# Relativistic corrections to Schrodinger's equation

1. Jul 7, 2011

### jfy4

Hi,

I was looking at the relativistic equation for energy $E^2=p^2+m_{0}^2$ and was wondering about a methodology I took with it. Making the substitution $E^2=\gamma^2 m_{0}^2$ then
$$p^2=m_{0}^2(\gamma^2-1)$$
Considering only the $\gamma-1$ factor, this can be expanded
$$\gamma-1=\frac{v^2}{2}+\frac{3v^4}{8}+\frac{5v^6}{16}+....$$
I then made the identification with $E=\hat{E}$,
$$\hat{E}=i\hbar\frac{\partial}{\partial t}$$
which yields the following from before
$$p^2=m_{0}^2\left(\left(\frac{i\hbar}{m_0}\frac{ \partial }{\partial t}\right)^2-1\right)$$
which led me to again identify
$$\gamma-1=\frac{i\hbar}{m_0}\frac{\partial}{\partial t}-1=\frac{v^2}{2}+\frac{3v^4}{8}+\frac{5v^6}{16}+....$$
If I make the appropriate substitution for $v=\hat{v}$
$$\hat{v}=\frac{i\hbar}{m_0}\nabla$$
and write the sum on the right as
$$\sum_{n=1}^{\infty}\frac{\left(1+(-1)^n\right) \left(\frac{1}{2} (-1+n)\right)!}{2 \sqrt{\pi } \frac{n}{2}!}\hat{v}^n$$
Then I end up with
$$i\hbar\frac{\partial}{\partial t}\psi=m_0\psi+\sum_{n=1}^{\infty}\frac{\left(1+(-1)^n\right) \left(\frac{1}{2} (-1+n)\right)!}{2 \sqrt{\pi } \frac{n}{2}!}m_{0}^{1-n}\left(i\hbar\nabla\right)^n\psi$$
Now for n=1, this equals
$$i\hbar\frac{\partial\psi}{\partial t}=m_0\psi$$
which seems manifestly true (EDIT: by which I mean, there are no kinetic energy terms on the right), and for n=2
$$i\hbar\frac{\partial\psi}{\partial t}=m_0\psi-\frac{\hbar^2}{2m_0}\nabla^2\psi$$
which seems Schrodinger-esque. So I was wondering... for n=4 (which is the next time a term is added on)
$$i\hbar\frac{\partial\psi}{\partial t}=m_0\psi-\frac{\hbar^2}{2m_0}\nabla^2\psi+\frac{3\hbar^4}{8m_{0}^3}\nabla^4\psi$$
is the last term on the right a relativistic correction to the Schrodinger-esque equation above it?

Thanks,

Last edited: Jul 7, 2011
2. Jul 7, 2011

### dextercioby

Yes, the lowest correction should have a 4th derivative, since in operator terms, it's $p^4$ which is actually $p p p p$ so it's a derivative of a derivative in total 4 times.

3. Jul 7, 2011

### Matterwave

I believe that that's correct; however, as you can see, that equation is going to be extremely difficult to solve in general.

That's why usually, different approaches are taken to incorporate SR into QM.

4. Jul 7, 2011

### jfy4

Yes, while Mathematica has no problem doing the stationary states, it refuses to solve anything with time dependence... I'm glad I could follow my nose though and get here. Thanks for comments.

5. Jul 7, 2011

### Bill_K

I'm curious how you would solve for a stationary state. Since the equation is fourth order you will have four linearly independent solutions, and will need to specify four boundary conditions.

6. Jul 7, 2011

### jfy4

There are still four constants $c_1, c_2, c_3, c_4$. Here was my procedure.

