I once overhead some spack-tard on the subway tell his friend that an antisymmetric wavefunction for two particles is necessarily the difference of products of one-body wavefunctions.
The moron had no idea that even though one can construct a basis of such states, an arbitrary state will be a linear combination of such basis states. If the particles interact, then even the energy eigenstates will not (in general) be in the "difference of products" form.
(augsav forgot about B3ta, Wed 5 Sep 2012, 1:03, 6 replies)

ah hah! Saw through your little trick
you used the terms 'necessarily' and 'if' as a conditional in the same testing hypothesis. Also you can explain all the rest away in terms of 'well the measurements didn't match our predicted results based on the theory but there's always some fudging room available based on Casimir fluctuations. Let's get it peer reviewed and coincidentally fund our research for another 12 months.
(the zord of damocles 7F 22 78, Wed 5 Sep 2012, 1:36, closed)

Yes,
that's right.
(username failed moderation, Wed 5 Sep 2012, 9:30, closed)

...er, yeah
...alright.
(Penguinbert killed Ken Dodd's dad's dog, Wed 5 Sep 2012, 13:31, closed)

One moment.
I'm currently ploughing my way through "Quantum Mechanics" by A. Rae, and I've only got as far as the chapter on one-dimensional wave functions.

Can I reserve this space for a witty reply sometime in March? How about the 15th?
( Amish Information Systems hates misplaced, commas, Wed 5 Sep 2012, 14:07, closed)

You forgot to say
anything about how you almost died from holding in your gales of laughter.
(apeloverage committed the vile act of onanism on, Wed 5 Sep 2012, 16:13, closed)

i did wee a little if that helps.

(augsav forgot about B3ta, Wed 5 Sep 2012, 16:15, closed)