Possible Outcomes of Measuring Spin-1 Particle in Different Axes

[DanAbnormal]

Homework Statement

A spin-1 particle is measured in a stern gerlach device, set up to measure $$S_{z}$$. What are the possible outcomes?

In this case, the outcome is zero. The same particle is measured by a second deviced which measures $$S_{x}$$. What are the possible outcomes of this measurement and their respective probabilities?

Homework Equations

$$S_{x}$$

$$S_{z}$$

The Attempt at a Solution

So for the first part, acting $$S_{z}$$ on the orthonormal basis for a spin 1 particle ($$m_{s}$$= -1, 0, 1) gives the possible outcomes -$$\hbar$$, 0 and $$\hbar$$

In this case, the measurement is zero, which means the particle is in the state with $$m_{s}$$ = 0.

Now for this next part, making a measurement of the $$S_{x}$$ observable of this particle, do I need to diagonalise the $$S_{x}$$ matrix, or is it enough to act $$S_{x}$$ straight onto the vector for $$m_{s}$$ = 0? If I do the latter, it gives a superposition state of the $$m_{s}$$ = 1 and $$m_{s}$$ = -1 vectors, both multiplied by $$\hbar$$/$$\sqrt{2}$$, and this isn't an eigenfunction of $$S_{x}$$... is it?

I was trying to think about this physically. If the first measurement gives a value of zero for spin along the z axis then this means spin must be aligned either along the y-axis or x axis. So upon making the second measurement, the possible outcomes for spin along the x-axis must be zero (if the spin is along y) or the either plus/minus the component of spin along the x-axis.
So the main question is, are the possible outcomes for the second measurement 0, and whatever value I get for acting $$S_{x}$$ onto the vector for $$m_{s}$$ = 0, or do I need to diagonalise the $$S_{x}$$ to get the true eigenvalues/eigenvectors of $$S_{x}$$? As I understood, all spin measurements are made relative to the z axis...

Im just a bit confused is all.

Cheers

 

[kuruman]

Your answer to the first question is correct.

Second question
Once the measurement is made, the particle is in state (0,1,0) or |0> according to your suggested orthonormal set. What do you get when you operate on this state with J_x?

 

[DanAbnormal]

Your answer to the first question is correct.

Second question
Once the measurement is made, the particle is in state (0,1,0) or |0> according to your suggested orthonormal set. What do you get when you operate on this state with J_x?

When I act J_x onto |0> I get the added superposition of |1> and |-1> [(1,0,0) and (0,0,1)] both weighted by $$\hbar$$/2

But |0> is not an eigenvector of J_x is it? so what meaning does this action have? If the measurement in the z-axis defined the particle to be in state |0> in that axis, then does this not imply the spin of the particle must be aligned in either the x or y axes?

 

[kuruman]

When I act J_x onto |0> I get the added superposition of |1> and |-1> [(1,0,0) and (0,0,1)] both weighted by $$\hbar$$/2

But |0> is not an eigenvector of J_x is it?

It is not. However, regardless of what component you measure, the result of the measurement can only be -1, 0 or +1. What differs from orientation to orientation with each measurement is the probability of getting one of the three results. So in this particular case, what are the probabilities of getting -1, 0 or +1?

Incidentally, the weights of each state must be pure numbers, hbar should not be in the picture.

 

[DanAbnormal]

It is not. However, regardless of what component you measure, the result of the measurement can only be -1, 0 or +1. What differs from orientation to orientation with each measurement is the probability of getting one of the three results. So in this particular case, what are the probabilities of getting -1, 0 or +1?

Incidentally, the weights of each state must be pure numbers, hbar should not be in the picture.

Well what I did was obtain the eigenvalues of the Sx matrix (without the hbar/(root 2) element ), which turned out to be 0, and +/- root 2
I then obtained the eigenvectors, for these, which when acted on by Sx gives 0, hbar/2, and -hbar/2 as possible outcomes, and the probability I got for the +/- hbar/2 eigenvalues were each 1/2, and the 0 eigenvalue was associated with a probability of zero.

I still don't understand the difference between obtaining eigenvectors of the Sx matrix and finding outcomes from these, and acting Sx straight onto the z-axis |0> state.
Im really sorry!

Is my method correct though?

 

[kuruman]

Once you make the first measurement, you know that your particle is in the eigenstate |0> or column vector (0,1,0). After the first measurement you make a second measurement and measure S_x. The possible outcomes of this second measurement are one of the eigenvalues of S_x, which you can get by diagonalizing S_x, namely -1, 0 or +1. Note that, after the second measurement, your particle is in one of the eigenstates of S_x and that you can obtain these eigenstates from the diagonalization process. Now then, the probability that you get a given one of the three eigenvalues after the second measurement is the square of the inner product of the original state (0,1,0), expressed as a row vector, with the (column) eigenvector of S_x corresponding to that particular eigenvalue.

 

[vela]

I still don't understand the difference between obtaining eigenvectors of the Sx matrix and finding outcomes from these, and acting Sx straight onto the z-axis |0> state.

If a system is in the state $\vert \psi \rangle$, the amplitude of finding it in the state $\vert \phi \rangle$ is $A = \langle \phi \vert \psi \rangle$, and the probability is then given by the modulus of the amplitude squared, P=|A|². In this problem, $\vert \phi \rangle$ is an eigenstates of S_x and $\vert \psi \rangle$ is the m=0 eigenstate of S_z. Making the measurement has nothing to do with applying the operator S_x to a state. The only reason you need it is to find its eigenvalues and eigenvectors.

 

[DanAbnormal]

Oh thanks guys! That makes sense to me now, considering all the maths I've just done. I was following a procedure just not totally getting it. But do now!