Determine the state |n> given results and probabilities [QM]

[roguetechx86]

Homework Statement

In a spin-$\frac{1}{2}$ system all particles are in the state $|\psi\rangle$. 3 experiments performed and are separate, the results are as follows:

Particle in state $|\psi\rangle$, measured $S_z$ = $\frac{\hbar}{2}$, with P=1/4
Particle in state $|\psi\rangle$, measured $S_x$ = $\frac{\hbar}{2}$, with P=7/8
Particle in state $|\psi\rangle$, measured $S_y$ = $\frac{\hbar}{2}$, with P=$\frac{4+\sqrt{3}}{8}$

Determine $|\psi\rangle$ in the $S_z$ basis.

Homework Equations

$S_z = \frac{\hbar}{2}\left[\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right]$, $S_x = \frac{\hbar}{2}\left[\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right]$, $S_y = \frac{\hbar}{2}\left[\begin{array}{cc} 0 & -i \\ i & 0\end{array}\right]$
$S = S_x \hat{i} + S_y \hat{j} + S_z \hat{k}$

where $S_z = \frac{\hbar}{2}\sigma_z$

$\sigma = \sigma_x \hat{i} + \sigma_y \hat{j} + \sigma_z \hat{k}$

The Attempt at a Solution

So, if we are given $P_z$ = 1/4, I would think this implies that
$$|\psi\rangle = \frac{1}{2}|+z\rangle + \frac{\sqrt{3}}{2}|-z\rangle\qquad \blacksquare$$
but I don't think this is correct or the whole thing, as I think that $|\psi\rangle$ must satisfy all basis. Also I am not sure that $S_x, S_y$ results are in their $|+x\rangle, |+y\rangle$ respectively, such that the $S_x$ measurement is
$$|\psi\rangle = \sqrt{\frac{7}{8}}|+x\rangle + \sqrt{\frac{1}{8}}|-x\rangle$$
or are written as $|\pm x\rangle, |\pm y\rangle$ in the $S_z$ basis

Do I need determine the $|\pm x\rangle, |\pm y\rangle$ states in the $S_z$ basis of the one mentioned earlier?

Should I try to find the eigenstate such that in 3-D Euclidean space
$$\frac{\hbar}{2}\left[\sigma_x \cos\phi + \sigma_y \sin\phi\right]\left[\begin{array}{c} \langle +z|\mu\rangle \\ \langle -z|\mu\rangle\end{array}\right] = \mu\frac{\hbar}{2}\left[\begin{array}{c} \langle +z|\mu\rangle \\ \langle -z|\mu\rangle\end{array}\right]$$

I don't want the answer only some direction as to how to proceed, as I am lost.

 

[TSny]

Welcome to PF!:D

So, if we are given $P_z$ = 1/4, I would think this implies that
$$|\psi\rangle = \frac{1}{2}|+z\rangle + \frac{\sqrt{3}}{2}|-z\rangle\qquad \blacksquare$$.

You need to allow for possible phase differences in the different components. You can adjust the overall phase of the wavefunction so that the coefficient of $|+z\rangle$ is real and equal to 1/2, but you can't assume that the phase of the $|-z\rangle$ coefficient is simultaneously real.

 

[roguetechx86]

Thanks, feels good to join!

That clue, was extremely helpful, such that
$$|\psi \rangle = \exp(i \phi_+)\left(\frac{1}{2}|+z \rangle + \frac{\sqrt{3}}{2}\exp(i\phi) |+z \rangle \right)$$
where $\phi = \phi_- - \phi_+$. If ignoring the overall phase and rewriting the probabilities of
$$|\langle +x | \psi \rangle |^2 \to S_z basis$$
$$|\langle +y | \psi \rangle |^2 \to S_z basis$$
you can solve $\phi = \frac{\pi}{6}$.

I didn't write the complete solution online because going through the motions really helped understanding the problem!

 

[TSny]

Looks good!

 

[paranormal]

I would approach this problem by first understanding the concept of spin states in quantum mechanics. In a spin-\frac{1}{2} system, the spin of a particle can have two possible states, denoted as |+\rangle and |-\rangle. These states correspond to the spin being aligned with the positive or negative direction of a chosen axis, respectively.

In this problem, we are given the probabilities of measuring the spin of a particle in the state |\psi\rangle along three different axes, S_z, S_x, and S_y. To determine the state |\psi\rangle in the S_z basis, we need to find the coefficients of |+\rangle and |-\rangle in the state |\psi\rangle.

First, let's consider the result for S_z. The probability of measuring S_z = \frac{\hbar}{2} is P_z = 1/4. This means that the state |\psi\rangle must have a coefficient of 1/2 for |+\rangle and a coefficient of \sqrt{3}/2 for |-\rangle. However, this does not uniquely determine |\psi\rangle, as there are infinitely many states that satisfy this condition. So we cannot say that |\psi\rangle = \frac{1}{2}|+\rangle + \frac{\sqrt{3}}{2}|-\rangle.

Next, let's consider the results for S_x and S_y. The probabilities for these measurements are P_x = 7/8 and P_y = \frac{4+\sqrt{3}}{8}, respectively. These probabilities correspond to the coefficients of |+\rangle and |-\rangle in the states |\psi\rangle and i|\psi\rangle, respectively. So we can write the following equations:

P_x = \frac{7}{8} = |\langle +|\psi\rangle|^2 \implies |\psi\rangle = \sqrt{\frac{7}{8}}|+\rangle + \sqrt{\frac{1}{8}}|-\rangle
P_y = \frac{4+\sqrt{3}}{8} = |\langle +|i\psi\rangle|^2 \implies i|\psi\rangle = \sqrt{\frac{4+\sqrt{3}}{8}}|+\rangle + \sqrt{\frac{4-\sqrt{3}}{8}}|-\rangle

Now, we can use these equations to find the coefficients of