Someone argues: "Because gives

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Wave mechanics in one dimension · The formalism of quantum mechanics

The Structure of Quantum Mechanics, Ch The oscillator as an algebra

Problem

Someone argues: "Because [a^,a^†]=1[\hat a,\hat a^\dagger]=1 gives a^a^†=a^†a^+1\hat a\hat a^\dagger = \hat a^\dagger\hat a + 1, the order in which a^\hat a and a^†\hat a^\dagger are written in the Hamiltonian is a free relabelling with no physical content. So if I define H^′=ℏω a^a^†\hat H' = \hbar\omega\,\hat a\hat a^\dagger instead of the usual H^=ℏω(a^†a^+12)\hat H=\hbar\omega(\hat a^\dagger\hat a+\tfrac12), I am just choosing where the zero of energy sits, and the ground-state energy of H^′\hat H' must come out as −ℏω/2-\hbar\omega/2, the mirror image of the usual +ℏω/2+\hbar\omega/2." ((a)) Using only [a^,a^†]=1[\hat a,\hat a^\dagger]=1 and boundedness of the spectrum of H^′\hat H' below, find the actual spectrum of H^′=ℏω a^a^†\hat H' = \hbar\omega\,\hat a\hat a^\dagger. ((b)) For ω=3.5×1013 rad s−1\omega = 3.5\times10^{13}\ \mathrm{rad\,s^{-1}}, give the numerical value in eV of the ground-state energy of H^′\hat H', and say in one sentence why the quoted argument is wrong.

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