Quantum physics: Probability of first state using dirac notation

Question

I am struggling with this question. Let a two-dimensional Hilbert space H has orthonormal basis vectors |A⟩ and |B⟩. Consider a vector in H ⊗ H: $ |v⟩ = α_1 |AA⟩ + α_2 |AB⟩ + α_3 |BA⟩ + α_4 |BB⟩$ .

What is the probability to find the first system in state $\frac{1}{\sqrt{2}} (|A⟩ + |B⟩)$?

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I have tried defining the state $\frac{1}{\sqrt{2}} (|A⟩ + |B) = |a⟩$ and then using dirac notation to get $⟨a||v⟩ = |\frac{1}{\sqrt{2}}(α_1 + α_2) + \frac{1}{\sqrt{2}}(α_3 + α_4)|^2$. But this assumes that the braket operation acts only on the first state which I am not sure about, as well as just being wrong.

Provided solution says that you should define |a⟩ and then apply it like this (Not explaining what it does, only gives answer after it):

$|⟨a| ⊗ ⟨A||v⟩|^2 + |⟨a| ⊗ ⟨B||v⟩|^2$

Can someone explain to me what this operation does and how it works?

Answer 1

Instead of $$ |a\rangle \otimes|A\rangle $$ I will more briefly write $|a\,A\rangle$ etc.

Your combined system is in state $v\in H\otimes H\,.$ Saying that the first subystem is in state $$ |a\rangle =\frac{|A\rangle +|B\rangle }{\sqrt{2}} $$ is the same as saying that the combined system is in one of the orthogonal states $|a\,A\rangle$ or $|a\,B\rangle\,.$

The probability that this is the case is clearly the sum of the two probabilites (since $|a\,A\rangle$ and $|a\,B\rangle$ are orthogonal). This probability is obviously

\begin{align} \big|\big\langle a\,A\,\big|\,v\big\rangle\big|^2+ \big|\big\langle a\,B\,\big|\,v\big\rangle\big|^2\,. \end{align}