In the paper "Unusual Transitions Made Possible by Superoscillations", the author begins by solving for a coherent state \begin{equation}|\alpha\rangle\end{equation} such that

\begin{equation} \left\langle B_{x}(\mathbf{r}, t)\right\rangle=F(z-c t)=\left\langle\alpha\left|B_{x}(\mathbf{r}, t)\right| \alpha\right\rangle \end{equation} For an arbitrary function F(z-ct). Where: \begin{equation} \mathbf{B}(\mathbf{r}, t)=\nabla \times \mathbf{A}(\mathbf{r}, t) \end{equation} \begin{equation} \mathbf{A}(\mathbf{r}, t)=\sum_{\mathbf{k} p} \sqrt{\frac{2 \pi \hbar c^{2}}{V \omega_{k}}}\left(a_{\mathbf{k}, p} \hat{\mathbf{e}}_{\mathbf{k}, p} e^{i\left(\mathbf{k} \cdot \mathbf{r}-\omega_{k} t\right)}+a_{\mathbf{k}, p}^{\dagger} \hat{\mathbf{e}}_{\mathbf{k}, p}^{*} e^{-i\left(\mathbf{k} \cdot \mathbf{r}-\omega_{k} t\right)}\right) \end{equation}

And the unit vectors are chosen such that the magnetic field is parallel to the $x$ unit vector.

Calculating the curl of the above explicitly given that $a_{\mathbf{k}^{\prime}, p^{\prime}}\left|\alpha_{\mathbf{k}, p}\right\rangle=$ $\delta_{\mathbf{k}, \mathbf{k}^{\prime}} \delta_{p, p^{\prime}} \alpha_{\mathbf{k}, p}\left|\alpha_{\mathbf{k}, p}\right\rangle$ , the author concludes:

\begin{equation} \begin{aligned} &\left\langle\alpha_{k, p}\left|B_{x}(\mathbf{r}, t)\right| \alpha_{k, p}\right\rangle= \sqrt{\frac{8 \pi \hbar \omega_{k}}{V}}\left(\operatorname{Re}\left[\alpha_{k}\right] \sin \left(k z-\omega_{k} t\right)+\operatorname{Im}\left[\alpha_{k}\right] \cos \left(k z-\omega_{k} t\right)\right) \end{aligned} \end{equation}

Then the author points out that $F(z-ct)$ can be represented by a Fourier transform: \begin{equation} F(z-c t)=\sum_{k}\left(A_{n} \cos \left(k_{n} z-\omega_{k_{n}} t\right)+B_{n} \sin \left(k_{n} z-\omega_{k_{n}} t\right)\right) \end{equation} \begin{equation} A_{n} \quad=\quad \frac{2}{V^{1 / 3}} \int d z F(z) \cos \left(k_{n} z\right) \end{equation} \begin{equation} B_n=\frac{2}{V^{1 / 3}} \int d z F(z) \sin \left(k_{n} z\right) \end{equation}

Up to here I'm pretty much good with the derivation. The point which I'm not clear on is that the author states that from the previous two equations, it directly follows that: \begin{equation} |\alpha\rangle=\prod_{k>0}\left|i \sqrt{\frac{V^{1 / 3}}{2 \pi \hbar \omega_{k}}} \tilde{F}(k)\right\rangle \end{equation}

Which is a product of coherent states which satisfies the wish that the expectation value of the magnetic field operator on the state is $F(z-ct)$. I feel like I may be missing some fundamental bit of physics which makes this follow logically but I'm not sure what that bit of physics is. For example, is it generally true that if an expectation value on a coherent state is a plane wave, the expectation value on a product of such coherent states is a sum of such plane waves?

I would appreciate if anyone could help me see what I'm missing here and understand the physics and mathematics here a bit better.

1 R. Ber and M. Schwartz, Unusual Transitions Made Possible by Superoscillations, (2015).


1 Answer 1


Let's just work with a scalar field $\phi$ and ignore normalization factors, to cut down on the amount of notational baggage.

