Pseudoentanglement Ain’t Cheap

Denote the symplectic forms on $V$ and $W$ by $\omega_{V}$ and $\omega_{W}$, respectively. Let $A\coloneqq V\oplus W$, where $V\oplus W=\{(v,w):v\in V,w\in W\}$. Define the form $\omega_{A}$ on $A$ by $\omega_{A}(a_{1},a_{2})=\omega_{V}(v_{1},v_{2})+\omega_{W}(w_{1},w_{2})$ where $a_{i}=v_{i}+w_{i}$. We will prove that $(A,\omega_{A})$ is symplectic.

It is obvious that $A$ is a vector space, so it remains to prove that $\omega_{A}$ is a symplectic form. Recall from Definition 2.3 that we must show that $\omega_{A}$ is bilinear, alternating, and non-degenerate. It is clear that $\omega_{A}$ is bilinear because it is the sum of two bilinear forms. For $a=(v,w)\in A$, we have $\omega_{A}(a,a)=\omega_{V}(v,v)+\omega_{W}(w,w)=0$, so $\omega_{A}$ is alternating.

Finally, we prove that $\omega_{A}$ is non-degenerate. Suppose we have an element $a=(v,w)$ such that for all $a^{\prime}\in A$ we have $\omega_{A}(a,a^{\prime})=\omega_{V}(v,v^{\prime})+\omega_{W}(w,w^{\prime})=0$. Now choose a $v^{\prime}\in V$ such that $[v,v^{\prime}]=0$ (one must exist since $V$ is symplectic). $\omega_{A}(a,a^{\prime})=\omega_{V}(v,v^{\prime})+\omega_{W}(w,w^{\prime})=0$ by assumption, and, because $\omega_{V}(v,v^{\prime})=0$, it follows that $\omega_{W}(w,w^{\prime})=0$ for all $w^{\prime}\in W$. Since $\omega_{W}$ is non-degenerate, $w=0$. A similar argument shows that $v=0$. Therefore, $\omega_{A}$ is non-degenerate. ∎

Recall that $W_{C(x)}W_{C(y)}=(-1)^{[C(x),C(y)]}W_{C(y)}W_{C(x)}$. Suppose that $W_{C(x)}=(-1)^{c_{1}}CW_{x}C^{\dagger}$ and $W_{C(y)}=(-1)^{c_{2}}CW_{y}C^{\dagger}$. We have

Thus $[C(x),C(y)]=[x,y]$. ∎

Let $A^{\prime}\subseteq A$ be any set of size $\dim(\mathrm{Weyl}(\ket{\psi})_{A})$. By known techniques, one can find a Clifford circuit $C$ acting only on $A$ that maps the Paulis in $\mathrm{Weyl}(\ket{\psi})_{A}$ to $\mathcal{Z}_{A^{\prime}}$.⁴⁴4An explicit algorithm for computing this $C$ can be found in [GIKL23a, Section 3]. As a consequence, this $C$ behaves as $C\ket{\psi}=\ket{x}_{A^{\prime}}\ket{\phi}_{[n]\setminus A^{\prime}}$, where $\ket{x}_{A^{\prime}}$ is a computational basis state on $A^{\prime}$ and $\ket{\phi}_{[n]\setminus A^{\prime}}$ is an arbitrary state on the remaining qubits. Because $C$ is local to $A$, it does not affect the entanglement entropy across the partition. Furthermore, it is clear that the qubits $\ket{x}_{A^{\prime}}$ are unentangled from the rest of the system. As such, only the qubits in $A\setminus A^{\prime}$ can exhibit entanglement across the partition, and there are $\lvert A\rvert-\lvert A^{\prime}\rvert=\lvert A\rvert-\dim\left(\mathrm{Weyl}(\ket{\psi})_{A}\right)$ many such qubits. So, the entanglement entropy across the partition is bounded above by $\lvert A\rvert-\dim\left(\mathrm{Weyl}(\ket{\psi})_{A}\right)$. ∎

