In classical mechanics, adding a constant to the potential changes nothing. In quantum mechanics, this just shifts the energy and multiples the wavefunction with a phase term.
But now suppose I use the potential $$V(x)=(1/2)m \omega^2x^2 – (1/2)\hbar\omega$$ instead of $$V(x)=(1/2)m \omega^2x^2 $$ for the harmonic oscillator. Then the energy spectrum will be $$E_n=n\hbar\omega$$ instead of $$E_n=(n+1/2)\hbar\omega.$$ This would mean that the harmonic oscillator can have a vanishing ground state energy depending on the potential that we choose. But the non-vanishing ground state energy of the harmonic oscillator is the source of the zero point energy of vacuum. So does that mean adding a constant to potential changes the system in quantum mechanics?
Quantum Mechanics – Vanishing Zero Point Energy in Harmonic Oscillator
harmonic-oscillatorhilbert-spacepotential energyquantum mechanicsschroedinger equation
Best Answer
The right way to interpret energy in the quantum vacuum is not known. Especially, its gravitational influence is not known.
If we say that the energy $(1/2)\hbar \omega$ is in fact present in the vacuum for every field mode, then we get an enormous energy density, and this leads to predictions about cosmic expansion which are way off from the observations. Such a model predicts an expansion accelerating at rates many orders of magnitude higher than is observed. On the other hand the Casimir effect can be calculated correctly by taking the zero point energy as $(1/2)\hbar \omega$ for each mode. So there is a real puzzle here and no-one knows the answer.