How to take partial inner product between tensor product states and GHZ state?

Question

I am trying to solve some problems in which 3 people (Alice, Bob and Charlie) share 3 photons entangled in the state $|GHZ\rangle$ and Alice and Bob perform some joint measurement on $|GHZ\rangle$. I am required to find what the probability of measuring some other state $|\psi_{AB}\rangle$ is and and onto what state Charlies photon gets projected, assuming measurement in a basis that includes the state $|\psi_{AB}\rangle$.

I think I can solve this by taking the partial inner product $\langle\psi_{AB}|GHZ\rangle$ to yield some vector $|\phi\rangle \in \mathbb V_C$, where the squared magnitude of $|\phi\rangle$ gives the probability of measuring it and normalising $|\phi\rangle$ gives the state that Charlies photon is projected onto.

To do this, I therefore need to be able to take the inner product of a vector in $\mathbb V_A \otimes \mathbb V_B$ with a vector in $\mathbb V_A \otimes \mathbb V_B \otimes \mathbb V_C$. I understand how to take a partial inner product between two vectors when the first vector is a local vector, but I am unsure of how to do it in cases like these when both vectors exist in tensor product spaces. I have studied my textbook for a while, but could not understand entirely the method, but have attempted the first part of the question, where $|\psi\rangle = |\Psi^-\rangle$ with my understanding of how the partial inner product works. Is this correct, and if not, what have I misinterpreted?

$$\begin{align}\langle \Psi^-|GHZ\rangle &= \frac{1}{2}(\langle HV| - \langle VH|)(|HHH\rangle + |VVV\rangle)\\ &=\frac{1}{2}(\langle HV|HHH\rangle - \langle VH|HHH\rangle + \langle HV|VVV\rangle - \langle VH|VVV\rangle)\\ &=\frac{1}{2}(|zero\rangle - |zero\rangle + |zero\rangle - |zero\rangle)\\ &= |zero\rangle \end{align}$$

Where I am getting from line 2 to 3 because, since $|H_AV_B\rangle$ is perpendicular to $|H_AH_B\rangle$, $\langle H_AV_B|H_AH_BH_C\rangle = 0|H_C\rangle$ Therefore, the probability of Alice and Bob measuring $|\Psi^-\rangle$ is $0$

Answer 1

This all looks correct, other than your notation $|0\rangle$. The overlaps for tensor products are simply composed from the overlaps for each of the subspaces.

In general, you can think of the operation $\langle \psi^-|GHZ\rangle$ as really being a shorthand for $$ \langle \psi^-|GHZ\rangle=\left(\langle \psi^-|\otimes \mathbb{I}_C\right)|GHZ\rangle, $$ where $\mathbb{I}_C$ is the identity operator on Charlie's subspace. So you are correct in doing calculations like \begin{align} \langle HV|GHZ\rangle&=\frac{1}{\sqrt{2}}(\langle HV\otimes\mathbb{I}_C)\left(|HHH\rangle+|VVV\rangle\right)\\ &=\frac{1}{\sqrt{2}}\left[\left(\langle H|H\rangle_A\right) \left(\langle V|H\rangle_B\right) \left(\mathbb{I}_C|H\rangle_C\right) + \left(\langle H|V\rangle_A\right) \left(\langle V|V\rangle_B\right) \left(\mathbb{I}_C|V\rangle_C\right)\right]\\ &=\frac{1}{\sqrt{2}}\left[\left(1\right) \left(0\right) \left(|H\rangle_C\right) + \left(0\right) \left(1\right) \left(|V\rangle_C\right)\right]=0. \end{align} The final result is not a state in Charlie's Hilbert space, but a lack of a state - this process has in some sense annihilated any state that Charlie has. A similar result is found from the overlap with the state $|VH\rangle$, as you correctly showed.