Take any nonzero $e_{1}\in S$. Because $\dim(S)>v$, there exists some $f_{1}\in S$ such that $[e_{1},f_{1}]=1$. Let $W_{1}$ be the span of $e_{1}$ and $f_{1}$. We will prove that $S=W_{1}\oplus\left(W_{1}^{\perp}\cap S\right)$ is a direct sum. First, we argue that $W_{1}\cap\left(W_{1}^{\perp}\cap S\right)=\{0\}$.⁵⁵5In fact, because $W_{1}$ is symplectic, even $W_{1}\cap W_{1}^{\perp}=\{0\}$. Take $z\in W_{1}$. Since $W$ is the span of $e_{1}$ and $f_{1}$, we can write $z=\alpha e_{1}+\beta f_{1}$. If $z$ is also in $W^{\perp}\cap S\subseteq W^{\perp}$, then $0=[x,z]=\beta$ and $0=[y,z]=\alpha$, so $z=0$. Next, we prove that any $v\in S$ can be written as a sum of $w_{1}\in W_{1}$ and $w_{1}^{c}\in W_{1}^{\perp}\cap S$. Clearly $w_{1}\coloneqq[v,e_{1}]f_{1}+[v,f_{1}]e_{1}\in W_{1}$, and define $w_{1}^{c}\coloneqq v+[v,e_{1}]f_{1}+[v,f_{1}]e_{1}$. It is easy to check $[e_{1},w_{1}^{c}]=[f_{1},w_{1}^{c}]=0$, so $w_{1}^{c}\in W_{1}^{\perp}$. Furthermore, if since $e_{1},f_{1}\in W_{1}\subset S$, $w_{1}^{c}\in S$ as well. It is clear that $v=w_{1}+w_{1}^{c}$.

Repeat this process to collect pairs $(e_{1},f_{1}),\dots,(e_{r},f_{r})$ until we have that $S\cap_{j=1}^{r}W_{j}^{\perp}$ doesn’t contain an $e_{r+1}$ and $f_{r+1}$ such that $[e_{r+1},f_{r+1}]=1$. Observe by linearity that any object in $x\in\cap_{j=1}^{r}W_{j}^{\perp}$ must have $[x,y]=0$ for all $y\in\bigoplus_{i=1}^{r}W_{i}$. Consequently, $S\cap_{j=1}^{r}W_{j}^{\perp}\oplus\langle e_{1},\ldots,e_{r}\rangle$ is an isotropic subspace of dimension $v+k-r$. Since all isotropic subspaces within $V$ must have dimension at most $v$,

Hence, $\bigoplus_{i=1}^{r}W_{i}$ has dimension at least $2k$. Since each $W_{i}$ is symplectic, their direct sum is symplectic by 2.9. ∎

Let $\{b_{i}\}_{i}$ be a basis for $\mathrm{Weyl}(\ket{\psi})_{B}$, and let $\{e_{i}\}_{i}$ be an extension such that, together, they span $\mathrm{Weyl}(\ket{\psi})$. Define the subspace $S\coloneqq\langle\{e_{i}\}_{i}\rangle$. Clearly $\dim(S)=\dim(\mathrm{Weyl}(\ket{\psi}))-\dim(\mathrm{Weyl}(\ket{\psi})_{B})$ and $S\cap\left(\mathbb{F}_{2}^{2n}\right)_{B}=\{0\}$. Define $e^{A}_{i}\in\left(\mathbb{F}_{2}^{2n}\right)_{A}$ to be $e_{i}$ except with the coordinates not in $A$ set to $0$.

Define $S^{A}\coloneqq\langle\{e^{A}_{i}\}_{i}\rangle$.

• Claim 3.5.

  $\dim(S^{A})=\dim(S)=\dim(\mathrm{Weyl}(\ket{\psi}))-\dim(\mathrm{Weyl}(\ket{\psi})_{B})$

  Proof.

  $\dim(S)\geq\dim(S^{A})$ is trivial, so we just need to argue that the vectors $\{e_{i}^{A}\}_{i}$ are linearly independent. For the sake of contradiction, assume they’re not, i.e., that there exists some set of indices $I\subseteq[\dim(S)]$ such that $\sum_{i\in I}e^{A}_{i}=0$. Note that $\sum_{i\in I}e_{i}\neq 0$ because is $\{e_{i}\}_{i}$ is a basis by construction. Therefore, $\sum_{i\in I}e_{i}$ will be zero on the coordinates of $A$, but not $B$. That is, $\sum_{i\in I}e_{i}\in\mathrm{Weyl}(\ket{\psi})_{B}$, a contradiction. We conclude that $\dim(S^{A})=\dim(\mathrm{Weyl}(\ket{\psi}))-\dim(\mathrm{Weyl}(\ket{\psi})_{B})$. ∎

