Thread: commutators

evaluate the commutator [-d^2 / dx^2 , x] f (x)

can anyone please check my work thanks.

[-d^2 / dx^2 , x] f (x) = -2f '(x) -xf "(x) + -xf "(x)

[-d^2 / dx^2 , x] f (x) = -2f '(x) = -2 (d/dx (f(x))

[-d^2 / dx^2 , x] = -2 d/dx

Originally Posted by myoplex11

evaluate the commutator [-d^2 / dx^2 , x] f (x)

can anyone please check my work thanks.

[-d^2 / dx^2 , x] f (x) = -2f '(x) -xf "(x) + -xf "(x)

[-d^2 / dx^2 , x] f (x) = -2f '(x) = -2 (d/dx (f(x))

[-d^2 / dx^2 , x] = -2 d/dx

There's a typo in the first line (the last term should be +xf''(x).) Otherwise it looks good.

-Dan