A certain partial diff. equ.

[fuhrerkeebs]

When Schrodinger derived his famous equation, he derived it with the electron in mind, as is obvious with his use of E=p²/2m+V. This is fine if you want to work with electrons, but I wanted to see what the equation would be like if you derived it for the photon, using E=pc. I worked through it and got the equation:

 

[math]\frac{d\psi}{dt} = -c \frac{d\psi}{dx}[/math]

 

Does anyone know how to find a square-integrable solution to this equation?

[MandrakeRoot]

For every [math]\sigma \in \mathbb{C}[/math], [math]\phi(t,x) = e^{\sigma(t - \frac{1}{c}x)}[/math] is a solution to your equation.

 

Mandrake

[Aeschylus]

The derivitaion does not come from considering electrons alone, but 'matter waves' (i.e. massive particles obeying the De Broglie relations as all masive particle should do) in general.

 

What I'd guess you'd get is this (if you look at the derivation you find for light waves you can stop before you have to actually postualte anything new):

 

[math]\frac{\partial^2 \Psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2\Psi}{\partial t^2}[/math]

 

(note: this is a partial differential equation as psi is a function of both x and t)

 

which is just the equation for a classical plane wave (and obeys the relationship, from De Broglie, E = pc) which has the solution:

 

[math]\Psi(x,t) = \Psi_0 e^{i(kx - \omega t)[/math]

 

 

and it's only the real part of Psi that we're interested in to give us th dispalcement (classically).

 

edited to add: I've just looked up the equation and does indeed take the form of the classical wave equation above.

[fuhrerkeebs]

Thanks guys! I found Mandrakes solution but I didn't think it was right because it was not square-integrable...until I remember that a free particle has equal probability of being anywhere. Aeschylus, I remembered the classical equation, but I was just having fun try to see what the answer would be...

[MandrakeRoot]

Well it depends a little what domain of the function you consider. You didnt specify so well i just gave you the first solution that came to mind. If you restrict t and x to some cube this is a perfectly square integrable solution.

 

Mandrake

[Severian]

What I'd guess you'd get is this (if you look at the derivation you find for light waves you can stop before you have to actually postualte anything new):

 

[math]\frac{\partial^2 \Psi}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2\Psi}{\partial t^2}[/math]

 

This is the massless Klein-Gordon Equation. If you put a mass in you would get the Klein-Gordon Eq:

 

[math]\frac{\partial^2 \Psi}{\partial t^2} = c^2 \frac{\partial^2\Psi}{\partial x^2} - c^4 m^2 \Psi[/math]

 

which is directly analogous to

 

[math]E^2=p^2c^2+m^2c^4[/math]

 

(of which E=pc is a special case).