6
$\begingroup$

If the energy of a photon

$E_{p}=hv$

And the energy of an electromagnetic wave is

$E_{w}\propto \hat{\mathbf B}^2$

What is the relationship between $E_{w}$ and $E_{p}$?

$\endgroup$
0

2 Answers 2

11
$\begingroup$

You only need to rewrite $\mathbf B$ and $\mathbf E$ in terms of field $A_{\mu}$ (here $\hbar = c = 1$), $$ \tag 1 \hat{\mathbf B} = [\nabla \times \hat{\mathbf A}], \quad \hat{\mathbf E} = -\frac{\partial \hat{\mathbf A}}{\partial t} - \nabla \hat{A}_{0}, $$ which is written as infinite "sum" of photons: $$ \tag 2 A_{\mu} = \sum_{\lambda} \int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2E_{\mathbf p}}}e_{\mu}^{\lambda}(\mathbf p )\left( \hat{a}_{\lambda}(\mathbf p )e^{-ipx} + \hat{a}_{\lambda}^{\dagger}(\mathbf p ) e^{ipx}\right). $$ After that you can easily obtain the relation between energies of sets of photons and "real" EM field: $$ \tag 3 \hat{H} = \int \hat{T}_{00}d^{3}\mathbf r = \int \frac{1}{2}\left( \hat{\mathbf B}^{2} + \hat{\mathbf E}^{2}\right)d^{3}\mathbf r. $$

If you need I'll derive it.

Tedious derivation

For simplicity you need Coulomb gauge $A_{0} = 0, (\nabla \cdot \mathbf A) = 0$ (eq. $(3)$ already implies that), polarization sum rule and orthogonality relations for polarization vectors, $$ \sum_{\lambda}e_{i}^{\lambda}(\mathbf p)e_{j}^{\lambda}(\mathbf p) = \delta_{ij}, \quad (\mathbf e_{\lambda}(\mathbf p) \cdot \mathbf e_{\lambda'}(\mathbf p)) = \delta_{\lambda\lambda'}. $$ and commutation relations $$ [\hat{a}_{\lambda}(\mathbf p), \hat{a}^{\dagger}_{\lambda {'}}(\mathbf k)] = \delta_{\lambda \lambda {'}}\delta (\mathbf p - \mathbf k), \quad [\hat{a}_{\lambda}(\mathbf p ), \hat{a}_{\lambda{'}}(\mathbf k)] = 0. $$ First let's calculate $(1)$ by using $(2)$ ($E_{\mathbf p} = p_{0}$): $$ \hat{\mathbf E}(x) = -\partial_{0}\hat{\mathbf A}(x) = i\sum_{\lambda}\int \frac{d^{3}\mathbf p}{\sqrt{2 (2 \pi )^{3}}}\mathbf e_{\lambda}(\mathbf p) \sqrt{E}_{\mathbf p}\left( \hat{a}_{\lambda}(\mathbf p )e^{-ipx} - \hat{a}^{\dagger}_{\lambda}(\mathbf p )e^{ipx}\right), $$ $$ \hat{\mathbf B}(x) = [\nabla \times \hat{\mathbf A}] = i\sum_{\lambda}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2E_{\mathbf p}}}[\mathbf p \times \mathbf e_{\lambda}(\mathbf p)]\left( \hat{a}_{\lambda}(\mathbf p )e^{-ipx} - \hat{a}^{\dagger}_{\lambda}(\mathbf p )e^{ipx}\right). $$ Then $$ \int d^{3}\mathbf r \hat{\mathbf E}^{2} = -\sum_{\lambda , \lambda '}\int \frac{d^{3}\mathbf r d^{3}\mathbf p d^{3}\mathbf k}{(2 \pi )^{3}2}\sqrt{E_{\mathbf p}E_{\mathbf k}}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(\mathbf k )) \times $$ $$ \times \left( \hat{a}_{\lambda}(\mathbf p )e^{-ipx} - \hat{a}^{\dagger}_{\lambda}(\mathbf p )e^{ipx}\right) \left( \hat{a}_{\lambda {'}}(\mathbf k )e^{-ikx} - \hat{a}^{\dagger}_{\lambda {'}}(\mathbf k )e^{ikx}\right) = \left| \frac{1}{(2\pi )^{3}}\int d^{inx}d^{3}\mathbf r = \delta (\mathbf n) e^{in_{0}x_{0}}\right| = $$ $$ = -\sum_{\lambda , \lambda {'}}\frac{1}{2}\int d^{3}\mathbf p d^{3}\mathbf k \sqrt{E_{\mathbf p}E_{\mathbf k}}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(\mathbf k )) \times $$ $$ \times \delta (\mathbf p + \mathbf k)\left( e^{ix_{0}(k_{0} + p_{0})}\hat{a}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(\mathbf k) + e^{-ix_{0}(k_{0} + p_{0})}\hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(\mathbf k) \right) + $$ $$ +\sum_{\lambda , \lambda {'}}\frac{1}{2}\int d^{3}\mathbf p d^{3}\mathbf k \sqrt{E_{\mathbf p}E_{\mathbf k}}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(\mathbf k )) \times $$ $$ \times \delta (\mathbf p - \mathbf k) \left( e^{ix_{0}(k_{0} - p_{0})}\hat{a}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(\mathbf k) + e^{-ix_{0}(k_{0} - p_{0})}\hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(\mathbf k)\right) = $$ $$ = -\frac{1}{2}\sum_{\lambda , \lambda {'}}\int d^{3}\mathbf p E_{\mathbf p}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(-\mathbf p )) \left( e^{2ip_{0}x_{0}}\hat{a}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(-\mathbf p) + e^{-2ix_{0}p_{0}}\hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(-\mathbf p) \right) + $$ $$ \tag 4 + \frac{1}{2}\sum_{\lambda , \lambda {'}}\int d^{3}\mathbf p E_{\mathbf p}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(\mathbf p )) \left( \hat{a}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(\mathbf p) + \hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(\mathbf p) \right). $$ The same thing with $\int d^{3}\mathbf r\hat{\mathbf B}^{2}$ by using relation $$ ([\mathbf p \times \mathbf e_{\lambda}(\mathbf p)] \cdot [\mathbf k \times \mathbf e_{\lambda {'}}(\mathbf k)]) = (\mathbf p \cdot \mathbf k)(\mathbf e_{\lambda}(\mathbf p) \cdot \mathbf e_{\lambda {'}}(\mathbf k)) - (\mathbf p \cdot \mathbf e_{\lambda}(\mathbf p)) (\mathbf k \cdot \mathbf e_{\lambda {'}}(\mathbf k)) = $$ $$ = (\mathbf p \cdot \mathbf k)(\mathbf e_{\lambda}(\mathbf p) \cdot \mathbf e_{\lambda {'}}(\mathbf k)) $$ can give $$ \int d^{3}\mathbf r \hat{\mathbf B}^{2} = $$ $$ = \frac{1}{2}\sum_{\lambda , \lambda {'}}\int d^{3}\mathbf p E_{p}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(-\mathbf p )) \left( e^{2ip_{0}x_{0}}\hat{a}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(-\mathbf p) + e^{-2ix_{0}p_{0}}\hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(-\mathbf p) \right) $$ $$ \tag 5 + \frac{1}{2}\sum_{\lambda , \lambda {'}}\int d^{3}\mathbf p E_{p}(\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(\mathbf p )) \left( \hat{a}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(\mathbf p) + \hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(\mathbf p) \right). $$ So after summation of $(4), (5)$ you will get that $$ \hat{H} = \frac{1}{2}\sum_{\lambda , \lambda {'}}\int d^{3}\mathbf p (\mathbf {e}_{\lambda }(\mathbf p ) \cdot \mathbf {e}_{\lambda {'}}(\mathbf p )) E_{\mathbf p} \left(\hat{a}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda {'}}(\mathbf p) + \hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}_{\lambda {'}}(\mathbf p) \right) = $$ $$ \tag 6 \frac{1}{2}\sum_{\lambda}\int d^{3}\mathbf p E_{\mathbf p}\left(\hat{a}_{\lambda}(\mathbf p) \hat{a}^{\dagger}_{\lambda}(\mathbf p) + \hat{a}^{\dagger}_{\lambda}(\mathbf p) \hat{a}_{\lambda}(\mathbf p) \right) = \sum_{\lambda}\int d^{3}\mathbf p E_{\mathbf p}\left( \hat{a}_{\lambda}^{\dagger}(\mathbf p)\hat{a}_{\lambda}(\mathbf p) + \delta (0)\right). $$ Eq. 6 implies "representation" of the energy of EM field as sum of energies of photons ($E_{\mathbf p} = \omega_{\mathbf p}$), because $\int d^{3}\mathbf p \hat{a}_{\lambda}^{\dagger}(\mathbf p)\hat{a}_{\lambda}(\mathbf p)$ refers to the particles number operator.

