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NOTE: The 'correct' solution in this post is actually incorrect; see Cye Waldman's correct solution.

I am asked to solve the following integral for $E_n$, which appears in the study of the quartic anharmonic oscillator:

$$\int_{-E_n^{\frac{1}{4}}}^{E_n^{\frac{1}{4}}}\frac{dx}{\sqrt{E_n-x^4}} = 2\int_0^{E_n^{\frac{1}{4}}}\frac{dx}{\sqrt{E_n-x^4}} = \left(n+\frac{1}{2}\right)$$

The equality follows because the integrand is even. I now use the substitution $u=E_n-x^4$ to obtain:

$$dx=-\frac{du}{4(E_n-u)^\frac{3}{4}}$$

And the integral now becomes:

$$\frac{1}{2} E_n^{-\frac{3}{4}} \int_{0}^{E_n}u^{-\frac{1}{2}}(1-\frac{u}{E_n})^{-\frac{3}{4}}du$$

Making another substitution $v=\frac{u}{E_n}$, we obtain:

$$\frac 1 2 E_n^{-\frac{1}{4}} \int_0^1 v^{-\frac{1}{2}}(1-v)^{-\frac{3}{4}} \, dv$$

This is now a Beta function, and thus we obtain:

$$\frac{1}{2} E_n^{-\frac{1}{4}} \int_0^1 v^{-\frac{1}{2}}(1-v)^{-\frac{3}{4}}\,dv = \frac{1}{2} E_n^{-\frac{1}{4}} \ B(\frac{1}{2},\frac{1}{4}) = (n+\frac{1}{2})$$

Thus, rearranging for $E_n$, and writing the Beta function using Gamma functions, we obtain:

$$\implies E_n = \left(\frac{\Gamma(1/2) \Gamma(1/4)}{2 \Gamma(3/4) (n+\frac{1}{2})} \right)^4$$

However, the solution quoted by the question-setter, is:

$$E_n = \left(\frac{3\Gamma(3/4)^2}{\sqrt{2 \pi}}(n+\frac{1}{2}) \right)^{4/3}$$

This is drastically different. The identity:

$$\Gamma(1/4) \Gamma(3/4) = \sqrt{2} \pi$$

Is given, but this obviously would not resolve the discrepancy. I'm reasonably certain the error lies in one of my variable changes, but I cannot find where.

I am further inclined to think my answer is incorrect, as if the energy $E_n$ were indeed:

$$\implies E_n = \left(\frac{\Gamma(1/2) \Gamma(1/4)}{2 \Gamma(3/4) (n+\frac{1}{2})} \right)^4$$

Then $\displaystyle \lim_{n \to \infty} E_n = 0$, and the potential is bounded with a finite binding energy equal to the $n=0$ energy: $E_0 = \left(\frac{\Gamma(1/2) \Gamma(1/4)}{2 \Gamma(3/4) (\frac{1}{2})} \right)^4$. In other words, a particle could 'escape' the $x^4$ potential well, which does not make intuitive sense as it increases to infinity on both sides.

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  • $\begingroup$ One thing I note is your bounds are not correct after your first substitution $\endgroup$ – Triatticus Apr 23 '17 at 21:56
  • $\begingroup$ Apologies; it was a typing error. This has now been corrected, but the question still stands. $\endgroup$ – CrossProduct Apr 23 '17 at 22:00
  • $\begingroup$ I have gone through this much as you did, except that I started with the transform $x^4=at$. I end up with the same result as you did. Your problem is much worse than the gamma identity. You/we have the n-term in the denominator, while the so-called correct solution has it in the numerator. So I ask, who says that is the correct solution. $\endgroup$ – Cye Waldman Apr 23 '17 at 23:18
  • $\begingroup$ Thanks for your efforts; ff two different approaches using different substitutions have yielded the same answer then I am indeed inclined to think that perhaps the 'correct' solution is not so! However, if the $\frac{1}{(n+1/2)^4}$ dependence in my/our answer is correct, it would imply that the energy levels of a quartic potential well decrease (rapidly) with increasing n, which doesn't make intuitive sense - it would imply that a particle could escape the $x^4$ potential with binding energy equal to $E_0$ ($n=0$ energy). I've added this argument to my original post. $\endgroup$ – CrossProduct Apr 23 '17 at 23:34
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    $\begingroup$ From the context and from the right-hand side of your first equation, it appears that you're applying Bohr-Sommerfeld quantization to the Hamiltonian $H=p^2+x^4=E$. But in that case the appropriate integral to quantize is the classical action $\oint p\,dx=\oint \sqrt{E-x^4}\,dx$, not the classical period. $\endgroup$ – Semiclassical Apr 24 '17 at 0:02
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I disagree with correct solution in the OP and will present my own analysis for your scrutiny. First, let's reduce the clutter and set $a=E_n$. We seek to find $a$ such that

$$2\int_0^{a^{1/4}}\sqrt{a-x^4}dx=\left(n+\frac{1}{2}\right)$$

Now let $x^4=at$ or $x=(at)^{1/4}$, then

$$ dx=\frac{a^{1/4}}{4}t^{-3/4}dt\\ \sqrt{a-x^4}=\sqrt{a}\sqrt{1-t}\\ x=a^{1/4}\to t=1 $$

