I am working on a problem in the context of an infinite symmetric potential barrier around the origin, with barriers at $x=- \frac{a}{2}$ and $x=\frac{a}{2}$. The wave function is a symmetric triangle-function:

$$ \Psi (x) = \begin{cases} 0 & x < - \frac{a}{2} \\ \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right) & -\frac{a}{2} \leq x \leq \frac{a}{2} \\ 0 & x > \frac{a}{2} \end{cases} $$

I must calculate the probability of measuring a particle in the above eigenstate in the ground state (when $n=1$). The way to attack this problem is to calculate the coefficient $c_1$, and then square it. Here follows my (wrong) attempt.

$$ c_n = \int\limits_{-\infty}^{\infty} \psi_n^*(x)\Psi (x)\,\mathrm dx $$

I understand from the above expression that the $\psi_n^* (x)$ is just the "regular" solution to the particle in a box. Given this symmetric potential, and that $n=1$ (odd $n$, even function), I must set

$$ \psi _1^* (x) = \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a} $$

The expression for $c_n$ becomes

$$ c_n = \int\limits_{-a/2}^{a/2} \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a} \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx \\ $$

Simplifying gives

$$ c_n = \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx $$

This can be solved by partial integration, setting $u'=\cos \left( \frac{\pi x}{a} \right)$ and $v= \left( \frac{a}{2} - \lvert x \rvert \right)$. Hence, $u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $ and $v'=-1$. The integral itself then becomes

$$ \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx = \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right)\left( \frac{a}{2} - \lvert x \rvert \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) (-1)\,\mathrm dx $$

Cleaning up

$$ \mbox{Integral} = \frac{a^2}{2 \pi} \sin \left( \frac{\pi x}{a} \right) - \lvert x \rvert \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right) - \frac{a^2}{\pi ^2} \cos \left( \frac{\pi x}{a} \right) $$

Now we just need to insert the limits. However, we see quite easily from the above expression that upon inserting $x=\pm \frac{a}{2}$, the last term vanishes, since $\cos ( \pm\pi / 2 = 0)$. The sine terms will equal $1$ when the limit is positive and $-1$ when the limit is negative. Hence, we are left with

$$ \mbox{Integral} = \left[ \frac{a^2}{2 \pi} - \frac{a}{2} \frac{a}{\pi} \right] - \left[ \frac{a^2}{2 \pi}(-1) - \lvert \frac{-a}{2} \rvert \frac{a}{\pi}(-1) \right] = \frac{a^2}{2 \pi} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi} $$

Plugging this into the original expression

$$ c_n = \frac{2 \sqrt{6}}{a^2} \cdot \frac{a^2}{\pi} = \frac{2 \sqrt{6}}{\pi} \approx 1.56 $$

So, you see the problem. My $c_n$ is larger than one, which will give me a rather un-physical probability. Am I wrong in this derivation in my assumption for $\psi _n^* (x)$? If not, then where?

Wolfram Alpha gives me that

$$ \mbox{Integral} = \frac{2a^2}{\pi ^2} \\ \Rightarrow c_1 = \frac{2 \sqrt{6}}{a^2} \cdot \frac{2a^2}{\pi ^2} = \frac{4 \sqrt{6}}{\pi ^2} $$

Using this gives me a probability of $c_1^2 \approx 0.986$, which at least has physical meaning. Also, logic tells me that there is a very high probability of finding the particle in the ground state, so this further backs up WA's answer.

  • $\begingroup$ How do you get from Wolfram Alpha's expression $c_1 = \frac{2a^2}{\pi ^2}$ to $c_1^2 = 0.986$ - what value of $a$ are you using and where does this $a$ come from? $\endgroup$
    – Philipp
    May 14, 2014 at 17:01
  • $\begingroup$ $a$ is the length of the potential well. I have made an error; I see that now! The WA expression for $c_1$ should be multiplied by the constant that was factored out from the integral. I will update my question! $\endgroup$
    – Yoda
    May 14, 2014 at 17:03

1 Answer 1


I think your tactics using partial integration from the start is the problem because $\lvert x \rvert$ is not differentiable at $x=0$ so working with its derivative is not quite kosher (and $v' = -1$ is certainly wrong since for $x < 0$ the gradient of $-\lvert x \rvert$ is $1$).

I suggest you go about it step by step:

\begin{align} c_1 &= \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right) dx \\ &= \frac{2 \sqrt{6}}{a^2} \biggl( \frac{a}{2} \underbrace{\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{1} } - \underbrace{\int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{2} } \biggr) \end{align}

Now, integral $I_1$ is easy:

\begin{align} I_{1} &= \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx \\ &= \left[ \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right) \right]_{-a/2}^{a/2} = \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) + \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) = \frac{2a}{\pi} \end{align}

For integral $I_2$ I suggest that you use that

\begin{equation} \lvert x \rvert = \begin{cases} x & x \geq 0 \\ -x & x \leq 0 \end{cases} \end{equation}

and split up the integral into two integrals

\begin{align} I_{2} &= \int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx \\ &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx \end{align}

Now use integration by parts with $u'=\cos \left( \frac{\pi x}{a} \right)$ ($\Rightarrow u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $) and $v= x$ ($\Rightarrow v'= 1 $) to solve the integral

\begin{align} \int x \cos \left( \frac{\pi x}{a} \right) dx &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) dx \\ &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \ , \end{align}


\begin{align} I_{2} &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx \\ &= \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{0}^{a/2} - \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{-a/2}^{0} \\ &= \frac{a^2}{2 \pi} - \frac{a^2}{\pi^2} - \frac{a^2}{\pi^2} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi} - \frac{2 a^2}{\pi^2} \end{align}

Plugging this into the equation for $c_1$ gives

\begin{align} c_1 &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a}{2} I_{1} - I_{2} \right) \\ &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a^{2}}{\pi} - \frac{a^2}{\pi} + \frac{2 a^2}{\pi^2} \right) \\ &= \frac{4 \sqrt{6}}{\pi^2} \end{align}

and that is indeed what Wolfram Alpha gives you.

  • $\begingroup$ Hmm. The text does not explicitly say that the function is normalized; I just assumed that it was. That square root fooled, because it looks like a normalizing constant. But still, WA gets another answer than mine, and that one looks (to me) to be the more likely correct answer. $\endgroup$
    – Yoda
    May 14, 2014 at 18:11
  • $\begingroup$ @AndersMB I checked it again and have to correct my first comment: the wavefunction is indeed normalized. $\endgroup$
    – Philipp
    May 14, 2014 at 20:13
  • $\begingroup$ I just redid my integration based on your notes. Thanks a lot! I never thought that I had to treat $\lvert x \rvert$ in a special way! $\endgroup$
    – Yoda
    May 15, 2014 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.