Electron Wavefunction After a Photon–Electron Collision

Short answer: you generally cannot write a single pure-state wavefunction for the electron after the collision unless you condition on the detected outgoing photon. The photon and electron scatter into an entangled state. If you do not observe the photon, the electron is left in a mixed state (use a density matrix, not a single wavefunction).

Setup

Initial (asymptotic “in”) state:

  • Photon plane wave \lvert \gamma:\mathbf{k},\varepsilon \rangle
  • Electron wavepacket \lvert \psi_e^{\mathrm{in}} \rangle = \displaystyle \int d^3p\, \phi_{\mathrm{in}}(\mathbf{p})\, \lvert \mathbf{p} \rangle

The scattering operator S produces the entangled “out” state:

    \[ \lvert \Psi_{\mathrm{out}} \rangle = \sum_{\varepsilon'} \int d^3k'\, d^3p'\; \mathcal{M}_{\varepsilon'\varepsilon}\!\big(\mathbf{p},\mathbf{k} \to \mathbf{p}',\mathbf{k}'\big)\, \phi_{\mathrm{in}}(\mathbf{p})\; \lvert \gamma:\mathbf{k}',\varepsilon' \rangle \otimes \lvert \mathbf{p}' \rangle, \]

Energy–momentum conservation is built into the matrix element \mathcal{M}.

Electron State Conditioned on Detecting the Outgoing Photon (\mathbf{k}', \varepsilon')

Project onto \langle \gamma:\mathbf{k}',\varepsilon' \rvert and renormalize. Let the momentum transfer be \mathbf{q} = \mathbf{k} - \mathbf{k}'. Then the electron’s conditional momentum-space wavefunction is

    \[ \boxed{ \phi_{\mathrm{out}}^{(\mathbf{k}',\varepsilon')}(\mathbf{p}') \;\propto\; \mathcal{M}_{\varepsilon'\varepsilon}\!\big(\mathbf{p}'-\mathbf{q},\mathbf{k}\to \mathbf{p}',\mathbf{k}'\big)\; \phi_{\mathrm{in}}(\mathbf{p}'-\mathbf{q}) } \]

Intuition: the packet is shifted by the recoil \mathbf{q}, weighted by the Compton/Thomson matrix element.

Position-Space Picture

    \[ \boxed{ \psi_{\mathrm{out}}^{(\mathbf{k}',\varepsilon')}(\mathbf{r}) \;\propto\; \Big[\mathcal{A}_{\varepsilon'\varepsilon}(\hat{\mathbf{k}} \to \hat{\mathbf{k}}',\,\mathbf{p})\Big]\; e^{\,i\,\mathbf{q}\cdot \mathbf{r}}\; \psi_{\mathrm{in}}(\mathbf{r}) } \]

That is, the electron wavefunction acquires a plane-wave phase factor e^{i\mathbf{q}\cdot\mathbf{r}} (a momentum kick) times an angular/polarization-dependent scattering amplitude \mathcal{A}.

Useful Limits

Thomson (Low-Energy) Limit: \hbar\omega \ll m_e c^2

Here \mathcal{M} is nearly momentum-independent and \mathcal{A} \propto r_e\,(\varepsilon'\!\cdot\!\varepsilon) with the classical electron radius r_e. Then

    \[ \phi_{\mathrm{out}}(\mathbf{p}') \propto \phi_{\mathrm{in}}(\mathbf{p}' - \mathbf{q}), \qquad \psi_{\mathrm{out}}(\mathbf{r}) \propto e^{i\mathbf{q}\cdot \mathbf{r}} \,\psi_{\mathrm{in}}(\mathbf{r}). \]

Compton Regime (Relativistic Photon Kinematics)

The detected photon energy fixes the recoil via the Compton formula

    \[ \lambda' - \lambda = \lambda_C (1 - \cos\theta), \qquad \lambda_C = \frac{h}{m_e c}, \]

and \mathcal{M} reduces to the Klein–Nishina amplitude, delivering the usual angular/polarization factors. The momentum-shift expression above still holds, with \mathcal{M} supplying the correct relativistic weights.

If You Do Not Detect the Photon

You must trace out the photon degrees of freedom:

    \[ \rho_e^{\mathrm{out}} = \mathrm{Tr}_{\gamma}\!\big[\, \lvert \Psi_{\mathrm{out}} \rangle \langle \Psi_{\mathrm{out}} \rvert \, \big] = \sum_{\varepsilon'} \int d^3k'\; \lvert \psi_e^{(\mathbf{k}',\varepsilon')} \rangle \langle \psi_e^{(\mathbf{k}',\varepsilon')} \rvert, \]

which is generally a mixed state—not representable by a single wavefunction.

Takeaway

  • The collision creates entanglement between the photon and electron.
  • Conditioned on a specific scattered photon (\mathbf{k}',\varepsilon'), the electron’s wavefunction is its initial packet shifted by the recoil \mathbf{q}=\mathbf{k}-\mathbf{k}' and weighted by the appropriate scattering amplitude.
  • Unconditioned, the electron is in a mixed state, so a density matrix is required.