Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The boost transformations are not unitary unlike rotations, the boost generators are not Hermitian. When this induces 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 transformation corresponding to boosts in the Hilbert space?

share|improve this question
Related : physics.stackexchange.com/q/56024 –  user38249 Feb 13 '14 at 15:12
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. –  Luboš Motl Feb 13 '14 at 15:52
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 ... –  WetSavannaAnimal aka Rod Vance Feb 15 '14 at 1:33
... 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. –  WetSavannaAnimal aka Rod Vance Feb 15 '14 at 1:41
... Also the Stone-von Neumann theorem, Wigner's theorem and Wigner's classification are important concepts you are walking into. –  WetSavannaAnimal aka Rod Vance Feb 15 '14 at 1:43

1 Answer 1

up vote 14 down vote accepted

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.

share|improve this answer
Hi Luboš, is there an easy way to get the hacek accent on the s in your name in MS-Word (on a US/English keyboard - I guess you're writing on a Czech one, yes?) –  WetSavannaAnimal aka Rod Vance Feb 15 '14 at 1:50
Try copy-paste. Yes, in Czechia, we have a Czech keyboard where "š" appears instead of "3" which is obtained as shift-3 and # is elsewhere, or with right alt, and so on. In tex, it is "backslash v space {s}", $\v s$, for some reason, it isn't allowed here. You can add a Czech keyboard in MS Windows. I am myself using a modified Czech American Lumo keyboard, downloadable from motls.blogspot.com/2011/09/… - it's as American as you can get, gives Czech characters more or less at usual Czech places, and allows 100 other characters by easy combos –  Luboš Motl Feb 15 '14 at 5:53
Thanks muchly. I was hoping there was an easier way: non-English letters STILL aren't that well supported in a supposedly globalised world! –  WetSavannaAnimal aka Rod Vance Feb 15 '14 at 6:45
Hi, they're hard to type on keyboards but most web pages etc. run on Unicode system to encode the characters which allows all of the alphabets etc. of the world! There are ways to type characters on English keyboard, too. Alt with 0161 pressed, release alt, could work on your keyboard (0160 is capital s-caron) although it types something else on mine LOL. –  Luboš Motl Feb 15 '14 at 13:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.