Quantum-entangled macroscopic objects

Question

As a thought experiment, what would happen if you had two macroscopic metal cubes, 1 kg each, and they are entangled atom-for-atom? Each atom in cube A would be entangled to its corresponding atom in the same position in cube B. And let's say they are far apart (1 light-year) so they can't interact with each other.

What would happen if you accelerate cube A, would cube B also start accelerating? Would there be anything at all interesting about their behavior?

Answer 1

Background You imagine two interesting states. The represent the difference between microscopically coherent state and macroscopically quantum state(entanglement). First let me describe your state more clearly, then tell you what I mean.

When you entangle the atoms, atom-by-atom, you are entangling a property of the atom. There are multiple ways to do this depending on the material. We can first imagine, it's ionicly bounded and the ions are bosons. Each Ion(atom), with position $r_i$ has a ground state in the material $\psi_0(r)$ and is located at a lattice site $R$_i: $\psi_{\alpha=0}(r_i-R_i)$. We can hide the spin, electronic degrees of freedom in the index $\alpha$. Since they are bosons the ground state of the solid will be $\prod_i\psi_0(r_i-R_i)$.

Assuming the atoms in material $A$ and different (distinguishable) from the atoms in material $B$, the ground state of both materials together will be:

$\prod_{i}\psi^A_0(r_{i,A}-R_{i,A})\psi^B_0(r_{i,B}-R_{i,B})$

Notice this is a product state and the atoms are not entangled at all. An experiment trying to create the states you propose would try to cool to this ground state first. They would then try to entangle the atoms, using a given excited state $\psi_{\alpha\neq1}$. This could be an excited electron state, a vibration state, or a spin flip in the nuclei or electrons. Lets arbitrarily label this excited state with $\alpha=1$

Now if just two atoms are entangled outside the solid in the bell state, they would have a wave function:

$\psi^A_0(r_{0,A})\psi^B_1(r_{0,B})+\psi^A_1(r_{0,A})\psi^B_0(r_{0,B})$

The states you described

The first state you described, entangled atom for atom, each pair of atoms at a given site $R$ are entangled. We would write this wave function as:

$\prod_i [\psi^A_0(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})+\psi^A_1(r_{i,A}-R_{i,A})\psi^B_0(r_{i,B}-R_{i,B})]$

The second state you described is:

$\prod_i \psi^A_0(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})+\prod_i\psi^A_1(r_{i,A}-R_{i,A})\psi^B_0(r_{i,B}-R_{i,B})$

This is called the GHZ state and is macroscopcially entangled. If you measure 1 atom in solid A you know the property of all the atoms in solid $B$. For the first state (the macroscopically coherent state), measuring one atom in solid $A$ only tells you about the atom it is entangled with in solid $B$. The solids know a macroscopically more amount of information about each other in the macroscopically entangled state. This state is indicative of macroscopic quantum mechanics similar to Schrodinger cat, while the macroscopic coherent state, only has quantum properties at the 2 particle level.

The macroscopically entangled state is much more fragile. If the enviorment manages to measure the state of one atoms(say a atom in $B$ and finds it in state $1$), then the state collapses into an unentangled state:

$\prod_{i}\psi^A_0(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})$

while if the particle is measured in the macroscopically coherent state, it only collapses that atom:

$\prod_{i\neq j} [\psi^A_0(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})+\psi^A_1(r_{i,A}-R_{i,A})\psi^B_0(r_{i,B}-R_{i,B})]\psi^A_0(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})$

This correspondence between fragility and macroscopicisty has been formalized and is used to test theories of quantum gravity which hypothesize a collapse mechanism.

Effects of actions on one solid

As Stéphane Rollandin pointed out, certain actions on solid $A$ have no effect on the state of solid $B$. So accelerating one solid won't effect the other. What it does is change the state of the atoms in one solid. Your two states go to:

$\prod_i [\psi^A_2(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})+\psi^A_3(r_{i,A}-R_{i,A})\psi^B_0(r_{i,B}-R_{i,B})]$

and

$\prod_i \psi^A_2(r_{i,A}-R_{i,A})\psi^B_1(r_{i,B}-R_{i,B})+\prod_i\psi^A_3(r_{i,A}-R_{i,A})\psi^B_0(r_{i,B}-R_{i,B})$

Notice the state of the $B$ atoms is still the same, it's just entangled with different states of the $A$ atoms. The action made on solid $A$ can only be detected, by statistics. To see this, suppose the states $2$ and $3$ are:

$\psi^A_2(r_{i,A}-R_{i,A}) = \psi^A_0(r_{i,A}-R_{i,A}) + \psi^A_1(r_{i,A}-R_{i,A})$

$\psi^A_3(r_{i,A}-R_{i,A}) = \psi^A_0(r_{i,A}-R_{i,A}) - \psi^A_1(r_{i,A}-R_{i,A})$

Measuring state an A atom in state $1$ produces a different state then it did before. The macroscopic state collapses to:

$\prod_{i}\psi^A_1(r_{i,A}-R_{i,A})[\prod_i\psi^B_1(r_{i,B}-R_{i,B})-\prod_i\psi^B_0(r_{i,B}-R_{i,B})]$

Now solid $B$ is directly in a cat state superposition of all atoms in two different states. This is a dramatic change in the statistics compared to before. This dramatic change is the key to violating bell inequality expiremnets which test classical locality.

See if you can figure out what happens to the macroscopicly coherent state.