Partial inner product on Hilbert space

Question

I'm trying to work through Schumacher and Westmoreland's Quantum Processes, Systems, and Information, and in their discussion on conditional states (6.4) they introduce the partial inner product which has me quite stumped.

Consider a composite quantum system AB consisting of two subsystems A and B. The subsystems are described by Hilbert Spaces $\mathcal{H}^{(\mathrm{A})}$ and $\mathcal{H}^{(\mathrm{B})}$ respectively, and AB is described by $\mathcal{H}^{(\mathrm{AB})} = \mathcal{H}^{(\mathrm{A})} \otimes \mathcal{H}^{(\mathrm{B})}$. Let $|\alpha^{(\mathrm{A})}\rangle \in \mathcal{H}^{(\mathrm{A})}$ and $|\Phi^{(\mathrm{AB})}\rangle \in \mathcal{H}^{(\mathrm{AB})}$ be arbitrary vectors. Then $|\phi_\alpha^{(\mathrm{B})}\rangle = \langle \alpha^{(\mathrm{A})} | \Phi^{(\mathrm{AB})}\rangle$ (the partial inner product) is defined to be a vector in $\mathcal{H}^{(\mathrm{B})}$ such that, for all $|\beta^{(\mathrm{B})}\rangle \in \mathcal{H}^{(\mathrm{B})}$,

$$ \langle \beta^{(\mathrm{B})} | \phi_\alpha^{(\mathrm{B})}\rangle = \langle \alpha^{(\mathrm{A})}, \beta^{(\mathrm{B})} | \Phi^{(\mathrm{AB})}\rangle.$$

I am first asked to show the existence and uniqueness of $|\phi_\alpha^{(\mathrm{B})}\rangle$. I think I understand how to do this, but I would appreciate any comments on it nonetheless.

Then I am asked to show that, given $|\Psi^{(\mathrm{AB})}\rangle \in \mathcal{H}^{(\mathrm{AB})}$ and an orthonormal A-basis $\{ |a^{(\mathrm{A})}\rangle \}$,

$$ |\Psi^{(\mathrm{AB})}\rangle = \sum_a |a^{(\mathrm{A})}\rangle \otimes \langle a^{(\mathrm{A})} |\Psi^{(\mathrm{AB})}\rangle, $$

and have no idea how to even begin.

In general, if you have any advice for how to think about the partial inner products (and conditional states as well), it would be greatly appreciated. Thanks!

Answer 1

These kinds of manipulations are way easier with explicit components over some fixed basis. To show how this works, define orthonormal bases $|a_n\rangle$ and $|b_m\rangle$ for $\mathcal H^\mathrm{(A)}$ and $\mathcal H^\mathrm{(B)}$, and let $|\alpha^{(\mathrm{A})}\rangle \in \mathcal{H}^\mathrm{(\mathrm{A})}$ and $|\Phi^{(\mathrm{AB})}\rangle \in \mathcal{H}^\mathrm{(\mathrm{AB})}$, which can be decomposed as $$ |\alpha^{(\mathrm{A})}\rangle =\sum_{n}\alpha_n|a_n⟩ \quad\text{and}\quad |\Phi^{(\mathrm{AB})}\rangle =\sum_{nm}\Phi_{nm}|a_n⟩\otimes|b_m⟩. $$ Intuitively, you'd expect the partial inner product $⟨\alpha^\mathrm{(A)}| \Phi^\mathrm{(AB)}⟩$ to contract along the $n$ index of those two vectors, and that is exactly what you do: you just define the vector in $\mathcal H^{(B)}$ with components $\sum_n \alpha_n^*\Phi_{nm}$, i.e. you define \begin{align} ⟨\alpha^\mathrm{(A)}|\Phi^\mathrm{(AB)}⟩:= |\phi_\alpha^\mathrm{(B)}⟩ & := \sum_m \left(\sum_n \alpha_n^*\Phi_{nm}\right)|b_m⟩ \\& = \sum_m \left(\sum_n ⟨\alpha^\mathrm{(A)}|a_n⟩⟨a_n,b_m|\Phi^\mathrm{(AB)}⟩\right)|b_m⟩ . \end{align} That's all well and good, since you can always define whatever you want, but now the problem is making it useful. So, suppose that you have some arbitrary $$ |\beta^\mathrm{(B)}⟩ = \sum_m \beta_m |b_m⟩ \in \mathcal H^\mathrm{(B)}, $$ and you want to check the desired property: \begin{align} ⟨\beta^\mathrm{(B)}|\phi_\alpha^\mathrm{(B)}⟩ & = \sum_m \beta_m^*\times \left(\sum_n \alpha_n^*\Phi_{nm}\right) \\ & = \sum_{nm} \beta_m^*\alpha_n^*\Phi_{nm} \\ & = \sum_{nm} ⟨\alpha^\mathrm{(A)}|a_n⟩⟨\beta^\mathrm{(B)}|b_m⟩⟨a_n,b_m|\Phi^\mathrm{(AB)}⟩ \\ & = \sum_{nm} ⟨\alpha^\mathrm{(A)},\beta^\mathrm{(B)}|a_n,b_m⟩⟨a_n,b_m|\Phi^\mathrm{(AB)}⟩ \\ & = ⟨\alpha^\mathrm{(A)},\beta^\mathrm{(B)}|\Phi^\mathrm{(AB)}⟩, \end{align} as desired.

