- #1
tunafish
- 11
- 0
Hi everybody!
Two question for you:
1) Take a free particle, moving in the x direction.
Its (time indipendent) wave function, in terms of the momentum is [itex]\psi(x)=\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}[/itex].
Now, i know the momentum of the particle: p.
So i should not know anything about its position, according to the uncertainty principle.
Effectively i don't get any information on "where" the particle is looking at the wave function; but the fact that the [itex]\psi[/itex] depends on the position does mean something particular?2) Take an eigenstate of the hamiltonan operator [itex]|E\rangle[/itex].
Now, they say that this wave function, in terms of the momentum is:[tex]\psi_p=\langle p|E\rangle[/tex] or, expressed with position eigenstates:[tex]\psi_x=\langle x|E\rangle[/tex]But those are just numbers, so it should't be like this instead:[tex]\psi_p=\int |p\rangle\langle p|E\rangle\;dp[/tex]and [tex]\psi_x=\int |x\rangle\langle x|E\rangle\;dx[/tex]??
Thanks for your help!
Two question for you:
1) Take a free particle, moving in the x direction.
Its (time indipendent) wave function, in terms of the momentum is [itex]\psi(x)=\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}[/itex].
Now, i know the momentum of the particle: p.
So i should not know anything about its position, according to the uncertainty principle.
Effectively i don't get any information on "where" the particle is looking at the wave function; but the fact that the [itex]\psi[/itex] depends on the position does mean something particular?2) Take an eigenstate of the hamiltonan operator [itex]|E\rangle[/itex].
Now, they say that this wave function, in terms of the momentum is:[tex]\psi_p=\langle p|E\rangle[/tex] or, expressed with position eigenstates:[tex]\psi_x=\langle x|E\rangle[/tex]But those are just numbers, so it should't be like this instead:[tex]\psi_p=\int |p\rangle\langle p|E\rangle\;dp[/tex]and [tex]\psi_x=\int |x\rangle\langle x|E\rangle\;dx[/tex]??
Thanks for your help!