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 $(\langle A| + \langle B|) |C\rangle = \langle A|C\rangle + \langle B|C\rangle$


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.

  • 1
    $\begingroup$ Your question is just about elementary mathematics actually! Everything relies upon this pair of identities where $a$ and $b$ are complex numbers, $(a+b)^* = a^*+ b^*$ and $(a^*)^* = a$. $\endgroup$ – Valter Moretti Aug 7 '16 at 15:04

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?

  • $\begingroup$ I had bolded the wrong line, but after reading your writing I can understand now, the line which is bolded now makes sense since ⟨C|(|A⟩+|B⟩)=[(⟨A|+⟨B|)|C⟩]∗ can be true. I adjusted to bold the right line. $\endgroup$ – Phil Aug 8 '16 at 2:01
  • $\begingroup$ The bolded line ⟨C|(|A⟩+|B⟩)=[(⟨A|+⟨B|)|C⟩]∗ must be true because if I factor them out then <C|A> + <C|B> = <A|C>* + <B|C>* which is true according to Axiom ( <A|B> = <B|A>* ) and therefore <A|C>* + <B|C>* = [(<A|+<B|)|C>]* $\endgroup$ – Phil Aug 8 '16 at 2:20

The line you have bolded just assumes: |D> = |A> + |B>

Thus, taking adjoint on both sides: <D| = <A| + <B|

Plugging this into the identity: <C | D> = <D | C>* gives your desired result.'

  • $\begingroup$ Thanks! I didn't realize this transformation could be done, presumably this is adjoint of matrix. $\endgroup$ – Phil Aug 8 '16 at 11:40
  • $\begingroup$ Please up-vote the answer if it helped you. Thanks. $\endgroup$ – user115625 Aug 9 '16 at 3:44

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