# Does a random local unitary destroy entanglement

There is a massive debate raging about whether the wavefunction is an aspect of reality or just a way of handling the data you have about a system. This question is in that vein.

A local unitary on a qubit is an operator, $O$ on a two dimensional hilbert space, for which there exists an adjoint, $O^{\dagger}$, such that $O O^{\dagger} = I$.

Suppose you have an entagled pair of qubits. Next, suppose you perform a Uniformly random, local unitary, which you have no knowledge of, on one of the qubits. Does this destroy the entanglement?

I would say that the entanglement is destroyed if the value $E= Tr(\rho_a ln(\rho_a))$ departs from its original value, of say 1, towards 0. $\rho_a$ is the density matrix of the $a$ qubit.

Note: My first guess is that for states like this: $\psi = | 00 \rangle + |11\rangle$ the Von Neumann entropy, $E$ is 1. If you apply a random unitary to any one qubit, you are just randomizing an already completely random state, so there is no effect, so, no, you cannot destroy the entanglement with a random local unitary.

• A uniformly random, local unitary what? Also, define your notion of "destroying" entanglement. – ACuriousMind Sep 12 '14 at 14:46
• Why does this question have anything to do with the reality of the wave function? – Martin Sep 12 '14 at 14:58
• @Martin, I guess, the idea is that if you can destroy a joint state by a random local, you are destroying it just by not knowing about something. However, I am thinking of density matrices which handle classical probabilities which can safely be seen as information based. – Ben Sprott Sep 12 '14 at 15:02
• What do you mean by local? If what you mean by local is, consider a two particle Hilbert space $\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$ and define a local unitary to be of the form $U=U_A \otimes U_B$ then you can never change the entanglement entropy or the Schmidt number of a state $\rho$. It amounts to a change of basis. I suppose I should infer you mean the subsystems are spatially local and one should choose a unitary which acts on both simultaneously. – SM Kravec Sep 12 '14 at 16:19
• @SMKravec Hi, yes, by local, I mean applied to one qubit or another, or, as you wrote, a product of operators on the subsystems. So $U_A$ is local to $A$. Its not just a change of basis, however. I am considering the possibility of a random change of basis, for which there is absolutely no information for any observer. – Ben Sprott Sep 12 '14 at 16:59

I'd say: Since entanglement is correlation, randomly changing one part of the system is like adding random noise and should destroy the correlation.

But let's make this a bit more formal. You have a system $\rho\in\mathcal{B}(\mathcal{H}_A\otimes \mathcal{H}_B)$ where $\mathcal{H}_{A,B}$ are the Hilbert spaces of qubit A and B (just $\mathbb{C}^2$). Now you apply a random (according to which measure? The Haar measure? Some other measure? There is a debate about which measure is physical - the Haar measure is not it) unitary to one part of the system. If we take the Haar measure, it seems that you do the following (hopefully I'm not misunderstanding anything or erring in my calculations):

$$\rho\mapsto \int (1\otimes U) \rho (1\otimes U)^{\dagger}\,dU$$

This defines you a quantum channel $T$. The quantum channel would "destroy entanglement", if any measure of mixed entanglement would be decreasing. The entanglement would be completely destroyed, if the final state is always separable. Such channels are called "entanglement breaking channels". You think that this channel should be entanglement breaking. If it is, it suffices to show this for the maximally entangled Bell state (you can get any other pure state from a Bell state by a linear transformation on one of the two systems, i.e.: given a pure state $|\psi\rangle\in\mathcal{H}_A\otimes \mathcal{H}_B$ then $|\psi\rangle=(X\otimes 1)|\Omega\rangle$ where $|\Omega\rangle$ is a Bell state).

Now let's calculate this. Note that $\int UAU^{\dagger}\,dU=tr(A)1$. Then

$$T(|\Omega\rangle\langle \Omega|) =\sum_{ij} |i\rangle\langle j| \otimes \int U|i\rangle \langle j|U^{\dagger}\,dU=1/tr(1)$$

Hence your outcome is diagonal and as such separable ($T(|\Omega\rangle\langle \Omega|)=\sum_{ij} |ij\rangle\langle ij|/n$ (with the normalization $n$). Hence, yes, this would be an entanglement breaking channel.

• Neat stuff! Thanks Martin. Still, what about my initial intuition. The state $\rho_A$, after the trace, is $0.5|0 \rangle + 0.5|1\rangle$. A random unitary on this would not change it. The VN entropy would not change. – Ben Sprott Sep 12 '14 at 17:09
• The operation $T$ given here is a random local unitary operation on the 2nd subsystem. It is the convex combination of unitaries in a particular range performed on the state, therefore qualifies as mixing or noise. A genuine local unitary operation does not alter the Schmidt coefficients of a pure bipartite state, hence the entanglement is unchanged. However, local measurements do decrease entanglement. – updraft Mar 6 '18 at 13:42

Suppose $\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$ and $|\psi \rangle \in \mathcal{H}$ is a pure state which is entangled between the subsystems. Define a 'local' unitary to be $U=U_A \otimes U_B$. Then $U$ cannot change the entanglement entropy independent of what it is, random or not.

Expand $|\psi \rangle = \sum_{j,k} a_{j,k} |j_A \rangle \otimes |k_B \rangle$ where $\{|j \rangle\} ,\{ |k \rangle\}$ form orthonomal basis of $\mathcal{H}_A$ and $\mathcal{H}_B$ respectively. $|\psi \rangle$ has a Schmidt decomposition expressed as $|\psi \rangle = \sum_i \lambda_i |i_A \rangle \otimes |i_B \rangle$ where $\sum_i \lambda_i^2=1$. It follows the entanglement entropy of this state, independent of whether I trace out $A$ or $B$ is $S=-\sum_i \lambda_i^2 \log(\lambda_i^2)$

The claim is that acting the $U$ defined above cannot change any of these coefficients. $U|\psi \rangle = \sum_{j,k} a_{j,k} U_A|j_A \rangle \otimes U_B|k_B \rangle \equiv \sum_{j,k} a_{j,k} |j'_A \rangle \otimes |k'_B \rangle$ where by unitarity $\{|j' \rangle\} ,\{ |k' \rangle\}$ are still independent orthonormal bases. Then do the Schmidt decomposition using this new basis; nothing about the $\lambda_i$ changes.