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I often hear my teachers say that the negative norm states break unitarity. And I can also read this elsewhere, such as at this place

In this gauge the relation between unitarity and gauge invariance is thus obvious: by breaking the gauge invariance we thus break the unitarity, since the negative norm states enter the game.

Whose unitarity is broken? How to break it?

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I asked Mark Srednicki about this, and he told me that it's not really correct to say that negative-norm states break unitarity, because negative-norm states don't exist by the definition of the inner product. It's often a convenient calculational trick to formally expand your state space so that it's no longer a Hilbert space by adding in negative-norm ghosts, and the presence of physical states formally appearing to couple to ghosts indicates the presence of a quantum anomaly that prevents you from consistently quantizing your theory. But this is just a calculational trick to see the anomaly - the anomaly is real, the ghosts aren't.

In particular, you can always in principle see the existence of the anomaly without introducing ghosts. For example, the usual explanation for the fact that bosonic string theory can only be formulated in 26 dimensions is that that's the only number of dimensions in which the ghosts decouple. But we can alternatively work in light-cone gauge with only positive-norm states, and we find that only in 26 dimensions do the Lorentz generators close. This is another way to see the anomaly that doesn't require any mention of ghosts.

Mark also said that another reason it's incorrect to say that ghosts "break unitarity" is that they really just prevent you from consistently quantizing your theory at all - there's no reason to specifically single out unitarity as being broken.

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Unitarity is the idea that states in a (quantum mechanical) Hilbert space are always normalized (have length one), $\langle \psi|\psi\rangle = 1$. Normally the evolution of a physical system (in quantum theory) is described by a unitary operator acting on the state vector in Hilbert space. For example, the evolution of a system in time is given by $e^{i H t} |\phi\rangle$, where the exponent of the Hamiltonian $e^{i H t}$ is the unitary operator describing how the system evolves throughout time.

When the evolution is non-unitary the state will now longer be normalized. This is bad because the probabilities of all the possible outcomes added will now no longer add up to one. For the same reason, when you obtain a non-normalized state, or a negative norm state, the evolution of the system has been nonunitary.

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  • $\begingroup$ Can you give some references? $\endgroup$ – lewton Mar 7 '17 at 17:07
  • $\begingroup$ Wikipedia? en.wikipedia.org/wiki/Unitarity_(physics) (see the references to other articles there) $\endgroup$ – JgL Mar 7 '17 at 17:08
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    $\begingroup$ I'd say that negative-norm states break the probabilistic interpretation of QM, which is sometimes loosely called unitarity, beause unitary transformations preserve probabilities. Do you agree, @JgL? $\endgroup$ – Prof. Legolasov Mar 8 '17 at 10:44
  • $\begingroup$ I agree with @SolenodonParadoxus. I don't see what the existence of negative-norm states has to do with unitarity. First of all, all (nonzero) states in the Hilbert are required to have positive norm by the definition of inner product, so what do negative-norm states even mean? If they don't live in the Hilbert space, then they don't have a norm at all. Second of all, unitarity is a property of the time-evolution operator and has nothing to do with the states themselves. I agree that if a normalized state time-evolves into a negative-norm state then that violates unitarity, ... $\endgroup$ – tparker May 8 '17 at 18:16
  • $\begingroup$ ... but only because the evolution of a normalized state into another state with any norm other than 1 violates unitarity. There's nothing particularly bad about negative-norm states in this regard. $\endgroup$ – tparker May 8 '17 at 18:17
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Unitarity means that the time evolution is implemented by a unitary operator $U(t)=e^{-itH}$ (in units where $\hbar=1$) on a Hilbert space. If $H$ is implemented as a Hermitian linear operator on a vector space with an indefinite inner product, this is guaranteed only if some subspace of physical states is both a Hilbert space and invariant under time evolution. Since in a Hilbert space all $\psi^*\psi$ are nonnegative, this implies that no physical state $\psi$ can have $\psi^*\psi<0$. Thus all such states are unphysical. This is the precise content of the statement ''negative norm states break unitarity''

In a general inner product space, $U(t)$ will still preserve all inner products and hence all norms. But in general there is no useful subspace consisting only of positive norm states and zero. Only the latter qualifies as a Hilbert space. Note that linear combinations of positive norm states may have negative norm; so the condition of being positive on a subspace large enough to accommodate a representation of all relevant observables rather than just a subset is very restrictive.

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  • $\begingroup$ (a) Is the plus sign supposed to be an equals sign? And (b) why can't you have a unitary time-evolution operator that preserves both positive and negative norms of states? $\endgroup$ – tparker May 14 '17 at 13:29
  • $\begingroup$ I corrected the equation. - In a general inner product space, $U(t)$ will still preserve all inner products and hence all norms. But in general there is no useful subspace consisting only of positive norm states and zero. Only the latter qualifies as a Hilbert space. Note that linear combinations of positive norm states may have negative norms; so the condition of being positive on a subspace rather than just a subset is very restrictive. $\endgroup$ – Arnold Neumaier May 15 '17 at 7:47

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