Lets say we have a complex vector $\vec{z} \!=\!(1\!+\!2i~~2\!+\!3i~~3\!+\!4i)^T$. Its scalar product $\vec{z}^T\!\! \cdot \vec{z}$ with itself will be a complex number, but if we conjugate the transposed vector we get $\overline{\vec{z}^T}\!\! \cdot \vec{z}$ *(this is a inner product right?)* and a positive real number as a result:

\begin{align*}

\overline{\vec{z}\,^{T}}\! \cdot \vec{z}&=\begin{pmatrix}1-2i&2-3i&3-4i\end{pmatrix} \begin{pmatrix}1+2i\\2+3i\\3+4i\end{pmatrix} =\\

&=\begin{pmatrix}(1-2i)(1+2i) + (2-3i)(2+3i) + (3-4i)(3+4i)\end{pmatrix} = \\

&= \begin{pmatrix}(1-2i+2i+4) + (4+6i-6i+9) + (9-12i+12i+16)\end{pmatrix} = \\

&= (5 + 13 + 25) = 43

\end{align*}

**1st question:**

I know that ket $\left|z\right\rangle$ is a vector of a Hilbert space and i know that $\vec{z}$ is the same as $\left|z\right\rangle$. But what about $\overline{\vec{z}^T}$? Is it equal to a bra $\left\langle z\right|$ ?

**2nd question**

Notation $\overline{\vec{z}^T}$ means we have to conjugate & transposethe a vector $\vec{z}$. Can this notation be swapped with a dagger $\dagger$ *(afterall this is an operation named conjugate transpose)*?

**3rd question:**

From all of the above it seems logical to ask if this equality holds $\left|z\right\rangle^\dagger = \left \langle z \right|$ ?

## Best Answer

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