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Probability of Finding System in a State Given a Particular Basis
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[QUOTE="TSny, post: 4303109, member: 229090"] Your calculation looks correct to me. I think 5/9 is the answer. |ω1> and |ω2> are just two possible quantum states of the system. They do not need to be eigenstates of any particular operator. If the system is in state |ω1> then the probability of measuring the system to be in state |ω2> is just |<ω2|ω1>|[SUP]2[/SUP] assuming normalized states. The "inner product" <ω2|ω1> is similar to finding the scalar product ##\vec{a}\cdot\vec{b}## of two ordinary vectors in 3D space. You can pick any orthonormal set of basis vectors ##\hat{i}##, ##\hat{j}##, ##\hat{k}##, and express ##\vec{a}## and ##\vec{b}## as ##\vec{a} = a_1\hat{i}+a_2\hat{j}+a_3\hat{k}## ##\vec{b} = b_1\hat{i}+b_2\hat{j}+b_3\hat{k}## and find ##\vec{a}\cdot\vec{b} = a_1b_1+a_2b_2+a_3b_3## Similarly, if we have any orthonormal basis |δ1>, |δ2> ,|δ3>, |ω1> = a[SUB]1[/SUB]|δ1> + a[SUB]2[/SUB]|δ2> +a[SUB]3[/SUB]|δ3> |ω2> = b[SUB]1[/SUB]|δ1> + b[SUB]2[/SUB]|δ2> +b[SUB]3[/SUB]|δ3> and <ω2|ω1> = b[SUB]1[/SUB][SUP]*[/SUP] a[SUB]1[/SUB] + b[SUB]2[/SUB][SUP]*[/SUP] a[SUB]2[/SUB] + b[SUB]3[/SUB][SUP]*[/SUP] a[SUB]3[/SUB] [/QUOTE]
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Probability of Finding System in a State Given a Particular Basis
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