What is the Correct Value of the Square of a Momentum Operator?

[hnicholls]

know I'm missing something obvious.

for a momentum operator p = -iħ d/dx

if I square the -iħ part I get (+1)ħ²

but I believe the correct value (as in the kinetic energy of the Hamiltonian) is

-ħ/2m d²/dx².

how is the value of the term -ħ/2m where the square of -i = +1?

Thanks!

 

[jedishrfu]

i * i = -1 and so -i * -i = -1 * i * -1 * i = 1 * i * i = i * i = -1

 

[charlemagne94]

Kinetic energy is given by
K = (1/2)*m*v^2 = (p^2)/(2m).
(-ih)^2 is actually equal to -h^2, because (-i)*(-i) = -1.
Hence, K = (p^2)/2m = (-h^2/2m)*(d^2/dx^2)

 

[hnicholls]

i * i = -1 and so -i * -i = -1 * i * -1 * i = 1 * i * i = i * i = -1

this suggests, i x i = -1 and -i x -i = -1 x i x -1 x i = 1 x -1 = -1

right?