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I'm a beginner of QFT. Ref. 1 states that

[...] The Lorentz group $SO(1,3)$ is then essentially $SU(2)\times SU(2)$.

But how is it possible, because $SU(2)\times SU(2)$ is a compact Lie group while $SO(1,3)$ is non-compact?

And after some operation, he says that the Lorentz transformation on spinor is complex $2\times2$ matrices with unit determinant, so Lorentz group becomes $SL(2,\mathbb{C})$. I'm confused about these, and I think there must be something missing.

References:

  1. L.H. Ryder, QFT, chapter 2, p. 38.
Qmechanic
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    This is essentially a duplicate of http://physics.stackexchange.com/q/28505/2451 , http://physics.stackexchange.com/q/99283/2451 , and the answers given therein. – Qmechanic Apr 13 '14 at 12:18
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    Related Math.SE post: https://math.stackexchange.com/q/1109369/11127 – Qmechanic May 15 '17 at 06:46

3 Answers3

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Here's my two cents worth.

Why Lie Algebras?

First I'm just going to talk about Lie algebras. These capture almost all information about the underlying group. The only information omitted is the discrete symmetries of the theory. But in quantum mechanics we usually deal with these separately, so that's fine.

The Lorentz Lie Algebra

It turns out that the Lie algebra of the Lorentz group is isomorphic to that of $SL(2,\mathbb{C})$. Mathematically we write this (using Fraktur font for Lie algebras)

$$\mathfrak{so}(3,1)\cong \mathfrak{sl}(2,\mathbb{C})$$

This makes sense since $\mathfrak{sl}(2,\mathbb{C})$ is non-compact, just like the Lorentz group.

Representing the Situation

When we do quantum mechanics, we want our states to live in a vector space that forms a representation for our symmetry group. We live in a real world, so we should consider real representations of $\mathfrak{sl}(2,\mathbb{C})$.

A bit of thought will convince you of the following.

Fact: real representations of a Lie algebra are in one-to-one correspondence (bijection) with complex representations of its complexification.

That sounds quite technical, but it's actually simple. It just says that we can have complex vector spaces for our quantum mechanical states! That is, provided we use complex coefficients for our Lie algebra $\mathfrak{sl}(2,\mathbb{C})$.

When we complexify $\mathfrak{sl}(2,\mathbb{C})$ we get a direct sum of two copies of it. Mathematically we write

$$\mathfrak{sl}(2,\mathbb{C})_{\mathbb{C}} = \mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C})$$

So Where Does $SU(2)$ Come In?

So we're looking for complex representations of $\mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C})$. But these just come from a tensor product of two representations of $\mathfrak{sl}(2,\mathbb{C})$. These are usually labelled by a pair of numbers, like so

$$|\psi \rangle \textrm{ lives in the } (i,j) \textrm{ representation of } \mathfrak{sl}(2,\mathbb{C}) \oplus \mathfrak{sl}(2,\mathbb{C})$$

So what are the possible representations of $\mathfrak{sl}(2,\mathbb{C})$? Here we can use our fact again. It turns out that $\mathfrak{sl}(2,\mathbb{C})$ is the complexification of $\mathfrak{su}(2)$. But we know that the real representations of $\mathfrak{su}(2)$ are the spin representations!

So really the numbers $i$ and $j$ label the angular momentum and spin of particles. From this perspective you can see that spin is a consequence of special relativity!

What about Compactness?

This tortuous journey shows you that things aren't really as simple as Ryder makes out. You are absolutely right that

$$\mathfrak{su}(2)\oplus \mathfrak{su}(2) \neq \mathfrak{so}(3,1)$$

since the LHS is compact but the RHS isn't! But my arguments above show that compactness is not a property that survives the complexification procedure. It's my "fact" above that ties everything together.

Interestingly in Euclidean signature one does have that

$$\mathfrak{su}(2)\oplus \mathfrak{su}(2) = \mathfrak{so}(4)$$

You may know that QFT is closely related to statistical physics via Wick rotation. So this observation demonstrates that Ryder's intuitive story is good, even if his mathematical claim is imprecise.

Let me know if you need any more help!

