Undergrad Motivating the form of a bra vector

[LightPhoton]

TL;DR

How can we write that the amplitude coefficient of bra vector is complex conjugate of ket vector?

Consider a general ket $|\psi\rangle$, expressed in the $|\alpha_i\rangle$ basis:

$$
|\psi\rangle = \sum \langle \alpha_i | \psi \rangle |\alpha_i\rangle,
$$

where$\langle \alpha_i | \alpha_j \rangle = \delta_{ij}$,
with $\delta_{ij}$ being the Kronecker delta function, and $\langle \alpha_i | \psi \rangle$ being the probability amplitudes.

Now, how do we construct the bra vector $\langle \psi |$?

Starting from the normalization condition:

$$
\langle \psi | \psi \rangle = 1,
$$

we can expand $\langle \psi |$ as:

$$
\langle \psi | = \sum \langle \psi | \alpha_i \rangle \langle \alpha_i |.
$$

Substituting this into the normalization:

$$
\langle \psi | \psi \rangle = \bigg( \sum \langle \psi | \alpha_i \rangle \langle \alpha_i | \bigg) \bigg( \sum \langle \alpha_i | \psi \rangle |\alpha_i\rangle \bigg),
$$

which simplifies to:

$$
\langle \psi | \psi \rangle = \sum \langle \psi | \alpha_i \rangle \langle \alpha_i | \psi \rangle = 1.
$$

Since the right-hand side (RHS) is a real number, the left-hand side (LHS) must also be real:

$$
\langle \psi | \alpha_i \rangle \langle \alpha_i | \psi \rangle \in \mathbb{R}.
$$

From this, how do we conclude that:

$$
\langle \psi | \alpha_i \rangle = \langle \alpha_i | \psi \rangle^* \tag{1}
$$

In classical physics, dot products are commutative, so $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ and both are real. Analogously, if $\langle \psi | \alpha_i \rangle \in \mathbb{R}$, we could conclude $\langle \psi | \alpha_i \rangle = \langle \alpha_i | \psi \rangle$.

However, in the general case where these are complex numbers, we can use $1$.

But is this the only justification for these rules? Or can we expand more?

 

[PeroK]

You need two hashes to delimit inline latex.

[Note -- OP's LaTeX fixed now by the Mentors]

 

[marcusl]

It's by the definition of the inner product. In linear algebra, the dot product of complex vectors is
$$a^Hb$$
where H denotes the Hermitian transpose (complex conjugate transpose). This carries directly to the bra-ket inner product <a|b>. You could also have deduced this from your normalization expression for unit complex vectors,
$$\langle \psi | \psi \rangle = 1$$
which requires a complex conjugation to be true.