Question:
Suppose we have a state $ | \psi \rangle $ that is a linear combination of two other states $ | \psi \rangle = a | 1 \rangle + b | 2 \rangle $, where $ a $ and $ b $ are non-zero, but otherwise unspecified, and $ | 1 \rangle $ and $ | 2 \rangle $ are orthonormal ($ \langle 1 | 2 \rangle = \langle 2 | 1 \rangle = 0 $, $ \langle 1 | 1 \rangle = \langle 2 | 2 \rangle = 1 $). What is $ \langle 1 | \psi \rangle $ ?
Solution:
This is in fact easy, we have:
$ \begin{eqnarray*} \langle 1 | \psi \rangle &=& \langle 1 | (a | 1 \rangle + b | 2 \rangle) \rangle \\ &=& \langle 1 | a | 1 \rangle + \langle 1 | b | 2 \rangle \\ &=& a \langle 1 | 1 \rangle + b \langle 1 | 2 \rangle \\ &=& a \end{eqnarray*} $
So that is it - consider this is a warm up for week 2.
Suppose we have a state $ | \psi \rangle $ that is a linear combination of two other states $ | \psi \rangle = a | 1 \rangle + b | 2 \rangle $, where $ a $ and $ b $ are non-zero, but otherwise unspecified, and $ | 1 \rangle $ and $ | 2 \rangle $ are orthonormal ($ \langle 1 | 2 \rangle = \langle 2 | 1 \rangle = 0 $, $ \langle 1 | 1 \rangle = \langle 2 | 2 \rangle = 1 $). What is $ \langle 1 | \psi \rangle $ ?
Solution:
This is in fact easy, we have:
$ \begin{eqnarray*} \langle 1 | \psi \rangle &=& \langle 1 | (a | 1 \rangle + b | 2 \rangle) \rangle \\ &=& \langle 1 | a | 1 \rangle + \langle 1 | b | 2 \rangle \\ &=& a \langle 1 | 1 \rangle + b \langle 1 | 2 \rangle \\ &=& a \end{eqnarray*} $
So that is it - consider this is a warm up for week 2.
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