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Consider the operator $S^+ = \sum_{i=1}^L S^+_i$ acting on a spin-chain of spin-1/2 particles. Denote the half-chain Von Neumann entanglement entropy of a state $|\psi\rangle$ by $\mathbb{S}[|\psi\rangle]$. (For simplicity in notation in the following, take $\mathbb{S}[0] = 0$.)

Consider the following example. Define the state $|\Omega\rangle$ as having all spins down: $|\Omega\rangle = \otimes_{i=1}^L|\downarrow\rangle$. Then $\mathbb{S}[|\Omega\rangle] =0$ because $|\Omega\rangle $ is a product state, while $\mathbb{S}[S^+|\Omega\rangle] = \ln(2)$, as can be seen by a quick Schmidt decomposition by hand. Notice in particular that

$$\mathbb{S}[S^+|\Omega\rangle] - \mathbb{S}[|\Omega\rangle] = \ln(2)$$


After playing around a little with numerics, I have the following conjecture: $$\max_{|\psi\rangle \in \mathscr{H}} \left( \mathbb{S}[S^+|\psi\rangle] - \mathbb{S}[|\psi\rangle] \right)= \ln(2)$$

That is, $S^+$ can only increase the entropy of a state by $\ln(2)$ and no more. Similarly, I conjecture that $$\max_{|\psi\rangle \in \mathscr{H}} \left( \mathbb{S}[(S^+)^n|\psi\rangle] - \mathbb{S}[|\psi\rangle] \right)=\mathbb{S}[(S^+)^n|\Omega\rangle] - \mathbb{S}[|\Omega\rangle]$$

Are these conjectures correct? How can I prove these conjectures?


A small piece of supporting evidence (not near a proof) for the first conjecture is that it is easy to check that all product states $|p\rangle$ in the $S^z$-basis obey $\mathbb{S}[S^+|p\rangle] \leq \ln(2)$, as the resulting state's Schmidt decomposition has at most two states. Another small piece of supporting evidence, when I feed in random states for $|\psi\rangle$, the entanglement entropy decreases relative to the entropy of the random state. However, this is just supporting evidence that is far from a statement about all possible states in the Hilbert space.

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  • $\begingroup$ Only saw this now: The point is that the operator S+ has Schmidt rank 2 (as an operator), and thus, it can increase the Schmidt rank of any state by at most a factor of 2. However, what happens to the Schmidt coefficients can be rather different, as you correctly note in your answer. $\endgroup$ May 2 at 22:03
  • $\begingroup$ @NorbertSchuch Thanks, that adds clarity to what's happening. $\endgroup$
    – user196574
    May 2 at 22:34
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My conjectures in my question were incorrect. Since $S^+$ can destroy certain states, it's possible to begin with a low entanglement entropy state that gains a high entropy after being acted upon by $S^+$.

For example, consider (for some large $L$) a state $$|\psi\rangle = \sqrt{.999999999} \otimes_{i=1}^L|\uparrow\rangle_i + \sqrt{.000000001} (\text{scrambled, normalized superposition of a massive number of states}).$$ The entropy of this initial state arising from the second term is suppressed by the tiny coefficient. However, the action of $S^+$ will destroy the first term, removing the suppression of the second term after normalizing. The second term after being acted upon by $S^+$ will still have some large amount of entanglement, perhaps smaller than the second term had before, but still a much larger amount than the full initial state. Thus, the entanglement of the final state will be much larger than the initial state, easily exceeding $\ln(2)$.

I will see if I can formulate a similar question in the context of unitary operators, which cannot annihilate states.

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