# Lorentz transforming a momentum eigenstate

Tags:
1. Oct 20, 2015

### terra

Let's take a quantum state $\Psi_p$, which is an eigenstate of momentum, i.e. $\hat{P}^{\mu} \Psi_p = p^{\mu} \Psi_p$.
Now, Weinberg states that if $L(p')^{\mu}\,_{\nu}\, p^{\nu} = p'$, then $\Psi_{p'} = N(p') U(L(p')) \Psi_{p}$, where $N(p')$ is a normalisation constant. How to construct an unitary operator like that, or how to even see that such an operator is to be found?

Some thoughts.
I define $\Lambda^{\mu}\,_{\nu} \, k^{\mu} = p^{\mu}$. Component by component:
\begin{align*}
\Lambda_{00} &= \gamma \\
\Lambda_{0i} = \Lambda_{i0} &= \gamma \beta_i \\
\Lambda_{ij} = \Lambda_{ji} &= (\gamma - 1) \frac{\beta_i \beta_j }{ \beta^2 } + \delta_{ij}.
\end{align*}
Now let's try $U(\Lambda) \Psi_{k,\sigma} := \sum_{\rho} D^{(j)}\,_{\rho,\sigma}\,(\Lambda) \Psi_{k, \rho}$, where $D(\Lambda)$ is Wigner's D-matrix, which gets different representations depending on the range of the state's quantum numbers $\sigma$: $\sigma = j, j-1, ..., -j+1, -j$, altogether $2j+1$ possible values.

If we take $\Psi_{k,\sigma}$ to be a scalar particle for which $\sigma = j = 0$, $D^{(0)}\,_{0,0}\,(\Lambda) = 1$ so that $\Psi_{p,0} = N(p) \Psi_{k,0}$. In Dirac's notation, $\Psi_{k} = | \boldsymbol{k} \rangle = \int d^3 \boldsymbol{x} \, | \boldsymbol{x} \rangle \langle \boldsymbol{x} | \boldsymbol{k} \rangle = \int d^3 \boldsymbol{x} \, | \boldsymbol{x} \rangle e^{-i \boldsymbol{k} \cdot \boldsymbol{x}}$, and $\Psi_p = N(p) \Psi_k = \int d^3 \boldsymbol{x} \, | \boldsymbol{x} \rangle N(p) e^{-i \boldsymbol{k} \cdot \boldsymbol{x}}$. For these to be equal, I could choose $N(p) = \int d^3 \boldsymbol{q} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{q} + i \boldsymbol{k} \cdot \boldsymbol{q}}$, so that
\begin{align*}
N(p) \Psi_{k} &= \int d^3\boldsymbol{q} \, \int d^3\boldsymbol{x} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{q} + i \boldsymbol{k} \cdot \boldsymbol{q} - i \boldsymbol{k} \cdot \boldsymbol{x}} |\boldsymbol{x} \rangle \\
&= \int d^3\boldsymbol{x} \, \delta^{(3)}(\boldsymbol{p} - \boldsymbol{q}) e^{-i \boldsymbol{k} \cdot \boldsymbol{x}} |\boldsymbol{x} \rangle \\
&= \int d^3\boldsymbol{x} \, e^{-i \boldsymbol{p} \cdot \boldsymbol{x}} |\boldsymbol{x} \rangle \\
&= \Psi_{p,0}
\end{align*}
However, this choice for $N(p)$ also depends on $\boldsymbol{k}$...

Last edited: Oct 20, 2015
2. Oct 21, 2015

### Gavin Gong

$U(\Lambda) \Psi_{k,\sigma} := \sum_{\rho} D^{(j)}\,_{\rho,\sigma}\,(\Lambda) \Psi_{k, \rho}$,
this line is not correct, unless you assume that
$\Lambda^{\mu}\,_{\nu} \, k^{\mu} = k^{\mu}$,
so the result is
$\Psi_{k,0} = N(p) \Psi_{k,0}$.
and $N(p)=1$
actually N(p) is just a constant to fix the different normalization conventions between states, and it isn't so important here.

the logic here is: if you assume the physical states have a lorentz symmetry, there must be a representation on the hilbert space.

if you want to construct the explict form of $U(\Lambda)$,you can refer to 2.4 of weinberg's book.

Last edited: Oct 21, 2015
3. Oct 30, 2015

### terra

That's right, sorry, I was sloppy adding that.
Weinberg shows that
$$U(\Lambda) \Psi_{p,\sigma} = \sum_{\rho} D^{(j)}_{\rho \sigma}(W(\Lambda,p)) \Psi_{\Lambda p, \rho}$$
where $W(\Lambda,p)$ is such that $W^{\mu}_{\nu} k^{\nu} = k^{\mu}$ so such transformations form a little group for every 'class' of momentum eigenstates. Division to 'classes' is obtained by noticing which quantities stay invariant under every proper orthocronous Lorentz transformation. Three physical 'classes' are obtained: the vacuum for which $p^2 = 0, p^0=0$, a massless particle for which $p^2 = 0, p^0 > 0$ and a massive particle $p^2 < 0, p^0 > 0$.
Weinberg then goes on to explaining how to obtain a $2j+1$ dimensional irreducible representation from the generators: angular momentum operators. On the other hand he gives a possible decomposition of Wigner rotation $W$ into boosts and rotations that operate on four-vectors. But he doesn't seem to connect these two types of objects, $W$ and $D(W)$. That is, how to obtain a representation for $W = L^{-1}(\Lambda p) \Lambda L(p)$ in $2j+1$ dimensions?

Actually, as for massive particles $W$ is a spatial rotation, I should indeed be able to construct a representation for the rotations. But, I will not know how will it be related to boosts performed on four vectors. That is, I know $D(\alpha, \beta, \gamma)$ and a way to write $W(\Lambda,p)$ but not $D(W(\Lambda, p))$.

Last edited: Oct 30, 2015