# What is the matrix representation of the momentum operator (generator of translations) that is used in the commutators of the Poincaré Group?

So the commutators of the Poincareé group are given by

Where the $P$'s are the generators of translation (linear momentum operators), the $J$'s are the generators of rotations (Angular Momentum operators), $K$'s are boost generators, $H$ is the energy.

The boost generators are

\begin{eqnarray} \textbf{K}=\left\{ \left(\begin{array}{cccc} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0 & i & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \end{array}\right) \right\} \end{eqnarray}

And the AM operators are

\begin{eqnarray} \textbf{J}=\left\{ \left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & -i & 0 & 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \right\} \end{eqnarray}

It's fairly straightforward to derive the above from the boost matrices and the rotation matrices, respectively, but I'm pretty confused about what the $P$ matrices are and how to derive them. I'm sure I'm overlooking something simple, but none of the texts I have seem to do this explicitly. Can anyone help?

• This is a quite valid question, but have you taken a look at the wikipedia article on translations? I think it pretty much says everything you need. Check it at en.wikipedia.org/wiki/Translation_%28geometry%29 Jun 30, 2016 at 16:27

The boost generators are

First of all, note that You've chosen specific representation of Lie algebra generators of Poincare group, which is vector-like matrix representation. There are many representations in general (below I'll write about them).

In Your question, You've chosen the matrix representation of Poincare group algebra generators in pseudo-euclidean space. The translation transformation is $$x_{a}\to x_{a}+b_{a}$$ Note that it isn't linear transformation in Minkowski space-time, so it can't be represented in terms of matrices.

It can be, however, made linear, if we embed Minkowski space-time into fictive 5-dimensional space-time with extra coordinate $x_{5}$. Then Poincare group transformations now are in matrix form: with $x = x_{\mu}, a = a_{\mu}, \Lambda = \Lambda_{\mu\nu}$ we have $$\begin{pmatrix} x{'} \\ x_{5}{'} \end{pmatrix} = \begin{pmatrix} \Lambda & a \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ x_{5}\end{pmatrix}$$ Now You can get matrix representation of translation operator. Try to do that.

As for different representation, it is possible to represent generators in terms of differential operators. Namely, for group transformation $$x_{\alpha} = f_{\alpha}(a,x)$$ corresponding differential operator $\hat{X}$ is $$\hat{X}_{i} = \left(\frac{df^{\alpha}(x, a)}{da_{i}}\right)_{a = 0}\partial_{\alpha}$$ These generators, as can be shown, preserve Lie group algebra. In terms of these operators, Poincare group generators $J_{\mu\nu},P_{\mu}$ are $$\hat{P}_{\mu} = i\partial_{\mu},\quad \hat{J}_{\mu\nu} = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})$$ (an edit)

What object do they act on, the wave-function?

These expressions are derived from generators, which You've obtained from the definition of Poincare group transformations of ordinary 4-vector. Getting the generators for "wave function" transformations is completely different story.

Briefly, suppose You have the world with Poincare symmetry. On the level of quantum mechanics this means that the modulus of scalar product $\langle \kappa |\psi\rangle$ of states $|\kappa\rangle |\psi\rangle$ of system is invariant under Poincare group transformations $|\psi\rangle \to |\psi'\rangle , \ |\kappa\rangle \to |\kappa'\rangle$: $$|\langle \psi |\kappa \rangle|^{2} = |\langle \psi{'} |\kappa{'} \rangle|^{2}$$ By Wigner theorem, this means that Poincare group transformations are realized linerly and unitary (for simplicity, here I omitted discussion about anti-linear anti-unitary case): $$|\psi{'}\rangle = U(\Lambda , a)|\psi\rangle$$ $$U(\Lambda , a) = \text{id} + ia^{\mu}\hat{P}_{\mu} + \frac{i}{2}\omega^{\mu \nu}\hat{J}_{\mu \nu},$$ where $\omega_{\mu \nu}, a_{\mu}$ are parameters of Poincare group transformation, while hermitean operators $\hat{P}_{\mu}, \hat{J}_{\mu \nu}$ are called Poincare group generators.

Note that in general their explicit form depends on representation. In general, for $|\psi \rangle$ being irreducible representation of the poincare group, $|\psi\rangle = |p, \sigma\rangle$ (with $p$ being momentum and $\sigma$ being the label of so-called little group of $p$), we can use following correspondence $$|\psi\rangle \to \hat{a}^{\dagger}_{\sigma} \to \hat{\Psi}_{A},$$ where $\hat{a}^{\dagger}_{\sigma}(\mathbf p)$ is the creation operator of the state with given momentum $p$ and spin projection (helicity) $\sigma$, $\hat{\Psi}_{A}$ is the creation-destruction field with (in general) spinor indices $A$ (it may be 4-vector operator, Dirac spinor operator, and etc). The number and structure of indices, and hence the transformation law of field under Lorentz group, are determined from the value of spin (helicity) of the representation. In general, $$\hat{J}_{\mu \nu} = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu}) + \hat{M}_{\mu \nu},$$ where $\hat{M}_{\mu \nu} = (M_{\mu \nu})_{A}^{\ B}$ is the matrix of generators of given finite dimensional representation of the Lorentz group. This determines transformation law of the "wave function" $\hat{\Psi}$ (precisely, the transformation of coefficients near $\hat{a}$, $\hat{a}^{\dagger}$ in the expansion of $\hat{\Psi}$).

The $\hat{P}$ is obviously the linear momentum operator with $\hbar = 1$, but what are the $\hat{J}$'s?

$\hat{P}_{\mu}$ is called 4-momemtum operator, while $\hat{J}_{\mu \nu}$ is called angular momentum operator. The reason of this is that they frequently can be obtained from classical stress-energy and angular momentum tensors (which are obtained from Noether theorem) by using correspondence rules. Correspondingly, $\hat{J}_{i} = \frac{1}{2}\epsilon_{ijk}\hat{J}^{jk}$ is called angular momentum operator, while $\hat{K}^{i} = -\hat{J}^{0i}$ is called boost operator. In classical limit $\hat{K}^{i}$ is associated with the vector of center of energy. This can be seen by using differential operators for 4-vector representation of the Poincare group.

• Thanks for this! Can you say a little more about the last line (or maybe point me somewhere)? The $\hat{P}$ is obviously the linear momentum operator with $\hbar=1$, but what are the $\hat{J}$'s? A few will be the AM operators, but the others are what? The "boost generators"? What object do they act on, the wave-function? Jul 3, 2016 at 14:00
• @quantum_loser : I've added some description. Jul 6, 2016 at 13:07