- #1
amjad-sh
- 246
- 13
Hello all;
Sorry if my questions are stupid or weird, but I'm still a beginner in quantum mechanics. So please be patient with me!
I'm reading shankar text of quantum mechanics,and I reached the part related to momentum operator and the momentum wave function in position basis.
He derived that [itex]\langle x | p \rangle=\frac{1}{√2πħ} e^{ipx}[/itex] which means that it is a plane wave.
so [itex]\langle x | p \rangle[/itex] can't be normalized and we can't associate with them a sensible probability distribution as the probability density will be uniform in all space.
But prof shankar then said:"But since the plane waves are eigenfunctions of P does it means that well-defined states of momentum do not exist ? yes,in strict sence".
my first question is:1)does he mean by saying"well-defined states of momentum do not exist" that we can't measure the momentum of a particle in a specific position?I know that we can't but what he means by this?
2)does this mean that |p> does not belong to Hilbert space?
Now prof shankar continued:"however there do exist states that are normalizable to unity(i.e correspond to proper vectors)and come arbitrarily close to having a precise momentum".
3) does this mean that for proper states we can detect momentum which is very close to precise momentum?
while for improper states it is impossible to detect momentum?
then prof shankar said:"thus a particle coming out from an accelerator with some advertised momentum ,say 500 Gev/c,is in a proper normalizable state(since it is located in our laboratory)and not in a plane wave state correponding to [itex]| p=500Gev/c \rangle[/itex].
4) I didn't get this at all, if someone can simplify it to me.
Thanks in advance!
Sorry if my questions are stupid or weird, but I'm still a beginner in quantum mechanics. So please be patient with me!
I'm reading shankar text of quantum mechanics,and I reached the part related to momentum operator and the momentum wave function in position basis.
He derived that [itex]\langle x | p \rangle=\frac{1}{√2πħ} e^{ipx}[/itex] which means that it is a plane wave.
so [itex]\langle x | p \rangle[/itex] can't be normalized and we can't associate with them a sensible probability distribution as the probability density will be uniform in all space.
But prof shankar then said:"But since the plane waves are eigenfunctions of P does it means that well-defined states of momentum do not exist ? yes,in strict sence".
my first question is:1)does he mean by saying"well-defined states of momentum do not exist" that we can't measure the momentum of a particle in a specific position?I know that we can't but what he means by this?
2)does this mean that |p> does not belong to Hilbert space?
Now prof shankar continued:"however there do exist states that are normalizable to unity(i.e correspond to proper vectors)and come arbitrarily close to having a precise momentum".
3) does this mean that for proper states we can detect momentum which is very close to precise momentum?
while for improper states it is impossible to detect momentum?
then prof shankar said:"thus a particle coming out from an accelerator with some advertised momentum ,say 500 Gev/c,is in a proper normalizable state(since it is located in our laboratory)and not in a plane wave state correponding to [itex]| p=500Gev/c \rangle[/itex].
4) I didn't get this at all, if someone can simplify it to me.
Thanks in advance!