Simple question about compatible observables

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  • #1
etotheipi
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Hey, sorry if this is a dumb question but I wondered if someone could clarify. If you have two Cartesian coordinate systems ##\mathcal{F}## and ##\mathcal{F}'## whose origins coincide except their ##x## axes point in opposite directions (i.e. ##\hat{x} = -\hat{x}'##), then the spin operators along their respective positive ##x## directions would seem to be related via. ##\hat{S}_x = - \hat{S}_{x'}##. AFAIK they're compatible because ##[\hat{S}_x, \hat{S}_{x'}] = 0##.

Does that mean that if you measure the observable corresponding to ##\hat{S}_x## to be a definite ##S_x##, then this automatically fixes ##S_{x'} = -S_x##? That's to say, does a single measurement cut it, or do you still need to actually measure the spin in each coordinate system individually?

Sorry if this doesn't make sense 😜
 
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  • #2
DrClaude
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Does that mean that if you measure the observable corresponding to ##\hat{S}_x## to be a definite ##S_x##, then this automatically fixes ##S_{x'} = -S_x##? ¨
Yes.
 
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  • #3
PeroK
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Hey, sorry if this is a dumb question but I wondered if someone could clarify. If you have two Cartesian coordinate systems ##\mathcal{F}## and ##\mathcal{F}'## whose origins coincide except their ##x## axes point in opposite directions (i.e. ##\hat{x} = -\hat{x}'##), then the spin operators along their respective positive ##x## directions would seem to be related via. ##\hat{S}_x = - \hat{S}_{x'}##. AFAIK they're compatible because ##[\hat{S}_x, \hat{S}_{x'}] = 0##.

Does that mean that if you measure the observable corresponding to ##\hat{S}_x## to be a definite ##S_x##, then this automatically fixes ##S_{x'} = -S_x##? That's to say, does a single measurement cut it, or do you still need to actually measure the spin in each coordinate system individually?

Sorry if this doesn't make sense 😜
It's fundamentally the same measurement. The orientation of the axis in terms of up and down is not physically significant. For example, measuring the position or momentum in the x-direction is the same physical measurement as measuring them in the "-x" direction. The choice of what is ##\pm## is not a physical difference.
 
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  • #4
Nugatory
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Does that mean that if you measure the observable corresponding to ##\hat{S}_x## to be a definite ##S_x##, then this automatically fixes ##S_{x'} = -S_x##? That's to say, does a single measurement cut it
Yes.
Note that every eigenstate of ##S_x## is an eigenstate of ##S_{x’}## and vice versa which is the mathematical way of saying what you probably already knew, that it’s fundamentally the same measurement.
 
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etotheipi
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Awesome, thanks for the speedy replies :smile:
 
  • #6
etotheipi
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Now I'm getting confused about something else. In our first coordinate system, the Pauli vector is expressed$$\vec{S} = S_x \hat{x} + S_y \hat{y} + S_z \hat{z}$$and here the components are operators. Then, suppose another coordinate system is rotated by ##\theta## around the ##\hat{z}## axis, so that ##\hat{x} = \cos{(\theta)} \hat{x}' - \sin{(\theta)} \hat{y}'##, ##\hat{y} = \cos{(\theta)} \hat{y}' + \sin{(\theta)} \hat{x}'##, and then$$\vec{S} = (\cos{(\theta)} S_x + \sin{(\theta)} S_y) \hat{x}' + (\cos{(\theta)} S_y - \sin{(\theta)} S_x) \hat{y}' + S_z \hat{z}$$which means that our new operators are ##S_{x'} = \cos{(\theta)} S_x + \sin{(\theta)} S_y##, ##S_{y'} = \cos{(\theta)} S_y - \sin{(\theta)} S_x## and ##S_{z'} = S_z##.

For this to be of any use, we need to be able to put the operators into a matrix representation w.r.t. some basis of the space of states. This is the slightly confusing part. Let's take ##\theta = \pi##, just to keep things simple. We arbitrarily choose a basis for the Hilbert space of states, say ##B = \{ |s_z = +\frac{1}{2} \rangle, |s_z = -\frac{1}{2} \rangle \}##. First question; would I be right in saying that the following states satisfy$$|s_x = +\frac{1}{2} \rangle = |s_{x'} = -\frac{1}{2} \rangle, \quad \quad|s_x = -\frac{1}{2} \rangle = |s_{x'} = +\frac{1}{2} \rangle$$i.e. that the actual states of the system don't care about the coordinate system?

Secondly, how do we find the matrix representation in the new system? If the representation w.r.t. the basis ##B## in the space ##\mathcal{H}## is$$
S_x =
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}

\quad

S_y =
\begin{pmatrix}
0 & -\frac{i}{2} \\
\frac{i}{2} & 0
\end{pmatrix}

\quad

S_y =
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & -\frac{1}{2}
\end{pmatrix}$$then is it just a case of going like $$
S_{x'} =
\begin{pmatrix}
0 & \frac{1}{2}\cos{\theta} - \frac{i}{2}\sin{\theta} \\
\frac{1}{2}\cos{\theta} + \frac{i}{2}\sin{\theta} & 0
\end{pmatrix}
$$ $$
S_{y'} =
\begin{pmatrix}
0 & -\frac{i}{2} \cos{\theta} - \frac{1}{2} \sin{\theta} \\
\frac{i}{2} \cos{\theta} - \frac{1}{2} \sin{\theta} & 0
\end{pmatrix}
$$ $$S_{z'} =
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & -\frac{1}{2}
\end{pmatrix}$$and these matrices are applied in exactly the same way to the column representations of the states in ##\mathcal{H}##? Sorry for the long post o_O
 
  • #7
PeroK
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I think the simplest way to work this out is to define the vector operator
$$\mathbf{S} = (S_x, S_y, S_z)$$
And the operator that measures the spin of the system about an axis defined by the unit vector ##\mathbf{\hat r}## is:
$$S_{\mathbf{\hat r}} = \mathbf{S} \cdot \mathbf{\hat r} = r_xS_x + r_yS_y + r_zS_z$$
It's then just some algebra to calculate the eigenvectors of ##S_{\mathbf{\hat r}} ## in the original z-basis. Once you have this, you can look at the effect of active and passive transformations.
 
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  • #8
etotheipi
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I think the simplest way to work this out is to define the vector operator
S=(Sx,Sy,Sz)
And the operator that measures the spin of the system about an axis defined by the unit vector r^ is:
Sr^=S⋅r^=rxSx+rySy+rzSz
Aight, so for a sanity check, if ##\hat{r} = \cos{\theta} \hat{x} + \sin{\theta} \hat{y}## then ##S_{\hat{r}} = \vec{S} \cdot \hat{r} = \cos{\theta}
\begin{pmatrix}

0 & \frac{1}{2} \\

\frac{1}{2} & 0

\end{pmatrix} +

\sin{\theta}
\begin{pmatrix}

0 & -\frac{i}{2} \\

\frac{i}{2} & 0

\end{pmatrix}
##
and this should indeed correspond to what I had as the ##\hat{x}'## axis in the above, so sanity remains in tact!

Thanks, okay I'll think of it like this in the future.
 

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