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I'm trying to find out how to quantize EM field. It seems like $\vec{A}$ and $\vec{E}$ are it's canonical coordinates. For example: $$\mathfrak{H} = \frac12E^2 + \frac12(\nabla\times A)^2$$ $$H = \int \limits_V \mathfrak{H} dV$$ Thus $$\frac{\partial \mathfrak{H}}{\partial E} = E = -\frac{\partial A}{\partial t} $$ But there is a problem for me: how to obtain functional derivative $\frac{\delta H}{\delta A}$? I'm sure there should be: $$\frac{\delta H}{\delta A} = [\nabla \times [\nabla \times A]] = \frac{\partial E}{\partial t},$$ as it is used in London's Equation for superconductors.

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  • $\begingroup$ Try writing vectorial expressions component-by-component using the fully anti-symmetric tensor $\epsilon_{ijk}$. $\endgroup$ – Dvij Mankad Feb 27 at 16:10
  • $\begingroup$ Also, what is $H$? Hamiltonian? Then you should have a volume integral there, or Hamiltonian density, in which case you need to use a different kind of derivative to get the time-evolution of canonical variables. $\endgroup$ – Cryo Feb 27 at 16:46
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If all you're asking is how to take the functional derivative $\delta H/\delta \vec{A}$, here's how: First, take the variation $\vec{A} \to \vec{A} + \delta \vec{A}$, plug it into $H$, and discard all terms not linear in $\delta \vec{A}$ to obtain $\delta H$. This becomes: $$ \delta H = \int (\nabla \times \vec{A}) \cdot (\nabla \times \delta \vec{A}) \, dV. $$ Next, integrate by parts, using the identity $\vec{v} \cdot (\nabla \times \vec{w}) = \nabla \cdot (\vec{v} \times \vec{w}) + \vec{w} \cdot (\nabla \times \vec{w})$: $$ \delta H = \int \left[ \nabla \cdot \left( (\nabla \times \vec{A}) \times \delta \vec{A} \right) + \delta \vec{A} \cdot \left( \nabla \times (\nabla \times \vec{A}) \right) \right] \, dV. $$ The first term can be written as an integral over the boundary of the volume in question; assuming that the variation $\delta \vec{A}$ vanishes on this boundary (which is a typical assumption), then this term goes away. And since by definition, the functional derivative $\delta H/\delta \vec{A}$ is defined by $$ \delta H = \int \left( \frac{\delta H}{\delta \vec{A}} \right) \cdot \delta \vec{A} \, dV $$ for all variations $\delta \vec{A}$, we conclude that $$ \frac{\delta H}{\delta \vec{A}} = \nabla \times (\nabla \times \vec{A}). $$

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You have a partial derivative, you show what you differentiate with respect to, but it is not clear what you keep fixed. Can you please specify all your independent variables?

The reason I ask for clarity is that under normal Lagrangian treatment the $A^j$ and $\nabla_i A^j$ are treated as independent variables, in which case:

$\left(\frac{\partial \nabla_i A^j}{\partial A^k}\right)_{\dots\, \nabla_s A^p} \equiv 0$


Ok so have a very standard situation. First things first, derivatives of vector potential are independent quantities from vector potential itself. Think of it this way. Given the value of the vector potential in one point $\mathbf{A}\left(\mathbf{r},t\right)$, can you tell me its gradient? No, you will need values of the neighbouring points, and more neighbouring points etc. Fundamentally this goes back to how you get the Euler-Lagrange equations by minimizing action.

Next, the Hamiltonian. Given Lagrangian,

$L=\int d^3 r \mathcal{L}\left(\dot{A}^i, \nabla_j A^i\right) = \frac{1}{2}\int d^3 r \left( \epsilon_0 \dot{A}^2 - \frac{1}{\mu_0}\left(\mathbf{\nabla}\times\mathbf{A}\right)^2\right)$

one can show that the quantity that is conserved in time, the Hamiltonian is:

$H=\frac{1}{2} \int d^3 r \left( \epsilon_0 \dot{A}^2 + \frac{1}{\mu_0}\left(\mathbf{\nabla}\times\mathbf{A}\right)^2\right)$

Next you need to decide how to proceed. A standard way to go is to limit the size of the space you are considering, and to look at Fourier transforms of the vector potential

$\mathbf{\tilde{A}}\left(\mathbf{k},t\right)=\frac{1}{\left(2\pi\right)^3}\int d^3 r \exp\left(i \mathbf{k}.\mathbf{r}\right) \mathbf{A}\left(\mathbf{r}, t\right)$.

This will convert the integral in the expression for the Hamiltonian into a sum (because you have assumed that the space under consideration is finite). You then end up with a Hamiltonian that is a sum of many simple harmonic oscillators, so one can easliy quantize. If you want to go back to real space, simply do the inverse fourier transform on the vector potential operator. This approach can be found in any book on quantum optics, i.e. Loudon's "The Quantum Theory of Light".

If you do not want to take the fourier transform it gets tricky, I don't have time to write it up now, so I will leave it.

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  • $\begingroup$ I think $A$ and $E$ are independent variables. That what I'm asking for: are there any accurate methods to quantize the EM field? $\endgroup$ – Flammifer Feb 27 at 16:59
  • $\begingroup$ What about $\mathbf{A}$ and $\mathbf{\nabla}\times\mathbf{A}$? I can show you a way of quantizing the field, but you need to be more specific. What is your Lagrangian? This matters because, for example, in presence of polarization density ($\mathbf{P}$), the canonical momentum density for the field is $\boldsymbol{\pi}=\epsilon_0 \mathbf{\dot{A}}-\mathbf{P}$, i.e. not the electric field. $\endgroup$ – Cryo Feb 27 at 17:09
  • $\begingroup$ $$\mathfrak{L} = \frac{1}{8 \pi}\int \limits_V (E^2 - B^2 ) dV $$ as it is in Landau-Lifshiz Course, Field Theory Polarisation remains constant everywhere $\endgroup$ – Flammifer Feb 27 at 17:21

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