Substituting and rearranging we get

$$\frac{a^{3/4}}{2}\int_0^1 t^{-3/4}(1-t)^{1/2}dt=\left(n+\frac{1}{2}\right)$$

Introducing the complete beta function,

$$B(\nu,\mu)=\int_0^1 t^{\nu-1}(1-t)^{\mu-1}dt=\frac{\Gamma(\nu)\Gamma(\mu)}{\Gamma(\nu+\mu)}$$

Clearly, $\nu=1/4$ and $\mu=3/2$ and we can then show that

$$a=\left[ \frac{2\left(n+\frac{1}{2}\right)\Gamma(7/4)}{\Gamma(1/4)\Gamma(3/2)}\right]^{4/3}$$

At the OP's suggestion, we can substitute

$$ \Gamma(7/4)=(3/4)\Gamma(3/4)\\ \Gamma(3/2)=\sqrt{\pi}/2\\ \Gamma(1/4)\Gamma(3/4)=\pi\sqrt{2} $$

and demonstrate that

$$E_n=\left[ \frac{2\Gamma(3/4)^2\left(n+\frac{1}{2}\right)}{\pi\sqrt{2\pi}}\right]^{4/3}$$

So, it would appear that the original correct solution was missing a factor of $\pi$ in the denominator.

The solution I present here has been validated numerically for $n\in\mathbb{R^+}$, i.e., not just $n\in\mathbb{Z}$.

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  • $\begingroup$ Indeed: The identity $\Gamma(1/4) \Gamma(3/4) = \sqrt{2} \pi$ was incorrectly quoted as $\Gamma(1/4) \Gamma(3/4) = \sqrt{2\pi}$ by the question setter. Thank you for clearing this up! $\endgroup$ – CrossProduct Apr 25 '17 at 15:07
  • $\begingroup$ Your welcome. Don't forget to tip the help. $\endgroup$ – Cye Waldman Apr 25 '17 at 16:02
  • $\begingroup$ I'll just make a small editorial comment here. If you leave the result in the original form I derived it, a smart person can see that the ratio of gamma terms arises from the beta function. The alternative expression just looks like magic (where did that come from?). $\endgroup$ – Cye Waldman Apr 25 '17 at 16:14
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The original integral is quoted incorrectly. For a quartic potential, the WKB approximation integral should read:

$$\int_{-E_n^{\frac{1}{4}}}^{E_n^{\frac{1}{4}}}\sqrt{E_n-x^4} \ dx = (n+\frac{1}{2})$$

Evaluation of this integral in terms of Beta functions yields the answer given by Cye Waldman.

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  • $\begingroup$ I'm having a problem with the correct solution in that I tested it numerically and it doesn't work. I derived the solution to the revised integral and I get $E_n=\left[\frac{2(n+1/2)\Gamma(7/4)}{\Gamma(1/4)\Gamma(3/2)}\right]^{4/3}$. This has been verified numerically for $n=1:100$. $\endgroup$ – Cye Waldman Apr 24 '17 at 19:49
  • $\begingroup$ I've worked through it again, and using the revised integral posted in my answer, I am indeed still obtaining the answer as quoted (exactly). $\endgroup$ – CrossProduct Apr 24 '17 at 23:34
  • $\begingroup$ Interesting, have you tried a numerical integration? I will double check my work tomorrow. However, just ran the beta integral on WolframAlpha and it agrees with my $\Gamma(7/4)/(\Gamma(1/4)\Gamma(3/2))$. $\endgroup$ – Cye Waldman Apr 25 '17 at 0:26
  • $\begingroup$ I've not tried a numerical integration no - I seem to still obtain the exact answer as posted. I'll post my working in around two day's time or so. $\endgroup$ – CrossProduct Apr 25 '17 at 1:05
  • $\begingroup$ Just for curiosity, where does the correct solution come from? $\endgroup$ – Cye Waldman Apr 25 '17 at 4:09

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