I'll pose my question in the form of a statement, and you can tell me if there's anything wrong with what I say:

Every Hilbert space which describes a system that obeys the "fundamental symmetries of the universe" (namely (a) invariance under time translation, (b) space translation, (c) rotations and (d) boosts), must contain the operators $H$, $\mathbf{P}$, $\mathbf{J}$ and $\mathbf{K}$, which are the corresponding generators of the Lie groups corresponding to the symmetries (a), (b), (c) and (d).

I think there's something wrong in there, because when I take for example the 2 dimensional Hilbert space of a qubit (describing for example the spin of an electron), I can't think of what would be the "momentum operator" for that space. Would that mean that the spin state of the electron is not invariant under space translations?

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    $\begingroup$ Your statement is a matter of definition as the statement "has a symmetry" is meaningless without an operation defining the posed symmetry. Your counterexample is not one because the 2d qubit does not have the proposed symmetries (except time translation). $\endgroup$ Feb 12, 2021 at 19:33
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    $\begingroup$ I think it s a theorem that any unitary representation of Poincare requires a infinite dimensional space. On the other hand, the Hilbert space for the qubit is just an approximation of the entire Hilbert space of the system, for example, the case of only considering the spin degree of freedom of an electron and not its translational motion. So, it should not be a surprise that if you consider only the spin then you can't have a translation generator. $\endgroup$
    – AndresB
    Feb 12, 2021 at 20:20

1 Answer 1


The two-dimensional Hilbert space of a qubit is simply not including any information about where the qubit is in space, so it is meaningless to talk about moving the qubit in space in this context. However, you can always make the description bigger.

For example, if your qubit is a spin $1/2$ particle, its state in space would not be described by two numbers, representing the amplitudes for up or down, but a pair of wavefunctions $(\psi_\uparrow(\mathbf{x}), \psi_\downarrow(\mathbf{x}))$, representing the amplitude to be up or down in each specific point in space. Then you can define translations. This is essential to, e.g. describe how the Stern-Gerlach experiment works. (However, you still don't really have true Lorentz boosts, since this description is nonrelativistic. To get boosts in the picture, you would have to go to relativistic quantum mechanics.)

This kind of thing happens all the time in physics. For example, in high school fluid dynamics you might hear a lot about the pressure and density and velocity of a liquid, such as in Bernoulli's principle. But why doesn't the temperature of the fluid ever get mentioned? The answer is that when fluid move at subsonic speeds in small setups, the internal thermal energy doesn't really do anything; it's just "carried along for the ride", so it gets dropped from the description. Of course you can always put it back if you need it, such as when describing supersonic flow, or large-scale motion like convection in the atmosphere.

  • $\begingroup$ Atmospheric physics is all about the „temperature field” and the speed of air masses is clearly subsonic. $\endgroup$
    – DanielC
    Feb 12, 2021 at 20:08
  • $\begingroup$ @DanielC Good point, thanks, I added that! $\endgroup$
    – knzhou
    Feb 12, 2021 at 20:09

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