Edward Hughes
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    When we do quantum mechanics, we want our Hilbert space to be a representation of the symmetry group. Hilbert space is complex, so we want complex, not real, representations. Your statement about real representations being in bijection is false. Complex representations are. – Peter Kravchuk Oct 07 '17 at 02:35
  • @PeterKravchuk I'm not sure what you mean by 'Hilbert space is complex'? $(\mathbb{R}^\mathfrak{n},\circ)$ is a Hilbert space for any cardinal number $\mathfrak{n}$, as are projective real spaces etc. – Alec Rhea Oct 07 '17 at 02:53
  • @AlecRhea the physical Hilbert space in quantum mechanics is complex as in "over $\mathbb{C}$." – Peter Kravchuk Oct 07 '17 at 08:50
  • @PeterKravchuk But $\mathbb{C}$ is isomorphic to $\mathbb{R}^2$ as an algebraically closed field, so any vector space doodads we build over $\mathbb{C}^n$ (resp. $P\mathbb{C}^n$, like the Bloch sphere) could be equivalently built over $\mathbb{R}^{2n}$ (resp. $P\mathbb{R}^{2n}$). – Alec Rhea Oct 07 '17 at 09:29
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    @AlecRhea, $\mathbb{R}^2$ is not a field unless you endow it with $\mathbb{C}$ structure, what is your point? – Peter Kravchuk Oct 07 '17 at 16:00
  • @PeterKravchuk The point is that anything built over $\mathbb{C}$ can be built over $\mathbb{R}^2$ where $(a,b)+(c,d)=(a+b,c+d)$ and $(a,b)\times(c,d)=(ac-bd,ad+bc)$ with no a-priori reference to $\mathbb{C}$, so your original critique that "Hilbert space is complex not real" is incoherent. This post is fine as far as I can tell -- all finite dimensional complex structures are real structures of twice the 'dimension' from the perspective of model theory. I don't know about your claim that the bijection is nonexistent, but a reference would be appreciated. – Alec Rhea Oct 08 '17 at 05:05
  • @AlecRhea what you said is essentially equivalent to "anything built over $\mathbb{C}$ can be built over $\mathbb{C}$", because your $\mathbb{R}^2$ is $\mathbb{C}$. In quantum mechanics, the Hilbert space has $\mathbb{C}$ (and not anything else) as the base field. This is a meaningful statement, and if you don't understand it, I encourage you to ask a question on this site. Comments are not for this sort of discussion. – Peter Kravchuk Oct 08 '17 at 23:28
  • @PeterKravchuk No, the $\mathbb{R}^2$ I've given is $\mathbb{R}^2$, and from the perspective of model theory I can develop a theory of $\mathbb{R}^2$ and a theory of $\mathbb{C}$ with entirely different vocabularies which you would then want to identify as 'the same theory' even though they aren't if we're using the term theory in a precise, model-theoretic sense. I suspect that all critiques in your original comment come from a misunderstanding of this point -- something which is in bijection with representations over $\mathbb{C}^n$ will be in bijection with real $\mathbb{R}^{2n}$ reps. – Alec Rhea Oct 09 '17 at 02:20
  • @PeterKravchuk It is completely correct and coherent to say that 'Hilbert space has $\mathbb{R}^2$ as a base field in QM' with the binary operations I've given, and then move forward from there with no need to create a new number system, just a manipulation of ordered pairs of our existing number system. The fact that we can't trivially do the same thing for $\mathbb{R}^3$ ($\mathbb{R}^4$ has the quaternions) might end up being related to something interesting and physical. – Alec Rhea Oct 09 '17 at 02:25
  • @PeterKravchuk Unless I'm mistaken, $\mathbb{C}$ is desirable in QM because it is complete in its canonical metric and algebraically closed, both of which have many physical consequences that we want. We can recover the latter of these properties by looking at larger, nonstandard models of $\mathbb{C}$ and the former by looking at a 'closest finite complex number' metric in this model, so I think there is some good reason not to just ascribe the nice properties of $\mathbb{C}$ as a base field to the fact that 'it is $\mathbb{C}$'. – Alec Rhea Oct 09 '17 at 03:53
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    But the question was about the group, not the Lie algebra. On the group level there are subtle differences! – Arnold Neumaier Nov 22 '17 at 10:19
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    The group version is handled in https://physics.stackexchange.com/a/28651/7924 – Arnold Neumaier Nov 22 '17 at 13:43
  • As stated, your "Fact" is misleading at best. It should read: "complex representations of a real (semisimple) Lie algebra are in one-to-one correspondence with complex representations of its complexification". Probably you mean that, as a few lines later you write "provided we use complex coefficients". Cf. my recent answer on math, https://math.stackexchange.com/a/3258221/96384, which tried to sort out this confusion. – Torsten Schoeneberg Jul 19 '19 at 16:56
  • And a minor point, it would prevent some confusion if once you introduce the Lorenz Lie algebra, you state explicitly that $\mathfrak{sl}_2(\mathbb C)$ is viewed here as a real Lie algebra, of dimension $6$. (Which explains why its complexification, which you correctly state later as $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$, but now as complex Lie algebra, is $6$-dimensional over $\mathbb C$.) – Torsten Schoeneberg Jul 19 '19 at 17:00
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    And, sorry, but your sentence "We live in a real world, so we should consider real representations of $\mathfrak{sl}_2(\mathbb C)$" is nonsense and highly misleading on many levels, please delete it. Really, the world is more complex than that. – Torsten Schoeneberg Jul 19 '19 at 17:02
  • Yes, can someone justify what this means "We live in a real world, so we should consider real representations of 2(ℂ)" ? – wonderich Jul 27 '20 at 15:17
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Firstly, what book is this? It will help greatly if I can reference it myself.

