0
$\begingroup$

Say there's a positron and an electron which are occupying quantum states and are entangled. There's two observers, Alice and Bob. Assume Alice measured spin of positron along $z-$axis and obtained $+z$ spin. Now entangled electron will have $-z$ spin, if Bob tried to measure it.

Now my question is that, what if, Alice made second measurement of positron along $x-$axis, and obtained $+x$ spin? What would happen if Bob measured spin of electron along $x-$axis after Alice's second measurement?

Would that guarantee entangled electron to have $-x$ spin? Or was quantum entanglement broken already at first measurement and Bob's electron still has $-z$ spin and hence result is $+x$ $50/-x$ $50$?

$\endgroup$
1
$\begingroup$

Yes, entanglement is destroyed/broken/ruined, etc. as soon as you measure the spin of one of the particles. Let's see why.


tl;dr

The measurement of an operator defined over the Hilbert space of a subsystem of the entangled system destroys the entanglement. This can be understood as follows: the eigenstates of an operator defined over the Hilbert space of a subsystem are states in the Hilbert space of the subsystem, and since the measurement of such an operator would bring the subsystem into one of the eigenstates of the operator, the post-measurement subsystem will have a separate (independent) state living in its Hilbert space (namely, the eigenstate of the said operator). This in itself means that the entanglement of the subsystem with the rest of the system is destroyed because, in an entangled state, a subsystem cannot have its separate (independent) state (in other words, no state in the Hilbert space of the subsystem can describe the subsystem when it's entangled to the rest of the system).


Brief summary of what is an entangled state

For two particles who would individually live in Hilbert spaces $\mathcal{H}_\mathcal{A}$ and $\mathcal{H}_\mathcal{B}$ respectively, their composite system lives in the Hilbert space $\mathcal{H}_\mathcal{A}\otimes\mathcal{H}_\mathcal{B}$. Now, there are two types of states in this composite Hilbert space:

  • Product states which are of the form $\vert\psi_A\rangle\otimes\vert\psi_B\rangle$ where $\vert\psi_{A, B} \rangle\in\mathcal{H}_{\mathcal{A},\mathcal{B}}$.

  • Entangled states which are not of the form described above. For example, a state like $\frac{1}{\sqrt{2}}\big[\vert \phi_1\rangle\otimes\vert\xi_1\rangle+\vert \phi_2\rangle\otimes\vert\xi_2\rangle\big]$ where $\vert\phi_{1,2}\rangle\in\mathcal{H}_\mathcal{A}$, $\vert\xi_{1,2}\rangle\in\mathcal{H}_\mathcal{B}$ is not of the form $\vert\psi_A\rangle\otimes\vert\psi_B\rangle$.

In particular, the key thing to notice is that there is no separate state of the subsystem $\mathcal{A}$ or $\mathcal{B}$ in the composite system iff the composite system is described by an entangled state. This is because if either subsystem had a separate state then any composite system involving the subsystem would have a state which would be of the form $\vert \psi_{\text{sub-system}}\rangle\otimes\vert\psi_{\text{rest of the composite system}}\rangle$. This can be shown more rigorously but I suppose this intuitive exposure is convincing.

Why measurement of spin over one of the subsystems destroys the entanglement?

OK, so, let's consider an operator over one of the subsystems, say $\mathcal{B}$. Such an operator, by definition, is of the form $\mathcal{I}_\mathcal{A}\otimes\hat{\mathcal{O}}_\mathcal{B}$ which means that it acts as an identity operator over the Hilbert space of subsystem $\mathcal{A}$ and as some non-trivial operator $\hat{\mathcal{O}}_\mathcal{B}$ over the Hilbert space of subsystem $\mathcal{B}$. Clearly, this means that the measurement of such an operator over an entangled state would project the entangled state $\vert\psi\rangle$ to $\mathbb{P}_b\vert\psi\rangle$ with a probability $\langle \psi\vert\mathbb{P}_b\vert\psi\rangle$ where $\mathbb{P}_b=\mathbb{I}_\mathcal{A}\otimes \vert b\rangle\langle b\vert$; here $\vert b\rangle$ are the eigenstates of the operator $\hat{\mathcal{O}}_\mathcal{B}$ and they naturally live in $\mathcal{H}_\mathcal{B}$.

