In tensor notation. A state vector $|uv\rangle$ is a tensor product(non entangled states) if and only if there is $A\in E_1(u)$ and $B\in E_2(v)$ such that $A\otimes B$.

So by postulate of quantum mechanics, the state vector after the measurement(one part measurement by a projector) is $P_u(i)=(|u_i\rangle\langle u_i|\otimes \mathbf{1})/n$ where $n$ is a normalization factor.

Therefor the state after measure $|u'v'\rangle_{after}=P_u(i)|uv\rangle=|u_i\rangle\otimes |v\rangle\cdot \lambda$ where $|u'v'\rangle_{after}$ could be written as the tensor product of $|u_i\rangle=A$ and $B=|v'\rangle=|v\rangle\cdot \lambda$. Thus $|u'v'\rangle_{after}$ became a tensor product, thus a non entangled states.

However,(a false statement) consider $P_u(U)=(\mathbf{1}\otimes\mathbf{1})$(taking "measure" to all of basis vector $|u_i\rangle\in E_1(u)$). It's obvious that $P_u(U)|uv\rangle=|uv\rangle$.

Thus after the "measure", $|uv\rangle$ is still an entangled states.

By analogy, suppose $z$ is a states encountered a massive degeneracy, possibly infinite, then $P_u(z)|uv\rangle=$ some vectors. However, there is no reason to be sure that the vector after the measurement is a tensor product, or say a non entangled states, i.e. there is no definite requirement for the existence of $A$ and $B$ such that $A\otimes B=|u'v'\rangle$ when there is degeneracy.

My questions are:

  1. Does entanglement must be destroyed after the projection operator?(with degeneracy, of course.)

  2. Does all the entangled states have to be destroyed after the measurement?

  3. If not, prove the condition for which entanglement can be preserved. Further, prove the condition for which entanglement must be destroyed.(Under projection operator, and in general cases.)

  • $\begingroup$ Why do you claim $|u'v'\rangle_{after}=P_u(i)|uv\rangle=|u_i\rangle\otimes |v\rangle\cdot \lambda$? What is $\lambda$? In general when you have an entangled state, performing a measurement on one qubit will affect the other one. $\endgroup$ – bRost03 Sep 27 '18 at 2:35
  • $\begingroup$ my guess is that $\lambda$ is one of the amplitudes (selected by the measurement process) in the Schmidt decomposition of the entangled state in the bipartite system. $\endgroup$ – wcc Sep 27 '18 at 7:02
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    $\begingroup$ This is not a real answer, but related to your question 2. Measurement does not always produce a non-entangled state at the end. In fact, entanglement by measurement (!) is a common experimental method to generate highly entangled state. Google the keyword "entanglement by measurement" or look at this example (arxiv.org/pdf/1508.03056.pdf), where a highly entangled state of many atoms is generated after a detection of a single photon from the atomic cloud. $\endgroup$ – wcc Sep 27 '18 at 7:07

You must distinguish between collective measurement and separabale measurement. I think that I can answer to all of your questions starting from a simple example. Let suppose that we have a pair of qubits, i.e., the state space is $\mathcal{H}=\mathcal{H}_2\otimes\mathcal{H}_2$ where $\mathcal{H}_2$ is a $2$-dimensional space with basis vectors $|0\rangle$ and $|1\rangle$. You can find a basis of entangled states in $\mathcal{H}$, e.g., the Bell states: $$|\phi^+\rangle = \frac{|00\rangle+|11\rangle}{\sqrt{2}} \quad;\quad|\phi^-\rangle = \frac{|00\rangle-|11\rangle}{\sqrt{2}}$$ $$|\psi^+\rangle = \frac{|01\rangle+|10\rangle}{\sqrt{2}} \quad;\quad|\psi^-\rangle = \frac{|01\rangle-|10\rangle}{\sqrt{2}}$$ Such states form an orthonormal basis for $\mathcal{H}$, and thus they can be used as a projective measurement.

Note that all of these states are entangled. Therefore, independently from the initial state, the output after the measurement will always be entangled.

