# Probability of measuring a particle in the ground state: having trouble with the integration

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.

• 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? – Philipp May 14 '14 at 17:01
• $a$ 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! – Yoda May 14 '14 at 17:03

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

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

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}

so

\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.

• 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. – Yoda May 14 '14 at 18:11
• @AndersMB I checked it again and have to correct my first comment: the wavefunction is indeed normalized. – Philipp May 14 '14 at 20:13
• 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! – Yoda May 15 '14 at 7:58