Unification of Weak and Electromagnetic forces I'm curious how the weak and em forces relate to one another. My knowledge in this area is weak (ha) but I am interested in an explanation for how these two forces relate to one another, and how / why (and when) they are able to be considered 'part of the same force' (electroweak).
I understand there's a "mixing angle" involved though this means little to me; would also like to understand how the photon relates to W/Z bosons (I expect they are related in more ways than simply all of them being bosons)...
This definitely has been asked and there are answers on this SE, but I can't make sense of the technical answers here (ones that rely on the language of QM heavily) so am after a more layperson explanation if that's at all possible...
 A: Instead of explaining how the weak and electromagentic interactions unify (a bottom-up approach), I will try to showcase this from the other direction, explaining how the electroweak interaction gives rise to the weak and electromagetic ones after symmetry breaking. I will include some minor technicalities, but try to render them as insubstatial to the general argumentation as possible. Of course some simplifications are made and my language is not 100% precise, the conclusions are correct though, and I try to balance the loss in precision against the gain in intuition.
What is the electroweak interaction?
While we say there is only one electroweak interaction that is not entirely true. The electroweak theory still includes two interactions that have nothing to do with each other before symmetry breaking. However, neither of these directly relate to either the weak or the electromagetic interactions we observe at low energies.
These two interactions are the so-called $SU(2)_L$, where $L$ is for left-handed and the $U(1)_Y$ with $Y$ for "Hypercharge". An interaction of $SU(2)$ type has three force-carrying bosons, which are often called $W_1, W_2, W_3$, that can interact among themselves, i.e., a $W_1$ can "feel" $W_2$s and $W_3$s but not other $W_1$s. These are related to the $W$ bosons we know in the weak interaction, but bear with me... A $U(1)$ interaction only has a single gauge boson, usually called $B$, which can not "feel" other $B$s, nor do the $W$ and $B$ talk to each other.
This already fully characterizes the electroweak interactions without symmetry breaking. There are 3 $W$ type bosons that interact among themselves and a separate $B$ type boson that does not talk to anyone.
What do other fields do?
Now we need to give these bosons something to interact with besides themselves, we need to introduce other fields. For the $SU(2)$ type interaction we will focus on the doublet. A doublet consists of two fields that are tied together. We usually write them as a vector 
$$\begin{pmatrix} h_1 \\ h_2 \end{pmatrix}.$$
How do the bosons interact with this? This is a bit more technical, but I'll include a tl;dr at the end of the section.
The $W$ bosons act on this vector as matrices. They can be written as
$$ W_1 \propto \frac{1}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad W_2 \propto \frac{1}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad W_3 \propto \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}.$$
Now, $W_1$ and $W_2$ have a very similar structure. Therefore, we combine them in the following way (I will not include normalization matters; while they are important, they make things look more complicated than they are!):
$$ W^+ = W_1 + i W_2 \propto  \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$
$$ W^- = W_1 - i W_2 \propto \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$
For these matrices, in addition to $W_3$, it is easy to see what they do to our doublet:
$$ W^+ \begin{pmatrix} h_1 \\ h_2 \end{pmatrix} \propto \begin{pmatrix} h_2 \\ 0 \end{pmatrix} $$
$$ W^-\begin{pmatrix} h_1 \\ h_2 \end{pmatrix} \propto  \begin{pmatrix} 0 \\ h_1 \end{pmatrix} $$
$$W_3 \begin{pmatrix} h_1 \\ h_2 \end{pmatrix} \propto   \frac{1}{2} \begin{pmatrix} h_1 \\ - h_2 \end{pmatrix}$$
So, the $W^+$ takes the lower component of our doublet and shifts it into the upper component, the $W^-$ takes the upper component and shift it into the lower component and the $W_3$ just gives a minus sign to the lower component and gives a factor 1/2.
The $B$ boson does something slightly different. It multiplies both components with the same number, the Hypercharge of the doublet, which has to be assigned:
$$ B \begin{pmatrix} h_1 \\ h_2 \end{pmatrix} \propto  Y \begin{pmatrix} h_1 \\ h_2 \end{pmatrix} $$
tl;dr: The bosons $W_1$ and $W_2$ shift around the components of the doublet. This is seen most easily by introducing $W^+$ and $W^-$, which replace $W_1$ and $W_2$. $W_3$ multiplies the upper component with 1/2 and the lower component with -1/2. $B$ multiplies both components with the same number $Y$, which has to be given externally. Also, there are interaction strengths $g$ and $g'$ involved which I no longer display explicitly.
Symmetry breaking and the Higgs field
The Higgs field is a doublet with hypercharge $Y = \frac{1}{2}$. Due to the specific form of its potential (keyword "Mexican Hat potential"), it obtains a vacuum expectation value in its lower component:
$$ H = \begin{pmatrix} H_1 \\ H_2 \end{pmatrix} \rightarrow \begin{pmatrix} 0 \\ v \end{pmatrix} $$
This vacuum expectation value (vev) makes all fields massive that do anything at all to it. If you look at the paragraph above, you'll find that (NOTE: This part is simplified a bit too much. I give a more thorough explanation at the end of the post.)


