# How to convert a ket vector into a bra vector? [closed]

Suppose I have been given a ket vector $$|i\rangle = \frac{1}{\sqrt{2}}|u\rangle + \frac{i}{\sqrt{2}}|d\rangle$$ and I want to find the corresponding bra vector. How can I do that?

My attempt : My guess is that the corresponding bra vector would be something like : $$\langle i| = \frac{1}{\sqrt{2}}\langle u| + \frac{i}{\sqrt{2}} \langle d|$$ But I am not sure about it. Any help/hint is appreciated

Thanks in advance :)

• @Jakob That is an answer. It's not appropriate to put answers in comments. It is possible that it will be deleted. Jul 29, 2021 at 14:36
• Hint: have you tested your guess by computing $\langle i|i\rangle$?
– J.G.
Jul 29, 2021 at 14:44
• @J.G. ah I see, that's not correct. The correct one should be $$\langle i| = \frac{1}{\sqrt{2}}\langle u| - \frac{i}{\sqrt{2}} \langle d|$$? Jul 29, 2021 at 14:47

Bras are the Hermitian adjoint of kets; thus, as said by @Jacobs, you have to take the Hermitian adjoint of $$|i\rangle$$, which means taking the transpose of the complex conjugate. For numbers it's just equal to the complex conjugate. For example,
$$\frac{1}{\sqrt{2}}|u\rangle\to\langle u|\frac{1}{\sqrt{2}}$$ since the coefficient is real.