What is the intuition behind the independence of $X_2-X_1$ and $X_1+X_2$, $X_i \sim N(0,1)$? I was hoping someone could propose an argument explaining why the random variables  $Y_1=X_2-X_1$ and $Y_2=X_1+X_2$,   $X_i$ having the standard normal distribution, are statistically independent. The proof for that fact follows easily from the MGF technique, yet I find it extremely counter-intuitive.
I would appreciate therefore the intuition here, if any.
Thank you in advance.
EDIT: The subscripts do not indicate Order Statistics but IID observations from the standard normal distrubution. 
 A: The result works for $(X_1,X_2)$ jointly normal (i.e. with correlation, $-1<\rho<1$), with common $\sigma$. 
If you know a couple of basic results, this is about all you need:
$\quad\quad\quad$ 
dobiwan's approach is essentially fine - it's just that the result is more general than the case dealt with there.
A: The result you claim to be true is not true in general, not even for the case when all that is known is that $X_1$ and $X_2$ are normal random variables with identical variance, but the result does hold for the usual interpretation of the condition you stated later:

The subscripts do not indicate Order Statistics but observations from the standard normal distrubution.

The usual interpretation of the last few words in this statement is, of course, that $X_1$ and $X_2$ are independent
(normal) random variables, and hence jointly normal random variables.

For jointly normal random variables with identical variance, it is
  true that $X_1+X_2$ and $X_1-X_2$ are independent (normal) random
  variables (with, in general, unequal variances), and the intuitive explanation for this is best given in Glen_b's answer. For your special case of
  $X_1$ and $X_2$ being independent as well, dobiwan's answer, which you
  have accepted, is simplest, and indeed reveals that any rotation of
  the axes, not just by the $\pm \frac{\pi}{4}$ implicit in the transformation $(X_1,X_2)\to (X_1+X_2, X_1-X_2)$, will yield independent random variables.


What can be said in general? In everything I say below, please bear in
mind that $X$ and $Y$ have the same variance, no matter what other
properties might be attributed to them.
If $X$ and $Y$ are any random variables (note: not necessarily
normal) with identical variance, then
$X+Y$ and $X-Y$ are uncorrelated random variables (that is, they have
zero covariance). This is because
the covariance function is bilinear:
$$\begin{align}
\operatorname{cov}(X+Y, X-Y) &= \operatorname{cov}(X,X) - \operatorname{cov}(X,Y) + \operatorname{cov}(Y,X) - \operatorname{cov}(Y,Y)\\
&= \operatorname{var}(X) - \operatorname{cov}(X,Y) + \operatorname{cov}(X,Y) - \operatorname{var}(Y)\\
&= 0.
\end{align}$$
Here we have used the fact that $\operatorname{cov}(X,X)$ is just 
the variance $\operatorname{var}(X)$ of $X$ (and similarly
for $Y$) and, of course,
$\operatorname{cov}(Y,X) = \operatorname{cov}(X,Y)$.   Note that
this result holds when $X$ and $Y$ are (marginally) normal random variables
but not necessarily jointly normal random variables. 
(If you are not familiar with this notion of marginal normality
not being the same as joint normality, see 
this great answer by cardinal).
In the special case when $X$ and $Y$ are jointly  normal
(but not necessarily independent) normal random variables,
so are $X+Y$ and $X-Y$ jointly normal, and since their covariance
is $0$, $X+Y$ and $X-Y$ are independent random variables. 
A: This is standard normal distributed data:

Notice that the distribution is circulary symmetric.
When you switch to $Y_1 = X_2 - X_1$ and $Y_2 = X_1 + X_2$, you effectively rotate and scale the axis, like this:

This new coordinate system has the same origin as the original one, and the axis are orthogonal.  Due to the circulary symmetry, the variables are still independent in the new coordinate system.
A: I first argue for general identically distributed $X_1,X_2$ that the conditional mean of $Y_1$ conditional on $Y_2$ is constant $0$. Based on this, I argue that the covariance of  $Y_1,Y_2$ is 0. Then, under normality, zero covariance implies independence.
The conditional mean
Intuition: $X_1+X_2=y$ does not imply anything about which component contributed more to the sum (e.g., $X_1=x, X_2 = y-x$ is as likely as $X_1 = y-x, X_2=x$). Thus, the expected difference must be 0.
Proof: $X_1$ and $X_2$ have identical distribution and $X_1+X_2$ is symmetric with respect to the indexing. Thus, for symmetry reasons, the conditional distribution $X_1 \mid Y_2  = y$ must be equal to the conditional distribution $X_2 \mid Y_2 = y$. Hence, the conditional distributions also have the same mean, and 
\begin{equation}
\mathbb{E}(Y_1 \mid Y_2 = y) = \mathbb{E}(X_1 - X_2 \mid X_1+X_2 = y) \\ = \mathbb{E}(X_1 \mid X_1+X_2 = y) - \mathbb{E}(X_2 \mid X_1+X_2 = y)= 0.
\end{equation}
(Caveat: I did not consider the possibility that the conditional mean might not exist.)
Constant conditional mean implies zero correlation/covariance
Intuition: correlation measures how much $Y_1$ tends to increase when $Y_2$ increases. If observing $Y_2$ never changes our mean  of $Y_1$, $Y_1$ and $Y_2$ are uncorrelated. 
Proof: By definition, covariance is 
\begin{equation}
Cov(Y_1,Y_2) = \mathbb{E}\left[\left(Y_1 - \mathbb{E}(Y_1)\right)\left(Y_2 -\mathbb{E}(Y_2) \right)\right]
\end{equation}
to this expectation, we apply the law of iterated expectations: take the expectation of the conditional expectation conditional on $Y_2$:
\begin{equation}
= \mathbb{E}\left[\mathbb{E}\left[\left(Y_1 - \mathbb{E}(Y_1)\right)\left(Y_2 -\mathbb{E}(Y_2) \right)  \mid  Y_2\right]\right] = \mathbb{E}\left[(Y_2 - \mathbb{E}(Y_2))\mathbb{E}\left[Y_1 - \mathbb{E}(Y_1) \mid Y_2\right] \right].
\end{equation}
Recall that the conditional mean was shown to be independent of $Y_2$ and thus the expression simplifies as
\begin{equation}
= \mathbb{E}\left[(Y_2 - \mathbb{E}(Y_2))\mathbb{E}\left[Y_1-\mathbb{E}(Y_1)\right]\right]
\end{equation}
but the inner expectation is $0$ and we get
\begin{equation}
= \mathbb{E}\left[(Y_2 - \mathbb{E}(Y_2))\times0\right] = 0.
\end{equation}
Independence
Just by assuming identical distributions for $X_1,X_2$, it was shown that $Y_1$ and $Y_2$ are uncorrelated. When $X_1,X_2$ are jointly normal (for example, iid. normal as in the question), their linear combinations $Y_1,Y_2$ are also jointly normal and thus uncorrelatedness implies independence. 
