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.]

  • 3
    $\begingroup$ What exactly is your question? $\endgroup$ Commented Aug 27, 2015 at 22:43
  • $\begingroup$ 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. $\endgroup$
    – suissidle
    Commented Aug 27, 2015 at 23:42
  • $\begingroup$ Related, if not an answer: physics.stackexchange.com/questions/173776/… $\endgroup$
    – Rococo
    Commented Sep 13, 2021 at 0:08

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – suissidle
    Commented Aug 28, 2015 at 8:12
  • $\begingroup$ @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. $\endgroup$
    – Timaeus
    Commented Aug 28, 2015 at 15:05
  • $\begingroup$ @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. $\endgroup$
    – Timaeus
    Commented Aug 28, 2015 at 15:14

I'll answer by jotting down a few basic results in quantum information. For convenience let's adopt the notation $$ | \uparrow \rangle \equiv |0\rangle,\qquad | \downarrow \rangle \equiv |1\rangle . $$ For example, the singlet state is $(|01\rangle - |10\rangle)/\sqrt{2}$.

Let's also define the single-qubit operators \begin{eqnarray} X &\equiv& | 0 \rangle\langle 1 | + | 1 \rangle\langle 0 |,\\ Z &\equiv& | 0 \rangle\langle 0 | - | 1 \rangle\langle 1 |,\\ Y &\equiv& Z X. \end{eqnarray} $X$ and $Z$ are Pauli operators, and $Y$ is $i$ times a Pauli operator.

The notation $ZX$ means a product of operators acting on a single qubit. But we also want to write tensor products of operators acting on different qubits. For that purpose we add a subscript, so for example $Z_1 X_2$ means a $Z$ acting on the first qubit and an $X$ acting on the second qubit.

Now we can notice that the singlet state is an eigenstate of $Z_1 Z_2$ with eigenvalue $-1$: $$ Z_1 Z_2 \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) = \frac{1}{\sqrt{2}}(-|01\rangle + |10\rangle) = -\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) $$ and it is also an eigenstate of $X_1 X_2$ with eigenvalue $-1$ (exercise for the reader).

How about the state $(|00\rangle + |11\rangle)$? (I am not bothering with normalization now). It is an eigenstate of $Z_1 Z_2$ and of $X_1 X_2$ with eigenvalue $1$ in each case.

You can now explore the other Bell states.

The thing to notice is, of course, that these entangled states are not eigenstates of any single-qubit operator. So to measure a state and be sure that after the measurement the state is entangled, you have to measure two different two-qubit observables, in such a way that you do not also measure any single-qubit observables whose measurement would destroy the entanglement. The experimental method to measure an observable such as $Z_1 Z_2$ without measuring either $Z_1$ or $Z_2$ on its own is whatever a good experimentalist would like to propose, but the following is one good way.

To measure $Z_1 Z_2$ we may combine two-qubit operators with single-qubit measurements. To measure $Z_1 Z_2$ (without measuring either of $Z_1$ or $Z_2$ on its own) you can, for example, perform controlled-not from each qubit to a third qubit (prepared initially in $|0\rangle$) and then measure that third qubit. To measure $X_1 X_2$ you can use that $X = H Z H$ where $H$ is the Hadamard transform. It follows that you can adopt the same method as for $Z_1 Z_2$ but with $H$ operators suitably inserted.

All this is a few notes on basic ideas in quantum information which could also be found in any introductory course notes or textbook.


Finally, let me take issue with one statement in the question, where it says "In order to be 100% sure to have an entangled spin state, one would have to measure it". This is first a bit muddled, because if you have a single system in a state that is unknown to you, then performing a measurement can never give 100% certainty about what state it was in before the measurement was enacted, no matter what kind of state or observable is involved. This is because if the state was unknown then you cannot be 100% sure that your measurement did not disturb it. So the only way to get the kind of certainly one is asking for here is either to prepare the state yourself, or to have many systems which some oracle has guaranteed are all in the same state. The former is the way to do it. So to be 100% sure to have an entangled spin state, one would first prepare two spins in a known unentangled state by projective measurement, and then drive them through a known, controlled, unitary evolution to the entangled state you wanted to prepare.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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