$(\mathbb{F}_{2}^{2n})_{A}$ is a symplectic subspace, and each $e^{A}_{i}\in(\mathbb{F}_{2}^{2n})_{A}$. Therefore, $S^{A}$ is a subspace of a $2\lvert A\rvert$-dimensional symplectic subspace. By Lemma 3.3, there must exist some symplectic subspace $T^{A}\subseteq S^{A}$ of dimension at least $2\left(\dim\left(S^{A}\right)-\lvert A\rvert\right)=2\left(\dim\left(\mathrm{Weyl}(\ket{\psi})\right)-\dim\left(\mathrm{Weyl}(\ket{\psi})_{B}\right)-\lvert A\rvert\right)$. Let $\{t_{i}^{A}\}_{i}$ be a symplectic basis of $T^{A}$. We can express each basis element as $t_{i}^{A}=\sum_{j}\alpha_{j}e^{A}_{j}$ for some setting of $\alpha_{j}\in\{0,1\}$. Define $t_{i}\coloneqq\sum_{j}\alpha_{j}e_{j}$, and observe that their span defines a subspace $T\subseteq\mathbb{F}_{2}^{2n}$ that shares the same dimension as $T^{\prime}$. We can then similarly define $t^{B}_{i}\in\left(\mathbb{F}_{2}^{2n}\right)_{B}$ to be $t_{i}$ except with the coordinates not in $B$ set to $0$. Observe that $t_{i}=t^{A}_{i}+t^{B}_{i}$. By the linearity of the symplectic product and the fact that $\mathrm{Weyl}(\ket{\psi})$ is isotropic,

Therefore, $\{t_{i}^{B}\}$ is also a symplectic basis for a symplectic subspace of $\left(\mathbb{F}_{2}^{2n}\right)_{B}$.

Using a Clifford circuit $C^{A}$ acting locally on the qubits in $A$, we can perform the symplectic map that takes $\{t_{i}^{A}\}$ to the symplectic basis of $\left(\mathbb{F}_{2}^{2n}\right)_{A^{\prime}}$ where $A^{\prime}\subset A$ and $\lvert A^{\prime}\rvert=\dim(T)$.⁶⁶6Again, this mapping is described in detail in [GIKL23a, Section 3]. Note that $C^{A}(t_{i}^{B})=t_{i}^{B}$. Using a second Clifford circuit $C^{B}$ acting locally on the qubits in $B$, we can take $\{t_{i}^{B}\}$ to the symplectic basis of $\left(\mathbb{F}_{2}^{2n}\right)_{B^{\prime}}$ where $B^{\prime}\subset B$ and $\lvert B^{\prime}\rvert=\lvert A^{\prime}\rvert=\dim(T)$. Note $C^{B}(C^{A}(t_{i}^{A}))=C^{A}(t_{i}^{A})$.

• Claim 3.6.

  $C^{B}(C^{A}(T))$ is a Lagrangian subspace of $\left(\mathbb{F}_{2}^{2n}\right)_{A^{\prime}\sqcup B^{\prime}}$.

  Proof.

  It is clear from the actions of $C^{A}$ and $C^{B}$ that each $C^{B}(C^{A}(t_{i}))$ is a member of $\left(\mathbb{F}_{2}^{2n}\right)_{A^{\prime}\sqcup B^{\prime}}$. Furthermore, $\dim\left(C^{B}(C^{A}(T))\right)=\dim(T)$, which is half the dimension of $\left(\mathbb{F}_{2}^{2n}\right)_{A^{\prime}\sqcup B^{\prime}}$. Finally, since $C^{B}\left(C^{A}\left(\mathrm{Weyl}(\ket{\psi})\right)\right)$ is isotropic, so too must $C^{B}(C^{A}(T))$ as a subset of $C^{B}\left(C^{A}\left(\mathrm{Weyl}(\ket{\psi})\right)\right)$. ∎

We conclude that the state of the qubits indexed by $A^{\prime}\sqcup B^{\prime}$ is a stabilizer state $\ket{\phi}$ of $2\dim(T)$ qubits that is unentangled from the rest of the system.

Our last step is to prove that $C^{B}(C^{A}(T))_{B^{\prime}}=\{0\}$, which will imply that the entanglement across $(A^{\prime},B^{\prime})$ is $\dim(T)/2$ by [FCY⁺04, Eq. 1]. First recall that $S\cap\left(\mathbb{F}_{2}^{2n}\right)_{B}=\{0\}$ by construction, which implies $T\cap\left(\mathbb{F}_{2}^{2n}\right)_{B}=\{0\}$ because $T\subseteq S$. Next, we note that $C^{A}$ has no effect on $\left(\mathbb{F}_{2}^{2n}\right)_{B}$ since its action is local to $A$. Furthermore, $C^{B}$ simply permutes $\left(\mathbb{F}_{2}^{2n}\right)_{B}$ (and cannot map elements into $(\mathbb{F}_{2}^{2n})_{B})$. Hence,

Finally, since $\left(\mathbb{F}_{2}^{2n}\right)_{B^{\prime}}\subseteq\left(\mathbb{F}_{2}^{2n}\right)_{B}$,

Because $C^{A}$ and $C^{B}$ act locally on $A$ and $B$ respectively, they do not change the entanglement between $A$ and $B$, thus completing the proof. ∎