# Why is a conjugate transpose of $|+\rangle$ a vector $1/\sqrt{2} (\langle0| + \langle1|)$?

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 $$\frac{1}{\sqrt2}(\langle0| + \langle1|)$$.

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

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 $$\frac{1}{\sqrt{2}}(\langle0|+\langle1|)$$