Simplification in Schrodinger derivation

[TheFerruccio]

This is not a homework question. This is not for a course. However, I got a warning for posting such questions elsewhere, so, I suppose I must post them here.

Homework Statement

The following is an excerpt of the derivation of the Schrodinger equation. After deriving the Klein-Gordon equation, the relativistic total energy is approximated to arrive at the Schrodinger equation.

Homework Equations

$E = mc^2\sqrt{1+\frac{p^2}{m^2c^2}}$
$\approx mc^2\left(1+\frac{1}{2}\frac{p^2}{m^2c^2}\right)$

The Attempt at a Solution

Well, frankly, I do not see how they went from the first step to the second step. Where did the 1/2 come from? How does the removal of the square root effectively approximate this? I am not seeing it. Was a conjugate used and multiplied somehow?

 

[TSny]

See Binomial Approximation

 

[Mute]

It may also help you to play around with some of these plots to get a feel for how well the approximations work. For example, letting $x = p/(mc)$, take a look at a comparison of the two curves, plotted with wolframalpha:

http://www.wolframalpha.com/input/?i=Plot[{Sqrt[1+x^2],1+x^2/2},{x,0,1}]

Play around with some other examples of the binomial approximation as well.

 

[TheFerruccio]

Thank you! This is perfect. I do not know how I managed to get this far without having heard of the binomial approximation, and having tutored math for years. I guess I am one of the lucky ten thousand.

 

[paranormal]

I can provide some explanation for the simplification in the derivation of the Schrodinger equation.

The first step in the derivation involves using the relativistic energy equation, which includes the square root term. This term accounts for the relativistic effects of momentum on the total energy of a particle. However, in the context of quantum mechanics, we are dealing with particles of very small masses and velocities, so the relativistic effects are negligible. Therefore, we can approximate the relativistic energy equation by dropping the higher order terms, which is where the 1/2 comes from in the second step.

To better understand this, let's look at the Taylor series expansion of the square root term:

\sqrt{1+\frac{p^2}{m^2c^2}} \approx 1 + \frac{1}{2}\frac{p^2}{m^2c^2} - \frac{1}{8}\frac{p^4}{m^4c^4} + ...

As you can see, the first term is just 1, and the second term is the one we are interested in, which is where the 1/2 comes from. The higher order terms are much smaller and can be neglected. So, by dropping these higher order terms, we effectively approximate the relativistic energy equation to the non-relativistic form, which is the second step in the derivation.

I hope this helps in understanding the simplification in the derivation of the Schrodinger equation.