[Physics] Beginner question, for Bra-Ket algebra for exercise 1.1 in Quantum Mechanics The Theoretical Minimum

complex numbershilbert-spacequantum mechanics

I am reading the book Quantum Mechanics The Theoretical Minimum and the first exercise is as follows with a given solution from this webpage. The problem I have is I don't understand how $\langle C| (|A\rangle + |B\rangle)=[ \langle A| + \langle B|] |C\rangle)^*$. I have bolded the line I do not understand below. How did the $(|A\rangle + |B\rangle)$ suddenly change into $(\langle A| + \langle B|)$ when the axiom $\langle B|A\rangle=\langle A|B\rangle^*$ is applied here?

I can accept the rest of the solution.

Exercise 1.1

a) using the axioms for inner products, prove $(\langle A| + \langle B|) |C\rangle = \langle A|C\rangle + \langle B|C\rangle$

Answer

Section 1.9.4 has 2 axioms for inner products:

  1. $\langle C| (|A\rangle + |B\rangle) = \langle C|A\rangle + \langle C|B\rangle$
  2. $\langle B|A\rangle=\langle A|B\rangle^*$

Well the first axiom is pretty like we are trying to ‘prove’ we just want to get all the $C$’s to the other side. The second axiom does this for simple bra-kets.

If we say $\langle C|D\rangle=\langle D|C\rangle^*$ from axiom 2 and that $|D\rangle=|A> + |B\rangle$
then replacing $D$ by the sum of $A$ and $B$: $\langle C| (|A\rangle + |B\rangle)=[ (\langle A| + \langle B|) |C\rangle]^*$

If we multiply out the left hand side. $\langle C|A\rangle +\langle C|B\rangle=[ (\langle A| + \langle B|) |C\rangle]^*$
and then take the complex conjugate of each term $\langle C|A\rangle^* +\langle C|B\rangle^*=[ (\langle A| + \langle B|) |C\rangle]^{* *}$

Then axiom 2 says the two items on the left become $\langle A|C\rangle +\langle B|C\rangle$ and can we assume the conjugating a conjugate gives the original? since conjugation is reversing the sign on the imaginary bit I think we can.

So $\langle A|C\rangle +\langle B|C\rangle=(\langle A| + \langle B|) |C\rangle^{ **}=(\langle A| + \langle B|) |C\rangle$
or $(\langle A| + \langle B|) |C\rangle=\langle A|C\rangle +\langle B|C\rangle$ as required.

Best Answer

Yes, there is a property of complex numbers that $(A^*)^* = A$, where $A^*$ is the complex conjugate of $A$. So

\begin{align}\langle C|A \rangle^* + \langle C|B \rangle^* &=( \{\langle A| + \langle B|\} |C \rangle)^{* *}\\ &= \langle A|C \rangle ^{* *} + \langle B|C \rangle^{* *}\\ &= \langle A|C \rangle + \langle B|C \rangle~.\end{align}

I think this was your doubt, correct?

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