From the definition of entanglement of pure states:

Let the first system be in state $|\psi\rangle_A$ and the second in $|\psi\rangle_B$ with eigenstates $a_n|a_n\rangle$ and $b_m|b_m\rangle$ respectively.

We can write the combined wavefunction as $$|\psi\rangle_{AB} = \sum_{n,m}c_{nm}|a_n\rangle|b_m\rangle$$

A non-separable or entangled state is one where the combined wavefunction cannot be expressed as a cartesian product of the two states $|\psi\rangle_{AB} \neq |\psi\rangle_A \otimes |\psi\rangle_B$.

As given on wikipedia, it is inseparable if for any vectors $[a_n],[b_m]$ at least for one pair of coordinates $a_{n},b_{m}$ we have $c_{nm}\neq a_{n}b_{m}$


How does it follow from this definition of entanglement that if we measure $|\psi\rangle_A$ then we know the state of $|\psi\rangle_B$. It is clear from the example of Bell's states, however, how do we show this is true generally for any $N\times M$ vector space?

  • 1
    $\begingroup$ Note that the correct typesetting for kets is as |\psi\rangle. $\endgroup$ Commented Apr 7, 2021 at 15:01
  • $\begingroup$ Does this answer your question? $\endgroup$ Commented Apr 7, 2021 at 15:15
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    $\begingroup$ Are you sure that this is even true in the general case? $\endgroup$
    – noah
    Commented Apr 7, 2021 at 15:27
  • $\begingroup$ @Jakob This answer shows that any one part measurement gives a seperable final state $\endgroup$ Commented Apr 7, 2021 at 15:42
  • $\begingroup$ The final state could be seperable but still not certain. $|a_1\rangle \otimes (|b_1\rangle + |b_2\rangle)$ is seperable but $\psi_b$ is still a superposition, for example $\endgroup$ Commented Apr 7, 2021 at 15:57

2 Answers 2


The question is not quite well-defined. Given a bipartite pure state $|\Psi\rangle$ shared by Alice and Bob, assuming Alice performed a measurement and found some $|\phi\rangle$, and assuming Bon knows the initial state $|\Psi\rangle$, then Bob oughts to describe his state as $(\langle \phi|\otimes I)|\Psi\rangle$. This is always some state, regardless of whether $|\Psi\rangle$ is entangled or not. In this sense, regardless of there being entanglement, measuring one side of a system "implies" the state on the other side. The same applies to general mixed states.

A more sensible interpretation of the question is as follows: is it always true that, knowing Alice's measurement result, we can predict the result of Bob's measurement?

The answer to this depends on the choices of measurement bases, as well as on the purity of the initial state.

  1. Let the shared state be a bipartite pure separable state $|\Psi\rangle$. Then there cannot be any correlation between Alice's and Bob's measurement results, regardless of their choices of measurement basis.

  2. Let the shared state be a bipartite pure non-separable state $|\Psi\rangle$. Then there is always some choice of measurement bases such that measurement results are fully correlated. This follows from the Schmidt decomposition: you can always find orthonormal bases $\{|u_i\rangle\}_i$ and $\{|v_i\rangle\}_i$ such that $|\Psi\rangle=\sum_i \sqrt{p_i}|u_i\rangle\otimes|v_i\rangle$ for some $p_i\ge0$ such that $\sum_i p_i=1$.

    If $|\Psi\rangle$ is maximally entangled, then any choice of measurement basis will result in full correlation. If the state is not maximally entangled, then bases need to be chosen appropriately: there is always a choice of measurement bases such that the measurement outcomes are not fully correlated.

  3. If the shared state is not pure, then everything becomes more complicated. You can have a separable bipartite mixed state $\rho$ that still gives you correlations between the measurement outcomes. As an example, if $\rho=I/2$ is the one-qubit maximally mixed state, then measurement outcomes are fully correlated if Alice and Bob measure in the computational basis.


I don't think it's generally true. If we're looking at 2 entangled electrons that come from an initial state

$$ \psi = \frac 1{\sqrt 2}(|0,0\rangle + |1,1\rangle)$$

(in the total angular momentum basis $|J,M\rangle$), then in terms of the spins, that is:

$$ \psi = \frac 1{\sqrt 2}[\frac 1{\sqrt 2}(|\uparrow\downarrow\rangle -|\downarrow\uparrow\rangle)+ |\uparrow\uparrow\rangle]$$

So if I measure the 1st spin to be down, then the other is up, but if I measure it up, then the other is twice as likely to be up than down.

  • $\begingroup$ Can you show that its not generally true? $\endgroup$ Commented Apr 7, 2021 at 15:38
  • $\begingroup$ In the last phrase, after the comma, shouldn't be "but if I measure it up"? $\endgroup$
    – Mark_Bell
    Commented Apr 9, 2021 at 10:40

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