# How to produce a given entangled state of two quantum bits?

I was watching Leonard Susskind's video series on quantum entanglement, where he looks at the spins of two electrons. In particular, there are entangled states of the form $$\alpha\left|\uparrow \downarrow \right\rangle + \beta\left|\downarrow \uparrow\right\rangle\tag {*})$$ A special case is the singlet state $$\left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow\right\rangle$$, whose spin can be measured to be 0 in any direction.

In order to be 100% sure to have an entangled spin state, one would have to measure it, but can entangled states be eigenvectors of Hermitian operators (= results of measurements) other than the trivial one? Can these operators be expressed as tensor products of measurements on each electron (“sigmas” and “taus” in the lectures)?

[I imagine the following method. Electron pairs shot through a Stern-Gerlach-type apparatus split into three families: $$\left|\uparrow \uparrow \right\rangle$$ $$\left|\uparrow \downarrow \right\rangle, \left|\downarrow \uparrow\right\rangle$$ $$\left|\downarrow \downarrow \right\rangle$$ (careful not to split the pairs!)

The $$0$$-spin fraction then contains the entangled pairs, with the coefficients of the linear combination (*) such that the spin is with 100% probability equal to 0 on the measured axis. One can then refine this fraction again along the remaining directions and end up with the singlet state.]

• What exactly is your question? – Norbert Schuch Aug 27 '15 at 22:43
• I guess (1) is my method correct? And (2) how to express this measurement in terms of Hermitian operators, if possible in terms of commuting operators, each on each particle. – suissidle Aug 27 '15 at 23:42

I'm not sure your question is as well posed as you think it is.

In order to be 100% sure to have an entangled spin state, one would have to measure it, but can entangled states be eigenvectors of Hermitian operators (= results of measurements) other than the trivial one?

If you know something is in one of various orthogonal states then in principle the official answer is that you can find out which one. But otherwise in general you can find out of something was in a particular state unless you have many copies (and by no cloning you can't make the copies you have to already have them). And doing a measurement doesn't help. He word measurement is seriously misleading. When you so the thing called a measurement you make something become an eigenvector of that measurement. But you can do it in such a way that if it already were an eigenvector of that operator that it doesn't change.

When you force it into becoming an eigenvector you force it to be in one of various subspaces. But the set of entangled states ... is ... not ... a ... subspace.

For instance $\left|\uparrow \downarrow \right\rangle-\left|\downarrow \uparrow\right\rangle$ is an entangled state as is $\left|\uparrow \downarrow \right\rangle+\left|\downarrow \uparrow\right\rangle$ but their sum is $\left|\uparrow \downarrow \right\rangle$ which is not. So there is not a subspace of entangled states onto which you could project.

You could project onto a specific entangled state.

Can these operators be expressed as tensor products of measurements on each electron ("sigmas" and "taus" in the lectures)?

No. And as for the pair thing, that's vague and a Stern-Gerlach is just going to separate states. If you want to measure the z component of the total spin you could break it into three subspaces the span of $\left|\uparrow \uparrow\right\rangle,$ the span of $\left|\downarrow \downarrow\right\rangle,$ and the span of $\{\left|\uparrow \downarrow \right\rangle,\left|\downarrow \uparrow\right\rangle\}.$ But this would not give you entangled states if they weren't already entangled. And the result would not tell you if they were entangled.

• Any vector is in a subspace (span of itself). As to $\left|\uparrow \downarrow \right\rangle-\left|\downarrow \uparrow\right\rangle$ and $\left|\uparrow \downarrow \right\rangle+\left|\downarrow \uparrow\right\rangle$, they're in the span of $\{\left|\uparrow \downarrow \right\rangle,\left|\downarrow \uparrow\right\rangle\}$, but it's now clear that some of these states aren't entangled. However, how should one think of a pair randomly selected among the middle fraction if not as being entangled? – suissidle Aug 28 '15 at 8:12
• @suissidle Almost every joint state is entangled. And being entangled is normal. In fact, measurements are when you entangle the measurement device with the system in a particular way. Things naturally become entangled. Being unentangled is basically having two things be independent. Are two things ever truly totally independent? It's possible that everything in the universe is entangled at least a little bit. And if you randomly pick a state in your space with every state equally likely then there is a 100% chance you get an entangled state. And 100% doesn't mean it has to happen. – Timaeus Aug 28 '15 at 15:05
• @suissidle It just means you don't expect to get get 1% to be unentangled PR even to get 0.5% or 0.00000001% to be unentangled. You could get one, but each one would be like a weird result you don't expect k. A reproducible way. So just stop thinking of entangled as weird, and stop thinking entangled or not entangled is meaningful distinction. You can have a pile of a million entangled states and a different pile of a million unentangled states that are so close to each other that you couldn't tell which pile was which because a million state wasn't enough. – Timaeus Aug 28 '15 at 15:14