Operator Algebra

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In Quantum Mechanics the commutator $[,]$ plays a fundamental role as we pointed out in the articles on Heisenberg's Microscope and the Uncertainty Principle. The Schrödinger Equation even has an analog, the Heiseberg equation of motion, which includes the commutator and can be solved purely algebraically. The same thing is true for the Harmonic Oscillator, that is solved much easier algebraically.

The operator or matrix algebra is fundamentally distinct from ordinary algebra in the sence that the operators $A,B$ and $C$, thought they obey the associative law for addition \begin{align} (A+B)+C=A+(B+C) \end{align} as well as the commutative law for addition \begin{align} A+B=B+A \end{align}

they in general DO NOT COMMUTE with each other for multiplication \begin{align} A \cdot B \not =B \cdot A \end{align} To shorten the notation we bring $B \cdot A$ on the LHS and introduce

the Commutator \begin{align} A \cdot B -B \cdot A := [A,B]\not =0 \end{align}

The associative law for multiplication however holds \begin{align} (A \cdot B) \cdot C=A \cdot (B \cdot C) \end{align}


Calculate the commutator of $\frac{\partial}{\partial x}$ and $x$ \begin{align} \frac{\partial}{\partial x}(x \psi(x))=\psi(x) + x \frac{\partial \psi(x)}{\partial x} \end{align} and \begin{align} x \frac{\partial}{\partial x} \psi(x)= x\frac{\partial \psi(x)}{\partial x} \end{align} so the commutator is \begin{align} \left[ \frac{\partial}{\partial x},x \right]\psi(x)= \psi(x) + x \frac{\partial \psi(x)}{\partial x}-x\frac{\partial \psi(x)}{\partial x}=\psi(x) \end{align} \begin{align} \left[ \frac{\partial}{\partial x},x \right]=1 \end{align}

This differnece $[A,B]\not =0$ turns out to bring with it a whole new mathematical structure termed operator algebra (or Heisenberg algebra) whose rules we will study now.

\begin{align} \left[ A, B \right]&= -\left[ B,A \right] \\ \tag{1} \end{align} \begin{align} \left[ A, B+C \right]&=\left[ A, B\right] +\left[ A, C\right]\\ \tag{2} \end{align} \begin{align} \left[ A+ B,C \right]&=\left[ A, C\right] +\left[ B, C\right]\\ \tag{3} \end{align} \begin{align} \left[ A, B \, C \right]&= ABC - BCA + BAC - BAC \\ \tag{4} \end{align} \begin{align} &= (AB-BA)C + B(AC-CA) \\ \tag{5} \end{align} \begin{align} &=\left[ A, B\right]C + B \left[ A, C\right]\\ \tag{6} \end{align} \begin{align} \left[ A B , C \right]&= A \left[ B,C\right] + \left[ A, C\right] B\\ \tag{7} \end{align}

Moreover, the commutator of two operators $[A,B]$ commutes with the operators \begin{align} \left[ \left[A, B\right],A \right]&=0\\ \left[ \left[A, B\right],B \right]&=0 \end{align} and the commutator of an operator raised to some power $n$ is \begin{align} \left[A^n, B\right]&= n A^{n-1} \left[A, B\right] \end{align} by (1) this is the same as \begin{align} \left[A^n, B\right]&=-\left[B, A^n\right] = - n A^{n-1} \left[B, A \right] \end{align} for a proof see K. F. Riley - Mathematical Methods for Physics and Engineering.

The exponential of an operator can be expanded as a Taylor Series \begin{align} e^A= \sum_{n=0}^{\infty} \frac{A^n}{n!} \end{align} and the product of two operator exponentials is defined as \begin{align} e^Ae^B= e^{A+B+[A,B]/2} \end{align}