Probabilities of measuring ##\pm \hbar/2## along ##\hat{n}##?

In summary, the probability of measuring ##+\frac \hbar 2## along ##\hat n## is given by:P(+\frac \hbar 2 \, \, \hat n) = 1/2 (cos (\theta/2) e^{i\theta /2} |+ \rangle_n + sin (\theta/2)e^{i\theta /2} |- \rangle_n)
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happyparticle
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Homework Statement
Probabilities of measuring ##\pm \hbar/2## along ##\hat{n}##?
Relevant Equations
##|+\rangle_n = cos(\theta/2)e^{-i\phi/2}|+\rangle +sin(\theta/2)e^{i\phi/2}|-\rangle##
##|-\rangle_n = sin(\theta/2)e^{-i\phi/2}|+\rangle -cos(\theta/2)e^{i\phi/2}|-\rangle##
Hi,

Given a spin in the state ##|z + \rangle##, i.e., pointing up along the z-axis what are the probabilities of measuring ##\pm \hbar/2## along ##\hat{n}##?

My problem is that I'm not sure to understand the statement. It seems like I have to find the probabilities of measuring an eigenvalue along ##\hat{n}##. What does that mean exactly? Is it the probability to measure ##\pm \hbar/2## in the ##n## basis?

I tried to find ## |+\rangle ## in the ##n## basis, which I think this is ##|z+\rangle## in the ##n## basis. I thought maybe it could help me.
I got ## |+\rangle = 1/2 (cos (\theta/2) e^{i\theta /2} |+ \rangle_n + sin (\theta/2)e^{i\theta /2} |- \rangle_n)##

I'm really confuse with this statement. I'm not sure to understand the difference between ##|z + \rangle## and ##|+\rangle##
 
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  • #2
happyparticle said:
My problem is that I'm not sure to understand the statement. It seems like I have to find the probabilities of measuring an eigenvalue along ##\hat{n}##. What does that mean exactly?
The question is this. Imagine you have a system that produces electrons in the ##\ket{z+}## state. You can confirm this by setting up a measurement aparatus to measure the spin component about the z-axis and finding that every electron produced by your system behaves in the same way, corresponding to a measurement of ##+\frac \hbar 2##. Note that the z direction is simply some direction you have chosen in your experimental set-up.

Now, you leave your electron production system in place and re-orient your measurement aparatus along an axis ##\hat n##, represented by azimuthal and polar angles ##\theta, \phi##.

Each electron will behave in one of two ways, corresponding to the measurements of ##\pm \frac \hbar 2##. The question is: what is the probability that the electron is measured to have ##+\frac \hbar 2##; and what is the probability that the electron is measured to have ##-\frac \hbar 2##?
 
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FAQ: Probabilities of measuring ##\pm \hbar/2## along ##\hat{n}##?

What does it mean to measure ##\pm \hbar/2## along ##\hat{n}##?

Measuring ##\pm \hbar/2## along ##\hat{n}## refers to the possible outcomes when measuring the spin component of a quantum particle, such as an electron, along a specified direction ##\hat{n}##. In quantum mechanics, spin measurements along any axis yield discrete values, typically ##+\hbar/2## or ##-\hbar/2## for spin-1/2 particles.

How do you calculate the probabilities of measuring ##\pm \hbar/2## along ##\hat{n}##?

The probabilities of measuring ##\pm \hbar/2## along a direction ##\hat{n}## are determined by the projection of the spin state onto the eigenstates of the spin operator along ##\hat{n}##. If the spin state is represented by a state vector ##|\psi\rangle##, the probabilities are given by the squares of the magnitudes of the projection coefficients: ##P(\pm \hbar/2) = |\langle \pm \hbar/2|\psi \rangle|^2##.

What is the role of the direction ##\hat{n}## in these measurements?

The direction ##\hat{n}## specifies the axis along which the spin component is measured. The orientation of ##\hat{n}## affects the eigenstates of the spin operator along this direction, thereby influencing the probabilities of obtaining ##+\hbar/2## or ##-\hbar/2## as measurement outcomes.

How does the initial spin state affect the measurement probabilities?

The initial spin state determines the likelihood of obtaining either ##+\hbar/2## or ##-\hbar/2## when measuring along ##\hat{n}##. For example, if the initial state is an eigenstate of the spin operator along ##\hat{n}##, the probability of measuring the corresponding eigenvalue is 1, and the probability of measuring the opposite eigenvalue is 0. For a general state, the probabilities are given by the projections onto the eigenstates of the spin operator along ##\hat{n}##.

Can you provide an example calculation for the probabilities of measuring ##\pm \hbar/2##?

Sure! Consider an initial spin state ##|\psi\rangle = \alpha|+\hbar/2_z\rangle + \beta|-\hbar/2_z\rangle##, where ##|+\hbar/2_z\rangle## and ##|-\hbar/2_z\rangle## are eigenstates of the spin operator along the z-axis

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