Let's consider a generic state $\vert \psi \rangle =\sum_{ij}c_{ij}\vert \phi^{A}_i\rangle\otimes\vert \xi^B_j\rangle$. This would be an entangled state iff $c_{ij}$ cannot be written as some $c_ic_j$. In any case, Let's see the action of the projection operator that we introduced. \begin{align} \mathbb{P}_b\vert\psi\rangle&=\Big(\mathbb{I}_\mathcal{A}\otimes \vert b\rangle\langle b\vert\Big)\Big(\sum_{ij}c_{ij}\vert \phi^{A}_i\rangle\otimes\vert \xi^B_j\rangle\Big)\\ &=\sum_{ij}~c_{ij}\vert \phi^{A}_i\rangle\otimes\vert b\rangle\langle b \vert \xi^B_j\rangle\\ &=\sum_{ij}~\langle b \vert \xi^B_j\rangle c_{ij}~\vert \phi^{A}_i\rangle\otimes\vert b\rangle\\ &=\bigg[\sum_{ij}\langle b \vert \xi^B_j\rangle c_{ij}\vert \phi^{A}_i\rangle\bigg]\otimes\vert b\rangle\\ \end{align} This is clearly of the form $\vert\psi_A\rangle\otimes\vert\psi_B\rangle$ and thus, we see that the action of the projection operator destroys the entanglement, or, in other words, a measurement of an operator defined over one of the subsystems would destroy the entanglement of the subsystem with the rest of the system.

Notice that we have implicitly assumed that the operator $\hat{\mathcal{O}}_\mathcal{B}$ is non-degenerate. If this isn't the case then, in general, we cannot say if its measurement would destroy the entanglement of not. For example, in the extreme case, a non-degenerate $\hat{\mathcal{O}}_\mathcal{B}$ can be taken to be the identity operator and it is clear that this wouldn't destroy the entanglement. In any case, this is not relevant to our example for the spin operator is not degenerate.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

$\newcommand{\ket}[1]{\lvert #1 \rangle}$Before the first measurement is made, the system is prepared in the state $$\alpha\ket{S_z^{(1)};+}\ket{S_z^{(2)};-}+\beta\ket{S_z^{(1)};-}\ket{S_z^{(2)};+}$$ where $|\alpha|^2+|\beta|^2=1$, and the positron and electron are particles $(1)$ and $(2)$ respectively. Then, Alice measures $S_z^{(1)}$, the spin of the positron in the $z$-direction, and get the value $+\frac{1}{2}$, meaning that the system is now in the state $$\ket{S_z^{(1)};+}\ket{S_z^{(2)};-}$$ so Bob would get $-\frac{1}{2}$ if he were to measure $S_z^{(2)}$, the spin of the electron in the $z$-direction. Now, we ask about measuring the spins in the $x$-direction. In order to address this, let us rewrite the positron's state in the $S_x^{(1)}$ basis. Doing that gives us the state $$\frac{1}{\sqrt2} \left(\ket{S_x^{(1)};+}+\ket{S_x^{(1)};-}\right) \ket{S_z^{(2)};-}$$ Here, we can see that Alice's measurement of $S_x^{(1)}$ will not affect the electron, as either outcome will leave the electron in the $\ket{S_z^{(2)};-}$ state; i.e. the positron and electron are not entangled. The same applies for Bob measuring $S_x^{(2)}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $S_x^{(1)}$ and $S_z^{(2)}$ are operators. What do you mean when you say that they are not entangled? $\endgroup$ – Dvij D.C. Jun 3 at 5:51
  • $\begingroup$ @DvijD.C. Sorry, I was a little confused about the terminology used with entanglement. I have edited. $\endgroup$ – Sandejo Jun 3 at 6:01
  • $\begingroup$ Ah, I see :) Much better! $\endgroup$ – Dvij D.C. Jun 3 at 6:01

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.