# Example of time-dependent factorization of a Hilbert space

In these notes on entanglement in QFT it is pointed out that in the Heisenberg picture the factorization of a Hilbert space is time-dependent (pages 18 and 19):

In the Schrödinger picture, it is clear that the state evolves from an unentangled to an entangled one, and more generally the entanglement entropy $$S(A)$$ evolves with time. In the Heisenberg picture, however, this is confusing: the state doesn't evolve, so how can it become entangled? The answer is that the factorization of the Hilbert space into $$\mathcal{H}_A\otimes \mathcal{H}_B$$ evolves! More precisely, instead of $$\mathcal{H}_{AB}=\mathcal{H}_A\otimes \mathcal{H}_B$$ we should have written $$\mathcal{H}_{AB}\simeq \mathcal{H}_A\otimes \mathcal{H}_B$$ where $$\simeq$$ means isomorphic. Normally we wouldn't bother with such a fine point, but in the Heisenberg picture the isomorphism is time-dependent. Whether a given state $$|\psi\rangle_{AB}\in \mathcal{H}_{AB}$$ is isomorphic to a tensor product $$|\psi'\rangle_A\otimes |\psi''\rangle_B\in \mathcal{H}_A\otimes \mathcal{H}_B$$ depends on the isomorphism and therefore on the time. Thus, in the Heisenberg picture, when factorizing a Hilbert space into $$\mathcal{H}_A\otimes \mathcal{H}_B$$ and computing $$\rho_A,S(A)$$, etc, we need to specify not only the subsystems $$A$$ and $$B$$ but also the time at which we are looking at them.

Now I confess I fail to see how that happens in practice.

I mean assuming that indeed there is one time-dependent isomorphis $$\mathcal{S}(t) : \mathcal{H}_{AB}\to \mathcal{H}_{A,t}\otimes \mathcal{H}_{B,t}$$ I have no problem in understanding that if $$\mathcal{S}(t_1)|\psi\rangle$$ is a tensor product it can be the case that $$\mathcal{S}(t_2)|\psi\rangle$$ is not, and hence the entanglement entropy evolves in time.

What I fail to see is why there is such a time-dependent isomorphism, why the factorization depends on time on the Heisenberg picture.

What would be a simple example that in the Heisenberg picture, the factorization of a Hilbert space into two parts is time-dependent so that we have the said time-dependent isomorphism?

In the Heisenberg picture, states do not evolve in time. Instead, the operators are defined to evolve with time. Here we have a situation where a state is unentangled (i.e. factorizable) at one time and entangled (i.e. non-factorizable) at some later time. Since the state itself doesn't change with time in the Heisenberg picture, something else must have changed.

There are two things that determine whether a state is factorizable: the state itself and the factorization of the Hilbert space. We know that the state went from being factorizable to being non-factorizable. Therefore, the factorization of the Hilbert space must have changed with time.

It's important to stress that entanglement is defined relative to a given factorization. For example, consider a system of three spin-1/2 particles in the state:

$$|\psi\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle)\otimes|\uparrow\rangle=\frac{1}{\sqrt{2}}(|\uparrow\downarrow\uparrow\rangle-|\downarrow\uparrow\uparrow\rangle)$$

This three-spin system lives in the Hilbert space $$\mathcal{H}^3=\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}$$, where $$\mathcal{H}$$ is the Hilbert space of a single spin-1/2 particle. There are different ways to factorize this Hilbert space into two pieces: you can, for example, separate it as $$\mathcal{H}^2\otimes\mathcal{H}$$, or you can separate it as $$\mathcal{H}\otimes\mathcal{H}^2$$. If you consider this as a system of (spins 1 and 2 together) + spin 3, then this state is separable. If you consider this as a system of spin 1 + (spins 2 and 3 together), then this state is entangled.*

Knowing this, how do we decide which factorization we're using? The answer is based on what you can observe about the system. If you have the ability to measure the spin of spin 1 separately, but you do not have this ability for spins 2 and 3, then you might choose the factorization $$\mathcal{H}\otimes\mathcal{H}^2$$ (this is similar to if you're studying a deuterium atom, for example, where you can measure the electron magnetic moment and the nuclear magnetic moment, but not the magnetic moment of the individual nucleons in the nucleus). Similarly, if you can measure spin 3 separately but can only measure spins 1 and 2 together, then you might choose the factorization $$\mathcal{H}^2\otimes\mathcal{H}$$. The point is that the choice of factorization is determined by a set of available operators.

Knowing that, we can finally compare what happens in the Schrodinger and Heisenberg pictures, in the following example: you have a system of two spin-1/2 particles which starts with their spins both pointing up. You apply a particular time-independent Hamiltonian which, after a certain amount of time, evolves into the maximally-entangled spin singlet state.

In the Schrodinger picture:

The Hamiltonian and the tensor-products of lower-dimensional operators that it's constructed from are time-independent. Therefore, the regions corresponding to the domains of each single-spin operator are also time-independent, so the factorization itself is time-independent. The state evolves, under these operators, from a separable member of this factorized Hilbert space to an inseparable member of this factorized Hilbert space.

In the Heisenberg picture:

The state is time-independent. The operators evolve with time, which means that the domains of each of the components of the tensor product also evolve with time. Hence, the factorization itself (which defines the separation of the Hilbert space into the domains of the lower-dimensional operators) changes with time. The state does not move, but the regions of Hilbert space which are defined by the factorization to be separable move around, and the state therefore ends up not being covered by any of these regions.

*This argument was lifted from tparker's answer here: Is entanglement *not* intrinsic to state, but dependent on division into subsystems? (Susskind QM)

• Thanks for the answer. So let me see if I got your point. We have the space $\mathcal{H}_{AB}$ on which the operators $S_{A,i}$ and $S_{B,i}$ for $i=1,2,3$ act. All observables are built out of these. So once we say that "someone will measure $\mathbf{S}_A$ and someone will measure $\mathbf{S}_B$ this choice of "grouping" of observables determines the factorization? And then since in the Heisenberg pictures the operators change in time, so does the factorization?
– Gold
Jan 22, 2019 at 0:46