# On an appearance of the Lorentz group in QM.

The Lorentz group is the group of symmetries of SR; mathematically this is group, $SO(1,3)$.

There is a double covering of this group by $SL(2,C)$, that is we have the map

$SL(2,C) \rightarrow SO(1,3)$

Which is a 2-1 map, (and it turns out that this map can be explicitly given by the Pauli matrixes); and since this group is simply connected it is fact the universal cover.

Now, the kernel of this covering is just {I,-I}, hence we can identify the target of this map, SO(1,3) - the Lorentz group - with the quotient $SL(2,C)/{I, -I}$, and this is just the projective special linear group, $PSL(2,C)$.

Further, a two-state system in QM is modelled by a 2d complex Hilbert space $H$; but this is not quite true, since states are represented by lines in this space, and so it's the projective Hilbert space that we need to consider, ie $PH$. It's linear symmetry group (of determinant one), is just $PSL(2,C)$, that is the Lorentz group.

So here we have a natural way in which the Lorentz group, a group naturally associated with SR, appearing in QM.

Is there more to this than simple mathematical serendipity?

• That observation seems to be just Lie group numerology. – Qmechanic Aug 3 '17 at 19:55
• @qmechanic: I'm not so sure; consider this 'consider a CFT on the Riemann sphere, it has the möbius group as the conformal group, which is isomorphic to $PSL(2,C)$' – Mozibur Ullah Aug 3 '17 at 20:40

I don't think its as serendipitous as you think

If I understand you correctly, you're imagining the group action of $PSL(2, \mathbb{C})$ acting on your two-level system just as a $2 \times 2$ matrix on a vector in $\mathbb{C}^2$.

The thing is that, in quantum mechanics, you're usually not interested in all of the possible representations on your Hilbert space. You're just interested in the unitary ones that preserve the inner products. If you transform all of your states by the same unitary transformation they'll all have the same inner product. Therefore its impossible to detect that you've done the transformation at all.

There's another way to tell that the Lorentz group action on a 2 level system is nothing to write home about (in my opinion). Up to a phase (assuming proper normalization) a state in the two level system is specified by two real degrees of freedom. The Lorentz group has six. What are all those extra degrees of freedom even doing?

I think this is probably just one of those "embarrassment of riches" things that always comes up in representation theory.

• Well, maybe; however, I'll point out that Penroses twistors are based on the coincidence of $Spin(4,2)$ with $SU(2,2)$; I'm not claiming that my observation is up there with his, but I'd be surprised if no-one has noted this somewhere. – Mozibur Ullah Aug 4 '17 at 1:39
• Physically, we're interested in unitary transformations because they preserve probability, of course. – Mozibur Ullah Aug 4 '17 at 1:42

This is just a coincidence, which fundamentally comes from the fact that there just aren't very many nonisomorphic low-dimensional Lie groups. As the name implies, non-relativistic QM doesn't "know" anything about Lorentz invariance. Sure, there's a certain formal similarity between Minkowski spacetime and a two-level system in QM, but it's hard to see what useful insight could emerge from analogizing the entirety of spacetime in SR to a single qubit in QM. I think even most physicists would consider that cow too spherical.