Eigenvalue and Amplitude

Question

What is the relationship between an eigenvalue and an amplitude?

1. Suppose that $\hat{B}|\psi\rangle = \frac{1}{\sqrt{2}}|\psi_{1}\rangle + \frac{1}{\sqrt{2}}|\psi_{2}\rangle$ for some observable $\hat{B}$. I understand that $\frac{1}{\sqrt{2}}$ is the amplitude for each of $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$. Can we know the eignvalues of $\hat{B}$ with respect to $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ from this equation?

2. Also, suppose that $\hat{B}|\phi\rangle = k|\phi\rangle$. I understand that k is an eigenvalue of the eigenstate for $\hat{B}$. But what is the amplitude for $|\phi\rangle$?

Answer 1

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.

Answer 2

1. No. I think your equation is perhaps meant to be $$|\psi\rangle = \frac{1}{\sqrt{2}}|\psi_1\rangle+\frac{1}{\sqrt{2}}|\psi_2\rangle,$$ 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 $|\psi_1\rangle$ and $|\psi_2\rangle$ are eigenvectors of the $\hat B$ operator. Let's make the following definition: $$\hat B|\psi_1\rangle=b_1|\psi_1\rangle,$$ $$\hat B|\psi_2\rangle=b_2|\psi_2\rangle.$$ 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. $|\phi\rangle$ by itself doesn't have an amplitude. If the system is in the state $|\psi\rangle=|\phi\rangle$ then a measurement of $\hat B$ will yield a value of $k$ with probability $|1|^2=1$.

Answer 3

1. The states $\vert \psi_1\rangle$ and $\vert\psi_2\rangle$ will not be in general eigenstates of $B$.

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