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I am quite confused to the manifestation of an electric field on the surface of a metal due to the work function. Can anyone give me an intuitive answer behind this?

For example:

enter image description here

So I am studying surface states, and spin orbit coupling on such states, and the whole reason behind this coupling is this electric field. Now every book I read kind of gives this hand waving argument that it the field arises due to the work function of the material. I don't see the connection at all.

So I would like some clarification on this.

Thank you.

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    $\begingroup$ Please describe more fully what is confusing you so that we know exactly what needs explaining. Thanks! $\endgroup$ Commented Apr 17, 2016 at 1:33

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In real materials and also in jellium, the positive background charge distribution does not perfectly align with the negative charge distribution. Especially, at the surface, where the translation symmetry is broken, charge can accumulate and creates a surface dipole. http://images.slideplayer.com/22/6402875/slides/slide_4.jpg Here is a very heuristic derivation of potential shift due to surface dipole. (I don't care about prefactors, units or sings) Let $\delta'$ be a derivative of of Dirac delta function, representing the charge distribution of this surface dipole. Now, the Poisson equation is

$$ \frac{d^2 V(z)}{dz^2} = p \delta'(z), $$

$$ \frac{d V(z)}{dz} = \int dz p \delta'(z) = p \delta(z), $$

$$ V(z) = \int dz \frac{d V(z)}{dz} = p \Theta(z), $$

where $\Theta$ is the Heaviside step function. We see here that the surface dipole term causes a constant shift of the potential at the surface. In other words, the work function can differ between different surfaces (100,111 etc.) because the surface dipole induced work function shift.

You cannot of course create a perpetual motion machine by taking electrons from one surface and adding them to another, because of the macroscopic multipole moments outside the sample will make the electric field conservative.

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  • $\begingroup$ Hmm, so -- ok, I'm trying to get a picture in my head. We have electrons and protons on the surface, they create an electric dipole, with an electric field pointing into the vacuum -- like my picture above represents. In the bulk, there is no electric field because all the dipoles cancel out, but on the surface the symmetry is broken due to a lack of a layer of atoms above the surface to cancel out the surface dipole. And electrons are not ejected out because the layer below the surface doesn't provide enough energy equal to the work function to eject them? is that right... hmm... $\endgroup$
    – sci-guy
    Commented Apr 17, 2016 at 11:48
  • $\begingroup$ @renegade arxiv.org/ftp/cond-mat/papers/0608/0608089.pdf Here is an article, which studies different surfaces and their work functions. Figure 1 is a nice example how different surfaces (left and right) can have different potentials. Electric field is then a derivative of the potential. $\endgroup$ Commented Apr 17, 2016 at 14:15
  • $\begingroup$ Also, note that even spin-orbit coupling changes positions of energy levels (which is probably what you mean), I think that it might not be much caused by these small surface dipole fields, but spin orbit coupling is mostly due to 'huge' 'spherical' electric fields within each atom. (Correct me if I am wrong, I have not explicitly ever calculated spin-orbit couplings.) $\endgroup$ Commented Apr 17, 2016 at 14:18

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