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Quantum mechanics

A^,B^ \hat{A}, \hat{B} are two Hermitian operators. consider the state ϕ=(δA^iλδB^)ψ \phi = (\delta \hat{A} - i \lambda \delta \hat{B}) \psi where λ \lambda is real and ΔB2>0 \Delta B^2 > 0 . By taking the inner products on both sides and minimizing with respect to λ \lambda derive the uncertainty relation:
ΔA2ΔB214(i[A,B])2 \Delta A^2 \Delta B^2 \geq \frac{1}{4} (i \langle [A,B] \rangle)^2 where [A^,B^]=A^B^B^A^ [\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A} is the commutator of A^ \hat{A} with B^ \hat{B} and ΔA2=(δA)2=ψ,δA^δA^ψ \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.
Original post by JBKProductions
A^,B^ \hat{A}, \hat{B} are two Hermitian operators. consider the state ϕ=(δA^iλδB^)ψ \phi = (\delta \hat{A} - i \lambda \delta \hat{B}) \psi where λ \lambda is real and ΔB2>0 \Delta B^2 > 0 . By taking the inner products on both sides and minimizing with respect to λ \lambda derive the uncertainty relation:
ΔA2ΔB214(i[A,B])2 \Delta A^2 \Delta B^2 \geq \frac{1}{4} (i \langle [A,B] \rangle)^2 where [A^,B^]=A^B^B^A^ [\hat{A}, \hat{B}] = \hat{A} \hat{B} - \hat{B} \hat{A} is the commutator of A^ \hat{A} with B^ \hat{B} and ΔA2=(δA)2=ψ,δA^δA^ψ \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?
Original post by ben-smith
What's lower case delta?


The lower case delta is δA=ψ,δA^ψ \langle \delta A \rangle = \langle \psi, \delta \hat{A} \psi \rangle . Where δA^=A^A \delta \hat{A} = \hat{A} - \langle A \rangle and \langle A \rangle = \langle \psi, \hat{A} \psi \rangle
(edited 11 years ago)

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