# Two qubit Pauli expectation value of $\underset{U}{\mathbb{E}}[U^{\otimes 2} (P_1 \otimes P_2)^{\otimes 2} U^{*\otimes 2}]$

I want to find a value for the expression:

$$\underset{U}{\mathbb{E}}[U^{\otimes 2} (P_1 \otimes P_2)^{\otimes 2} U^{*\otimes 2}],$$

where $$U$$ is a two-qubit unitary operator chosen Haar randomly, $$P_1$$ and $$P_2$$ are two single qubit Pauli operators, and the expectation is taken over choices of $$U$$.

When both Paulis are identity, the expression is trivially $$1$$. However, I wasn't entirely sure of the other cases.

• is the second unitary supposed to be $U^{\dagger}$ instead of $U^*$? Jan 30 at 5:58
• Yes, thats’s right! Jan 30 at 6:58
• When you say that the expectation is equal to $1$, do you mean the identity? Or else, isn't the inside of the expectation a matrix? Jan 30 at 9:12
• I mean the identity. Jan 30 at 10:43
• Do you know already the answer to the simpler problem where you do not have $^{\otimes 2}$ everywhere? In general, of course, one does not know $E(X^2)$ from $E(X)$ and it may be even harder to find $E(X^{\otimes 2})$ from $E(X)$ Jan 30 at 16:41

We can prove a slightly more general statement, which works for any Hermitian "sandwiched" matrix $$P$$ in any dimension $$d$$... but only for $$t=2$$ in the number of copies :'(
Let us call $$V$$ this expectation, and let $$W$$ be a unitary matrix. We have: \begin{align*} W^{\otimes2}V &= \int_U\left(WUPU^\dagger\right)^{\otimes 2}\,\mathrm{d}\mu(U)\\ &= \int_U\left(UPU^\dagger W\right)^{\otimes 2}\,\mathrm{d}\mu(UW)\\ &= \int_U\left(UPU^\dagger W\right)^{\otimes 2}\,\mathrm{d}\mu(U)\\ &= VW^{\otimes 2}. \end{align*} Thus, $$V$$ commutes with every $$W^{\otimes 2}$$ for $$W\in \mathcal{U}(d)$$.
We now want to use Schur's Lemma. We know that the symmetric subspace $$\vee^2\mathbb{C}^d$$ and the antisymmetric one are both irreducible representations of the group action $$W\mapsto W^{\otimes 2}$$. Furthermore, the former has dimension $$\binom{d+1}{2}$$ and the latter $$\binom{d}{2}$$, which conveniently sums up to $$d^2$$. Thus, we know that $$V$$ can be written as: $$V=\lambda_1P_{\text{sym}}+\lambda_2P_{\text{antisym}}$$ where $$P_{\text{sym}}$$ and $$P_{\text{antisym}}$$ are the respective projectors onto these subspaces.
We can evaluate $$\lambda_1$$ by computing: $$\lambda_1=\langle00|V|00\rangle=\int_U(\langle0|UPU^\dagger|0\rangle)^2\,\mathrm{d}\mu(U).$$ Since $$P$$ is Hermitian, $$\langle0|UPU^\dagger|0\rangle$$ is real. Thus, we can write: $$\lambda_1=\int_U|\langle0|UPU^\dagger|0\rangle|^2\,\mathrm{d}\mu(U)$$ This is a known quantity and we have: $$\lambda_1=\frac{\mathrm{tr}^2(P)+d}{(d+1)d}$$ Note also that we have, once again using the fact that $$P$$ is Hermitian: $$\mathrm{tr}(V)=\lambda_1\binom{d+1}{2}+\lambda_2\binom{d}{2}=\mathrm{tr}^2(P).$$ Which gives us: $$\lambda_2=\frac{\mathrm{tr}^2(P)-d}{d(d-1)}.$$ If $$P$$ is a Pauli matrix, it simplifies a lot, since the identity is the only one with a non-nil trace. For $$P=I$$, we find $$\lambda_1=\lambda_2=1$$, which gives us $$V=P_{\text{sym}}+P_{\text{anti}}=I,$$ and for any other Pauli matrix this gives us $$\lambda_1=\frac{1}{d+1}$$ and $$\lambda_2=-\frac{1}{d-1}$$, which gives us $$V=\frac{1}{d+1}P_{\text{sym}}-\frac{1}{d-1}P_{\text{anti}}.$$