We can expand the field operator $\phi(\vec{r}, t)$ in terms of creation and annihilation operators as \begin{equation} \phi(\vec{r}, t) = \int \frac{d^3 k}{(2\pi)^3} \left( a_\vec{k} e^{i (\vec k \cdot \vec r - \omega_k t)} + a^\dagger_\vec{k} e^{-i (\vec k \cdot \vec r - \omega_k t)} \right) \end{equation} The usual definition of a coherent state $|\alpha_{\vec{k}}\rangle$ is that it is an eigenvalue of the annhilation operator \begin{equation} a_\vec{k} | \alpha_\vec{k}\rangle = \alpha_\vec{k} | \alpha_{\vec k}\rangle \end{equation} Your state is a product of coherent states \begin{equation} |\Omega\rangle = \prod_\vec{k} | \alpha_{\vec{k}}\rangle \end{equation} Now we will sandwich $\phi(\vec{r}, t)$ in between the $|\Omega\rangle$ states \begin{eqnarray} \langle \Omega | \phi(\vec{r}, t) | \Omega\rangle &=& \langle \Omega |\left[\int \frac{d^3 k}{(2\pi)^3} \left( a_\vec{k} e^{i (\vec k \cdot \vec r - \omega_k t)} + a^\dagger_\vec{k} e^{-i (\vec k \cdot \vec r - \omega_k t)} \right)\right] | \Omega\rangle \\ &=& \int \frac{d^3 k}{(2\pi)^3} \left( \langle \Omega | a_\vec{k} | \Omega\rangle e^{i (\vec k \cdot \vec r - \omega_k t)} + \langle \Omega | a^\dagger_\vec{k} | \Omega\rangle e^{-i (\vec k \cdot \vec r - \omega_k t)} \right) \end{eqnarray} Now let's consider evaluating one of these expectation values. To make it transparent what is going on, suppose that the product in $\Omega$ is only over 3 values of momentum: $\vec{p}, \vec{k}, \vec{q}$. (I'll leave it as an exercise to generalize this to arbitrary products over momenta). Then... \begin{eqnarray} \langle \Omega | a_\vec{k} | \Omega \rangle &=& \left(\langle \alpha_\vec{p} | \langle \alpha_\vec{k} | \langle \alpha_\vec{q} |\right) a_{\vec{k}} \left(| \alpha_{\vec{p}} \rangle | \alpha_\vec{k}\rangle | \alpha_{\vec{q}}\rangle \right) \\ &=& \langle \alpha_\vec{p} | \alpha_\vec{p} \rangle \langle \alpha_\vec{k} | a_\vec{k} | \alpha_\vec{k} \rangle \langle \alpha_\vec{q} | \alpha_\vec{q}\rangle \\ &=& \langle \alpha_\vec{k} | a_\vec{k} | \alpha_\vec{k} \rangle \\ &=& \alpha_{\vec{k}} \end{eqnarray} To go from the first line to the second line, I used the fact that the operator $a_\vec{k}$ is properly a tensor product $1 \times 1 \times \cdots \times a_{\vec{k}} \times 1 \times ... $, where $1$ is the identity operator, and the product is taken over the different factors of Fock space with different values of the momentum. To go from the second to the third, I assumed each of the coherent states is normalized. The go from the third line to the fourth, I used the definition of a coherent state.

Given this, we can now rewrite the field expansion sandwiched between $|\Omega\rangle$ states as \begin{eqnarray} \langle \Omega | \phi(\vec{r}, t) | \Omega\rangle &=& \int \frac{d^3 k}{(2\pi)^3} \left( \alpha_k e^{i (\vec k \cdot \vec r - \omega_k t)} + \alpha_{\vec{k}}^\star e^{-i (\vec k \cdot \vec r - \omega_k t)} \right) \end{eqnarray} which is just the inverse Fourier transform of the ordinary function (not an operator) $\alpha_\vec{k}$. In other words, we can represent a "classical-like" state of the field $\phi$ by using a state $|\Omega\rangle$ that is a product of coherent states in each Fourier mode.

With some additional bookkeeping involving indices and polarization tensors and curl operators (but no new quantum concepts), you can use the exact same logic to understand the paper's statements about coherent states of the EM field.

  • $\begingroup$ thank you. To clarify one thing, there is no upper or lower limit on the spectrum of the Fourier transform of the field state or the function, so long as they match, is that right? So your example with three momenta values is just as good a field state as a field state whose Fourier transform has 6 momenta values or 1000 and vice versa? It all basically just depends on what field state you want to deal with? $\endgroup$
    – Cody Payne
    Commented Mar 7, 2022 at 5:40
  • $\begingroup$ @CodyPayne Yes, I just chose three values of momenta to make the algebra more transparent. You could have an infinite number of values of momenta and the same logic would work. $\endgroup$
    – Andrew
    Commented Mar 7, 2022 at 12:42

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