Why Does a Measurement of A Yield ##\frac{\hbar}{2}## with Certainty?

[BOAS]

Homework Statement

Consider an electron described initially by $\psi = \frac{1}{\sqrt{10}} \begin{pmatrix} 1\\ 3 \end{pmatrix}$. A measurement of the spin component along a certain axis, described by an operator $\hat{A}$, has the eigenvalues $\pm \frac{\hbar}{2}$ as possible outcomes (as with any axis), and the corresponding eigenstates of $\hat{A}$ are $\psi_1 = \frac{1}{\sqrt{10}} \begin{pmatrix} 1\\ 3 \end{pmatrix}$ , $\psi_2 = \frac{1}{\sqrt{10}} \begin{pmatrix} 3\\ -1 \end{pmatrix}$.

(a) Explain without calculation why a measurement of A returns the result $\frac{\hbar}{2}$ with certainty.
(b) If the spin-z component were now to be measured, what would be the probabilities of getting $\frac{\hbar}{2}$ and $- \frac{\hbar}{2}$, respectively?
(c) If the spin-z component is now indeed measured, and subsequently A again, show that the probability of getting $\frac{\hbar}{2}$ in the second measurement is 41/50.

Homework Equations

The Attempt at a Solution

[/B]
For part (a) the initial state is the same as the eigenspinor with that corresponding eigenvalue, so when multiplying the state with the adjoint of the eigenspinor, of course we will get 1.

(b) For the spin z measurement $\psi_{z +} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\psi_{z-} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ and the corresponding probabilities are found by multiplying the state with the adjoint of the eigenspinor and then squaring. $P_{z+} = 1/10$ and $P_{z-} = 9/10$

(c) I'm not sure how to tackle this, how do the measurements of spin-z effect the measurements of A?

Al I can think of is that in order to get the same measurement of A, the z component of the spins must have remained unchanged. The sum of two measurements of unchanged spin-z (1/10 * 1/10 + 9/10 * 9/10 = 41/50)

Thank you for any help you can give

 

[Orodruin]

What states will the electron be in after measuring the z-component and with what probabilities. What are the probabilities of measuring hbar/2 in the A-direction for those states?

 

[BOAS]

What states will the electron be in after measuring the z-component and with what probabilities. What are the probabilities of measuring hbar/2 in the A-direction for those states?

The electron will either be in the spin up or spin down states with probabilities 1/10 and 9/10 respectively.

The probability of measuring hbar/2 in the A direction for the spin up state is 1/10.

The probability of measuring hbar/2 in the A direction for the spin down state is 9/10.

Therefore, the overall probability is the sum of the two, i.e 41/50