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In non-relativistic QM, does it make a difference if an energy shift is applied to the systems's Lagrangian or Hamiltonian?

AbdulQat
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  • Possible duplicate: http://physics.stackexchange.com/q/8359/2451 – Qmechanic Oct 14 '13 at 22:47
  • This question has not been asked before. If you believe it has already been answered, kindly point out how the answer differs from that posted, which has neither been confirmed nor disputed. – AbdulQat Oct 25 '13 at 20:03
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    Are you talking about (i) shifting the energy reference level as a matter of chosing another convention, or are you talking about (ii) performing/extracting external work on the system, i.e. working with a Hamiltonian with explicit time dependence? Please clarify(edit) the post accordingly, in order to get it re-opened. – Qmechanic Oct 25 '13 at 20:33
  • We appear to be separated by a couple of common languages, viz. English and physics. – AbdulQat Oct 25 '13 at 20:52

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Because it detracts from the potential, as the shift in $L$ increases it tends to increase $\langle T \rangle$ to the point that a molecule first implodes ($E=2 \langle T \rangle$) and then -- discounting nuclear fusion -- explodes ($E=2 \langle V \rangle$), consistent with the maintenance of $ E = \langle T \rangle + \langle V \rangle$.

In contrast to $L$, if $H$ is shifted enough, implosion is bypassed, and the system directly ionizes.

AbdulQat
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