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When a particle enters an excited state, the energy appears in its quantum wavefunction according to $E = h \nu$.

Does the $E$ in this equation also include kinetic energy, and rest mass energy? Or are they "held" somewhere else? Or are kinetic/mass energy a different framework for looking at the same phenomenon, and if so how do they relate?

Also, if the particle gathers enough excitation energy to escape a potential well, and therefore escapes that well, from which 'budget' is the necessary energy taken - kinetic, frequency or mass; and why?

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  • $\begingroup$ $E = \hbar \nu$ is for photons. Energy of quantum systems is generally more complicated, but the energy can come from a variety of degrees of freedom, kinetic and otherwise. $\endgroup$
    – DumpsterDoofus
    Commented Mar 4, 2014 at 23:25

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  1. The equation [now corrected] in the question is incorrect by a factor of 2pi, because h-bar should just be h.

  2. The equation is not limited to excited states.

  3. E is the relativistic energy from $E^2 = p^2c^2 + m^2c^4$, where m is the rest mass and p is momentum.

  4. While the equation was originally for photons, De Broglie extended it to all particles in his 1924 PhD thesis.

reference:

http://www.rpi.edu/dept/phys/Dept2/modern-physics/lecture-notes.d/6-Quantum-early.pdf

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  • $\begingroup$ Fixed (1). So by your point (3) you're saying that $E=h \nu$ is another way of looking at kinetic+rest energy. Does that include potential energy as well? $\endgroup$
    – Sideshow Bob
    Commented Mar 5, 2014 at 12:20
  • $\begingroup$ E is the relativistic energy from $E^2 = p^2c^2 + m^2c^4$, where m is the rest mass and p is momentum. Will edit answer to clarify, add reference. $\endgroup$
    – DavePhD
    Commented Mar 5, 2014 at 13:49

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