Single particle measurement in entangled states [closed]

Question

There is a system with two particles in a state given by the wave function $F(x_1,x_2)$. What does it mean to make a measurement on one of the particles? This is not defined among the axioms of quantum mechanics. If we measure quantity $A$ of particle 1, how does the wave function collapse? What could be the wave function after the measurement?

Answer 1

Let's say your system is represented by the entangled state $|F \rangle$. You measure particle 1 to be in state $|\varphi\rangle$. Mathematically, a measurement is described by projecting onto the state you measure, meaning \begin{equation} |F \rangle \rightarrow (|\varphi \rangle \langle \varphi | \otimes I_2)|F \rangle. \end{equation} $|\varphi \rangle \langle \varphi | $ is a projection operator only acting one the first particle, which is why the identity for particle 2 $I_2$ is included. If you want to write it in terms of wave functions i.e. position space, the new wave function is given by \begin{equation} F(x_1, x_2) \rightarrow\varphi(x_1)\int d x_1^\prime \varphi^\ast(x_1^\prime) F(x_1^\prime ,x_2), \end{equation} where $\varphi(x_1)$ is the wave function of the state you measured. These two formulas are the same, just written in different representations. You take the scalar product with respect to the first particle and leave the second particle untouched. Then you multiply by the state you measured. As you can see, the state after measurement is a product state and therefore not entangled anymore.

Answer 2

This is an important point, quite often overlooked in textbooks. In fact it implies a specific axiom in the rigorous formulation of QM regarding the allowed observables for a system of $N$ identical systems:

Since the $N$ systems are indistinguishable, the observables of the overall system must be invariant under the unitary representation of the permutation group of $N$ elements.

Starting from elementary yes-no observables represented by orthogonal projectors, in case of two indistinguishable particles, possible elementary observables can be of the form $$P\otimes Q + Q\otimes P\:.$$ The meaning of this elementary observable is:

one particle of the two has the property P and the other particle has the property Q.

For instance $P$ could be the particle stays here and $Q$ could mean the particle stays there.

According to this perspective, the elementary observable $R:= P\otimes I + I\otimes P$ means:

one particle of the two stays here

Regarding the post measurement state, no special postulates are necessary, instead. Let us assume the Luders von Neumann projection postulate and suppose that the initial state is, for instance for a pair of identical fermions, $$\Psi = \psi\otimes \psi’ - \psi’\otimes \psi\:.$$ Suppose that a detector, described by $R$ reveals a particle here. According to the above hypotheses, the post-measurement state-vector is $R\Psi$ up to normalisation, i.e., $$P\psi \otimes \psi’ + \psi \otimes P\psi’- P\psi’ \otimes \psi-\psi’ \otimes P \psi,$$ which is again anti symmetric.

In experiments like the Bell one, the particles are strictly separated and the detectors very localized, so that, if $P\psi\neq 0$, necessarily $P\psi’=0$. In that case if a particle is found here, the post measurement state reads $$P\psi \otimes \psi’ -\psi’ \otimes P \psi$$ as is assumed in standard computations.

Reference: my Springer book Spectral Theory and Quantum Mechanics (2nd English edition) 2017

Answer 3

Generally, "take a measurement of A" means "apply the operator $\hat A$", the initial state:

$$ F_i(x_1, x_2) = F(x_1, x_2) $$

becomes a final state:

$$ F_f = n_{A, i}\hat AF_i(x_1, x_2) $$

where $n_{A, i}$ is a normalization factor.

So without any basis state, it's difficult to say anything more.

Suppose the individual particle states can be expressed in a basis $B$ composed of eigenstates of $\hat A$, that is:

$$B = \{a\big | |a\rangle \}$$

(that is supposed to say the set of all states for all eigenvalues $a$--it may need help).

anyway, the key is that for a basis state (ket):

$$ \hat A|a\rangle = a|a\rangle $$

Now you can write the initial state as a sum of products of individual particle eigenstates. The product state is

$$ |a_1 a_2\rangle \equiv |a_1\rangle_1|a_2\rangle $$

just a product of particle 1 with eigenvalue $a_1$ and particle 2 with eigenvalue $a_2$, and sum is over all basis states:

$$ |F_i\rangle = \sum_{a_1, a_2}c_{1,2}|a_1 a_2\rangle $$

Since the operator acts on particle 1:

$$ \hat A|a_1 a_2\rangle = \hat A|a_1\rangle_1| a_2\rangle_2$$ $$ \hat A|a_1 a_2\rangle = a_1|a_1\rangle_1| a_2\rangle_2$$ $$ \hat A|a_1 a_2\rangle = a_1|a_1 a_2\rangle $$

Then:

$$ F_f = \hat AF_i = \sum_{a_1, a_2}a_1c_{1,2}|a_1 a_2\rangle $$

and

$$ n_{A, i} = \Big[\sum_{a_1, a_2} ||c_{1, 2}||^2 a_1^2 \Big]^{-\frac 1 2} $$

idk, seems kind of abstract. Do you have an examples in mind?

Answer 4

This is not defined among the axioms of quantum mechanics.

This is not the case. In fact, measurement works exactly the same as before once you understand what it means for an operator to be a single-particle operator. Consider an operator $\hat{A}$ that acts in the state space of particle 1. Then its eigenvectors $\lvert a_i \rangle$ are a basis for this state space (for convenience, we'll take the spectrum of $\hat{A}$ to be non-degenerate, i.e., $a_i \neq a_j$ for $i\neq j$.). Take any arbitrary basis $\lvert \psi_n \rangle$ for the state space of particle 2, in which case a basis for the state space of the combined two-particle system is $$ \lvert a_i, \psi_n\rangle = \lvert a_i\rangle \lvert\psi_n\rangle\,, $$ where the "product" on the right is really a tensor product.

This basis is an eigenbasis of the operator $\hat{A}$ because (a) it is a basis for the two-particle space, and (b) $$ \hat{A}\lvert a_i, \psi_n\rangle = \left(\hat{A}\lvert a_i\rangle\right) \lvert\psi_n\rangle = \left(a_i\lvert a_i\rangle\right) \lvert\psi_n\rangle =a_i\lvert a_i \rangle \lvert\psi_n\rangle\,. $$ Consider an arbitrary two-particle state $$ \lvert \psi \rangle = \sum_{i,n}c_{i,n}\lvert a_i \rangle \lvert\psi_n\rangle\,. $$ Then, the details of a measurement of $\hat{A}$ follow through exactly as before. The probability of getting $a_i$ as the outcome of the measurement is exactly the sum of the squares of the coefficients of the eigenvectors corresponding to eigenvalue $a_i$, given by $$ \Pr(A = a_i) = \sum_{n} |c_{i,n}|^2\,, $$ and the post-measurement state is the piece of the state that "lives" in the subspace spanned by the eigenvectors of $\hat{A}$ with eigenvalue $a_i$, given by $$ \lvert\psi_{\textrm{post-meas.}}\rangle = \mathcal{N}\sum_{n}c_{i,n}\lvert a_i \rangle \lvert\psi_n\rangle\,, $$ where $\mathcal{N}$ is a normalization factor (in fact, it's equal to $(\Pr(A = a_i))^{-1/2}$. (Note that we are no long summing over $i$ here.)