darkenlighten Posted April 13, 2010 Posted April 13, 2010 Whether I have taken a Quantum Mechanics class is not the point. The physical description of the wave is the wave function. First, it does matter because I'm afraid you are not understanding the difference between the wave function and its probability, they describe two different things. And I know the wave function can be the physical description of the wave. The values of the wave function are probability amplitudes, and with the wave function you get the probability distribution. How is a wave function not a probability wave? Just to be clear then, what is your meaning of a probability wave, to make sure we aren't arguing semantics? Now you seem to be changing the point on semantics, you originally said this though and that was what was dead wrong, The reason it does not have a definite location is because it has a probable location in all possible paths from A to B. And yea I did say that, because just because something can be represented by all possible states, doesn't mean it is in all possible states.
theoriginal169 Posted June 4, 2010 Posted June 4, 2010 photon have no momentum so it cant be wave function. also photon doesnt have mass and you cant use observables.
swansont Posted June 5, 2010 Posted June 5, 2010 photon have no momentum so it cant be wave function. also photon doesnt have mass and you cant use observables. Photons have momentum. p = E/c
theoriginal169 Posted June 5, 2010 Posted June 5, 2010 Photons have momentum. p = E/c in principle it seems that they have momentum but it is said that photons have no mass so it doesnt seem logical that they have momentum. Merged post follows: Consecutive posts mergedin principle it seems that they have momentum but it is said that photons have no mass so it doesnt seem logical that they have momentum. e=mc^2 the m is not the photons mass (m-m[math]_{0}[/math])c^2=e ???
swansont Posted June 5, 2010 Posted June 5, 2010 in principle it seems that they have momentum but it is said that photons have no mass so it doesnt seem logical that they have momentum. p=mv is not a relativistic equation E^2 = p^2c^2 + m^2c^4 E=pc for massless particles
theoriginal169 Posted June 5, 2010 Posted June 5, 2010 so what does m represents according to my knowledge, at equation e=mc^2 m represent the change of mass of high speed particles is it true? p=mv is not a relativistic equation E^2 = p^2c^2 + m^2c^4 E=pc for massless particles
swansont Posted June 5, 2010 Posted June 5, 2010 so what does m represents according to my knowledge, at equation e=mc^2 m represent the change of mass of high speed particles is it true? No, it's not. It's a misuse of the equation I gave, since E=mc^2 only holds for particles at rest. To apply it to particles in motion is to redefine what is meant by mass.
theoriginal169 Posted June 5, 2010 Posted June 5, 2010 (edited) No, it's not. It's a misuse of the equation I gave, since E=mc^2 only holds for particles at rest. To apply it to particles in motion is to redefine what is meant by mass. ok i got it is relativistic QM i am not good at it. so can E be plugged in to schodringer equation? Merged post follows: Consecutive posts mergedand 1 more question does uncertainty principle holds for photon? if it does,how does [math]\Delta[/math][math]_{x}[/math] be calculated. Edited June 5, 2010 by theoriginal169 Consecutive posts merged.
swansont Posted June 5, 2010 Posted June 5, 2010 (edited) You use the Klein-Gordon equation http://en.wikipedia.org/wiki/Klein–Gordon_equation The uncertainty in position is associated with the wavelength. Edited June 5, 2010 by swansont
theoriginal169 Posted June 6, 2010 Posted June 6, 2010 You use the Klein-Gordon equation http://en.wikipedia.org/wiki/Klein–Gordon_equation The uncertainty in position is associated with the wavelength. ty very much
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