- #1

jfy4

- 649

- 3

Hi,

I was looking at the relativistic equation for energy [itex]E^2=p^2+m_{0}^2[/itex] and was wondering about a methodology I took with it. Making the substitution [itex]E^2=\gamma^2 m_{0}^2[/itex] then

[tex]p^2=m_{0}^2(\gamma^2-1)[/tex]

Considering only the [itex]\gamma-1[/itex] factor, this can be expanded

[tex]\gamma-1=\frac{v^2}{2}+\frac{3v^4}{8}+\frac{5v^6}{16}+...[/tex]

I then made the identification with [itex]E=\hat{E}[/itex],

[tex]\hat{E}=i\hbar\frac{\partial}{\partial t}[/tex]

which yields the following from before

[tex]p^2=m_{0}^2\left(\left(\frac{i\hbar}{m_0}\frac{ \partial }{\partial t}\right)^2-1\right)[/tex]

which led me to again identify

[tex]\gamma-1=\frac{i\hbar}{m_0}\frac{\partial}{\partial t}-1=\frac{v^2}{2}+\frac{3v^4}{8}+\frac{5v^6}{16}+...[/tex]

If I make the appropriate substitution for [itex]v=\hat{v}[/itex]

[tex]\hat{v}=\frac{i\hbar}{m_0}\nabla[/tex]

and write the sum on the right as

[tex]

\sum_{n=1}^{\infty}\frac{\left(1+(-1)^n\right) \left(\frac{1}{2} (-1+n)\right)!}{2 \sqrt{\pi } \frac{n}{2}!}\hat{v}^n

[/tex]

Then I end up with

[tex]i\hbar\frac{\partial}{\partial t}\psi=m_0\psi+\sum_{n=1}^{\infty}\frac{\left(1+(-1)^n\right) \left(\frac{1}{2} (-1+n)\right)!}{2 \sqrt{\pi } \frac{n}{2}!}m_{0}^{1-n}\left(i\hbar\nabla\right)^n\psi

[/tex]

Now for n=1, this equals

[tex]

i\hbar\frac{\partial\psi}{\partial t}=m_0\psi

[/tex]

which seems manifestly true (EDIT: by which I mean, there are no kinetic energy terms on the right), and for n=2

[tex]

i\hbar\frac{\partial\psi}{\partial t}=m_0\psi-\frac{\hbar^2}{2m_0}\nabla^2\psi

[/tex]

which seems Schrodinger-esque. So I was wondering... for n=4 (which is the next time a term is added on)

[tex]

i\hbar\frac{\partial\psi}{\partial t}=m_0\psi-\frac{\hbar^2}{2m_0}\nabla^2\psi+\frac{3\hbar^4}{8m_{0}^3}\nabla^4\psi

[/tex]

is the last term on the right a relativistic correction to the Schrodinger-esque equation above it?

Thanks,

I was looking at the relativistic equation for energy [itex]E^2=p^2+m_{0}^2[/itex] and was wondering about a methodology I took with it. Making the substitution [itex]E^2=\gamma^2 m_{0}^2[/itex] then

[tex]p^2=m_{0}^2(\gamma^2-1)[/tex]

Considering only the [itex]\gamma-1[/itex] factor, this can be expanded

[tex]\gamma-1=\frac{v^2}{2}+\frac{3v^4}{8}+\frac{5v^6}{16}+...[/tex]

I then made the identification with [itex]E=\hat{E}[/itex],

[tex]\hat{E}=i\hbar\frac{\partial}{\partial t}[/tex]

which yields the following from before

[tex]p^2=m_{0}^2\left(\left(\frac{i\hbar}{m_0}\frac{ \partial }{\partial t}\right)^2-1\right)[/tex]

which led me to again identify

[tex]\gamma-1=\frac{i\hbar}{m_0}\frac{\partial}{\partial t}-1=\frac{v^2}{2}+\frac{3v^4}{8}+\frac{5v^6}{16}+...[/tex]

If I make the appropriate substitution for [itex]v=\hat{v}[/itex]

[tex]\hat{v}=\frac{i\hbar}{m_0}\nabla[/tex]

and write the sum on the right as

[tex]

\sum_{n=1}^{\infty}\frac{\left(1+(-1)^n\right) \left(\frac{1}{2} (-1+n)\right)!}{2 \sqrt{\pi } \frac{n}{2}!}\hat{v}^n

[/tex]

Then I end up with

[tex]i\hbar\frac{\partial}{\partial t}\psi=m_0\psi+\sum_{n=1}^{\infty}\frac{\left(1+(-1)^n\right) \left(\frac{1}{2} (-1+n)\right)!}{2 \sqrt{\pi } \frac{n}{2}!}m_{0}^{1-n}\left(i\hbar\nabla\right)^n\psi

[/tex]

Now for n=1, this equals

[tex]

i\hbar\frac{\partial\psi}{\partial t}=m_0\psi

[/tex]

which seems manifestly true (EDIT: by which I mean, there are no kinetic energy terms on the right), and for n=2

[tex]

i\hbar\frac{\partial\psi}{\partial t}=m_0\psi-\frac{\hbar^2}{2m_0}\nabla^2\psi

[/tex]

which seems Schrodinger-esque. So I was wondering... for n=4 (which is the next time a term is added on)

[tex]

i\hbar\frac{\partial\psi}{\partial t}=m_0\psi-\frac{\hbar^2}{2m_0}\nabla^2\psi+\frac{3\hbar^4}{8m_{0}^3}\nabla^4\psi

[/tex]

is the last term on the right a relativistic correction to the Schrodinger-esque equation above it?

Thanks,

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