Why is the imaginary unit i in the momentum operator definition?

[dpa]

why is complex number i involved in defining momentum operator p
I mean Px=-ih...
What has complex number to do with momentum.

I do get however that i in other cases of quantum mechanics has to do with euler's formula that comes from harmonic nature of wave.

 

[bhobba]

why is complex number i involved in defining momentum operator p
I mean Px=-ih...
What has complex number to do with momentum.

I do get however that i in other cases of quantum mechanics has to do with euler's formula that comes from harmonic nature of wave.

Its not a matter of definition - you can actually derive it from Galilaian invariance - you can find the derivation in Ballentine - Quantum Mechanics - A Modern Development. But aside from that the derivative part is a pure imaginary operator and real valued operators should - well be real - multiplying it by i makes it real.

Thanks
Bill

 

[Ken G]

And if you think about it from the perspective of kinetic energy, you can say the same thing like this. Confining a particle in a small region requires giving the particle kinetic energy, and it also requires getting the particle's wave function to have a negative second derivative (so you can get the wavefunction to be large in the confined region and small outside of it). Since KE ~ P², the momentum must inherit a meaning related to the square root of the kinetic energy. The square root of a second derivative is like d/dx, but a square root of a minus second derivative is like i d/dx.