Observable of a Part of a Composite Systems in QM [closed]

Question

Let $A$ and $B$ be two spin$\frac12$ particles. The state of the system composed by $A$ and $B$ is the tensor product of the state of the two particles. Each particle has a base for the relevant Hilbert space given by the two states $\left|\uparrow\right>,\left|\downarrow\right>$

The state of the composite system is therefore a 4-dimensional space with base vectors $\left|\uparrow\uparrow\right>,\left|\uparrow\downarrow\right>,\left|\downarrow\uparrow\right>,\left|\downarrow\downarrow\right>$

For example the singlet state of two entangled particles (the most popular example of entangled system) is the vector $\frac1{\sqrt2}\left(\left|\uparrow\downarrow\right>-\left|\downarrow\uparrow\right>\right)$ while the coefficients for the bases $\left|\uparrow\uparrow\right>$ and $\left|\downarrow\downarrow\right>$ are zero.

Question:

When we measure the spin of $A$ along $z$ while $B$ is far away, associated to this experiment/interaction with the system, there must be an observable associated to it.

If there is an observable then there must be an Hermitian self-adjoin operator associated to it and being the state space 4-dimensional, this operator shall be a $4\times4$ matrix.

What is this operator and what are its eigenstates?

I initially thought that two eigenstates of the operator must have been $\left|\uparrow\downarrow\right>$ and $\left|\downarrow\uparrow\right>$ because this are possible results of measurements on the singlet state. But I cannot think of an operator that covers also the case where $A$ and $B$ are not entangled and the measurement on $A$ should not affect $B$.

This is because each of the 4 base vectors of the space of states of the composite system is the tensor product of base vectors from each subsystem (see above) and therefore any possible eigenstate of an operator, in which the system will collapse after the measurement, will change $A$ under measurement but also the $B$ which is far away and should be left unchanged.

Answer 1

The $z$ spin observable for a single system is given by: $$\sigma_z=|\uparrow\rangle\langle\uparrow|-|\downarrow\rangle\langle\downarrow|.$$

The matrix for any observable $a$ on system A is given by $a_A\otimes I_B$; the corresponding observable on B is given by $I_A\otimes a_B$, where $I_A,I_B$ are the identity operators on the Hibert spaces for systems $A,B$ respectively. So the observables for $z$ measurements on $A$ and $B$ respectively are $\sigma_{Az}\otimes I_B$ and $I_A\otimes\sigma_{Bz}$.

An account of EPR experiments in terms of Heisenberg picture observables that explains how the correlations are created without the measurement on $A$ affecting $B$ or vice versa can be found in this paper:

https://arxiv.org/abs/quant-ph/9906007

Answer 2

There are such operators! However, before answering, consider a single Hilbert space rather than a composite one. Furthermore, suppose $\{ \vert i \rangle \}$ is an orthonormal set of basis vectors. Mainly, they satisfy

$$ \sum_{i} \vert i \rangle \langle i \vert = \textbf{1}, \hspace{0.5cm} \langle i \vert j \rangle = \delta_{ij} \tag{1}$$

Now consider the operator $\hat{P}_i = \vert i \rangle \langle i \vert$. It satisfies two crucial properties: $\hat{P}^2_i = \hat{P}_i$ and $\hat{P}^{\dagger}_i = \hat{P}_i$. Due to the first property, it is called a $\textbf{projection operator}$. The second property implies that it is an observable since it is hermitian. This operator, when acted on a state $\vert \psi \rangle$, $\textbf{projects}$ the state that is proportional to basis $\vert i \rangle$. These operators are used to describe measurements since (in the Copenhagen Interpretation) the wavefunction collapses to a single state. Furthermore, these measurements are called to be complete since $\sum_{i} \hat{P}_i = \textbf{1}$ by equation (1).

Now consider measuring the spin of a particle. Since the spin can take 2 values, the Hilbert space has dimension 2. We can denote the basis vectors as $\{ \vert\uparrow \rangle, \vert\downarrow \rangle\}$. The associated projection operators are $\hat{P}_{\uparrow} = \vert\uparrow \rangle \langle\uparrow \vert$ and $\hat{P}_{\downarrow} = \vert\downarrow \rangle \langle\downarrow \vert$. Clearly, the eigenvectors of these operators describe the state of the system. If the particle has spin up, it is an eigenvector of operator $\hat{P}_{\uparrow}$ and vice-versa.

Now, let's discuss the composite system. Before, label the individual spaces as $\mathcal{H}_1$ and $\mathcal{H}_2$. The composite system is given by $\mathcal{H}_{composite} = \mathcal{H}_1 \otimes \mathcal{H}_2 $. As you have said, the basis vectors are $\{ \vert\uparrow \rangle \otimes\vert\uparrow \rangle, \vert\uparrow \rangle \otimes\vert\downarrow \rangle, \vert\downarrow \rangle \otimes\vert\uparrow \rangle, \vert\downarrow \rangle \otimes\vert\downarrow \rangle \}$. Now any wavefunction living in this composite space can be written as a linear combination of these basis vectorsm as they span this space. Now if you want to make a measurement on this system you use these basis vectors as we did previously. However, say you want to make a measurement on $\mathcal{H}_1$ and leave $\mathcal{H}_2$ alone. To do this you need to use the following operator:

$$ \hat{P} = (\vert \uparrow \rangle \langle \uparrow \vert)_1 \otimes \textbf{1}_2 $$

where the subscripts refer to which space the operators act. To see why this is appropriate consider acting this on a physical wavefunction. The identity leaves the $\mathcal{H}_2$ alone but the projection operator in $\mathcal{H}_1$ puts the state in $\mathcal{H}_1$ to the state after the measurement.

I hope this helps.