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Quantum mechanics

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     \hat{A}, \hat{B} are two Hermitian operators. consider the state  \phi = (\delta \hat{A} - i \lambda \delta \hat{B}) \psi where  \lambda is real and  \Delta B^2 > 0 . By taking the inner products on both sides and minimizing with respect to  \lambda derive the uncertainty relation:
     \Delta A^2 \Delta B^2  \geq \frac{1}{4} (i \langle [A,B] \rangle)^2 where  [\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A} is the commutator of  \hat{A} with  \hat{B} and  \Delta A^2 = \langle (\delta A)^2 \rangle = \langle \psi, \delta \hat{A} \delta \hat{A} \psi \rangle
    I've been stuck on this question for a while and would appreciate any help. Thanks.
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    (Original post by JBKProductions)
     \hat{A}, \hat{B} are two Hermitian operators. consider the state  \phi = (\delta \hat{A} - i \lambda \delta \hat{B}) \psi where  \lambda is real and  \Delta B^2 > 0 . By taking the inner products on both sides and minimizing with respect to  \lambda derive the uncertainty relation:
     \Delta A^2 \Delta B^2  \geq \frac{1}{4} (i \langle [A,B] \rangle)^2 where  [\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A} is the commutator of  \hat{A} with  \hat{B} and  \Delta A^2 = \langle (\delta A)^2 \rangle = \langle \psi, \delta \hat{A} \delta \hat{A} \psi \rangle
    I've been stuck on this question for a while and would appreciate any help. Thanks.
    What's lower case delta?
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    ReputationRep:
    (Original post by ben-smith)
    What's lower case delta?
    The lower case delta is  \langle \delta A \rangle = \langle \psi, \delta \hat{A} \psi \rangle . Where  \delta \hat{A} = \hat{A} - \langle A \rangle and  \langle A \rangle = \langle \psi, \hat{A} \psi \rangle

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