In the Gupta-Bleuler formalism we have a problem with two states (scalar photons and longitudinal photons), because here $\langle \vec{k}_a|\vec{k}_b\rangle $ is negative or zero. However, I thought that only $|\langle \vec{k}_a|\vec{k}_b\rangle |^2$ corresponds to probabilities, a quantity which would be non-negative anyway.

What am I doing wrong?

  • 1
    $\begingroup$ You have : $||\vec k;\epsilon_\mu\rangle|^2 = \langle \vec k;\epsilon_\mu|\vec k;\epsilon_\mu \rangle = -\eta_{\mu \mu} \frac{1}{2|\vec k|} \delta(\vec 0)$. So, with a metrics $\eta = (1,-1,-1,-1)$, the state $|\vec k;\epsilon_0\rangle$ has a negative "norm". $\endgroup$
    – Trimok
    Jan 14, 2014 at 11:01
  • $\begingroup$ So that the norm is negative is a mathematical problem (which causes issues defining a proper Hilbert space), not a physical problem of negative probabilities. Is that correct? $\endgroup$
    – PPR
    Jan 14, 2014 at 12:12
  • 3
    $\begingroup$ The two are related. Consider states $|a\rangle$ and $|b\rangle$ which are not normed (to $1$). Suppose that the state $|a\rangle $ has an unitary evolution in time. The probability of finding the system later in a state $|b\rangle$ is : $p = \dfrac{|\langle a|b\rangle|^2}{\langle b|b\rangle \langle a|a\rangle}$. Now suppose that $|b\rangle$ has a positive norm, and $|a\rangle$ has a negative norm, then we see that $p <0$ $\endgroup$
    – Trimok
    Jan 14, 2014 at 13:46

1 Answer 1


You should think in terms of the norm of a state, and what happens here is that you have negative-norm states which you don't want, otherwise you can't construct a Hilbert space.

Copying the formula in Wikipedia, $$ \langle\vec{k}_a;\epsilon_\mu|\vec{k}_b;\epsilon_\nu\rangle=(-\eta_{\mu\nu}){1\over 2|\vec{k}_a|}\delta(\vec{k}_a-\vec{k}_b) $$ you see that for $\mu=\nu=0,$ taking the metric signature as $(+,-,-,-),$ you find negative norm states. Since this prevents you from constructing an Hilbert space, it doesn't make sense to speak of probability: namely if you define a probability as a norm, then you get a negative probability, but this is just rephrasing the issue.

Put in another way, $ \langle\vec{k}_a|\vec{k}_a \rangle$ would be both negative and the norm of a vector for some states, preventing a probabilistic interpretation, that's why you speak of negative probability. Note that indeed a Hilbert space guarantees that the probabilities are well defined.

I think another useful point of view on this story is that, once you implement the constraint $\epsilon \cdot k=0$ you are with 3 degrees of freedom, and you eliminate the last one by saying that your Hilbert space is the quotient of positive norm states over zero norm states, and this is really to be thought of as gauge equivalence, i.e. gauge degrees of freedom are redundancies, i.e. unphysical states.

  • 2
    $\begingroup$ Could you elaborate? It is indeed often stated that negative-norm states result in negative probabilities. Why so, if as the OP mentions, probabilities are always proportional to the absolute value of a norm? $\endgroup$
    – innisfree
    Jan 14, 2014 at 11:46

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