Unlike rotations, the boost transformations are non-unitary. Therefore, the boost generators are not Hermitian. When boosts induce transformations in the Hilbert space, will those transformation be unitary? I think no. If that is the case, what is the physical significance of such non-unitary transformations corresponding to boosts in the Hilbert space?

  • $\begingroup$ Related : physics.stackexchange.com/q/56024 $\endgroup$
    – user38249
    Feb 13, 2014 at 15:12
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    $\begingroup$ It's related but this question is about the relativistic case but the older debate - both question and answer - is fully about the non-relativistic Galilean case so it's not quite the same problem. $\endgroup$ Feb 13, 2014 at 15:52
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    $\begingroup$ DEar @Roopam. I have been following your probing questions with interest. As Luboš's answer shows in the details for your specific, this question is answered by recalling that the image of a group under a particular representation is in general different from the group itself. As you know from your other question, the image of $O(1,3)$ under the adjoint representation ${\rm Ad}$ is $O(1,3)$ itself. But this is not true of a general homomorphism. Nor indeed is it generally true even of ${\rm Ad}$ itself. You marked the question "representation-theory" so it sounds as though you are ... $\endgroup$ Feb 15, 2014 at 1:33
  • $\begingroup$ ... learning get your head around these ideas. You just need to keep on learning about representations and their properties. When something undergoes a spacetime transformation by a member of $O(1,3)$, the corresponding quantum state must undergo a unitary transformation (the object must end up in some state, after all!). WHen you are ready, see the Woit reference linked in my answer here or also this excellent reference by the inimitable John Baez arxiv.org/abs/0904.1556 was also a great help to me. $\endgroup$ Feb 15, 2014 at 1:41
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    $\begingroup$ ... Also the Stone-von Neumann theorem, Wigner's theorem and Wigner's classification are important concepts you are walking into. $\endgroup$ Feb 15, 2014 at 1:43

1 Answer 1


On the actual Hilbert space of a consistent relativistic quantum mechanical system, the Lorentz transformations including boosts actually are unitary – which also means that the generators $J_{0i}$ are as Hermitian as the generators of rotations $J_{ij}$.

We say that the Hilbert space forms a unitary representation of the Lorentz group.

What the OP must be confused by is the fact that the ordinary vector representation composed of vectors $(t,x,y,z)$ is not a unitary representation of $SO(3,1)$. The $SO(3,1)$ transformations don't preserve any positively definite quadratic invariant constructed out of the coordinates $(t,x,y,z)$. After all, we know that an indefinite form, $t^2-x^2-y^2-z^2$, is conserved by the Lorentz transformations. So on a representation like the vector space of such $(t,x,y,z)$, the generators $J_{0i}$ would end up being anti-Hermitian rather than Hermitian.

But if you take a Lorentz-invariant theory with a positive definite Hilbert space, like QED, the formula for $J_{0i}$ makes it manifest that it is a Hermitian operator, which means that $\langle \psi |\psi\rangle$ is preserved by the Lorentz boosts! The complex probability amplitudes for different states $c_i$ behave differently than the coordinates $t,x,y,z$ above.

Note that the (non-trivial) unitary transformations of $SO(3,1)$ are inevitably infinite-dimensional. Finite-dimensional reps may be constructed out of the fundamental vector representation above and they are as non-unitary as the vector representation. But that's not true for infinite-dimensional reps. For example, the space of one-scalar-particle states in a QFT is a unitary representation of the Lorentz group. For each $p^\mu$ obeying $p^\mu p_\mu=m^2$, and there are infinitely (continuously) many values of such a vector (on the mass shell), the representation contains one basis vector (which are normalized to the Dirac delta function). The boosts just "permute them" along the mass shell which makes it obvious that the positively definite form is preserved when normalized properly.

  • $\begingroup$ Which formula do you mean by the formula for $J_{0i}$? I am reading weinberg and in his exposition (pg59) he just assumes the $J_{\mu \nu}$ are hermitian so that U is unitary. I'd be curious if it can be formulated a different way. $\endgroup$ Feb 8, 2019 at 11:48
  • $\begingroup$ Dear @user27084, $J_{0i}$ is schematically $x_0 p_i - x_i p_0$ but in field theory, $p_\mu$ has to be written as an integral of the energy density or momentum density over space. Let me assume that you know how to write the total momentum or the total energy as an integral involving all the fields. So to get the boost generator, you just write down the same integrals that define $p_0$ or $p_i$ and add the $x_0$ or $x_i$ extra factor inside the integral - to get both terms in the difference that I started with. $\endgroup$ Feb 9, 2019 at 19:02
  • $\begingroup$ To be more precise, the formula that you got so far is just the orbital part of $J_{0i}$, the $L_{0i}$, if you wish. To get the full generator of the boost, you also need to add the terms responsible for the spin, the intrinsic angular momentum, of all the fields. All such extra terms are simple bilinear expressions in the field whose index structure is fully determined by the Lorentz covariance. About 1/2 of the QFT books contain explicit formulae for the form of $J_{\mu\nu}$ for any field theory they discuss. $\endgroup$ Feb 9, 2019 at 19:04
  • $\begingroup$ Note that all the terms are some integrals and the integrands contain factors such as $x_0$, $x_i$, energy density, or momentum density. And then the spin. If you look at all these factors, there is never any imaginary unit etc. which makes it almost manifest that all the generators, regardless of the indices' being spatial or temporal, are Hermitian. $\endgroup$ Feb 9, 2019 at 19:06

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