# Why is the single particle field state not on the quantum harmonic oscillator spectrum?

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Two kinds of waves can be present in quantum systems, which have corresponding frequencies. I will call the first "de Broglie" (quantum / probability) waves and frequencies, in simple systems they correspond to energy eigenstates. I will call the second "Maxwell" (classical / intensity) waves and frequencies, they correspond to coherent states.

There is a commonly taught correspondence between the states of a single momentum of a quantum field and the states of a simple harmonic oscillator (a second derivative plus a constant operating on either state is zero). This correspondence implies a similar solution space, in this case one which has creation and annihilation operators and coherent states.

My question concerns the energy spectrum of these solution spaces. The allowed energies of the harmonic oscillator are $h f_M (n+\frac{1}{2} )$ with n an integer. That is to say, the 'de Broglie' energies may be $(n+\frac{1}{2} ) h$ times the 'Maxwell' frequency. For an individual photon, $E = h f_{dB}$ . However, we know an individual photon is a 'de Broglie' state of the electromagnetic field. The corresponding 'Maxwell' frequency for all occupation states of the same momentum is the same as the 'de Broglie' frequency. Therefore we can also say $E = h f_M$ . However, this is not in the oscillator spectrum $h f_M ( n + 1/2 )$ . I would like to know why the two systems with dynamical equations of the same form do not have the same relationship between their 'Maxwell' frequencies and 'de Broglie' energies.

Apologies for the probably nonstandard terminology. I did not find a general naming convention for the two types of waves and their associated properties. asked May 16, 2018

I am not sure if I understand it correctly, but in QFT the constant 1/2 is considered as the zero-point energy and can be dropped out.

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