From: PointedEars@web.de   
      
   Thomas 'PointedEars' Lahn wrote:   
   > For example, if the quantum-electromagnetic field has an energy that is ℎ f   
   > (or ℎ c/λ, or ℎ k c) above its ground-state energy (which is also its   
   zero-   
   > point energy), then we say that there is a photon with frequency f (or   
   > wavelength λ, or wavenumber k, respectively): the photon is the quantum of   
   > that field (AISB).   
      
   Sorry, that's   
      
     E = p c = _ℏ_ k c = ℎ/(2π) (2π/λ) c = ℎ c/λ = ℎ f.   
      
     [Planck–Einstein relation]   
      
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   PointedEars   
      
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