Is there a reason why probability density is written as $\psi^*\psi$ instead of $\psi\psi^*$? As the title states, I see $|\psi|^2$ written as $\psi^*\psi$ instead of $\psi\psi^*$. Are both correct or is there a reason behind it? As far as I'm aware, the only time I see this sort of ordering being important is when ordering two operators. 
 A: Yes, so it looks like more general inner products. For example, if you have two real vectors $v$ and $w$, their dot product $v \cdot w$ can also be written as $v^Tw$ where the $T$ means 'transpose'. Similarly if the vectors are complex the inner product is $v^\dagger w$.
In general, you can either think of the dot product as an operation that takes two vectors and spits out a number ($v \cdot w$), or as the result of a row vector/"dual vector" ($v^T$) acting on a column vector ($w$). Note that order does matter here; even though $v \cdot w = w \cdot v$, $wv^T$ is not $v^T w$.
The $\psi^*\psi$ notation is the same. It happens to not be necessary in this case, though it is necessary in general; once you move onto Dirac notation, you'll want $\langle \psi | \psi \rangle$ and not $|\psi \rangle \langle \psi|$, and if you take the product of two different wavefunctions it matters which is conjugated.
A: As mentioned in other answers, in principle the order does not matter. However, it does matter once you evaluate certain operators such as differential operators, since e.g.
$$
\int \psi^*(x)\frac{\partial}{\partial x}\psi(x)
$$
is different from
$$
\int \psi(x)\frac{\partial}{\partial x}\psi^*(x)\ .
$$
For this reason, $\psi^*$ is generally placed on the very left and $\psi$ on the very right, even in cases where the order does not matter.
A: You're essentially evaluating a inner/dot product in a particular basis. So, the order doesn't matter for the same function dotted with itself. But if you choose two different functions and want to calculate their dot product density, you'll have a difference, namely why (a+ib)(c-id)$\neq$(c+id)(a-ib), because unless both complex numbers are the same, you won't get the same value always.
So, yes the order definitely matters. It looks prettier too :)
A: There is no particular reason, because it's obviously commutative. Personally, I guess it looks nicer that way.