I solved the following
$$\epsilon\psi(x)=m_0\psi(x)-\frac{\hbar^2}{2m_0}\frac{\partial^2}{\partial x^2}\psi(x)+\frac{3\hbar^4}{8m_{0}^3}\frac{ \partial^4}{\partial x^4}\psi(x)$$
in mathematica with the following code
Code (Text):

Simplify[DSolve[\[Epsilon] \[Psi][x] ==
m \[Psi][x] +
Sum[((1 + (-1)^n) (1/2 (-1 + n))!)/(2 Sqrt[\[Pi]] (n/2)!)
m^(1 - n) (I \[HBar])^n D[\[Psi][x], {x, n}], {n, 1,
4}], \[Psi][x], x]]

and the result was
$$\psi [x]\to e^{\sqrt{\frac{2}{3}} x \sqrt{\frac{m^2 \hbar ^2-\sqrt{-m^3 (5 m-6 \epsilon ) \hbar ^4}}{\hbar ^4}}} C[1]+e^{-\sqrt{\frac{2}{3}} x \sqrt{\frac{m^2 \hbar ^2-\sqrt{-m^3 (5 m-6 \epsilon ) \hbar ^4}}{\hbar ^4}}} C[2]+e^{\sqrt{\frac{2}{3}} x \sqrt{\frac{m^2 \hbar ^2+\sqrt{-m^3 (5 m-6 \epsilon ) \hbar ^4}}{\hbar ^4}}} C[3]+e^{-\sqrt{\frac{2}{3}} x \sqrt{\frac{m^2 \hbar ^2+\sqrt{-m^3 (5 m-6 \epsilon ) \hbar ^4}}{\hbar ^4}}} C[4]$$

does this clarify by what I meant by solve? Is something wrong with my procedure?

EDIT: I added a psi in front of m, thanks Matterwave.

Thanks,

Last edited: Jul 7, 2011
7. Jul 7, 2011

### Matterwave

That first m is not really important in your calculations. Simply move it to the left hand side and it just re-scales your E. (You're missing a psi next to the m in your last post.)

8. Jul 7, 2011

### Bill_K

Well there is a problem with it, and I'm not sure at the moment how to resolve it. The full relativistic relationship between E and p is E = √(m2c4 + p2c2). That, obviously, is single-valued: one p value for each E value. In terms of a plane wave solution ψ ~ exp(ikx - iωt)), one k value for each value of ω. The dispersion curve ω(k) climbs indefinitely.

But when you approximate E ≈ mc2 + p2/2m - 3/8 p4/m3, the approximate relationship is quartic. And it's a quartic that opens downward. Now the dispersion curve ω(k) climbs for a while, then reaches a maximum ωmax, turns over and heads back down and crosses the axis! For given energy, a free particle in this approximation will generally have two wavelengths, one nearly the same as in the nonrelativistic case, and one that's very large. To make it usable, one would have to spell out how to eliminate the spurious high-frequency solutions.

9. Jul 7, 2011

### dextercioby

Though valid, your concern bears little to no relevance for the problem at hand. We only care if the p^4 term's contribution to the ground state energy is small enough to grant us the permission to dismiss the square root by using a series expansion. Or perhaps I'm missing something obvious...

10. Jul 7, 2011

### qsa

11. Jul 7, 2011

### Bill_K

The fact that E turns negative for short wavelength means the ground state is unstable.

12. Jul 7, 2011

### Vanadium 50

Staff Emeritus
You have to decide what you are going to do with this. Adding the p4 term and treating it as exact will lead to nonsense as p gets large: but so would stopping at p2.

If one is interested in relativistic corrections, there is no problem in using this as a perturbation in 1st order perturbation theory. IIRC, for particle-in-a-box, one can even solve it to all orders this way. If you're trying to get close to relativistic quantum mechanics this way, it's kind of hopeless.

13. Jul 7, 2011

### jfy4

Just so we are on the same page, I would never use/even consider this practically for relativistic QM. Like in the OP, I just wanted to know if my methodology was marginal, and if the result was correct; in so far as, if one continued to take terms from the sum and tack them on would they correctly predict results for a relativistic free particle, in theory.

I'm not planning on making this a big thing if you got the impression I was somehow impressed by this result. It was part of an afternoon physics play time session. Don't draw it out.

14. Jul 7, 2011

### alxm

Looks like you've successfully derived the first term of the http://en.wikipedia.org/wiki/Breit_equation" [Broken].

Last edited by a moderator: May 5, 2017
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