Bras, the dual space and inner products

So, we saw the “ket” space. Let’s now talk about the bra space. This is “dual to” the ket space.

# Dual correspondence

Every ket in ket space has a bra in bra space.

$$ \boxed{|\alpha\rangle \overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha|} $$

Same with eigenkets and eigenbras:

$$ |a'\rangle, |a''\rangle, \dots \overset{\text{D.C.}}{\longleftrightarrow} \langle a'|, \langle a''|, \dots $$

and

$$ |\alpha\rangle + |\beta\rangle \overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha| + \langle\beta| $$

The bra space is some mirror of the ket space.

Note: the bra dual of $c|\alpha\rangle$ is $c^{\ast}\langle\alpha|$

In general,

$$ c_\alpha|\alpha\rangle + c_\beta|\beta\rangle \overset{\text{D.C.}}{\longleftrightarrow} c_\alpha^{\ast}\langle\alpha| + c_\beta^{\ast}\langle\beta| $$

you must take the complex conjugate of the coefficient.

# Inner product

Inner product $\to$ taken of a bra and a ket.

$$ \langle\beta|\alpha\rangle = \left(\langle\beta|\right)\cdot\left(|\alpha\rangle\right) $$

(bra $\cdot$ ket). This product in general is a complex no. We always take one vector from bra space $\times$ one vector from ket space.

Two fundamental properties:

1. $\langle\beta|\alpha\rangle = \langle\alpha|\beta\rangle^{\ast}$
2. $\langle\alpha|\alpha\rangle \geq 0$

$\langle\alpha|\alpha\rangle$ must be a real number (no complex part). Put $\beta = \alpha$ in (1) and you get $\langle\alpha|\alpha\rangle = \langle\alpha|\alpha\rangle^{\ast}$.

# Orthogonality

Two kets $|\alpha\rangle$ and $|\beta\rangle$ are orthogonal iff

$$ \langle\alpha|\beta\rangle = 0 $$

and then (from property 1) $\langle\beta|\alpha\rangle = 0 = \langle\alpha|\beta\rangle$.

# Normalization

The normalized ket $|\tilde\alpha\rangle$ is given by:

$$ |\tilde\alpha\rangle = \left(\frac{1}{\sqrt{\langle\alpha|\alpha\rangle}}\right)|\alpha\rangle $$

Normalized means

$$ \langle\tilde\alpha|\tilde\alpha\rangle = 1 $$

In general $\sqrt{\langle\alpha|\alpha\rangle}$ is called the norm of $|\alpha\rangle$.

Normalization is imp. before using kets for physical states, as $|\alpha\rangle$ and $c|\alpha\rangle$ represent the same state.

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