Square of a momentum operator

  • Thread starter hnicholls
  • Start date
  • #1
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Main Question or Discussion Point

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!
 

Answers and Replies

  • #2
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i * i = -1 and so -i * -i = -1 * i * -1 * i = 1 * i * i = i * i = -1
 
  • #3
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)
 
  • #4
49
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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?
 

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