*

*$W^+$ moves the vev to the upper component, therefore it becomes massive

*$W_3$ gives the vev a factor $- g' \frac{1}{2}$, therefore it becomes massive

*$B$ gives the vev a factor $g \frac{1}{2}$, therefore it becomes massive


But wait! Now all our bosons are massive! That's now what we see!
The Higgs field that triggers the change from elecroweak to electromagnetic + weak does interact with both, the $SU(2)$ bosons and the $U(1)$ boson, so what happens is that the two interactions mix.
If we look at the combination
$$ \gamma = \frac{1}{g'^2 + g^2} (g' B + g W_3), $$
where the fraction is just for normalization (so I can introduce the mixing angle properly!), we observe that is does nothing to the vev!
$B$ gives the vev a factor of $g \frac{1}{2}$ and $W_3$ gives the vev a factor of $-g' \frac{1}{2}$. If we add them up as above these factors compoensate and $ \gamma$ does nothing to the vev, it stays massless. This is what we call the photon. The photon is therefore not elementary, but is a mixture of the $B$ and $W_3$ bosons. The factor $\frac{g'}{g^2 + g'^2} = \cos(\theta_W)$ defines the mixing angle. It just describes how much of the photon comes from the $B$ boson, and how much comes from the $W_3$.
The other possible combination of $B$ and $W_3$
$$ Z = \frac{1}{g'^2 + g^2} (g' B - g W_3) $$
is called the $Z$ boson, it does give a factor to the vev and therefore obtains a mass.
Electric charge from weak isospin and hypercharge
Now that we understand how the photon emerges from the $B$ and $W_3$ bosons, we can understand some features of the electric charge:


*

*First of all, the $W^+$ and $W^-$ bosons are charged, since they "feel" the $W_3$ inside the photon.

*The photon itself is not charged. The $W_3$ does not interact with itself, neither does the $B$.

*The same arguments hold true for the $Z$ boson: It interacts with the $W^\pm$ because it is part $W_3$, but doesn't interact with the photon.

*The electric charge of fields that are not part of doublets (all the right-handed particles) is exactly their Hypercharge.

*The electric charge of fields that are part of a doublet that has Hypercharge $Y$ is  $Y + \frac{1}{2} $ for the upper component and $ Y - \frac{1}{2}$ for the lower component. This relation even has a name: It is the Gell-Mann-Nishijima formula.


The last point has a very direct implication: The charges of the upper and lower components of a doublet differ by exacly 1. This is observed between the electron and the neutrino but also between the up and down quark.
Summary
All in all the whole process can be summarized as follows:


*

*In the beginning, we had two unrelated interactions: The $SU(2)_L$ and $U(1)_Y$. $SU(2)$ has three force carriers, $U(1)$ only one.

*The Higgs field, which takes part in both interactions obtains a vacuum expectation value.

*Because of this, only one combination of the force carriers remains massless (and can therefore mediate a long-range force). This is the photon, which is singled out by the fact that it leaves the vev alone (and therefore it also leaves the Higgs boson alone! The Higgs has no electric charge!)

*All the other force carriers, the $W^\pm$ and $Z$ bosons become massive. Therefore, they can only mediate a short-range force which was dubbed "weak" (Physicists at the time could not discern between range and strength).


I hope this lenghty explaination was tangible enough. If there are suggestions on how to improve or what details to expand on, please comment!
More thorough explanation of how and why some bosons get a mass
The reason some of the gauge bosons get a mass is because they couple to the lower component of the Higgs boson doublet.
This coupling is given by the square of the Higgs' covariant derivative.
The following formula might seem overwhelming at first sight, but if you'll bear with me, things get more clear quickly!
$$ \vert D_\mu H \vert^2 \supset \vert  - i g B_\mu \tfrac{1}{2} H - i g' W^1_\mu H - i g' W^2_\mu H - i g' W^3_\mu H \vert^2$$
If we now 


*

*plug in the replacements from above ($W^\pm$, $\gamma$ and $Z$) and

*perform the square of this term (note, it is an absolute square, so a complex conjugation must be done) and then

*plug in the form of the vev ($H = (0, v)$)


we arrive at
$$ \frac{1}{2} g'^2 v^2 W^+ W^- + \frac{1}{2} \frac{g^2 + g'^2}{g g'} v^2 Z^2 = \frac{1}{2} m_W^2 W^+ W^- + \frac{1}{2} m_Z^2 Z^2. $$
This is exactly the kind of term that describes a mass for a vector boson.
Notes


*

*$W^+$ and $W^-$ are not independent. They are each other's antiparticles. This can be seen easily through the fact that $W^-$ is the hermitian conjugate of $W^+$. They therefore have a common mass term. One could also have written this in terms of $W_1$ and $W_2$, then the term would be $m_W^2 (W_1^2 + W_2^2)$.

*The interactions of the Higgs boson with these vector bosons comes from the very term $\vert D_\mu H \vert^2$. Therefore, the masses of the bosons and their interactions with the Higgs are completely correlated. This is one of the features that people at the LHC checked to make sure it really is the Higgs boson they discovered.

*There is nothing special about the lower component of the doublet. One could as well write it with the vev in the upper component or with the vev distributed among both components. This is a choice of gauge. The notation with the vev in the lower component has become standard, though, because in this notation mixing between $B$ and $W$ to give $\gamma$ and $Z$ is easiest to describe and the notion of electric charge looks most natural (it's not different in other gauges, but it might look strange to introduce it).

*There is a difference in how the electron interact with photons or $W$ and $Z$ bosons. This has to do with the fact that only the part of the electron which has left-handed chirality interacts with the $W_1, W_2, W_3$ bosons at all. This is a topic for another post, though.. this one is long enough as it is :)

