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Is there a simple mathematical way to prove that measurement destroys entanglement?

I can see that this is indeed true if I just take a specific measurement on an entangled state. What I am looking for is a mathematical proof that this is true in general.

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  • $\begingroup$ This may help $\endgroup$
    – epelaez
    Oct 7, 2021 at 18:50
  • $\begingroup$ Sorry, but which part says measurement destroys entanglement? $\endgroup$
    – user18492
    Oct 7, 2021 at 18:57
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    $\begingroup$ Not sure if there is a simple proof, but one approach would be to convince yourself that measurement is a certain type of quantum channel and the outputs of this type of channel cannot be entangled. $\endgroup$
    – Condo
    Oct 7, 2021 at 18:59
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    $\begingroup$ As currently stated, the claim isn't true. For example, measuring a Bell state in the Bell basis does not destroy entanglement. In fact, measurement may create entanglement as is the case when measuring a computational basis state in the Bell basis. $\endgroup$ Oct 7, 2021 at 19:16

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It is not true in general that measurement destroys entanglement. However, if the measurement basis consists of unentangled states and if the measurement outcome unambiguously identifies a specific basis element then this is indeed the case. Also, this type of measurement performed on a subsystem destroys its entanglement with other subsystems.

Projective measurement on a full system

Projective measurement can be described by a basis $|\psi_k\rangle$ labelled by a set of outcomes $\lambda_k$ for $k=1,\dots,d$ where $d$ is the dimension of the underlying Hilbert space. If $\lambda_k$ are all distinct, then the measurement outcome unambiguously identifies a specific basis element. In this case, the observable describing the measurement is said to be non-degenerate. When such a measurement is performed on the state $|\phi\rangle$ then the outcome $\lambda_k$ occurs with the probability $|\langle\phi|\psi_k\rangle|^2$ and the post-measurement state is $|\psi_k\rangle$.

If the measurement is non-degenerate, then the post-measurement state is one of the elements of the basis describing the measurement. Consequently, such measurement destroys entanglement if and only if the basis state $|\psi_k\rangle$ corresponding to the observed outcome $\lambda_k$ is unentangled. In particular, non-degenerate measurement is guaranteed to destroy entanglement if all the basis states $|\psi_k\rangle$ are unentangled.

On the other hand, if the measured observable is degenerate, then the post-measurement state may preserve preexisting entanglement.

Projective measurement on a subsystem

Even though the subsystems $A$ and $B$ of a composite system $AB$ in a pure state $|\psi_{AB}\rangle$ may be entangled with each other, the system $AB$ in a pure state is never entangled with a third system $C$. If the measurement is non-degenerate then, by arguments above, the post-measurement state is a pure state. Therefore, the non-degenerate measurement destroys entanglement of a system being measured with other systems. If the measurement is degenerate then it may preserve entanglement between the measured system and other systems.

In summary, measurement may destroy, preserve or create entanglement between subsystems that are jointly measured. If the measurement is non-degenerate then it destroys entanglement of the system with other systems not included in the measurement.

Example

Suppose that three subsystems $ABC$ are in a state where $A$ is not entangled with the other subsystems and $B$ and $C$ are entangled with each other, e.g. $|0_A\rangle\otimes(|0_B0_C\rangle+|1_B1_C\rangle)/\sqrt{2}$. If we measure the subsystems $AB$ in the Bell basis, then we create entanglement between $A$ and $B$ and destroy the entanglement between $B$ and $C$. Indeed, the post-measurement state of $AB$ is a Bell state and the post-measurement state of $C$ is a computational basis state.


$^1$ Measurement consisting of rank-1 projectors. Equivalently, measurement of an observable corresponding to a Hermitian operator with non-degenerate spectrum.
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  • $\begingroup$ That makes sense thanks !! $\endgroup$
    – user18492
    Oct 7, 2021 at 20:02
  • $\begingroup$ One extra question though, can you give an example of basis states which are entangled or rather, a measurement where after measurement, the entangled state remains entangled? $\endgroup$
    – user18492
    Oct 7, 2021 at 20:03
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    $\begingroup$ A good example is the Bell basis. Measurement in this basis will create (or preserve) entanglement since all states in the basis are (maximally) entangled. $\endgroup$ Oct 7, 2021 at 20:19
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    $\begingroup$ So basically if you measure in a bell basis the resulting states would be one of the basis and since this basis itself is entangled the resulting state would be entangled. Thanks a tonne :) $\endgroup$
    – user18492
    Oct 7, 2021 at 20:28
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    $\begingroup$ Your subsystem measurements destroy entanglement is not correct. I think you've implicitly assumed rank 1 projectors. If you allow larger rank projectors then it is not true anymore. E.g., larger rank projectors in the Schmidt basis will not necessarily destroy entanglement. $\endgroup$
    – Rammus
    Oct 6, 2022 at 8:51
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The question is not completely well-defined as stated, because the answer depends on what precisely you mean with "measurement", and what you assume the post-measurement states to be.

A possible interpretation is: consider a bipartite state $\rho$, and suppose some kind of measurement is performed on one part of it (and leaving the other untouched). The most general way to describe the process of performing some measurement, after which the state becomes something else conditionally to the measurement outcome, is via an entanglement breaking channel. For a proof that these always destroy entanglement, see this post.

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I will assume that you refer to measuring a state in the computational basis. Let us consider a state made of two quantum registers (not necessarily entangled). Without loss of generality, and assuming this state is pure for simplicity's sake, this state can be written as: $$\sum_{x,y}\alpha_{x,y}|x\rangle|y\rangle$$ with $\sum_{x,y}\left|\alpha_{x,y}\right|^2=1$. Without loss of generality, let us assume that you measure the first register and got the state $|x\rangle$. This means that the state collapsed to: $$\frac{1}{\sum\limits_y\left\|\alpha_{x,y}\right\|^2}\sum_y\alpha_{x,y}|x\rangle|y\rangle$$ In particular, you can write this state as: $$|x\rangle\otimes\left(\frac{1}{\sum\limits_y\left\|\alpha_{x,y}\right\|^2}\sum_y\alpha_{x,y}|y\rangle\right)$$ Thus, it is necessarily not entangled since it can be written as a tensor product between these two registers. Generally, it is even a common practice to ignore the registers that have been measured: as long as we know that the measurement on the first register yielded $x$, writing this state as: $$\frac{1}{\sum\limits_y\left\|\alpha_{x,y}\right\|^2}\sum_y\alpha_{x,y}|y\rangle$$ carries as much information as the previous equation. All in all, whatever the starting state is, measuring one of its registers destroys the entanglement on this register. Note however that this only applies to the register you've measured! For instance, let us consider the state: $$\frac{|000\rangle+|011\rangle+|110\rangle+|101\rangle}{2}$$ All these registers are entangled, but measuring the first register will yield either $$\frac{|00\rangle+|11\rangle}{\sqrt{2}}$$ or $$\frac{|10\rangle+|01\rangle}{\sqrt{2}}$$ In particular, the two remaining registers are still entangled with each other, though they are not entangled anymore with the first one.

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