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I have 2 vectors: $$|a\rangle=\begin{pmatrix} 2+i & 2+2i\end{pmatrix}$$ and $$ |b\rangle=\begin{pmatrix} 1+i & 1+2i\end{pmatrix}$$ So basically I need to show that $$\langle a| b\rangle=\langle b| a\rangle^*$$

Attempt of solution: So when I multiply matrices I get for left side: $$\langle a| b\rangle=\begin{pmatrix} 2-i \\ 2-2i \end{pmatrix}\begin{pmatrix} 1+i & 1+2i\end{pmatrix}=(2-i)(1+i)+(2-2i)(1+2i)=2+2i-i+1+2+4i-2i+4=9+3i$$ for the right side I have: $$\langle b| a\rangle=\begin{pmatrix} 1-i \\ 1-2i \end{pmatrix}\begin{pmatrix} 2+i & 2+2i\end{pmatrix}=(1-i)(2+i)+(1-2i)(2+2i)=2+i+2-2i+1+2i-4i+4=9-3i$$ But I expected to get (I have an answer given by professor) $\langle a| b\rangle=9-3i$ and$\langle b| a\rangle=9+3i$.

So I dont know, did I make a mistake, or professor?

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  • $\begingroup$ Even though i showed that left side is complex conjugate of right side, but still confused $\endgroup$
    – Kostya
    Commented May 23, 2020 at 23:35
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    $\begingroup$ Shouldn't your ket be column vectors and the bra be row vectors? $\endgroup$
    – Thormund
    Commented May 24, 2020 at 0:46
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    $\begingroup$ Thormund, you are right. Kostya Komless, bra is row vector, ket is column vector. However Kostya, your answer is correct. $\endgroup$
    – Hantarto
    Commented May 24, 2020 at 3:28
  • $\begingroup$ Okay, thanks to you people^) $\endgroup$
    – Kostya
    Commented May 24, 2020 at 7:24

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The bra should be a row vector and the ket should be a column vector. If you redo in that way, you will get the required answer.

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  • $\begingroup$ Thank you for an answer I will try to do that $\endgroup$
    – Kostya
    Commented May 24, 2020 at 7:26

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