0
$\begingroup$

Given the following photon field operators \begin{align} \mathbf{A}(\mathbf{r}) &=& \sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar}{2 \omega V\epsilon_0}} \left(\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{r}} + \bar{\mathbf{e}}^{(\mu)} {a^\dagger}^{(\mu)}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{r}} \right) \\ \mathbf{E}(\mathbf{r}) &=& i\sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar\omega}{2 V\epsilon_0}} \left(\mathbf{e}^{(\mu)} a^{(\mu)}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{r}} - \bar{\mathbf{e}}^{(\mu)} {a^\dagger}^{(\mu)}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{r}} \right) \\ \mathbf{B}(\mathbf{r}) &=& i\sum_{\mathbf{k},\mu} \sqrt{\frac{\hbar}{2 \omega V\epsilon_0}} \left((\mathbf{k}\times\mathbf{e}^{(\mu)}) a^{(\mu)}(\mathbf{k}) e^{i\mathbf{k}\cdot\mathbf{r}} - (\mathbf{k}\times\bar{\mathbf{e}}^{(\mu)}) {a^\dagger}^{(\mu)}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{r}} \right), \\ \end{align} how do these photon field operators change our understanding of the classical electromagnetic stress-energy tensor \begin{align} T^{\mu\nu} =\begin{bmatrix} \frac{1}{8\pi}(E^2+B^2) && S_x/c && S_y/c && S_z/c \\ S_x/c && -\sigma_{xx} && -\sigma_{xy} && -\sigma_{xz} \\ S_y/c && -\sigma_{yx} && -\sigma_{yy} && -\sigma_{yz} \\ S_z/c && -\sigma_{zx} && -\sigma_{zy} && -\sigma_{zz} \end{bmatrix} ? \end{align}

$\endgroup$
  • $\begingroup$ What do you mean by "change the understanding"?. By the promotion of fields to operators, conserved currents also become operators. I'm not sure what else you would want to know. Note that if that's a follow up to our comments on this, you've changed the paradigm from QM to QFT in this question. $\endgroup$ – ACuriousMind Jul 18 '14 at 15:47
  • $\begingroup$ Is the stress-energy tensor possibly an operator similar to how the field is treated as operator? $\endgroup$ – linuxfreebird Jul 18 '14 at 15:51
  • $\begingroup$ Well, it contains only expressions that are derived from the field, so its just a collection of sums and powers of the field operators. But operators are operators, so I still don't get your question. $\endgroup$ – ACuriousMind Jul 18 '14 at 15:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.