Commutator - Aetherwisp

The commutator is a value indicating how two elements related by a binary operation fail to satisfy the commutative property. In the case of multiplication, given two elements $A$ and $B$ , their commutator is defined as

[A,B] = AB - BA,

which equals zero if and only if the commutative property holds between $A$ and $B$ (and thus $AB = BA$ ).

The anticommutator is the same operation with the opposite sign:

\{A,B\} = AB + BA

Not to be confused with the identically denoted and often compared to Poisson brackets.

Properties#

The commutator has several properties:

$[A + B, C] = [A, C] + [B, C]$
$[AB, C] = ABC - CAB$
$[A, A] = 0$ , meaning an element always commutes with itself.
$[A, B] = -[B, A]$ , known as anticommutativity.
$[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0$ , known as the Jacobi identity.
$[A, B]^{+} = -[A, B]$ , where $^{+}$ denotes the adjoint.
$[T,c]=0$ where $T$ is an operator and $c$ is a Scalar constant.
$[T,f(T)]$ , meaning an operator always commutes with functions of itself.
If two operators $A$ and $B$ commute up to a constant $c$ , $[A, B] = c \hat{\mathbf{1}}$ , then the Baker-Campbell-Hausdorff formula holds: $e^{A + B} = e^{-[A,B]/2} e^{A} e^{B}$ . Note the additional scaling factor $e^{-[A,B]/2}$ compared to the usual case.

In the case of two operator exponentials $e^{-A}$ and $e^{A}$ acting on another operator $B$ , one can use nested commutators:

\begin{align} e^{A} B e^{-A} &= \sum_{k=0}^{\infty} \frac{1}{k!} \underbrace{[A, [A, \ldots [A, B] }_{k \text{ times}} \ldots]] \\ &= B + [A, B] + \frac{1}{2}[A, [A, B]] + \frac{1}{3!}[A, [A, [A, B]]] + \ldots \end{align}

This also holds for $e^{A} f(B) e^{-A}$ since $e^{A} f(B) e^{-A} = f(e^{A} B e^{-A})$. Just replace $B$ with $f(B)$ in the above formula. ### In quantum mechanics The commutator is of fundamental importance in quantum mechanics, as it allows expressing numerous concepts related to the [[Disuguaglianza di Heisenberg|uncertainty principle]]. The most important commutation relation is the so-called **canonical commutation relation**, which shows that the position operator and the momentum operator do not commute[^1]

[\hat{q}, \hat{p}] = i\hbar,

where $\hbar$ is the [[Planck constant|reduced Planck constant]]. > [!example] Proof > Consider the [[Spazi Lp|L^2]] [[Spazio di Hilbert|Hilbert space]] over $\mathbb{R}$. The commutator in this space is defined as $[\hat{q}, \hat{p}]: L^{2}(\mathbb{R}) = \mathcal{H} \to \mathcal{H}$. For a generic [[Funzione d'onda|wave function]] $\psi(x)$ (operators must act on *something*), we have > $$([\hat{q}, \hat{p}]\psi)(x) = (\hat{q}\hat{p}\psi)(x) - (\hat{p}\hat{q}\psi)(x) = x(\hat{p}\psi)(x) + i\hbar \partial_{x}(\hat{q}\psi)(x) = \ldots

remembering that operators always act on what is to their right. Continuing,

> thus > $$([\hat{q}, \hat{p}]\psi(x)) = i\hbar \psi(x)

which, being true for all $\psi(x) \in L^{2}(\mathbb{R})$ , allows us to conclude

In three dimensions, for the position vector $\mathbf{r}$ and the [[Linear momentum|momentum]] $\mathbf{p}$, we have

[r_{i}, p_{j}] = -[p_{i}, r_{j}] = i\hbar\delta_{ij}, \quad [r_{i}, r_{j}] = [p_{i}, p_{j}] = 0

with indices referring to the three spatial dimensions and $\delta_{ij}$ being the [[Kronecker delta]]. In other words, position always commutes with position; momentum always commutes with momentum; they commute between each other only in different coordinates, otherwise yielding $i\hbar$. For the [[Momento angolare quantistico|quantum angular momentum]] along the $z$-axis $L_{z}$, the following hold:

[L_{z}, x] = i\hbar y, \quad [L_{z}, y] = -i\hbar x, \quad [L_{z}, z] = 0

[L_{z}, p_{x}] = i\hbar p_{y}, \quad [L_{z}, p_{y}] = -i\hbar p_{x}, \quad [L_{z}, p_{z}] = 0

[^1]: Note that the commutator of two operators is itself an operator. To be precise, the equation is $[\hat{q},\hat{p}]=i\hbar \hat{\mathbf{1}}$, where $\hat{\mathbf{1}}$ is the [[Identity operator]].