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What is the motivation for complexifying a Lie algebra?

In quantum mechanical angular momentum the commutation relations

$$[J_x,J_y]=iJ_z, \quad [J_y,J_z] = iJ_x,\quad [J_z,J_x] = iJ_y$$

become, on complexifying (arbitrarily defining $J_{\pm} = J_x \pm iJ_y$)

$$[J_+,J_-] = 2J_z,\quad [J_z,J_\pm] = \pm 2J_z.$$

and then everything magically works in quantum mechanics. This complexification is done for the Lorentz group also, as well as in the conformal algebra.

There should be a unified reason for doing this in all cases explaining why it works, & further some way to predict the answers once you do this (without even doing it), though I was told by a famous physicist there is no motivation :(

Qmechanic
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bolbteppa
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2 Answers2

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Short answer: complexifications facilitate representation theory.

In physics, we typically want to find representations of a Lie algebra $\mathfrak g$, and often times determining the representations of its complexification $\mathfrak g_\mathbb C$ is easier. Moreover, we have the following theorem (see ref 1. Proposition 4.6) which tells us that determining the representations of the complexification allows us to determine the representations of the original algebra.

Theorem. Let $\mathfrak g$ be a real Lie algebra, and let $g_\mathbb C$ be its complexification. Every finite-dimensional complex representation $\pi$ of $\mathfrak g$ has a unique extension to a complex-linear representation $\pi_\mathbb C$ of $\mathfrak g_\mathbb C$ \begin{align} \pi_\mathbb C(X+iY) = \pi(X) + i\pi(Y) \end{align} for all $X,Y\in\mathfrak g$. Furthermore, $\pi_\mathbb C$ is irreducible as a representation of $\mathfrak g_\mathbb C$ if and only if $\pi$ it is irreducible as a representation of $\mathfrak g$.

Example. Angular momentum in QM

In the case of the angular momentum in quantum mechanics, what physics books are doing mathematically is attempting to find representations of $\mathfrak {su}(2)$ acting on the Hilbert space of a given physical system. The complexification of $\mathfrak{su}(2)$ is $\mathfrak{sl}(2,\mathbb C)$, and $\mathfrak{sl}(2,\mathbb C)$ has a nice basis $J_\pm, J_z$ which has no counterpart in $\mathfrak{su}(2)$ and which makes determining representations much easier. The structure relations in the $J_\pm, J_z$ basis allow one to use "raising" and "lowering" operators.

Example. Lorentz algebra

In relativistic quantum field theory, we look for representations of $\mathfrak{so}(1,3)$. It turns out, quite happily, that when we complexify this algebra, it splits into a direct sum of complexified angular momentum algebras: \begin{align} \mathfrak{so}(1,3)_\mathbb C \cong \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C), \end{align} and since we already know the representation theory of the complexified angular momentum algebra so well, this makes studying the representations of the Lorentz algebra easy.

References:

  1. Hall, Lie Groups, Lie Algebras, and Representations
joshphysics
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20

From a mathematical perspective, to develop Lie algebra representation theory most efficiently, we need the field $\mathbb{F}$ of the Lie algebra to be algebraically closed. See, e.g. Ref. 1, where this assumption is used already at the beginning of Chapter II.

The situation for Lie algebras is similar to when we in linear algebra try to diagonalize, say, a real normal matrix. Such a matrix is always diagonalizable in an orthonormal set of eigenvectors, but the eigenvectors and eigenvalues could be complex. Even for physical systems which are manifestly real in nature, such complex eigenvectors and complex eigenvalues are often useful concepts.

In more detail, for an $n$-dimensional Lie algebra $\frak{g}$, we would like something similar to a Chevaller-basis to exist. This means (among other things) that it should be possible to pick a Cartan subalgebra (CSA) $\frak{h}$ with generators $H_i$, $i=1,\ldots, r$; where $r$ is the rank of $\frak{g}$; and supplemented with basis elements $E_a$, $a=1, \ldots n-r$, $$ {\frak g}~=~{\rm span}_{\mathbb{F}} \left( \{ H_i | i=1,\ldots, r\} \cup \{ E_a | a=1,\ldots, n- r\}\right) ,$$ with the property that the Lie bracket $[E_a,H_i]$ is proportional to $E_a$. The $E_a$ plays the role of raising and lowering operators, or equivalently, creation and annihilation operators.

All finite-dimensional semisimple complex Lie algebras have a Chevaller-basis.

Example: The Lie algebra $sl(2,\mathbb{C})$: Think of $H_i$ as $J_3$, and $E_a$ as $J_{\pm}$.

From a physical perspective, weights in the facts that, e.g.

  1. quantum theory uses complex Hilbert spaces, cf. this Phys.SE post and links therein;

  2. the complex Lie group $SL(2,\mathbb{C})$ happens to be the (double cover of the) restricted Lorentz group $SO^{+}(3,1)$, cf. e.g. this Phys.SE post;

  3. one may speculate that it is easier to construct physically sensible theories based on the category of (complex) analytic functions rather than, say, the category of real smooth functions.

References:

  1. J.E. Humphreys, Intro to Lie Algebras and Representation Theory, Graduate texts in Math 9, Springer Verlag.
M. A.
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Qmechanic
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