Orthogonality on Inner Product (Quantum Mechanics also)

[RJLiberator]

Homework Statement

Consider a qubit in the state |v> ∈ ℂ^2. Suppose that a measurement of δn is made on the qubit. Show that the probability of obtaining the result "+1" in the measurement is equal to 0 if and only if |v> and |n,+> are orthogonal.

Homework Equations

Inner product axioms
|v>|w> are orthogonal if |v>|w> = 0

The Attempt at a Solution

First things first, that this is a if and only if proof.
a. Prob(+1) = 0 implies that |v> and |n,+> are orthogonal
b. |v> and |n,+> are orthogonal implies Prob(+1) = 0

Proof of b: Prob(+1) = | |n,+> |v> |^2 , but since |n,+> and |v> are orthogonal as assumed, by definition of orthogonality, this is equal to 0. And so Prob(+1) = 0.

Proof of a: Similarly, Prob(+1) = | |n,+> |v> |^2 and since this equals 0, we see that |n,+>|v> = 0. But this is precisely the definition of orthogonality.

My concern: My concern is mainly with the proof of a, it seems very weak. Would you consider this a proper proof?

 

[RJLiberator]

One little bump before the week starts. :)

 

[DrClaude]

To make the proof less "handwavy", I would start from the fact that $\left|\pm n \right\rangle$ form a complete basis, i.e., one can always write
$$
\left|\psi \right\rangle = a \left| +n \right\rangle + b \left|-n \right\rangle
$$
and deduce things about the values of $a$ and $b$.

Also, please note that the inner product is noted <v|w>, not |v>|w>. |v> is a column vector, and <v| a row vector.

 

[RJLiberator]

Thanks for the help here. That's a good idea and may make it a stronger proof.