However, this is never true if you measure only one part of the system. Indeed, If you take a generic bipartite state $$|\psi\rangle=\sum_{i,j}c_{ij}|i\rangle_A|j\rangle_B$$ And take, without loss of generality, a generic projective measurement $M$ on system $A$ (i.e., $M:\mathcal{H}_2\to\mathcal{H}_2$): $$M=\sum_k\lambda_k|\lambda_k\rangle\langle\lambda_k|$$ The extension $\tilde{M}$ of $M$ to the space $\mathcal{H}$, is given by: $$\tilde{M}=M\otimes I =\sum_k\lambda_k|\lambda_k\rangle\langle\lambda_k|\otimes \sum_h|\lambda_h\rangle\langle\lambda_h| = \sum_{k,h}\lambda_k|\lambda_k\rangle\langle\lambda_k|\otimes|\lambda_h\rangle\langle\lambda_h|$$

Which is the spectral decomposition of $\tilde{M}$. Therefore, any (one part) measurement on $|\psi\rangle$ would give a product state of the form $|\lambda_k\rangle|\lambda_h\rangle$.

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Here's a comment with respect to Stefano's answer. I'm still accepting his answer although it's not correct.

"A measurement always destory entanglement", that was if the oritinal particle $a$ and $b$ came from bases $E_A$ and $E_B$, and the projector $|ab><ab|$ is defined in terms of the states in space $E_A\otimes E_B$.

Notice that although $|\psi^+>$ seemed to be a "measurement", it was not a projection vector in $E_A\otimes E_B$.

In fact, the original entangled states in $E_A\otimes E_B$ is a linear combination of $c_1|\psi^+>+c_2|\psi^->$ (single states, and $|\psi^+>$ is neither an entangled states nor a product states.) in space created by $\{|\psi^+>,|\psi^->,|\phi^+>,|\phi^->\}$.

$|\psi^+>$ is a single states, not a tensor product. The basis has already changed. Although it's a measurement/projector in it's own space, it's not a measurement/projector in $E_A\otimes E_B$. Therefore, although the entanglment was preserved, the original statement was still true because there wasn't a measurement.

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  • $\begingroup$ Your point is not true because you have a wrong definition of entangled states. Let take a composite space $H=H_1\otimes H_2$. Then, a (pure) state $|\xi\rangle$ is called entangled iff for all $|\psi\rangle\in H_1, |\phi\rangle \in H_2$ we have $|\xi\rangle\ne|\psi\rangle\otimes |\phi\rangle$. It is also important to note that you may have combinations of entangled states which give a separable (non entangled) state, e.g., set $c_1=c_2=1/\sqrt{2}$ in your answer. $\endgroup$ – steg Oct 4 '18 at 14:33
  • $\begingroup$ @steg to be an entanglement, it has to be a measurelemnt. To be a measurement, you can find the eigenvector of some operator $H$, in this case, $H$ is fixed. I got what you are trying to say. but it's a matter of definition. The particle can be entangled in $H\otimes H$ by measurement outside $H\otimes H$. But then the wave collapse happend in say $A\otimes A$, not in space $H\otimes H$. That's how they say entanglement was destoried after the measurement. Entangled states in $H\otimes H$ is not an observable in $H\otimes H$.(because they are not eigenfunciton of $H\otimes H$.) $\endgroup$ – J C Oct 4 '18 at 14:42
  • $\begingroup$ The definition of entanglement does not require a measurement. It's a mathematical concept which has strong consequences (i.e., correlations) on the measurements, but the definition is measurement-independent. Take a look at this. In any case I cannot understand your point as there are some typos ($H$ is an operator or a space? What is $A$? What are $E_A$ and $E_B$? $\endgroup$ – steg Oct 4 '18 at 16:01
  • $\begingroup$ @steg The question is about after the measurement. Spin can be entangled in eigenvector space of spin. However, one take the measure in position space, which has nothing to do with spin. It's a measurement that collapse the wave(and destory the entanglement) in position space, but it had nothing to do with the entanglement in spin. The concept of measurement is with respect to eigvenvector and eigenvalues. It's the positulate of quantum mechanics in the textbook. $\endgroup$ – J C Oct 4 '18 at 16:06
  • $\begingroup$ Again, as I said to you (and also IamAstudent' answer), entanglement can exist, and can even be created after the measurement as it can be precisely defined mathematically. Obviously, since quantum mechanics needs to be reproducible, if you measure with the same observable a state after it has been collapsed, you'll obtain the same result (however, it can be entangled as I showed to you). What you're talking about is correlations of measurements which is a "consequence" of entanglement, but the two concepts are different. $\endgroup$ – steg Oct 5 '18 at 8:01

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