Eigenkets, base kets and the identity operator

# Eigenkets of observable operators

Let $\hat A$ be a Hermitian operator ($\hat A = \hat A^\dagger$). Hermitian operators turn out to be observables!!

Theorem: Eigenvalues of a Hermitian operator are real. Eigenkets of $\hat A$ corresponding to diff. eigenvalues are orthogonal.

Start with the eigenvalue equations:

$$ \hat A|a'\rangle = a'|a'\rangle \quad (1) \qquad \text{and} \qquad \hat A|a''\rangle = a''|a''\rangle $$

Taking the dual correspondence of the second one (and using $\hat A^\dagger = \hat A$):

$$ \langle a''|\hat A = a''^{\ast}\langle a''| \quad (2) $$

Now hit (1) with $\langle a''|$ from the left and (2) with $|a'\rangle$ from the right:

$$ \begin{aligned} \langle a''|\cdot(1): &\quad \langle a''|\hat A|a'\rangle = a'\langle a''|a'\rangle \\ (2)\cdot|a'\rangle: &\quad \langle a''|\hat A|a'\rangle = a''^{\ast}\langle a''|a'\rangle \end{aligned} $$

Subtracting:

$$ 0 = \left(a' - a''^{\ast}\right)\langle a''|a'\rangle $$

• If $a''$ and $a'$ are the same (and the ket is not NULL) $\implies$ we know $\langle a'|a'\rangle > 0$ $\implies a' = a'^{\ast}$ $\therefore$

$a'$ must be real

• If $a'$ and $a''$ are not the same, then $a' - a''^{\ast}$ cannot be 0 $\implies$ $\langle a''|a'\rangle = 0$ $(a' \neq a'')$, which proves the orthogonality.

We expect observables to have real eigenvalues. The theorem proves that for Hermitians.

It is conventional to normalize $|a'\rangle$ so that $|a'\rangle, |a''\rangle, \dots$ form an orthonormal set (orthogonal + normal):

$$ \langle a''|a'\rangle = \delta_{a''a'}, \qquad \delta_{a'a''} = \begin{cases} 1 & a' = a'' \\ 0 & a' \neq a'' \end{cases} $$

# Eigenkets as base kets

Recall that we said any ket $|\alpha\rangle$ can be formed by $|a'\rangle, |a''\rangle, \dots$ and so on (henceforth represented as $\lbrace|a'\rangle\rbrace$), or

$$ |\alpha\rangle = \sum_{a'} c_{a'}|a'\rangle $$

Let’s dive deeper into these basis vectors.

• We know that the normalized eigenkets of $\hat A$ (Hermitian) form a complete orthonormal set.

$$ |\alpha\rangle = \sum_{a'} c_{a'}|a'\rangle $$

where $|a'\rangle$ is an eigenket of $\hat A$ (Hermitian).

$$ \implies \langle a''|\alpha\rangle = \sum_{a'} c_{a'}\langle a''|a'\rangle $$

Since $\langle a''|a'\rangle = \delta_{a''a'}$, we get $\langle a''|\alpha\rangle = c_{a'}$ when $a' = a''$

$$ \boxed{\langle a'|\alpha\rangle = c_{a'}} $$

(So the expansion from the kets notes is unique, the coefficients are fixed by this.)

$$ \implies |\alpha\rangle = \sum_{a'}\langle a'|\alpha\rangle|a'\rangle = \sum_{a'}|a'\rangle\langle a'|\alpha\rangle $$

From associativity:

$$ |\alpha\rangle = \left(\sum_{a'}|a'\rangle\langle a'|\right)|\alpha\rangle $$

# The identity operator

This has to hold for any $|\alpha\rangle$, so

$$ \boxed{\sum_{a'}|a'\rangle\langle a'| = 1} $$

$\to$ must be true. This is an identity operator, and the relation is called “closure”.

Eg.

$$ \begin{aligned} \langle\alpha|\alpha\rangle &= \langle\alpha|\cdot\left(\sum_{a'}|a'\rangle\langle a'|\right)\cdot|\alpha\rangle \\ &= \sum_{a'}\langle\alpha|a'\rangle\langle a'|\alpha\rangle \\ &= \sum_{a'}|\langle\alpha|a'\rangle|^2 = \sum_{a'}|\langle a'|\alpha\rangle|^2 \end{aligned} $$

$\to$ Real no. But $\langle a'|\alpha\rangle = c_{a'}$

$$ \implies \sum_{a'}|c_{a'}|^2 = \sum_{a'}|\langle a'|\alpha\rangle|^2 = \langle\alpha|\alpha\rangle = 1 $$

if $\langle\alpha|\alpha\rangle = 1$ $\implies$ if $|\alpha\rangle$ is normalized.

# Meaning of the outer product

Coming back to $|a'\rangle\langle a'|$ $\to$ the outer product of two eigenkets of a Hermitian operator $\to$ must also be an operator.

$$ \left(|a'\rangle\langle a'|\right)|\alpha\rangle = |a'\rangle\cdot\langle a'|\alpha\rangle = c_{a'}|a'\rangle $$

$|a'\rangle\langle a'|$ selects that portion of $|\alpha\rangle$ parallel to $|a'\rangle$

$\implies |a'\rangle\langle a'|$ is called the projection operator along base ket $|a'\rangle$.

$$ \boxed{\Lambda_{a'} \equiv |a'\rangle\langle a'|} $$

Outer products like these get used to select or flip spin (eg. $|+\rangle\langle-|$ flips a down spin to up).

And we have,

$$ \sum_{a'}\Lambda_{a'} = 1 $$

$\to$ Identity.

Think of it as saying,

if you take a ket $|\alpha\rangle$ and take its projections in $\lbrace|a'\rangle\rbrace$, then adding all those projections will give back $|\alpha\rangle$.

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