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Vacuum state is the lowest possible quantum energy state but isn't this also the definition of the ground state?

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OP is asking whether "ground" and "vacuum" refer to the same thing in quantum mechanics. The answer is

No.

A ground state is a proper eigenvector of the Hamiltonian $|0\rangle$ such that any other state satisfies $$ \langle\psi|H|\psi\rangle\ge\langle 0|H|0\rangle\tag1 $$

A vacuum state is a ground state that is also invariant under the Poincaré Group, $$ U(a,\Lambda)|0\rangle=|0\rangle\tag2 $$ up to a phase (which can WLOG set to $1$), where $\{a,\Lambda\}\in\mathrm{ISO}(1,d-1)$ and $U$ is a unitary representation of this group.

In this sense, being a vacuum state is more restrictive than being just a ground state. For one thing, the former do only exist in Poincaré invariant systems, while the latter exist for any physically meaningful system (that is, whenever $H$ is bounded from below).

In fact, one could argue that the property $(2)$ is the one that is really important in QFT; the property $(1)$ is just accessory. If we could find a (normalisable) Poincaré invariant state, even if it was not a ground state, then most of the standard manipulations of QFT (the Feynman rules, the LSZ theorem, etc.) could be performed using that state. In other words, the fact that $|0\rangle$ is the state with the minimal energy is not really fundamental; but it being Poincaré invariant is. This is usually obscured in most books because they tend to use $(1)$ to simplify the analysis, but a careful investigation proves that that property can be lifted, and most of the results remain valid.

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    $\begingroup$ N.B.: for non-relativistic quantum-field-theories, one should replace Poincaré by Galilei (or Bargmann) in the definition of vacuum states. $\endgroup$ Oct 1, 2017 at 19:21
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We use the term ground state in quantum mechanics to describe the lowest-energy state of a particle or a system of particles.

On the other hand, the term vacuum state is used to describe the lowest energy state in a field theory, which typically is a state with no physical particles.

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  • $\begingroup$ The vacuum has no excitations, which means the expectation value of the field $\langle \phi \rangle = 0$. However you can still have a standard deviation from zero because $\langle \phi^2 \rangle \neq 0$. This is how Lamb shift, Casimir effect etc. come up. $\endgroup$
    – SuperCiocia
    Oct 1, 2017 at 2:31
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    $\begingroup$ @SuperCiocia That is not correct. $\langle \phi \rangle$ can be non-zero in the vacuum; this is true of the Higgs field in the Standard Model. $\endgroup$
    – gj255
    Oct 1, 2017 at 10:06
  • $\begingroup$ Ah, yes, that's correct. The Higgs field is the only example surely no? $\endgroup$
    – SuperCiocia
    Oct 1, 2017 at 10:38
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    $\begingroup$ In the Standard Model, yes. But non-zero expectation values of fields in the vacuum is characteristic of spontaneous symmetry breaking, which occurs in many models in high energy and condensed matter physics. $\endgroup$
    – gj255
    Oct 1, 2017 at 12:53
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The term "vacuum state" typically refers to the ground state of a weakly coupled second-quantized quantum field theory, in which the low-lying excitations can be interpreted as particles, so the ground state is the absence of excitations/particles. This terminology can be a bit misleading for a strongly-coupled system, as it implies that the ground state is simple and nothing interesting is going on, when in fact the ground state can already be highly entangled and difficult to understand.

So "vacuum state" is usually a special case of "ground state". But people who work a lot with QFT often get lazy and use the phrase "vacuum state" to refer to the ground state of any quantum-mechanical system, even if it's first-quantized or has a finite number of degrees of freedom.

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