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I'm having trouble understanding why is $\psi(\Lambda^{-1}p')$ the correct wave function in the Lorentz transformed frame $p' = \Lambda p$.

Suppose the state in frame $O$ is given by

$$ |\Psi\rangle = \int dp\, \psi(p)\, |p\rangle $$

then in frame $O'$ with $p' = \Lambda p$, the same state has representation

$$ |\Psi\rangle = \int dp'\, \psi'(p')\, |p'\rangle $$

where $|p'\rangle = U(\Lambda) |p\rangle =|\Lambda p\rangle$ and $\psi'(q') = \langle q' | \Psi \rangle$ (all quantities unprimed $p, q, . . .$ are in frame $O$, all quantities primed $p', q', . . .$ are in frame $O'$). Calculating now

$$ \langle q' | \Psi \rangle = \langle q | U^{\dagger}(\Lambda)|\Psi\rangle = \int dp\, \psi(p)\, \langle q | U^{\dagger}(\Lambda)\, |p\rangle = \int dp\, \psi(p)\, \delta(q - \Lambda^{-1} p) = \psi(\Lambda q) $$

where the last step follows since $q - \Lambda^{-1} p = 0$ gives $p = \Lambda q$. So all in all $\psi'(q') = \psi(\Lambda q)$ but instead we know the correct answer is $\psi'(q') = \psi(\Lambda^{-1} q')$. Where is the mistake in the reasoning?

Note: similar questions have been asked before but none addresses the issue I'm raising, namely, how to convince oneself that the function space transformation $\psi(\Lambda^{-1} q')$ arises in the way shown above.

konstam56
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1 Answers1

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For the mass $m>0$, spin $j$ representation of the (universal cover of the) Poincaré group, the theory of induced representations give the transformation law $$(u(A,a)\phi)(p) = e^{ip\cdot a} D^{(j)}(A_pAA_{\Lambda(A^{-1})p})\phi(\Lambda(A^{-1})p),\qquad \forall\phi\in L^2(\Omega_m^+,\text d\Omega_m^+)\otimes\mathbb C^{2j+1}$$ where $(A,a)\in SL_2(\mathbb C)\ltimes\mathbb R^n$, $D^{(j)}$ is the spin-$j$ irreducible representation of the little group $SU(2)$ (recall that I have assumed $m>0$ here for definiteness), $p\mapsto A_p$ is any section from the orbit $\Omega_m^+$ to Lorentz transformations in $SL_2(\mathbb C)$ (the exact details of these objects come from the Mackey-Wigner theory of induced representations) and $\text d\Omega_m^+$ is the invariant measure.

For spin $j=0$, this transformation law reduces to $$(u(A,a)\phi)(p) = e^{ip\cdot a}\phi(\Lambda(A^{-1})p),\qquad p\in\Omega_m^+,$$ and if we neglect the action of the translation semigroup this further reduces to $$(u(A,0)\phi)(p) = \phi(\Lambda(A^{-1})p),\qquad p\in\Omega_m^+.$$ Taking a Fourier transform to go from momentum space to position space, the transposed action is seen to be $$(u(A,0)\psi)(x) = \psi(\Lambda(A^{-1})x),\qquad x\in\mathbb R^4,$$ which is the special case of the more general relation $$(u(A,a)\psi)(x) = \psi(\Lambda(A^{-1})(x-a)),\qquad x\in\mathbb R^4.$$

Phoenix87
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