I just learned about $|0\rangle$ and siblings $|0_\gamma\rangle$ and $|0_\infty\rangle$ while studying coherent and squeezed states in a QM class, and I have a question about the meaning of "$|0\rangle$".

I get that it is the ground state of the QHO, with energy $\frac{1}{2}\hbar\omega$, no problem there. But the Prof in class proceeds to calling it the vacuum state", he "squeezes the vacuum state with the squeezing operator" (lovely!), but I fail to see that this "vacuum" is "empty" at all: there is a particle or a QHO there.

Why does the Prof call it the "vacuum" state? Is it because the lowest possible energy level of any field anywhere is in fact $\frac{\hbar}{2}$ via Heisenberg uncertainty? (that was not in the class, so I'm probably wrong here)


1 Answer 1


In quantum field theory the vacuum is the vector with no particles.

It is defined starting from the Fock space, that is the sum of spaces with any number $n\in \mathbb{N}$ of (identical particles). The space with zero particles is, roughly speaking, one-dimensional, and its basis vector is the vacuum.

In quantum mechanics, one can use the Fock space formalism but with a slightly different meaning, because the subspaces become the subspaces corresponding to a number $n$ of energy excitations of the harmonic oscillator. The vacuum is then the ground state, i.e. the state with minimal energy and no excitations.

  • $\begingroup$ Ground state wavefunction of the SHO is a gaussean. When you say minimal energy and no "excitation", do you mean that the particle is localized because the wavefunction is gaussean and does not oscillate? $\endgroup$
    – Draco_1125
    Jan 28, 2018 at 10:55

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