It passes by reference. Meaning that it won’t create a copy of the value passed.
See:
http://php.net/manual/en/language.references.php (See Adam’s Answer)
Usually, if you pass something like this:
$a = 5;
$b = $a;
$b = 3;
echo $a; // 5
echo $b; // 3
The original variable ($a) won’t be modified if you change the second variable ($b) . If you pass by reference:
$a = 5;
$b =& $a;
$b = 3;
echo $a; // 3
echo $b; // 3
The original is changed as well.
Which is useless when passing around objects, because they will be passed by reference by default.