How did Feynman do this approximation $\sqrt{A^2 + \mu^2\varepsilon^2} =A\biggl( 1+\frac{1}{2}\,\frac{\mu^2\varepsilon ^2}{A^2}\biggr)$?

Question

I was reading the 9th lecture of Feynman; here is the excerpt of my concern:

Let’s go back now to our particular example of the ammonia molecule in an electric field. Using the values for H11, H22, and H12 given in (9.14) and (9.15), we get for the energies of the two stationary states $$E_{I}=E_0+\sqrt{A^2+\mu^2 \varepsilon^2},\quad E_{II}=E_0-\sqrt{A^2+\mu^2\varepsilon^2} \tag{9.30}$$ [...] When the electric field is zero, the two energies are, of course, just $E_0±A.$ When an electric field is applied, the splitting between the two levels increases. The splitting increases at first slowly with $E$, but eventually becomes proportional to $E$. (The curve is a hyperbola.) For enormously strong fields, the energies are just $$E_I=E_0+μ\varepsilon =H_{11}\;,\;E_{II}=E_0−μ\varepsilon=H_{22}.$$

[...]How can we separate the two molecular states? One method is as follows. The ammonia gas is let out of a little jet and passed through a pair of slits to give a narrow beam.The beam is then sent through a region in which there is a large transverse electric field. The electrodes to produce the field are shaped so that the electric field varies rapidly across the beam. Then the square of the electric field $\mathbf{\varepsilon}\cdot \mathbf{\varepsilon}$ will have a large gradient perpendicular to the beam. Now a molecule in state $|I⟩$ has an energy which increases with $\varepsilon^2$, and therefore this part of the beam will be deflected toward the region of lower $\varepsilon^2.$ A molecule in state $|II⟩$ will, on the other hand, be deflected toward the region of larger $\varepsilon^2$, since its energy decreases as $\varepsilon^2$ increases.

Incidentally, with the electric fields which can be generated in the laboratory, the energy $\varepsilon$ is always much smaller than $A.$ In such cases, the square root in Eqs. (9.30) can be approximated by $$A\biggl( 1+\frac{1}{2}\,\frac{\mu^2\varepsilon ^2}{A^2} \biggr). \tag{9.32}$$ So the energy levels are, for all practical purposes, $$E_{I} =E_0+A+\frac{\mu^2\varepsilon^2}{2A} \; \\ \& \\ E_{II} =E_0-A-\frac{\mu^2\varepsilon^2}{2A}$$

I'm not understanding how Feynman made the square-root term of $(9.30)$ that is, $\sqrt{A^2 + \mu^2\varepsilon^2}$ equal to $(9.32)$ that is, $A\biggl( 1+\dfrac{1}{2}\,\dfrac{\mu^2\varepsilon ^2}{A^2}\biggr)$ ? How did he make the approximation?

Answer 1

Since for positive $x$ we have $\left(1+\frac{x}{2}\right)^2>1+x$, it follows that $\sqrt{1+x}<1+\frac{x}{2}$.

On the other hand, $$ 1+\frac{x}{2}-\sqrt{1+x} = \frac{\frac{x^2}{4}}{1+\frac{x}{2}+\sqrt{1+x}} < \frac{x^2}{8\sqrt{1+x}}$$ hence for small $x$ we have that $1+\frac{x}{2}$ is an excellent approximation for $\sqrt{1+x}$, and obviously: $$ \sqrt{A^2+\mu^2\varepsilon^2} = A\cdot \sqrt{1+\frac{\mu^2\varepsilon^2}{A^2}}\approx A+\frac{\mu^2 \varepsilon^2}{2A^2} $$ if $A^2$ is large compared to $\mu^2\varepsilon^2$.

Answer 2

Since $A > 0$, $$\sqrt{A^2 + \mu^2 \epsilon^2} = A \sqrt{1 + \frac{\mu^2 \epsilon^2}{A^2}}.$$

Then use the linear approximation to $\sqrt{x}$ near $1$: $$\sqrt{x} \approx 1 + \frac{1}{2} (x - 1),$$ when $x$ is close to $1$. (This comes from calculus; one of the most important reasons why the subject is so useful is the ability to use it to approximate functions.)

Here, you must assume that $\mu \epsilon$ is considerably smaller than $A$, so that $x = 1 + \frac{\mu^2 \epsilon^2}{A^2}$ is close to $1$. Then the approximation follows.

Answer 3

There is an extension of the binomial theorem that applies to a binomial raised to a non-integer power. As is mentioned in another question, the formula is

$$(a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2 + \frac{n(n-1)(n-2)}{3!}a^{n-3}x^3 + \ldots .$$

Setting $n = \frac12$, we get

\begin{align} \sqrt{a+x} = (a+x)^{1/2} &= a^{1/2} + \frac12 a^{-1/2}x + \frac{\frac12(-\frac12)}{2!}a^{-3/2}x^2 + \ldots \\ &= a^{1/2} + \frac{x}{2 a^{1/2}} - \frac{x^2}{8 a^{3/2}} + \ldots . \end{align}

If you also set $a = A^2$ and $x =\mu^2 \varepsilon^2$, you get

\begin{align} \sqrt{A^2+\mu^2 \varepsilon^2} &= A + \frac{\mu^2 \varepsilon^2}{2 A} - \frac{\mu^4 \varepsilon^4}{8 A^3} + \ldots \\ &= A \left(1 + \frac{\mu^2 \varepsilon^2}{2 A^2} - \frac{\mu^4 \varepsilon^4}{8 A^4} + \ldots \right).\\ \end{align}

If $\mu^2 \varepsilon^2$ is small compared to $A^2$, the first two terms of the series are a good approximation for the entire series, so we ignore $\frac{\mu^4 \varepsilon^4}{8 A^4}$ and all the terms after it.

This is a useful series to know about in physics, because it works just as well for $(a+x)^{-1/2}$ as for $(a+x)^{1/2}$; also, I seem to recall at some point (although I do not recall exactly for what; something in special relativity, I think), the third term became significant as well, which you would not likely guess if you did not know about the series.