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mysearch
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Hi,
In my ignorance of quantum mechanics, which I am trying to address, I had initially assumed that any resolution of the wave-particle duality of photon and electrons led to some sort of common probability equation. Therefore, I am trying to better understand how different dispersion relationships seem to lead to lead to different wave equations and how to interpret the implications. Up to now I had only understood dispersion as a wave attribute within optics, e.g. spectrum spread through a prism due to different propagation velocities, within a given media, associated with each frequency.
[1] [tex] A = A_0sin(kx-wt) [/tex]
Citing equation [1] as a classical starting point for a wave function, which leads to the following equation based on the 2nd derivative and the relationship [tex][c=w/k = f \lambda ] [/tex], where [c] is the speed of light.
[2] [tex] \frac { \delta^2 \Psi}{ \delta t^2} = c^2 \frac { \delta^2 \Psi}{ \delta x^2} [/tex]
So my first question is whether equation [2] is still a valid baseline for light, when being described as a wave and a photon particle within quantum mechanics?
The reason for asking this question is based on the apparent fact that a photon [c=w/k] and a particle have different dispersion relationships. In a pseudo-derivation of Schroedinger’s wave equation, which I won’t detail here, you are led to the dispersive relationship for a free particle of mass [m] that originates with deBroglie’s wavelength equation:
[3] [tex] w = \frac {\hbar k^2}{2m}[/tex]
This is also described as a dispersion relationship, but one which then seems to prevent the previous method, as used for light, of equating the 2nd derivatives with respect to [x] and [t]. However, the equivalent solution to equation [2] has the following form and was described as Schroedinger’s wave equation for a free particle, ignoring the issue of potential energy [V] for the moment.
[4] [tex] -\frac { \hbar ^2}{2m} \frac { \delta^2 \Psi}{ \delta x^2} = i \hbar \frac { \delta \Psi}{ \delta t} = E \Psi [/tex]
As such, I am assuming that equation [4] only applies to particles with rest mass [m] and therefore cannot be applied to a photon. So:
Is there another Schroedinger’s wave equation for photons?
How do I interpret the meaning of equation [3]?
Here is my current assumption, not a statement of fact:
In the case of a photon, the dispersion relationship [c=w/k] corresponds to the propagation velocity equalling the phase velocity [vp], while by re-arranging equation [3], the propagation velocity seems to correspond to the group velocity [vg]:
[5] [tex] v_g = \frac {dw}{dk} = \frac {\hbar k}{2m} [/tex]
On the basis of deBroglie’s wavelength [tex] \lambda = h/mv [/tex] being substituted back into [5]
[6] [tex] v_g = \frac {\hbar k}{2m} = \frac {h}{2m \lambda} = \frac {mv}{2m} [/tex]
I am slightly confused by the 2 in the denominator of [6], which comes from [3] and suggests that [vg =v/2] after the mass [m] is cancelled:
Does the 2 have anything to do with kinetic energy E=mv/2?
Have I just made a mistake?
Is mass [m] subject to relativistic effects, e.g
[tex]E^2=p^2c^2+m_0^2c^4[/tex] ?
Would appreciate correction of any wrong assumptions on my part. Thanks
In my ignorance of quantum mechanics, which I am trying to address, I had initially assumed that any resolution of the wave-particle duality of photon and electrons led to some sort of common probability equation. Therefore, I am trying to better understand how different dispersion relationships seem to lead to lead to different wave equations and how to interpret the implications. Up to now I had only understood dispersion as a wave attribute within optics, e.g. spectrum spread through a prism due to different propagation velocities, within a given media, associated with each frequency.
[1] [tex] A = A_0sin(kx-wt) [/tex]
Citing equation [1] as a classical starting point for a wave function, which leads to the following equation based on the 2nd derivative and the relationship [tex][c=w/k = f \lambda ] [/tex], where [c] is the speed of light.
[2] [tex] \frac { \delta^2 \Psi}{ \delta t^2} = c^2 \frac { \delta^2 \Psi}{ \delta x^2} [/tex]
So my first question is whether equation [2] is still a valid baseline for light, when being described as a wave and a photon particle within quantum mechanics?
The reason for asking this question is based on the apparent fact that a photon [c=w/k] and a particle have different dispersion relationships. In a pseudo-derivation of Schroedinger’s wave equation, which I won’t detail here, you are led to the dispersive relationship for a free particle of mass [m] that originates with deBroglie’s wavelength equation:
[3] [tex] w = \frac {\hbar k^2}{2m}[/tex]
This is also described as a dispersion relationship, but one which then seems to prevent the previous method, as used for light, of equating the 2nd derivatives with respect to [x] and [t]. However, the equivalent solution to equation [2] has the following form and was described as Schroedinger’s wave equation for a free particle, ignoring the issue of potential energy [V] for the moment.
[4] [tex] -\frac { \hbar ^2}{2m} \frac { \delta^2 \Psi}{ \delta x^2} = i \hbar \frac { \delta \Psi}{ \delta t} = E \Psi [/tex]
As such, I am assuming that equation [4] only applies to particles with rest mass [m] and therefore cannot be applied to a photon. So:
Is there another Schroedinger’s wave equation for photons?
How do I interpret the meaning of equation [3]?
Here is my current assumption, not a statement of fact:
In the case of a photon, the dispersion relationship [c=w/k] corresponds to the propagation velocity equalling the phase velocity [vp], while by re-arranging equation [3], the propagation velocity seems to correspond to the group velocity [vg]:
[5] [tex] v_g = \frac {dw}{dk} = \frac {\hbar k}{2m} [/tex]
On the basis of deBroglie’s wavelength [tex] \lambda = h/mv [/tex] being substituted back into [5]
[6] [tex] v_g = \frac {\hbar k}{2m} = \frac {h}{2m \lambda} = \frac {mv}{2m} [/tex]
I am slightly confused by the 2 in the denominator of [6], which comes from [3] and suggests that [vg =v/2] after the mass [m] is cancelled:
Does the 2 have anything to do with kinetic energy E=mv/2?
Have I just made a mistake?
Is mass [m] subject to relativistic effects, e.g
[tex]E^2=p^2c^2+m_0^2c^4[/tex] ?
Would appreciate correction of any wrong assumptions on my part. Thanks