Moreover, that property is enough to prove the uniqueness of $|\phi_\alpha^\mathrm{(B)}⟩$, since the linear functional $|\beta^\mathrm{(B)}⟩\mapsto ⟨\alpha^\mathrm{(A)},\beta^\mathrm{(B)}|\Phi^\mathrm{(AB)}⟩$ is continuous, and via the Riesz representation theorem there exists a unique vector $|\tilde \phi_\alpha^\mathrm{(B)}⟩ \in \mathcal H^{(B)}$ such that that functional is an inner product, i.e. $$ ⟨\tilde \phi_\alpha^\mathrm{(B)}|\beta^\mathrm{(B)}⟩ = ⟨\alpha^\mathrm{(A)},\beta^\mathrm{(B)}|\Phi^\mathrm{(AB)}⟩; $$ taking the conjugate we ge that the $|\tilde \phi_\alpha^\mathrm{(B)}⟩$ we were guaranteed is the $|\phi_\alpha^\mathrm{(B)}⟩$ we found earlier, and that concludes the proof.

So again, to remark on the structure of the proof I just gave: I started with an explicit coordinate representation, which I then used to perform an explicit coordinate calculation of an invariant property, that then turned out to be strong enough to stand on its own feet. This structure is remarkably common $-$ turn to any textbook on vector calculus and you'll see plenty of similar proofs. (Which is, of course, no surprise, since it's the same fundamental linear-algebraic structures that are of interest.)

The proof of the completeness relation is similar and it is exactly what you've already posted in your answer.

Answer 2

I think I might have found a solution myself, actually, so let me write that up here.

Let $\{ |b^{(\mathrm{B})}\rangle \}$ be an orthonormal B-basis. Then $\{ |a^{(\mathrm{A})}, b^{(\mathrm{B})} \rangle \}$ is the product basis for $\mathcal{H}^{(\mathrm{AB})}$. By completeness, the identity operator on $\mathcal{H}^{(\mathrm{B})}$ is given by

$$ \hat{I}^{(\mathrm{B})} = \sum_{b} | b^{(\mathrm{B})} \rangle \langle b^{(\mathrm{B})} |. $$

Applying this to the second factor in the expression for $|\Psi^{(\mathrm{AB})}\rangle$ yields:

\begin{align} |\Psi^{(\mathrm{AB})}\rangle &= \sum_a |a^{(\mathrm{A})}\rangle \otimes \hat{I}^{(\mathrm{B})} \langle a^{(\mathrm{A})} |\Psi^{(\mathrm{AB})}\rangle \\ &= \sum_a |a^{(\mathrm{A})}\rangle \otimes \left( \sum_{b} | b^{(\mathrm{B})} \rangle \langle b^{(\mathrm{B})} | \langle a^{(\mathrm{A})} |\Psi^{(\mathrm{AB})}\rangle \right) \\ &= \sum_a |a^{(\mathrm{A})}\rangle \otimes \left( \sum_{b} | b^{(\mathrm{B})} \rangle \langle a^{(\mathrm{A})}, b^{(\mathrm{B})} |\Psi^{(\mathrm{AB})}\rangle \right) \\ &= \sum_{a,b} |a^{(\mathrm{A})}\rangle \otimes | b^{(\mathrm{B})} \rangle \langle a^{(\mathrm{A})}, b^{(\mathrm{B})} |\Psi^{(\mathrm{AB})}\rangle \\ &= \sum_{a,b} |a^{(\mathrm{A})}, b^{(\mathrm{B})} \rangle \langle a^{(\mathrm{A})}, b^{(\mathrm{B})} |\Psi^{(\mathrm{AB})}\rangle \end{align}

And this equality holds by completeness on $\mathcal{H}^{(\mathrm{AB})}$.