one conductor waveguide cannot support these modes. This is because the transverse field lines, in the absence of free charge, can only end on conductors. If there is only one conductor at a different potential from the point at infinity and that conductor is contained in a compact region, this means that, far from the conductor, the field must asymptote to a $1/r$ variation (the potential for a 2D problem asymptotes to a $\log r$ variation. Such a field would contain infinite energy. A second conductor allows the field lines to end and have a dipole variation far from the conductors.
