Does entangled states must become non entangled states after the measurement?

Question

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.)

Answer 1

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 $|\lambda_k\rangle|\lambda_h\rangle$.

Answer 2

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.