Confusion regarding working around with the mathematics of operators acting on the whole tensor product state

Question

In 'Quantum Information by Stephen Barnett' (Page 95), we have:

$$ P(m, l)=\left|{ }_a\langle l|\otimes\langle m|\hat{U}| \psi\rangle \otimes| A\rangle_a\right|^2=\left\langle\psi\left|\hat{\pi}_{m l}\right| \psi\right\rangle $$ where $\{| l\rangle_a\}$ spans $| A\rangle_a$ and $\{| m\rangle\}$ spans $| \psi\rangle$. $\hat{U}$ is some overall unitary evolution that acts on the state $| \psi\rangle \otimes| A\rangle_a$.

With this, how do I obtain the following result mathematically? $$ \hat{\pi}_{m l}={}_a\left\langle A\left|\hat{U}^{\dagger}\right| m\right\rangle \otimes|l\rangle_{a a}\langle l| \otimes\langle m|\hat{U}| A\rangle_a $$

The thing is that the unitary acts on the tensor product state, therefore, in general, entangling it. I don't know how to mathematically work with an operator that acts on a tensor product state that itself doesn't have any mentioned decomposition.

Any help would be highly appreciated.

Answer 1

You actually already have the result: one can always define an operator $$\hat{\pi}_{ml}\equiv (\mathbb{I}_s\otimes \langle A|_a) \hat{U}^\dagger_{sa} (|m\rangle_s\otimes |l\rangle_a)(_a\langle l|\otimes \vphantom{a}_s\langle m|)\hat{U}_{sa}(|A\rangle_a\otimes \mathbb{I}_s),$$ where I just added some parentheses, added a subscript $s$ for the system Hilbert space, added the subscript $sa$ to the unitary because it acts on both, and added identity operations for the system. We simply match the parts for each subspace to obtain $$\hat{\pi}_{ml}=\mathbb{I}_s (\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a)^\dagger |m\rangle_s \vphantom{a}_s\langle m|(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a) \mathbb{I}_s=(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a)^\dagger |m\rangle_s \vphantom{a}_s\langle m|(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a) .$$ The operator in parentheses $(\vphantom{a}_a\langle l|\hat{U}_{sa}|A\rangle_a)$ is just an operator on the system.

As to the properties of $\hat{\pi}_{ml}$, one would have to prove more things, but that is not the focus of this question.

Why did we define it like this? We simply expanded the absolute square $|\langle a|B|c\rangle|^2=\langle a|B|c\rangle\langle c|B^\dagger|a\rangle=\langle c|B^\dagger|a\rangle\langle a|B|c\rangle$ to find $$\left|(_a\langle l|\otimes \vphantom{a}_s\langle m|)\hat{U}_{sa}(|A\rangle_a\otimes |\psi\rangle_s)\right|^2=(\langle\psi|_s\otimes \langle A|_a) \hat{U}^\dagger_{sa} (|m\rangle_s\otimes |l\rangle_a)(_a\langle l|\otimes \vphantom{a}_s\langle m|)\hat{U}_{sa}(|A\rangle_a\otimes |\psi\rangle_s).$$

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Now, your main confusion probably comes from not being used to operators that act on two systems. So let's just be clear and say that any such operator can be decomposed as $$\hat{M}_{sa}=\sum_{ijkl}m_{ijkl}|i\rangle_s \vphantom{a}_s\langle j|\otimes |k\rangle_a\vphantom{a}_a\langle l|.$$ Making this operator unitary, or choosing whether it entangles two systems or not, depends on the coefficients $m_{ijkl}$, but such a decomposition always exists. This lets us compute things like, for $$\hat{U}_{sa}\equiv\sum_{ijkl}u_{ijkl}|i\rangle_s \vphantom{a}_s\langle j|\otimes |k\rangle_a\vphantom{a}_a\langle l|,$$ $$\vphantom{a}_a\langle l^\prime|\hat{U}_{sa}|A\rangle_a=\sum_{ijkl}u_{ijkl}|i\rangle_s \vphantom{a}_s\langle j| (\vphantom{a}_a\langle l^\prime|k\rangle_a\vphantom{a}_a\langle l|A\rangle_a).$$ This is just a new operator $$\hat{M}_s=\sum_{ij}m_{ij}|i\rangle_s\vphantom{a}_s\langle j|$$ acting on the system alone, with coefficients that depend on $l^\prime$ and $A$: $$m_{ij}=\sum_{kl}u_{ijkl} (\vphantom{a}_a\langle l^\prime|k\rangle_a\vphantom{a}_a\langle l|A\rangle_a).$$