Can you depict the photon's EM field? Light are made of waves in the electromagnetic field, so I wonder what is the shape of that wave. I mean what would be the EM field surrounding a photon in a particular instant?
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Or alternatively, how is the (electromagnetic) energy of the photon distributed in space? This would be equivalent to consider a energy density function U(x,y,z) of the photon. Or at least U(x,y) where x is the direction of the photon and y some perpendicular direction. So the density spread in the y direction would give an idea of the "width" of the photon.
More generally and informally, could consider whether the energy of the photon is all in a single point or distributed in some region, or if the energy has a location at all.
 A: First, the photon is a neutral particle, so it does not source an electromagnetic field the same way an electron does.
Second, a photon with energy $E$ has an associated frequency which you can work out from $E = hf$ (where $h$ is Planck's constant), and wavelength $\lambda = c/f$ (where $c$ is the speed of light). So, at a very loose level, you can think of a photon as being somewhat like a small-amplitude wave packet with frequency $f$ and wavelength $\lambda$.
However, this picture is in fact a lie. The reason is subtle and quantum mechanical. It turns out it is impossible to simultaneously measure the number of photons and the phase of the electromagnetic field. There is an uncertainty principle relating phase and number of the electromagnetic field, in a way that is quite analogous to the more famous uncertainty principle relating position and momentum of a particle.
Therefore, if you want to picture one photon (the number is equal to one so we can't know the phase) you have to imagine a wave with a certain wavelength, but kind of smeared out in space so that it has no phase (ie, there is no definite location of the peaks and troughs of the wave). This may sound like utter nonsense, but such is quantum mechanics. [Technical aside: What is really going on is that the photon's state is a superposition over all possible phases.]
For what it is worth, my picture of a photon is a little ball with a wave drawn on it and a color corresponding to that wavelength. I think most physicists give up on having a clear visualization and learn to trust the math.
A: Short answer:
Asking "what does a photon look like across space and time?" is easy. More often that not people answer this by saying that a photon is a particle that has a some probability to be found across space and time. This probability wave can in fact be shaped in any kind of way, the only rule for it is that is has to be related to whatever created the photon. (answering like this, of course, doesn't actually answer the question though, as you asked that the e-field of a photon looks like.)
Asking "what does a photon's electric field" look like is a much more difficult question to answer "intuitively".  A single photon has a very unique probability distribution of measuring a certain electric field value, which looks like this:

The only thing that changes when you move it in time in space is that it gains a global phase of $e^{i (kx- \omega t)}$. But in quantum mechanics the square of the probability is measured, and this term disappears completely. Many argue that the "phase" of a single photon is non-existent or ill-defined because you can't measure it.
Long answer:
Photons are quantum mechanical, so to explain why this is the case, you need quantum mechanics.
In my PhD I use a detector that can directly measure the electric field of a photon, and we use it to figure out collect statistics to recover the quantum state of our light. To quote my thesis (which is a work in progress):

In classical physics, the free electric field is said to have energy:

\begin{equation}
    u_{EM} \propto  E^2 + c^2 B^2 \end{equation}

This looks remarkably similar to the energy of a harmonic oscillator,
which has the form:

\begin{equation}
    u_{HO} =  \frac{1}{2m} p^2 + m \omega^2 x^2 \end{equation}

so in the case of the $\frac{1}{2m} = 1$ and $c^2 = m \omega^2$, these
two equations look remarkably similar:

\begin{align*} H_{EM} &=   E^2 + c^2 B^2 \\
    H_{HO} &=   p^2 + c^2 x^2  \end{align*}

If [E, B] = $i \hbar$, then these two systems are compeletely
isomorphic. This would mean that all our understanding of how
particles in harmonic oscillator potentials behave could be directly
mapped to this new problem with $x \rightarrow B$ and  $p \rightarrow
> E$.  In the same way that position and momentum become ``conjugate
variables," the electric and magnetic can loosely be thought to be
conjugate variables.
The energy eigenstates have the form:
$$H|n\rangle=\left(N+\frac{1}{2}\right) \hbar \omega|n\rangle$$
Now normally to find the energy eigenfunctions we project our states
in the position basis $\langle x |n \rangle =  \psi_n(x)$. But with
our substitution that $x \rightarrow B$, this becomes: $\langle B |n
\rangle =  \psi_n(B)$.  Meaning that the value of the amplitude of the
magnetic field is a quantum operator equivalent to the position of a
harmonic oscillator. The electric field is generally more interesting
for us, and we identify that with $p \rightarrow E$, this becomes:
$\langle E |n \rangle =  \psi_n(E)$. So we can immediately look at the
energy eigenfunctions for the (which is equivalent to the ``momentum
quadrature'') harmonic oscillator:

\begin{align}
    \psi_{n}(p)&=\frac{1}{(\pi \hbar m \omega)^{1 / 4}} \frac{1}{\sqrt{2^{n} n !}} H_{n}\left(\frac{p}{\sqrt{\hbar m \omega}}\right) e^{-p^{2} / 2 \hbar m \omega} \nonumber \\
    \therefore \psi_{n}(E)&=\frac{1}{(\pi \hbar m \omega)^{1 / 4}} \frac{1}{\sqrt{2^{n} n !}} H_{n}\left(\frac{E}{\sqrt{\hbar m \omega}}\right) e^{-E^{2} / 2 \hbar m \omega} \nonumber \\ \label{hermiteE} \end{align}

