How Can Uncertainty Product DeltaADeltaB Be Zero in Quantum Mechanics?

[Gray]

Homework Statement

Let C|+-> = +-|+->, and consider a state |psi> = cosT|+> + sinT|->. Find T such that the product of uncertainties, deltaAdeltaB, vanishes (i.e. becomes zero).

*Note: +- means plus or minus repectively.

Homework Equations

[A,B] = iC
In a previous question I proved deltaAdeltaB>=1/2|<psi|C|psi>| using the Schwarz inequality and some other stuff.

The Attempt at a Solution

So we want 1/2|<psi|C|psi>| = 0.
i.e. |<psi|CcosT|+> + <psi|CsinT|->| = 0.

Then I assumed <psi| = +|cosT + <-|sinT similarly to the psi ket.

So
|<+|(cosT)^2.C|+> + <-|sinT.C.cosT|+> + <+|cosT.C.sinT|-> + <-|(sinT)^2.C|->| = 0

Then use C|+-> = +-|+-> and similarly I assumed <+-|C = <+-|+- for the C bra.

So
|<+|(cosT)^2|+> + <-|sinTcosT|+> + <+|cosTsinT|-> + <-|(sinT)^2|->| = 0
|(cosT)^2 + sinTcosT<-|+> + sinTcosT<+|-> + (sinT)^2| = 0
|(cosT)^2 - sinTcosT<+|-> + sinTcosT<+|-> + (sinT)^2| = 0
|(cosT)^2 + (sinT)^2| = 0
which is clearly nonsense.

Are my assumptions incorrect? Am I not allowed to convert the ket formalism to the bra formalism in this manner?

 

[Gray]

I came up with a different way to do the problem using <psi| = -|cosT + <+|sinT which gives me
|2sinTcosT| = 0
|sin2T| = 0 by double angle formula
T = n.Pi/2 , n an integer

But I still don't know if what I'm doing is right (actually I suspect it's wrong).

 

[Gray]

I posted this over three days ago. How long does it usually take? Surely someone can help with a second year QM problem?

 

[drizzle]

your work seems fine in your first post, but I doubt the signs of the products in the pre-last
equation, try first to find out what you come up with in this:
<epsi|epsi>=1=...