It is highly likely that when he says $\mbox{SO}(1,3)$ [or $\mbox{SO}(3,1)$!] that he means $\mbox{SO}(1,3)_\uparrow$, which is absolutely not the same! But most people are very lazy about this.

Here you're picking out the region of $\mbox{O}(1,3)$ path-connected to the identity element, where $\mbox{O}(1,3)$ consists of four disconnected regions, labeled by

$$ \det(L) = \pm1$$

and

$$ L^{00} > 1 \space \mbox{ or } \space L^{00} < -1 $$

Then we have

$$ (\mbox{SU}(2) \times \mbox{SU}(2))/ \mathbb{Z_2} \simeq \mbox{SO}(4) $$

You can show this by considering the action of each of $\mbox{SU}(2)$ and $\mbox{SO}(4)$ on 2-complex-dimensional and 4-dimensional vectors respectively. You'll find that

$$ (x^1)^2 + (y^1)^2 + (x^2)^2 + (y^2)^2 = 1 $$

and

$$ (x^1)^2 + (x^2)^2 + (x^3)^2 + (x^4)^2 = 1 $$

respectively, up to normalisation. Here we have to quotient out $\mathbb{Z_2}$ since we want only those $U \in \mbox{SU}(2) $ which have

$$ \det(U) = 1 $$

Then $\mbox{SO}(4)$ is to Euclidean space as $\mbox{SO}(1,3)$ is to Minkowski space (using the Russian metric). This is why he says that $\mbox{SO}(1,3)$ is essentially $\mbox{SU}(2) \times \mbox{SU}(2)$, but shys away of declaring that the former is the latter (which would be an incorrect statement).

Next, you need to recall that $\mbox{SL}(2,\mathbb{C})$ is the real part of the complexification of $\mbox{SU}(2) \times \mbox{SU}(2)$. That is, $\mbox{SL}(2,\mathbb{C})$ is a double cover of $\mbox{SU}(2)$. This is because, when you complexify, and then take the real parts, you get two copies of $\mbox{SU}(2)$. Think of the way when we 'complexify' $\mathbb{R}$ and get $\mathbb{C}$, and we know that we can always write, for $z \in \mathbb{C}$

$$ z = x + i y $$

where $x, y \in \mathbb{R}$. So if we took the real parts of $\mathbb{C}$ we would get two copies of $\mathbb{R}$, the $x$ and the $y$, $$ \mathbb{C} \simeq \mathbb{R^2} $$

We can do the same thing with Lie Algebras, since they are just vector spaces after all, just like $\mathbb{R}$ and $\mathbb{C}$ are (maybe a little less trivial however!).

DanielC
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Flint72
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Perhaps it also helps understanding real forms and complexification, not having to look at representations at this time ...

SL2(C) as the complexification of SU(2), which is the compact real form, hence exhibiting a "doubling" construction allowing to look at a decomposition (Re & Im) and project etc.

Alternatively SL2(C) has two real forms, depending on the signature of the Killing form: SL2(R) (non-compact with signature (2,1)) and SU(2) (signature (0,2)); then it is clear the the complexification of the real non-compact form SL(2;R) is the non-compact SL(2;C) (not changing signature), helping understand why the same complexification of SU(2), with negative definite Killing form "turns into" the "same" Lie algebra and group (over complex numbers as coefficients there is only one signature).

The PSL2(C)=SO(3,1) is indeed better understood as Mobius Transformations correspond to Lorentz Transformations (especially for Physicists). Or looking at "doubling C" (H=CXC spinors / Quantum Computing picture) and decompose as H=R X (R^3, x) (H as a Lie algebra central extension - trivial). These are deep connections i would like to understand better too ...

Lucian Ionescu
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