[Physics] Vector $\vec{z}$ and its conjugate transpose $\overline{\vec{v}^\top}$ – is it the same as $\left|z\right\rangle$ and $\left\langle z \right|$

hilbert-spacenotationquantum mechanics

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:

\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

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

From Wiki:

For a finite-dimensional vector space, using a fixed orthonormal basis, the inner product can be written as a matrix multiplication of a row vector with a column vector:

$ \langle A | B \rangle = A_1^* B_1 + A_2^* B_2 + \cdots + A_N^* B_N = \begin{pmatrix} A_1^* & A_2^* & \cdots & A_N^* \end{pmatrix} \begin{pmatrix} B_1 \\ B_2 \\ \vdots \\ B_N \end{pmatrix}$

Based on this, the bras and kets can be defined as:

$\langle A | = \begin{pmatrix} A_1^* & A_2^* & \cdots & A_N^* \end{pmatrix}$

$ | B \rangle = \begin{pmatrix} B_1 \\ B_2 \\ \vdots \\ B_N \end{pmatrix}$

and then it is understood that a bra next to a ket implies matrix multiplication.

The conjugate transpose (also called ''Hermitian conjugate'') of a bra is the corresponding ket and vice-versa:

$\langle A |^\dagger = |A \rangle, \quad |A \rangle^\dagger = \langle A |$

because if one starts with the bra

$\begin{pmatrix} A_1^* & A_2^* & \cdots & A_N^* \end{pmatrix},$ then performs a complex conjugation, and then a matrix transpose, one ends up with the ket

$\begin{pmatrix} A_1 \\ A_2 \\ \vdots \\ A_N \end{pmatrix}$

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