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- Homework Statement
- Prove that the equality sign in the generalized uncertainty relation holds if the state in question satisfies ##\Delta A|\alpha\rangle = \lambda \Delta B|\alpha\rangle## with ##\lambda## purely imaginary.

- Relevant Equations
- ##\langle(\Delta A)^2\rangle \langle(\Delta B)^2\rangle \geq \frac{1}{4}|\langle \alpha |[A,B]|\alpha\rangle|^2##

##\langle \alpha |[A,B]|\alpha\rangle = \langle \alpha |\Delta A \Delta B - \Delta B \Delta A|\alpha\rangle##

It's easy to show that ##[\Delta A, \Delta B] = [A,B]##. I'm specifically having issues with evaluating the bra-ket on the RHS of the uncertainty relation:

##\langle \alpha |[A,B]|\alpha\rangle = \langle \alpha |\Delta A \Delta B - \Delta B \Delta A|\alpha\rangle##

The answer is supposed to be ##-2\lambda \langle \alpha |(\Delta B)^2|\alpha\rangle##, but I don't know how to evaluate the bra-ket to reach it; do I apply ##\Delta A \Delta B## and ##\Delta B \Delta A## each to both the ket and the bra, or just the ones that are immediately associated with them, i.e. not the one separated by the minus sign? If it's the former what do I do with e.g. the term ##\Delta A \Delta B|\alpha \rangle##? The trick was supposed to be to notice that ##\langle\alpha| \Delta A = -\lambda \langle\alpha|\Delta B##, but I seem to lack a more fundamental understanding. Any help is appreciated.

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