Question

Show that ifA and Bare matrices which don't commute, then ${\mathbf{e}}^{\left(A+B\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{A}}{\mathbf{e}}^{\mathbf{B}}$ , but if they do commute then the relation holds. Hint: Write out several terms of the infinite series for ${\mathbf{e}}^{\mathbf{A}}{\mathbf{e}}^{\mathbf{B}}$ , and ${\mathbf{e}}^{(A+B)}$and, do the multiplications carefully assuming that anddon't commute. Then see what happens if they do commute

Short answer

The statement${\mathrm{e}}^{(\mathrm{A}+\mathrm{B})}={\mathrm{e}}^{\mathrm{A}}{\mathrm{e}}^{\mathrm{B}}$ holds only if the matrices A and B commute, otherwise it does not hold.

Step-by-step solution

Step 1: The Taylor expansion of exponential function:

A Taylor series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like $\mathbf{x}\mathbf{,}{\mathbf{x}}^{\mathbf{2}}\mathbf{,}{\mathbf{x}}^{\mathbf{3}}$etc.

The Taylor expansion for the first couple of terms of the exponential functions explicitly. The left-hand side is

$\mathrm{I}+(\mathrm{A}+\mathrm{B})+\frac{1}{2!}{\left(\mathrm{A}+\mathrm{B}\right)}^2+\frac{1}{3!}{(\mathrm{A}+\mathrm{B})}^3...$

And the right-hand side is equal to

$\left(1+\mathrm{A}+\frac{1}{2!}{\mathrm{A}}^2+\frac{1}{3!}{\mathrm{A}}^3+...\right)\left(1+\mathrm{B}+\frac{1}{2!}{\mathrm{B}}^2+\frac{1}{3!}{\mathrm{B}}^3+...\right)$

$=1+\mathrm{A}+\mathrm{B}+\frac{1}{2!}{\mathrm{A}}^2+\frac{1}{2!}{\mathrm{B}}^2+\mathrm{AB}+\frac{1}{2!}{\mathrm{A}}^2\mathrm{B}+\frac{1}{2!}{\mathrm{AB}}^2+\frac{1}{3!}{\mathrm{A}}^3+\frac{1}{3!}{\mathrm{B}}^3...$

Given Parameters:

Two matrices A and B are given.

It needs to be verified that the given matrices don’t commute if ${\mathrm{e}}^{\left(\mathrm{A}+\mathrm{B}\right)}={\mathrm{e}}^{\mathrm{A}}{\mathrm{e}}^{\mathrm{B}}$ and commute if this relation holds.

Expand the Taylor expansion:

The square of the sum of anticommuting matrices is equal to ${\mathrm{A}}^2+\mathrm{AB}+\mathrm{BA}+{\mathrm{B}}^2$and if commute, then it is ${\mathrm{A}}^2+2\mathrm{AB}+{\mathrm{B}}^2$.

Find out the cubic terms.

$$\begin{gathered} {\left(\mathrm{A}+\mathrm{B}\right)}^3=\left({\mathrm{A}}^2+\mathrm{AB}+\mathrm{BA}+{\mathrm{B}}^2\right)\left(\mathrm{A}+\mathrm{B}\right) \\ ={\mathrm{A}}^3+{\mathrm{A}}^2\mathrm{B}+\mathrm{ABA}+{\mathrm{AB}}^2+\mathrm{BAA}+\mathrm{BAB}+{\mathrm{B}}^2\mathrm{A}+{\mathrm{B}}^3 \\ ={\mathrm{A}}^3+3{\mathrm{A}}^2\mathrm{B}+3{\mathrm{AB}}^2+{\mathrm{B}}^3 \end{gathered}$$

Hence, the given matrices anticommute if the relationship does not hold and vice-versa.