Eigenvalue and Amplitude

Eigenvalues are a property of operators and amplitudes are a property of states (or the decomposition of a state in a given basis). An operator can be applied to any state and a state can be acted on by any operator. There is no relationship between the two.

1. No. I think your equation is perhaps meant to be $$|psirangle = frac{1}{sqrt{2}}|psi_1rangle+frac{1}{sqrt{2}}|psi_2rangle,$$ without $hat B$ in there. This is the state of the system expanded in some base. If we are talking about the $hat B$ basis then $|psi_1rangle$ and $|psi_2rangle$ are eigenvectors of the $hat B$ operator. Let's make the following definition: $$hat B|psi_1rangle=b_1|psi_1rangle,$$ $$hat B|psi_2rangle=b_2|psi_2rangle.$$ Now $b_1$ and $b_2$ are the eigenvalues of these eigenvectors of $hat B$. With this information we know the following:

• A measurement of the observable $hat B$ on our system will give the numerical value $b_1$ with probability $|frac{1}{sqrt{2}}|^2=frac{1}{2}$ and $b_2$ again with probability $|frac{1}{sqrt{2}}|^2=frac{1}{2}$. In other words we have equal chance of getting each outcome. The fact that the total probability is equal to $1$ tells us that this state is normalised.

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2. $|phirangle$ by itself doesn't have an amplitude. If the system is in the state $|psirangle=|phirangle$ then a measurement of $hat B$ will yield a value of $k$ with probability $|1|^2=1$.

1. The states $vert psi_1rangle$ and $vertpsi_2rangle$ will not be in general eigenstates of $B$.

2. If $vert phirangle$ is an eigenstate of $B$ then any multiple of $vertphirangle$ is also an eigenstate, so the question does not make sense. The norm of $vert phirangle$ does not depend on this state being eigenstate of any operator or on the magnitude of any eigenvalues of an operator.