We know that any of the Bell states has the special property that the second qubit will be found in a state which is predictable based on measuring one of the qubits. But assuming any other pair of entangled qubits (any other entangled state apart from the Bell states) how does measuring the first qubit affects the probability of finding the second qubit in a specific state after measurement? Can we deduce any property(ties) for any entangled pair of qubits?

Obviously, there is a general property that measuring a quantum system impacts other systems entangled with it forthwith. But here I mean some property which is more specific and more calculational based. For instance, if the first qubit after measurement collapses in zero state then the probability of finding the second qubit in zero state will be $n$ where $n$ is calculated through some computations.


  • $\begingroup$ Just note that there are four Bell states, e.g. $\frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)$. In this case when you measure state $x$ in the first qubit, the second one is in state $1-x$. Other two Bell states are $\frac{1}{\sqrt{2}}(|01\rangle - |10\rangle)$ and $\frac{1}{\sqrt{2}}(|00\rangle - |11\rangle)$. $\endgroup$ – Martin Vesely Jan 12 at 20:01
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    $\begingroup$ @MartinVesely Thanks. You are correct. What I mentioned was just one of those four Bell states, but my question is about any other entangle pairs. For more clarification, I edited the part of the question at which you pointed. $\endgroup$ – Coder Jan 12 at 20:12
  • $\begingroup$ What you describe is a property of any pure quantum states, not only entangled ones. Are you talking about pure states, or arbitrary (mixed) states? $\endgroup$ – Norbert Schuch Jan 12 at 20:20
  • $\begingroup$ @Coder I don't understand your comment. The Bell states are pure states. For any pure state (Bell states, entangled states, unentangled states), the state of B is completely determined after measuring A. For non-pure states, it isn't - except for special cases and measurement bases. $\endgroup$ – Norbert Schuch Jan 12 at 21:39
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    $\begingroup$ @Coder For pure entangled states, once you measure one part the state of the other part is completely determined. The same is true for pure unentangled states. (This holds for projective measurements.) $\endgroup$ – Norbert Schuch Jan 12 at 22:22

Suppose you have 2-qubit state $|\psi\rangle_{AB}$. Entangled or not, you can always write it as $$|\psi\rangle_{AB}=a_0|0\rangle_A|\psi_0\rangle_{B}+a_1|1\rangle_A|\psi_1\rangle_{B}$$ If you measure qubit $A$ in state $|0\rangle$, then the qubit $B$ is in state $|\psi_0\rangle$, and you can compute the probability of qubit $B$ being in state $|0\rangle$ as $|\langle 0|\psi_0\rangle|^2$.

As you can see, the probability tells you nothing about possible entanglement of the state $|\psi\rangle_{AB}$.

If you have an unknown state, there is no way to tell whether the state is entangled or not by doing a single measurement.

If you have a known state, you can detect entanglement by computing concurrence. For a pure state $$|\psi\rangle =a_{00}|00\rangle+a_{01}|01\rangle+a_{10}|10\rangle+a_{11}|11\rangle$$ concurrence is $$C(|\psi\rangle)=2|a_{00}a_{11}-a_{01}a_{10}|$$ If concurrence is zero, the state is not entangled, otherwise it is entangled.

  • $\begingroup$ Thanks for your answer. What I infer from your explanation is that if the state is pure then it is possible to detect whether or not it is entangled otherwise it is not possible, is this conclusion correct? $\endgroup$ – Coder Jan 13 at 8:57
  • $\begingroup$ Knowing that 2-qubit state is pure makes entanglement detection simpler, but you still need multiple measurements on an ensemble of identical states. $\endgroup$ – kludg Jan 13 at 9:15
  • $\begingroup$ about your last remark, note that not all entangled states can be used to observe Bell violations, see arxiv.org/abs/quant-ph/0612147 $\endgroup$ – glS Jan 13 at 10:28
  • $\begingroup$ @gIS Interesting, but do 2-qubit entangled states that can't be used to observe violation of Bell's inequalities exist? $\endgroup$ – kludg Jan 13 at 11:09
  • $\begingroup$ @kludg Regarding your comment Knowing that 2-qubit state is pure makes entanglement detection simpler, but you still need multiple measurements on an ensemble of identical states. This is what I am looking for, could you please introduce any tutorial or paper which explains about, your own explanation will be appreciated as well, thanks. $\endgroup$ – Coder Jan 14 at 22:35

There is a more direct characteristic that makes the state of an entangled pair of qubits distinct from a non-entangled pair (which is also known as a separable state).

When two qubits are not entangled, the state of the one qubit can be described without any knowledge of the state of the other qubit. Mathematically we can write:

\begin{equation} |\psi\rangle_{AB,\mathrm{sep}} = |\psi\rangle_{A}|\phi\rangle_{B}. \end{equation} That is to say, the first qubit is in the state $|\psi\rangle_{A}$, and the second qubit is in the state $|\phi\rangle_{B}$; these states do not have any correlation whatsoever. A measurement on the first qubit does not give any information about the second qubit (and vice-versa). Since the state of the first and second qubit is completely separated, we also refer to such a system as a separable state.

Entangled qubits do not permit the above description. The state of the one qubit is correlated with the state of the other qubit. Since there are only two orthogonal states for a single qubit, we can always write the complete state as a sum of two correlated states on the first and second qubit:

\begin{equation} |\psi\rangle_{AB,\mathrm{gen}} = a_{1}|\psi_{1}\rangle_{A}|\phi_{1}\rangle_{B} + a_{2}|\psi_{2}\rangle_{A}|\phi_{2}\rangle_{B}, \end{equation}

meaning that when the first qubit is in state $|\psi_{1}\rangle_{A}$, we know the second qubit to be in state $|\phi_{1}\rangle_{B}$, and likewise for $|\psi_{2}\rangle_{A}$ and $|\phi_{2}\rangle_{B}$.

Note that if $|\phi_{1}\rangle_{B} = |\phi_{2}\rangle_{B}$ we end up with just a separable state $\big(a_{1}|\psi_{1}\rangle_{A} + a_{2}|\psi_{2}\rangle_{A}\big)|\phi_{2}\rangle_{B}$. So, our general expression gives us an entangled state whenever this is not the case:

\begin{equation} |\psi\rangle_{AB,\mathrm{ent}} = a_{1}|\psi_{1}\rangle_{A}|\phi_{1}\rangle_{B} + a_{2}|\psi_{2}\rangle_{A}|\phi_{2}\rangle_{B}, \end{equation} when $|\psi_{1}\rangle_{A} \not= |\psi_{2}\rangle_{A}$ and $|\phi_{1}\rangle_{B} \not= |\phi_{2}\rangle_{B}$.

Moreover, when $\langle\psi_{0}|\psi_{1}\rangle = 0 $ and $ \langle\phi_{0}|\phi_{1}\rangle = 0$ (they are orthogonal to each other) we have what we call a maximally entangled state.


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