...
While the E-field of the vacuum being a Gaussian distribution probably
isn't very suprising, the E-field distribution of a single photon
might be.
This probability distribution is very unique, as compared to the
vacuum state. We will see very soon that by measuring the E-field of
our quantum light, if the light follows this distribution, we can
conclude that our states are pure single photons (the Fock state,
$|1\rangle$). As seen in figure 1.4, while the
probability of measuring $E = 0$ is exactly zero, the average value of
the E-field is exactly zero ($\langle E \rangle = 0$).

Basically, in summary, a "single photon" is a very special state that has this very specific probability distribution of measuring the value of the electric field. Experimentally, photons are sent into a special detector that can measure the electric field directly (unlike normal detectors which would measure energy/photon Number).
Now if you're wondering how this actually reproduces classical waves..that's a good question. The average value of the e-field of any state with a well-defined photon number state is actually zero.
Copying and pasting more stuff I've wrote:

A photon number state of
$|1000\rangle$ has zero average value of the electric field. You
couldn't push an ion even with the electric field of a
$|10^{10000}\rangle$ photon state! (Of course, such a large state
would likely probably push your ion very far, but you wouldn't be able
to tell which direction it would move in.)
So, if an infinite numbered Fock  state can't produce a nonzero $\langle E\rangle$, how does classical light even exist? The answer, believe it
or not, is quantum superposition. When light is in a quantum
superpositions of different
Fock states, then it can produce light with a non-zero  $\langle E \rangle$. For example consider the superposition $|\psi \rangle = c_0|0\rangle + c_1|1\rangle$:

\begin{align*}    \langle \psi | E |\psi \rangle &= \Big( \langle0|c_0^* + \langle 1|c_1^* \Big) E \Big(c_0|0\rangle + c_1|1\rangle \Big) \\
    &= |c_0|^2\langle0|E |0\rangle + c_1^*c_0\langle 1| E |0\rangle+ c_0^*c_1\langle 0| E |1\rangle + |c_1|^2\langle1|E |1\rangle \\
    &= c_1^*c_0\langle 1| E |0\rangle+ c_0^*c_1\langle 0| E |1\rangle \end{align*}

where we identify that the    $\langle0|E |0\rangle$ and $\langle1|E |1\rangle$ terms disappear as discussed earlier.
The easiest way to work out $\langle 0| E |1\rangle$ is to write the
E-field operator in terms of a and $a^\dagger.$ The annhilation and
creation operators are the same as in the Harmonic Oscillator (again,
with the substitutions $B, E \rightarrow X, P$):

\begin{align*} \hat{a} &=\sqrt{\frac{m \omega}{2 \hbar}} \hat{B}+\frac{i}{\sqrt{2 m \omega \hbar}} \hat{E} \\ \hat{a}^{\dagger} &=\sqrt{\frac{m \omega}{2 \hbar}} \hat{B}-\frac{i}{\sqrt{2 m \omega \hbar}} \hat{E} \end{align*}

inverting this transformation:

\begin{align*} B &=\sqrt{\frac{\hbar}{2 m \omega}}\left(a+a^{\dagger}\right) \\ E &=-i \sqrt{\frac{m \hbar \omega}{2}}\left(a-a^{\dagger}\right) \end{align*}

which indicates allows us to easily solve for $\langle E \rangle$:

\begin{align*}
    \langle \psi | E |\psi \rangle &= 2\sqrt{\frac{\hbar}{2 m \omega}} ( c_1^*c_0 + c_0^*c_1) \\
    &= 2\sqrt{\frac{\hbar}{2 m \omega}} ( |c_1||c_0| e^{-i( \phi_1-\phi_0)} + |c_0||c_1| e^{i(\phi_1-\phi_0)}) \\
    &= 2\sqrt{\frac{\hbar}{2 m \omega}} ( |c_0| |c_1| \cos{\Delta\phi}) \end{align*}

where we have written our coefficients in a complex form, such that
$c_0 = |c_0|e^{i \phi_0}$ and $c_1 = |c_1|e^{i \phi_1}$ and $\Delta \phi = \phi_1 - \phi_0 $.

Things are already a bit long here, but at this point you can see that you get a nonzero mean E-field for certain values of $\phi$. This phase evolves in time proportional to its energy $e^{i \hbar \omega t}.$ This is how you can get classical-like light. To get perfectly classical-looking light, you need a "coherent state" which is an infinite superposition of this energy states. I can work out more if you're interested but this answer is already very long.
