# I Potential/Hamiltonian in MWI

1. Jul 2, 2017

### mieral

When a system gets entangled to the environment or to the observer and the world (or observer) is duplicated. What happens to the potential and Hamiltonian, does it gets duplicated or entangled too. I'd like an actual example of a system with a potential or Hamiltonian where I can analyze this. Most examples of MWI ignore the potential or Hamiltonian part so I can't see even one example to think of. Thank you.

2. Jul 2, 2017

### Staff: Mentor

No. Entanglement is a property of the quantum state of the system. The potential and Hamiltonian are what govern the dynamics of the system--how states evolve with time.

3. Jul 2, 2017

### mieral

Can you please give an example? I can't imagine how worlds can duplicate yet the setups can't duplicate. Remembering the setup is what hold the potential and Hamiltonian. So in 2 copies of the worlds. There are two setups. so why can't the potential or Hamiltonian get duplicated too. I need an actual system or setup to understand this and I can't think of any. Thanks!

4. Jul 3, 2017

### Staff: Mentor

Worlds don't duplicate in the sense you mean. There is only one quantum state for the system. The "many worlds" is an artifact of splitting up the system into subsystems. (Yes, that means the "many worlds interpretation" is misnamed.)

5. Jul 3, 2017

### Staff: Mentor

Suppose the spin of a single electron is measured by a measuring device, whose eigenstates are "up" and "down". (Think of it as a Stern-Gerlach apparatus oriented in the vertical direction.) The electron starts out in a superposition of those states; its state is $\vert e \rangle = a \vert u \rangle + b \vert d \rangle$, where $a$ and $b$ are complex numbers such that $|a|^2 + |b|^2 = 1$, and the $u$ and $d$ kets are the spin eigenstates. The measuring device starts out in a state we'll call "ready", or $R$, where it hasn't yet measured the spin, and the end states corresponding to the spin eigenstates are "measured up" and "measured down", which we write as $U$ and $D$ kets.

The evolution of the system as a whole then looks like this:

$$\Psi_\text{initial} = \vert e \rangle \vert R \rangle \rightarrow \Psi_\text{final} = a \vert u \rangle \vert U \rangle + b \vert d \rangle \vert D \rangle$$

Notice that the initial state is separable--it factors into a single state for the electron and a single state for the measuring device--while the final state is not; it is entangled, because it cannot be factored into a single state for the electron and a single state for the measuring device. But the system as a whole is still in a single state; there is no "duplication" of states. (Duplicating a quantum state is impossible; this is an important result called the "quantum no-cloning theorem".)

The "many worlds" comes in when people insist on calling each of the two terms in $\Psi_\text{final}$ a separate "world", because only each of those terms individually looks like a "classical" state of the sort people are used to (which basically means a separable state). But neither of those terms individually is the state of the system as a whole; only $\Psi_\text{final}$ as a whole is. But that state doesn't have an easy classical interpretation, because it's entangled, and the entanglement involves a measuring device.

Also note that the evolution I described above is caused by a single Hamiltonian, the Hamiltonian that describes the interaction of the electron's spin with the measuring device. There are not two Hamiltonians; the Hamiltonian doesn't split just because $\Psi_\text{final}$ has two terms. There is just one Hamiltonian, one interaction, and one state of the system.

(Note, btw, that $\Psi_\text{final}$ only has two terms because we wrote it down in a particular basis, the up/down basis. In principle, for any quantum state, there is some basis in which it is a basis state, and therefore can be written as a single term. But a basis in which $\Psi_\text{final}$ was a basis state and could be written as a single term would not correspond to any measuring device we can easily imagine.)

6. Jul 3, 2017

### mieral

Thanks for the enlightening post! But in the Schrodinger Cat version in many worlds, the cat is indeed duplicated into two or more worlds.. the dead and live cat or sick cat or other possible states. But then the cats have molecules and atoms and they are composed of potentials and hamiltonians. So the potentials and hamiltonians inside the cat are duplicated too! unless you mean all the duplicated cats are comprised of entangled cat states that are not separable? What then serve as the potential or Hamiltonian that govern the dynamics of the cat systems or subsystems?

7. Jul 3, 2017

### Staff: Mentor

The cat is not duplicated. The situation is described the same way as PeterDonis's example with electron spin except that $\Psi_{final}$ is the sum of terms corresponding to a live cat and a dead cat instead of spin-up and spin-down.
In principle the Hamiltonian of the cat is a function of the individual positions and momenta of all the $10^{23}$ or so atoms that make up the cat, as well as the electromagnetic interactions between them. In practice there's no way we could actually write that down, let alone do any calculations with it.

8. Jul 3, 2017

### mieral

The cat is not duplicated. But the observer who got entangled with the spin up or spin down are duplicated in the sense there is now observer + spin up and observer + spin down. But is it only the observer who got duplicated into two? or does it include his universe too such that the billions and billions or trillions and trillions of trillions of galaxies got duplicated as well so the observer + spin up has its own universe and galaxies and the observer + spin down has its own universe and galaxies?

9. Jul 3, 2017

### Staff: Mentor

That is not duplicating the observer (if it were, it would equally well be duplicating the cat--or the electron, depending on which scenario we are talking about). It is entangling the observer's state with the electron's (or cat's) state. That's all it is. There is one quantum state for the system, and it is entangled.

Nothing gets duplicated. You have been told this repeatedly. Therefore all of your questions from this point on are meaningless, since they assume something is duplicated when nothing is.

10. Jul 3, 2017