My understanding of the inner product is that it multiplies a vector by the conjugate transpose, but I don't understand why the conjugate transpose of $|+\rangle$ is $\frac{1}{\sqrt2}(\langle0| + \langle1|)$.

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    $\begingroup$ What do you think the conjugate transpose should be? The conjugate transpose of $|0\rangle$ is $\langle0|$ and the conjugate transpose of $|1\rangle$ is $\langle1|$. So it follows that the conjugate transpose of $\frac{1}{\sqrt(2)}(|0\rangle + |1\rangle)$ is $\frac{1}{\sqrt(2)}(\langle0| + \langle1|)$. $\endgroup$ Sep 30, 2020 at 18:57
  • $\begingroup$ why is |+> = 1/(√2) * (|0⟩+|1⟩) exactly? $\endgroup$ Sep 30, 2020 at 19:37
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    $\begingroup$ It's a definition. $\endgroup$ Sep 30, 2020 at 20:14
  • $\begingroup$ sorry, I'm trying to understand. Why is it defined that way? I guess I don't know what I don't know here. $\endgroup$ Sep 30, 2020 at 20:45
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    $\begingroup$ It's a convention. Anytime you see $|+\rangle$ it means $\frac{1}{\sqrt(2) }(|0\rangle + |1\rangle)$ $\endgroup$ Sep 30, 2020 at 20:57

1 Answer 1


The complex conjugation flips the sign of the imaginary part of a complex number. Transposition exchanges the row and column co-ordinates of a value in a matrix. A vector can be thought of as a matrix with 1 column and a certain amount of rows. The conjugation of this takes it to it's row form, which for a vector $|\psi\rangle$, becomes $\langle\psi|$.

Now you have $\frac{1}{\sqrt{2}}|0\rangle+|1\rangle$.$\frac{1}{\sqrt{2}}$ is real, so complex conjugation does nothing. However, $|0\rangle+|1\rangle$ become $\langle0|+\langle1|$.

So overall, you get $\frac{1}{\sqrt{2}}(\langle0|+\langle1|)$


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