What is the difference between intervention and conditional distribution? Suppose that the distribution $P_{C,E}$ is entailed by a structural causal model $\ell$
$C := N_C$
$E := 4C + N_E.$
Given than $N_C$ and $N_E$ are iid $\mathcal{N}(0,1)$ and the graph $C\to E.$ Then:
$P_E^{do(C:=2)} = \mathcal{N}(8,1) = P_{E|C=2}$ 
$P_C^{do(E:=2)} = \mathcal{N}(0,1) \ne P_{C|E=2}.$ 
My focus is in the second intervention $P_C^{do(E:=2)}.$ The conditional distribution of $C$ given $E=2$ is different from the distribution of $C$ after intervening and setting $E$ to $2.$ But how so? I fail to see why. $C$ is not a function of $E.$ 
Source:
Elements of Causal Inference (3.2 page 34). Jonas Peters, Dominik Janzing and Bernhard Scholkopf.
R Code Snipped: 
C <- rnorm(300)
E <- 4*C + rnorm(300)
c(mean(E), var(E))
# [1] 0.1236532 16.1386767

# do(C:=2);
C <- rep(2,300)
E <- 4*C + rnorm(300)
c(mean(E), var(E))
# [1] 7.936917 1.187035

 A: It is true that $C$ is not a function of $E;$ moreover, as the causal diagram $C\to E$ clearly says, we are thinking of $C$ as a cause of $E.$ However, in a causal diagram, you can use the $\newcommand{\doop}{\operatorname{do}} \doop$ operator on any variable you like; doing so deletes all arrows going into that node. This is "graphical surgery", so to speak. If we are intervening on $E,$ we must delete all arrows going into $E,$ except from the exogenous variables $N_C,N_E,$ etc.; this produces the graph $C\; E,$ with no arrows at all. We simultaneously modify the structural equations so as to set $E=2.$ That is, the structural equations become
\begin{align*}
C&=N_C\\
E&=2.
\end{align*}
Hence, $C$ is distributed according to $N_C$ when we intervene on $E.$ So much for $P(C|\doop(E)).$
What about $P(C|E)?$ This is going to be a similar calculation as before, except that we don't do graph surgery. Graph surgery equals intervention equals the $\doop$ operator. The only relationship we have between $C$ and $E$ is the equation $E=4C+N_E.$ When doing mere conditionals (without the $\doop$ operator), it is perfectly permissible to use any of the structural model equations. But they will not be modified this time, because we're not intervening. Hence, you get
\begin{align*}
2&=4C+N_E\\
\frac{2-N_E}{4}&=C\\
\frac12-\frac{N_E}{4}&=C.
\end{align*}
In summary: the $\doop$ operator forces graph surgery and structural equation model alteration. Regular conditional probabilities do not.
