Step 1: Understanding the Concept:
This question asks for the property of an inner product with respect to scalar multiplication. An inner product is linear in its first argument and conjugate linear (or antilinear) in its second argument (by convention in physics and most modern mathematics).
Step 2: Key Formula or Approach:
The axioms defining an inner product \(\langle \cdot, \cdot \rangle\) on a vector space over a field F (which can be \(\mathbb{R}\) or \(\mathbb{C}\)) include:
1. Linearity in the first argument: \(\langle c\alpha + d\gamma, \beta \rangle = c\langle \alpha, \beta \rangle + d\langle \gamma, \beta \rangle\) for scalars c, d. A special case is \(\langle c\alpha, \beta \rangle = c\langle \alpha, \beta \rangle\).
2. Conjugate symmetry: \(\langle \alpha, \beta \rangle = \overline{\langle \beta, \alpha \rangle}\).
From these two axioms, we can derive the property for the second argument.
\(\langle \alpha, c\beta \rangle = \overline{\langle c\beta, \alpha \rangle} = \overline{c\langle \beta, \alpha \rangle} = \overline{c} \cdot \overline{\langle \beta, \alpha \rangle} = \overline{c}\langle \alpha, \beta \rangle\).
This shows the inner product is conjugate linear in the second argument.
Step 3: Detailed Explanation:
We want to evaluate \(\langle a\alpha, b\beta\rangle\).
We can apply the properties step-by-step.
First, use the linearity in the first argument to pull out the scalar \(a\):
\[ \langle a\alpha, b\beta \rangle = a \langle \alpha, b\beta \rangle \]
Next, use the conjugate linearity in the second argument to pull out the scalar \(b\). The scalar comes out as its complex conjugate, \(\overline{b}\).
\[ a \langle \alpha, b\beta \rangle = a (\overline{b} \langle \alpha, \beta \rangle) = a\overline{b}\langle\alpha, \beta\rangle \]
So, the final result is \(a\overline{b}\langle\alpha, \beta\rangle\).
Step 4: Final Answer:
The correct property is \(\langle a\alpha, b\beta\rangle = a\overline{b}\langle\alpha, \beta\rangle\), which corresponds to option (B).