#### ChrisVer

Gold Member

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I will repeat myself...

Once you measure the position with infinite precision , you have absolutely no clue what the momentum of the measured photon is...

If you tell me that the photon starts with some momentum [itex]p[/itex] and I measure it at point [itex]x_{0}[/itex] then you'll probably never know its momentum again- not until you measure it. It can still be [itex]p[/itex] of course, but it can as well be [itex]p+Δp[/itex]... In fact the momentum is totally undetermined, it spans the whole p-space.

Why does this happen?

Because the position and momentum quantum operators don't commute. And also because of that, if you measure the [itex]x[/itex] component of the photon's position, you will lose information about the [itex]p_{x}[/itex] component of momentum, not necessarily [itex]p_{y,z}[/itex] since they indeed commute with the operator you measure. Of course that's nonsence, because what you measure is a position vector, so you lose momentum's vector.

As I said this problem corresponds to finding the delta function's Fourier Transform.

Also before the photon collides on your measuring machine, you have absolutely no idea what its position is... In fact the [itex]Δx=L[/itex] where L is the distance between your source and your measuring machine (where the photon scatters or is absorbed to be measured). That's the natural way of choosing the x uncertainty in that thought experiment. What about p? If p is fully determined (that means you can somehow measure it before you measure the photon's position), then the photon's propagation is space is going to be like a wave - you won't know anything about the x.

Once you measure the position with infinite precision , you have absolutely no clue what the momentum of the measured photon is...

If you tell me that the photon starts with some momentum [itex]p[/itex] and I measure it at point [itex]x_{0}[/itex] then you'll probably never know its momentum again- not until you measure it. It can still be [itex]p[/itex] of course, but it can as well be [itex]p+Δp[/itex]... In fact the momentum is totally undetermined, it spans the whole p-space.

Why does this happen?

Because the position and momentum quantum operators don't commute. And also because of that, if you measure the [itex]x[/itex] component of the photon's position, you will lose information about the [itex]p_{x}[/itex] component of momentum, not necessarily [itex]p_{y,z}[/itex] since they indeed commute with the operator you measure. Of course that's nonsence, because what you measure is a position vector, so you lose momentum's vector.

As I said this problem corresponds to finding the delta function's Fourier Transform.

Also before the photon collides on your measuring machine, you have absolutely no idea what its position is... In fact the [itex]Δx=L[/itex] where L is the distance between your source and your measuring machine (where the photon scatters or is absorbed to be measured). That's the natural way of choosing the x uncertainty in that thought experiment. What about p? If p is fully determined (that means you can somehow measure it before you measure the photon's position), then the photon's propagation is space is going to be like a wave - you won't know anything about the x.

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