Why you are using h bar in heisenberg uncertainity principle?

Answer 1

Sometimes frequency is not in terms of Hertz (Hz) ,CYCLES PER SECOND, but instead it is expressed as RADIANS PER SECOND, which is angular frequency. Therefore a conversion factor must be used, which is 'h-bar' Recall the following ---------------------------- h-bar = h/(2*pi) where h is Planck's constant angular frequency, ω = 2*pi*ν where ν is frequency in Hertz. ---------------------------- So lets take Planck's relation: Energy (E) = Planck's Constant (h) * frequency( ν ) E = h * ν 1) If the frequency ( ν ) is in Hz, then just looking at the units, Planck's relation becomes E = h * ν = ( J-s ) * (1/s) = J ---> Expected unit for energy: Joule 2) If the frequency ( ν ) is in Radians per second, h must have a conversion factor to accommodate angular frequency. Again, if we look at Planck's relation using angular frequency, ω = 2*pi*ν E = h * ω = ( J-s) * [ (2*pi)/s ] = J * 2*pi ---> Not the expected unit for energy So we must use a reduced Planck constant, h-bar = h/2*pi to obtain Joules E = h * ω = [( J-s)/(2*pi)] * [ (2*pi)/s ) = J ---> Expected unit for energy: Joule