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I've read that the Zeeman effect can cause a single allowed energy level to split into two, one of which is higher in energy by some value \Delta E'. The other level is shifted down by the same amount.

My question is this: if an electron is in the ground state, the lowest orbital, can it shift even lower by \Delta E'?

Is it possible to go below 0 energy and have a negative energy orbital?

I'm not fully or formally educated on quantum systems. Thank you in advance for the help!

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  • $\begingroup$ My mistake. You're correct. $\endgroup$ – Will Brickner Mar 24 '16 at 21:57
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When describing the energies of bound states we can choose the zero point of the energy to be any value we want. This is because the absolute values of energy have no physical significance - all that matters is the difference in energies. Technically we say the energy has a gauge symmetry.

Typically we take the energy to be zero when the bound state has been dispersed to infinity. So for a hydrogen atom we take the energy to be zero when the proton and electron are separated by an infinite distance. As we bring the proton and electron together to form a hydrogen atom the energy decreases, so it becomes negative. Hence the energy of the hydrogen ground state is normally expressed as -13.6eV.

The Zeeman splitting can indeed lower the energy of the hydrogen ground state, but there's nothing exciting about this. It just means the energy would fall from -13.6eV to -13.6 - (a bit more) eV.

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  • $\begingroup$ Awesome! Thanks! I'm wondering, could one leverage the Zeeman effect to alter the absorption/emission spectrum of an atom? This is technically a separate question, sorry. $\endgroup$ – Will Brickner Mar 25 '16 at 23:43
  • $\begingroup$ @WillBrickner: the Zeeman effect is usually quite small, and it causes a splitting of lines in the spectrum rather than anything more dramatic. $\endgroup$ – John Rennie Mar 26 '16 at 5:59

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