If you were to ask the question of a Quantum Physicist, you'd get a very different answer from that got from anyone who thinks they understand computers...
Clearly data has weight - consider the number of statisticians who ask you to "consider the weight of data" when proposing their own spin on matters ...
Joking aside, I understand Sam's question in the following manner:
Given that a single data bit is essentially the sum of electron spins in the different ('doped') semi-conductor substrates implemented as multiple back-to-back NAND gates in a micro-chip; and given that the zero-or-one-ness of that data bit is based on the imbalance in numbers of electrons in each part of that circuit; how likely is it that the number of electrons required to hold it in the one-state differs from that number required to hold it in the zero-state?
[Recall that electrons have weight. It may be 1/1300+ the weight of a proton, which in turn is 1/(avogadro's number) of a gramme - which makes it absolutely miniscule - but it still weighs something]
Now, if we consider the manufacturing process of memory chips, such an imbalance in one bit will be down to different doping levels in the substrates - which will affect a whole region of bits in the same manner. So it becomes reasonable to ask - on the quantum scale - would the memory stick weigh fractionally more after the data is loaded.
Sam - the answer depends on the individual memory chip, on whether the formatted state is all-ones; all-zeroes; or 50-50 balance between zeroes and ones; AND on the nature of the data being stored.
For your particular memory stick - whether it weights more, less, or about the same, is difficult to ascertain to any degree of accuracy... Why not ask a quantum physicist to 'weigh' it before and after and find out. And please keep us 'posted'!
Hows that ??
Baz
:h
