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$|T_0\rangle = \frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle + | \downarrow\uparrow\rangle )$ is a triplet state, whose spin function has to be symmetric. $|S \rangle = \frac{1}{\sqrt{2}}(|\uparrow \downarrow\rangle - | \downarrow\uparrow\rangle )$ is the singlet state, whose spin function has to be antisymmetric.

So, my questions are following.

  • Whether it's $|T_0\rangle $ or $|S\rangle $, if you measure one spin to be up, is the other one always down?

  • In the case of $|T_0\rangle $ state, are the two spins pointing in the same direction but, upon measurement of one spin, the other spin flips itself in the opposite direction? Namely, the two spins of the $|T_0\rangle $ state are embedded in the equatorial plane of the Bloch sphere pointing in the same direction up until the Z projection of one spin and the other spin points in the opposite direction from whatever direction the measured spin was projected into.

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  1. Yes to your first question, as is clear from projecting onto either subspace with one of the spins up.
  2. I am not sure what you mean by the second question, as you can't factor out the two spins in the state and consider them separately. In both states nothing can be said about the individual spins before measurement, just that the system is in a superposition of the two basis states where they have opposite spin.
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  • $\begingroup$ Thank you for your answer. The 2nd question has to do with this paper link. They say that they can distinguish between T0 and S states by moving the left electron to the right dot. If it does tunnel, it's S. If it's T0, the electron won't tunnel to the right dot due to Pauli spin blockade. I wasn't sure why T0 won't tunnel if the electrons are antiparallel. $\endgroup$ – Blackwidow Apr 19 at 13:01
  • $\begingroup$ @Blackwidow I am not too sure about this specific experiment but of course the only thing that differs between your two states are the relative phases between the two eigenstates. I don't know the details of the experiment but I guess this "Pauli spin blockade" must be sensitive to this phase difference. Sorry for the very vague and non-specific answer $\endgroup$ – Joe Bentley Apr 19 at 13:11
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$\def\ket#1{|#1\rangle} \let\up=\uparrow \let\dn=\downarrow$ My answer to your second question is no. I can't talk about the experiment you've linked to as I didn't study that paper but I offer you a theoretical argument.

Consider the observables $S_{1z}$, $S_{2z}$. I don't know the notation you're more familiar with. I mean $S_{1z}$ is the $z$-component of spin angular momentum of first particle. E.g. neither $\ket S$ nor $\ket{T_0}$ are eigenkets of $S_{1z}$ (and not even of $S_{2z}$) although both are eigenkets of $S_{1z}+S_{2z}$ for eigenvalue 0 (note that $S_{1z}$ and $S_{2z}$ commute).

Yet both $\ket S$ and $\ket{T_0}$ are eigenkets of the product $S_{1z}\,S_{2z}$, with eigenvalue -1/4. The minus sign informs us that particles' spins are opposite. On the contrary, both $\ket{\up\up}$ and $\ket{\dn\dn}$ belong to eigenvalue +1/4.

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  1. Yes: for either state, if you measure both spins you will always obtain opposing outcomes.
  2. No: for these entangled states, trying to talk about separate, independent individual states for the two subsystems that preexist measurement will inevitably lead to confusion. This is a case where the parts do not describe the whole: you really can't think about it as two Bloch spheres (each in 3 dimensions). If you must have a Bloch sphere, there is an analogous 15-dimensional object in which the states $|T_0\rangle$ and $|S\rangle$ may be described, but it still does not allow you to make independent descriptions of the subsystems separately (before measurement).

Regarding the symmetry of $|T_0\rangle$: this refers to exchange symmetry (the state is the same if you swap the two subsystems), but does not imply that the two spins are pointing in the same direction. $|S\rangle$ is antisymmetric in the same sense: if you exchange the two subsystems, you pick up an overall minus sign. $|T_0\rangle$ and $|S\rangle$ are also parity symmetric and antisymmetric, respectively, which refers to their sign under spatial inversion.

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