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I'm reading Griffiths Introduction to QM and I'm having trouble understanding why you can't simultaneously measure the x,y and z components of spin. I know that the uncertainty principle prevents this but I still don't see why.

Griffiths' example is that if we have a particle in its up state, $\chi_+$ then we know the z-component of its spin is $\frac{\hbar}{2}$. If we then measure the x-component, then we're suddenly left with a 50-50 probability of the x-component being $\frac{\hbar}{2}$ or $-\frac{\hbar}{2}$. First off, why is it a 50-50 chance? If the state of the z-component is $\chi^z$ then $$ \chi^z=a\chi_+ ^z + b\chi_- ^z$$ and the x-component is $$ \chi^x =\dfrac{a+b}{\sqrt{2}}\chi_+ ^x +\dfrac{a-b}{\sqrt{2}}\chi_- ^x$$

If the z-component is in its upstate, does $\chi^z$ collapse to $$\chi^z = \chi_+ ^z$$

and so $a=1$ and then $b=1$. Therefore, there is a 50-50 chance of the x-component to be in its up or down state. Is this why it is 50-50 or am I understanding it wrong?

Next, if the particle is in its up state, then shouldn't the x-component also be in the x-component up state i.e $\frac{\hbar}{2}$ or does its up and down state 'reset' every time we measure? If it does reset, does it mean that once I measure for the x-component I lose knowledge about the z-component? So I'm left with a definite x-component but now I only have a 50-50 probability of knowing if it's spin up or down in the z-component? If it even does reset, what causes it? Is it just because of the uncertainty principle?

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  • $\begingroup$ Try computing the commutators of each spin operator. You can only measure things simultaneously if they commute..i.e. commutator of 2 operators = 0. Hope that helps! $\endgroup$ – user122066 Jul 5 '16 at 1:12
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Mutually non-commuting operators cannot have simultaneous eigenstates, namely the eigenstates of the former must by all means be expressed as a linear combinations of (all) the eigenstates of the latter. In the case at hand, given ${|+\rangle}_z$ as eigenstate of the operator $S_z$, the following must hold: $$ {|+\rangle}_z = c_1 {|+\rangle}_x + c_2 {|-\rangle}_x $$ and likewise for the other component ${|-\rangle}_z$, only with different coefficients. Exploiting the commutation relations and the $\mathfrak{su}(2)$ Lie-algebra one finds out that $c_1 = \pm c_2 = 1/\sqrt{2}$ (signs may be inverted though).

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In general you are getting it right: Non-commuting operators do not share eigenstates, thus measuring $S_x$ on an eigenstate of $S_z$ will result in a state that is not an eigenstate of $S_z$ anymore. The spin operators do not commute because they are defined via the Lie-algebra relation $[S_i, S_j] = i \hbar \varepsilon_{ijk} S_k$.

Next, if the particle is in its up state, then shouldn't the x-component also be in the x-component up state i.e $\hbar/2$ or does its up and down state 'reset' every time we measure?

The 'up' state is defined with respect to a direction, i.e. the $z$ 'up' state is not equivalent to the $x$ 'up' state. Think of the spin as a vector in a 3D cartesian space. Clearly, a vector pointing along the positive $z$ axis is not the same as the vector pointing along the positive $x$ axis.

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This might be easier to understand in terms of the Pauli matrices:

$\sigma_z=\pmatrix{0 &1\\1 &0}$, $\sigma_y=\pmatrix{0 &-i\\i &0}$, $\sigma_x=\pmatrix{1 &0\\0 &-1}$,

where the operators for $z, y$, and $x$, are given by $\hat{S_z}=\tfrac{\hbar}{2}\sigma_z$, $\hat{S_y}=\tfrac{\hbar}{2}\sigma_y$, and $\hat{S_x}=\tfrac{\hbar}{2}\sigma_x$, respectively.

The eigenvalues of the Pauli matrices are $\pm 1$ You can find the eigenvectors of these matrices to be

$\chi_z^+= \pmatrix{1\\0}\qquad \chi_z^-=\pmatrix{0\\1}$

$\chi_y^+=\tfrac{1}{\sqrt{2}}\pmatrix{1\\i} \quad \chi_y^-=\tfrac{1}{\sqrt{2}}\pmatrix{1\\-i}$

$\chi_x^+=\tfrac{1}{\sqrt{2}}\pmatrix{1\\1} \quad \chi_x^-\tfrac{1}{\sqrt{2}}\pmatrix{1\\-1}$,

where the "+" and "-" superscript distinguish between the two different eigenvectors.

So to take Griffith's example of system prepared in a spin-up state, i.e., $\pmatrix{1 \\0}$. We can express this state in terms of the $x$ eigenvectors:

$\pmatrix{1\\0}=\frac{1}{\sqrt{2}}\left [\chi_x^+ +\chi_x^- \right ]=\frac{1}{2}\pmatrix{1\\1}+\frac{1}{2}\pmatrix{1\\-1}$

If you now measure the $x$ component there is a $(1/\sqrt{2})^2$ chance of finding the $+\hbar/2$ or $-\hbar/2$ eigenvalue.

Once you measured a certain value the wave function collapses and will remain in the state you measured. So if you measured $-\hbar/2$ after measuring $\hat{S}_x$ you'll measure this value again once you apply $\hat{S}_x$ again. However, the state couples to the surrounding leading to a decoherence of the wave function. If you wait to long, the measurement will be stochastic again.

To see why the uncertainty principle prevents you form measuring all projections at the same time can be seen form the vector model of spin. As the total angular momentum is $\hbar\sqrt{j(j+1)}$ and the maximum value of the projection is $\hbar m_{j,\text{max}}=\hbar j$ and because $j<\sqrt{j(j+1)}$, the angular momentum does never point directly along an axis but precesses around. Therefore only one projection is defined.

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  • $\begingroup$ "However, the state couples to the surrounding leading to a decoherence of the wave function" is irrelevant to the argument. Unless other measurements are performed, if the state is in an eigenstate it will remain so indefinitely. $\endgroup$ – gented Jul 6 '16 at 10:21

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