$\endgroup$
32
  • 2
    $\begingroup$ Could you get it? $\endgroup$ Sep 14, 2014 at 14:47
  • $\begingroup$ This might be more useful to explain some of the terms here and maybe even calculate it! $\endgroup$
    – Kyle Kanos
    Sep 14, 2014 at 14:47
  • 2
    $\begingroup$ @J.D'Alembert : I afraid that in this cumbersome and non-interested math is possible to sink, but I've finished to derive the relation. $\endgroup$ Sep 14, 2014 at 16:56
  • 1
    $\begingroup$ Bravo! Nice answer, much more specific than I would have expected. +1 $\endgroup$
    – Danu
    Sep 14, 2014 at 17:22
  • 1
    $\begingroup$ @AndrewMcAddams Sorry that it took me this long to accept it my internet was disconnected and thank you for all the work that you put into your answer. $\endgroup$ Sep 14, 2014 at 17:59
5
$\begingroup$

An experimentalists answer:

If you divide the energy of the electromagnetic wave by hv you will have the number of photons that are building up the electromagnetic wave.

$\endgroup$
7
  • $\begingroup$ Isn't a macroscopic electromagnetic wave something more similar to a coherent state, rather than a Fock state? A coherent state has uncertain amount of photons. What does the number (Ew / hv) actually represent? $\endgroup$
    – mpv
    Sep 14, 2014 at 18:17
  • $\begingroup$ @mpv conservation of energy? $\endgroup$
    – anna v
    Sep 14, 2014 at 19:10
  • $\begingroup$ @mpv: the expectation value for the number of photons, I think. $\endgroup$ Sep 15, 2014 at 4:21
  • $\begingroup$ @HarryJohnston from the analysis by Andrew I would think it would be the expectation of the average number of photons. But conservation of energy is an exact relation, so I expect it will be the number of photons for the energy carried by the wave. Any uncertainty will come from the classical measurement of energy, imo, not from h*nu. $\endgroup$
    – anna v
    Sep 15, 2014 at 5:48
  • $\begingroup$ I think what happens is that if you measure the number of photons, that makes the frequency of the photons uncertain, in just the same way that measuring the position of a quantum particle makes its momentum uncertain. So if you do make this measurement the answer might be that there are fewer photons than you expected, but since the frequency distribution has changed you still have the same total energy. $\endgroup$ Sep 15, 2014 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.