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Continuing from the my previous 2-state system problem, I am told that the observable corresponding to the linear operator $\hat{L}$ is measured and we get the +1 state. Then it asks for the probability that +1 is measured again if the system is measured again at $t$. I would have thought that this probability is 1 since measurement projects the system into the first measured outcome such that successive measurements are consistent. But clearly I am mistaken, since it also asks to show that the probability = 0 for t = $\pi\hbar\over \Delta E$. 

Could anyone please explain to me what is going on? Thanks.

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The measurement does project the state, but unless $[H,L]=0$ it doesn't have to stay that way. – wsc Sep 2 '11 at 11:43
Thanks, wsc. Would you mind explaining that a little bit more? – justcurious Sep 2 '11 at 11:48
@justcurious the time evolution of an observable is $\frac{d}{dt}O=\frac{i}{\hbar}\left[H,O\right]+\frac{\partial}{\partial t}O$ just as in analytical mechanics where instead of the commurator, you take the poisson bracket. – luksen Sep 2 '11 at 14:11
@luksen, I fixed a typo in your comment - hope you don't mind. – David Zaslavsky♦ Sep 2 '11 at 17:46
@David, of course not, thanks, what was it (there no way to view revs of comment are there?) – luksen Sep 2 '11 at 17:49
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...since measurement projects the system into the first measured outcome...

That part is true...

...such that successive measurements are consistent.

...but that part isn't. Although a measurement of $\hat{L}$ does put the system into the eigenstate of $\hat{L}$ corresponding to the measured eigenvalue, it doesn't mean it's going to stay in that state - after all, the time evolution of the system (i.e. the Schroedinger equation, Heisenberg equation, evolution operator... however you want to describe it) doesn't stop just because you made a measurement.

Mathematically, when you make a measurement, you use the eigenstate as an initial condition for the evolution equation to figure out what happens in the future.

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