Operators, commutators and outer products

2026-10-19

# physics # quantum-mechanics

# Operators

Observables are represented as operators.

Obviously not all operators are observables.

An operator acts on a ket from the left:

$$ \hat X\cdot\left(|\alpha\rangle\right) = \hat X|\alpha\rangle $$

↳ another ket.

Two operators $\hat X$ and $\hat Y$ are equal iff

$$ \hat X|\alpha\rangle = \hat Y|\alpha\rangle $$

for any arbitrary ket in the ket space.

$\hat X$ is a NULL operator if $\hat X|\alpha\rangle = 0$ $\forall |\alpha\rangle$ in ket space.

# Associative and commutative

$$ \hat X + \hat Y = \hat Y + \hat X $$

and,

$$ \hat X + (\hat Y + \hat Z) = (\hat X + \hat Y) + \hat Z $$

Most operators are linear (except time reversal, forget that):

$$ \hat X\left(c_\alpha|\alpha\rangle + c_\beta|\beta\rangle\right) = c_\alpha\hat X|\alpha\rangle + c_\beta\hat X|\beta\rangle $$

# Operators acting on bras

For a bra, the operator acts from the right:

$$ \left(\langle\alpha|\right)\hat X = \langle\alpha|\hat X $$

↳ another bra.

Note for an operator $\hat X$, the ket $\hat X|\alpha\rangle$ and bra $\langle\alpha|\hat X$ are NOT duals. In fact:

$$ \hat X|\alpha\rangle \overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha|\hat X^\dagger $$

where

$\hat X^\dagger$ is the adjoint of $\hat X$

An operator $\hat X$ is Hermitian if $\hat X = \hat X^\dagger$

$$ \implies \hat X|\alpha\rangle \overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha|\hat X \quad \text{iff } \hat X \text{ is Hermitian} $$

# Multiplication of operators

$$ \hat X\hat Y \neq \hat Y\hat X $$

(order of application matters). However they are associative!

$$ \hat X(\hat Y\hat Z) = (\hat X\hat Y)\hat Z = \hat X\hat Y\hat Z $$

and

$$ \begin{aligned} \hat X\left(\hat Y\left(|\alpha\rangle\right)\right) &= \hat X\left(\hat Y|\alpha\rangle\right) = \hat X\hat Y|\alpha\rangle \\ \left(\left(\langle\alpha|\right)\hat X\right)\hat Y &= \left(\langle\alpha|\hat X\right)\hat Y = \langle\alpha|\hat X\hat Y \end{aligned} $$

note:

$$ (\hat X\hat Y)^\dagger = \hat Y^\dagger\hat X^\dagger $$

as,

$$ \hat X\hat Y|\alpha\rangle \overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha|\hat Y^\dagger\hat X^\dagger $$

# Commutator and anti-commutator

Since the order matters, it’s useful to have a way of measuring how much it matters. The commutator:

$$ [\hat X, \hat Y] \equiv \hat X\hat Y - \hat Y\hat X $$

and the anti-commutator:

$$ \lbrace\hat X, \hat Y\rbrace \equiv \hat X\hat Y + \hat Y\hat X $$

If $[\hat X, \hat Y] = 0$, the operators commute and the order doesn’t matter. Observables like that are called compatible.
$[\hat X, \hat Y] = -[\hat Y, \hat X]$ while $\lbrace\hat X, \hat Y\rbrace = \lbrace\hat Y, \hat X\rbrace$
Any product splits into the two: $\hat X\hat Y = \frac{1}{2}[\hat X, \hat Y] + \frac{1}{2}\lbrace\hat X, \hat Y\rbrace$

# Outer product

Can we multiply a bra and a ket? ↳ one way is the inner product $\langle\alpha|\beta\rangle$. The other way around:

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

This is called the outer product.

An outer product of a bra and ket is an operator $\to$ fundamentally diff. from $\langle\alpha|\beta\rangle \in \mathbb{C}$.

Now, we said that $|\beta\rangle\langle\alpha|$ is an operator. If this acts on a ket:

(through associativity) and since $\langle\alpha|\gamma\rangle \in \mathbb{C}$, it’s another ket.

That is why $|\beta\rangle\langle\alpha|$ is an operator. Also notice

how the state changes to $|\beta\rangle$ with a “new” constant $\langle\alpha|\gamma\rangle$ from $|\gamma\rangle$.

# Adjoint of an outer product

$$ \hat X|\alpha\rangle \overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha|\hat X^\dagger $$

$$ \begin{aligned} \implies \left(|\beta\rangle\langle\alpha|\right)|\gamma\rangle &\overset{\text{D.C.}}{\longleftrightarrow} \langle\gamma|\left(|\beta\rangle\langle\alpha|\right)^\dagger \\ &= \langle\gamma|\left(|\alpha\rangle\langle\beta|\right) \end{aligned} $$

Why?

$$ \begin{aligned} |\beta\rangle\langle\alpha|\gamma\rangle &\overset{\text{D.C.}}{\longleftrightarrow} \langle\alpha|\gamma\rangle^{\ast}\langle\beta| \\ &= \langle\gamma|\left(|\alpha\rangle\langle\beta|\right) \end{aligned} $$

as $\langle\alpha|\gamma\rangle = \langle\gamma|\alpha\rangle^{\ast}$.

Hence proved that if $\hat X = |\beta\rangle\langle\alpha|$ then $\hat X^\dagger = |\alpha\rangle\langle\beta|$.

# Illegal products

Illegal products

$|\alpha\rangle\cdot|\beta\rangle$ $\to$ illegal iff $|\alpha\rangle$ and $|\beta\rangle$ belong to the same vector space.
$|\alpha\rangle\otimes|\beta\rangle$ or $|\alpha\rangle|\beta\rangle$ $\to$ legal iff $|\alpha\rangle$ and $|\beta\rangle$ are kets from diff. vector spaces.

Eg. $|\alpha\rangle$ is from the vector space for $e^-$ spin and $|\beta\rangle$ is from the $e^-$ orbital angular momentum. Or for two diff. particles and so on.

# Sandwiching an operator

Also note:

$$ \underbrace{\left(\langle\beta|\right)}_{\text{bra}}\cdot\underbrace{\left(\hat X|\alpha\rangle\right)}_{\text{ket}} = \underbrace{\left(\langle\beta|\hat X\right)}_{\text{bra}}\underbrace{\left(|\alpha\rangle\right)}_{\text{ket}} = \langle\beta|\hat X|\alpha\rangle $$

since $\hat X$ can be first operated on the bra or the ket, and we’ll end up with a complex no.

Note:

$$ \langle\beta|\cdot\left(\hat X|\alpha\rangle\right) = \left[\left(\langle\alpha|\hat X^\dagger\right)\cdot|\beta\rangle\right]^{\ast} $$

$$ \boxed{\langle\beta|\hat X|\alpha\rangle = \langle\alpha|\hat X^\dagger|\beta\rangle^{\ast}} $$

If $\hat X$ is Hermitian, $\hat X = \hat X^\dagger$

$$ \implies \langle\beta|\hat X|\alpha\rangle = \langle\alpha|\hat X|\beta\rangle^{\ast} $$

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