- #1
etotheipi
The magnitude of the momentum ##p## satisfies ##p^2 = p_x^2 + p_y^2 + p_z^2## and this implies the operator equation ##\hat{p}^2 = \hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2##, so we can say that ##\hat{p}^2 = -\hbar^2 (\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2})##.
I had two questions. Firstly, is ##\hat{p}^2## equivalent to ##\hat{p^2}##, or does the latter not make any sense?
And secondly, we can say ##\hat{p}^2 \psi = p^2 \psi##, but does it make any sense to talk about the 'unsquared' version of the operator for the magnitude of the momentum, ##\hat{p}##? With the squared ##\hat{p}^2## the ##\psi## distributes into the bracket, but I'm not sure if it is valid to write ##\hat{p} \psi = p \psi##.
Thank you, and sorry if these are silly questions !
I had two questions. Firstly, is ##\hat{p}^2## equivalent to ##\hat{p^2}##, or does the latter not make any sense?
And secondly, we can say ##\hat{p}^2 \psi = p^2 \psi##, but does it make any sense to talk about the 'unsquared' version of the operator for the magnitude of the momentum, ##\hat{p}##? With the squared ##\hat{p}^2## the ##\psi## distributes into the bracket, but I'm not sure if it is valid to write ##\hat{p} \psi = p \psi##.
Thank you, and sorry if these are silly questions !