I have a system of two spin 1/2 particles in the z-direction given by $\phi=\frac{1}{\sqrt{2}} \Big( |->|+> +|->|-> \Big)$ where $+$ means up and $-$ means down. The first ket in each term designates the first particle and the second one is linked to the second particle. All the kets in the z-direction will be written without index from now on.

Let's say I measure $S_{1x}$ and I get the value $+\frac{\hbar}{2}$, which would mean that the state points towards the positive direction in the x axis. Then which would be the state after the measurement?

On the one hand, using the projectors and $|+>_x$ in terms of the eigenvectors of $S_{1z}$ ($|+>_x=\frac{|+>+|->}{\sqrt{2}}$), I obtain

$$ \Big(\frac{|+>+|->}{\sqrt{2}} \otimes I \Big) \Big(\frac{<+|+<-|}{\sqrt{2}} \otimes I \Big) \phi= \Big(\frac{|+>+|->}{\sqrt{2}}\otimes I \Big) \frac{1}{2} \Big(I \otimes |+> + I\otimes |-> \Big) = \frac{1}{2\sqrt{2}} \Big(|+>|+>+|->|+>+|+>|->+|->|-> \Big) $$

which, once normalized becomes

$$\frac{1}{2} \Big(|+>|+>+|->|+>+|+>|->+|->|-> \Big)$$

However, on the other hand, $S_{1x}=\frac{1}{2} \Big(S_{1+}+S_{1-} \Big)$, which applied to the initial state $\phi$ gives $$S_{1x} \phi = \frac{\hbar}{2\sqrt{2}} \Big(|+>|+>+|+>|-> \Big)$$

My question is why I'm not obtaining the same result with both methods, as well as which one is the correct method to find the state after the measurement.


Write your state as $$ \vert{\phi}\rangle=\frac{1}{\sqrt{2}}\vert{-}\rangle_1\left( \vert + \rangle_2+\vert -\rangle_2\right) $$ Since the eigenstate of spin-up along $\hat x$ is $$ \frac{1}{\sqrt{2}}\left( \vert + \rangle_1+\vert -\rangle_1\right) $$ your measurement ought to be (notwithstanding $\otimes 1$) $$ \hat \Pi_{x,1,+} = \frac{1}{2}\Bigl[ \vert + \rangle_1+\vert -\rangle_1\Bigr] \Bigl[_1\langle + \vert +{_1\langle} -\vert\Bigr] $$ which, when acting on $\vert -\rangle_1$ gives $$ \frac{1}{2}\left( \vert + \rangle_1+\vert -\rangle_1\right) $$ which is indeed the $\vert +\rangle_{x,1}$ state as desired (i.e. the outcome is $+x$). When acting on the full $\vert \phi\rangle$ you will get $$ \hat \Pi_{x,1,+}\vert\phi\rangle= \frac{1}{2\sqrt{2}}\Bigl[ \vert + \rangle_1+\vert -\rangle_1\Bigr]\Bigl[ \vert + \rangle_2+\vert -\rangle_2\Bigr]\, . $$ As expected, this is not a normalized state since the projector does not preserve the norm.

Your initial ket in the $1$ subspace $\vert -\rangle_z$ is not an eigenstate of $S_{x,1}$ and $S_{x,1}\ne \hat\Pi_{x,1,+}$ so I don't see why you want to use the action of $S_{x,1}$ in your problem. In particular, $S_{x,1}$ is hermitian but $\hat\Pi_{x,1,+}$ is not so there is no way their action can be equalled.

  • $\begingroup$ So what effect does the action of $S_{x,1}=S_{1+}+S{1-}$ have on the initial state if it's not the same as using the projector? I mean, if it doesn't give the state after the measurement, what's its physical meaning? Applying this operator definitely changes the state, after all. $\endgroup$ – Sky Apr 18 '17 at 7:57
  • $\begingroup$ Observables are not used to transform states. You compute their averages etc, and in some cases (like $S_x$) they have an interpretation as generators of infinitesimal transformations. Their eigenvalues and eigenvectors have important applications. $\endgroup$ – ZeroTheHero Apr 18 '17 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.