# Is there a relationship between the energy of a photon and the energy of an electromagnetic wave?

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 $E_{p}$?

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.

• Could you get it? Sep 14, 2014 at 14:47
• This might be more useful to explain some of the terms here and maybe even calculate it! Sep 14, 2014 at 14:47
• @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. Sep 14, 2014 at 16:56
• Bravo! Nice answer, much more specific than I would have expected. +1
– Danu
Sep 14, 2014 at 17:22
• @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. Sep 14, 2014 at 17:59