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?
Best Answer
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).