# Energy or a piecewise defined wave function in an Infinite Square well

So I'm asked to determine the most probable measured energy value of a particle in an infinite square well with wave function $$\psi(x)=\begin{cases} Ax, & 0< x<\frac{a}{4} \\[1em] A(\frac{a}{2}-x), & \frac{a}{4} They ask us to guess what the most probable energy is and justify. Then they ask to show mathematically what it is.

I claim that its most probable energy is that of a particle in the 1st exited states (as it looks similar to the 1st excited state). I have no idea how to go about calculating the energy for this wave function though. I imagine I use $$\left< H\right>$$, but how to actually go about doing that is escaping me at the moment.

Can anyone give me a hint or push in the right direction?

• Any wavefunction (including this one) is some linear combination of energy eigenfunctions, because they are complete. The square of the complex magnitude of the coefficients tells you the probability that you’ll measure that energy state. You know what all the energy eigenfunctions are. You can find the coefficients in the (infinite) linear combination by using the fact that the eigenfunctions are orthonormal. – G. Smith Sep 17 '19 at 19:44
• As a "how do I do this" question this is probably doomed by the homework policy. But there is a very solid conceptual question buried just below the surface. It ought to be possible to save the post by exposing the underlying issue. – dmckee --- ex-moderator kitten Sep 17 '19 at 19:56
• $\left< H\right>$ is the average energy. That won't tell you the most probable energy. – G. Smith Sep 17 '19 at 23:16
• I have no idea how to go about calculating the energy for this wave function. Your wavefunction doesn’t have one well-defined energy. It isn’t an energy eigenfunction. It is a superposition of an infinite number of energy eigenfunctions, so energy measurements on this state can produce an infinite number of different energies, each with some probability. – G. Smith Sep 17 '19 at 23:20

So we are given:

$$\psi(x)=\begin{cases} Ax, & 0< x<\frac{a}{4} \\[1em] A(\frac{a}{2}-x), & \frac{a}{4}

And we have to calculate the most probable energy. I'm assuming this is the initial wavefunction, $$\Psi(x, \ 0)$$ of the system. In order to calculate the probability of measuring some energy $$E_n$$, we have to project this wavefunction onto the energy eigenstate $$|E_n\rangle$$ of the infinite square well, in the position basis ($$\langle x | E_n \rangle$$) which, then take the modulus squared of that value to get the probability.

In order to find which energy is the most likely, we simply find this " discrete energy wavefunction", which is just the set of expansion coefficients $$c_n$$ when we expand $$\Psi(x, \ 0)$$ in terms of the stationary states.

I'm assuming you already solved the Schrödinger equation for the infinite square well to get the time-independent wavefunctions for each allowed energy:

$$\psi(x) \ = \ \sqrt{\frac{2}{a}} \sin \Big( \frac{n \pi x}{a} \Big)$$

So we then have:

$$c_n \ = \ \displaystyle\int_{0}^{a} \ \sqrt{\frac{2}{a}} \sin \Big( \frac{n \pi x}{a} \Big) \ \Psi(x, \ 0)$$ $$\Rightarrow \ A \sqrt{\frac{2}{a}} \Big( \displaystyle\int_{0}^{a/4} \sin \Big( \frac{n \pi x}{a} \Big) \ x \ + \ \displaystyle\int_{a/4}^{3a/4} \sin \Big( \frac{n \pi x}{a} \Big) (\frac{a}{2} \ - \ x) \ + \ \displaystyle\int_{3a/4}^{a} \sin \Big( \frac{n \pi x}{a} \Big) (x \ - \ a) \Big)$$

This integral is super nasty, so I just used Wolfram Alpha to solve it. We get:

$$c_n \ = \ \sqrt{\frac{2}{a}} A \frac{a^2}{\pi^2 n^2} \Big( 2\sin\Big(\frac{\pi n}{4}\Big) \ - \ 2\sin\Big( \frac{3\pi n}{4} \Big) \ + \ sin (\pi n ) \Big)$$

Ok, so now all we have to do is take the modulus squared of this, and we have a function telling us the probability of measuring each of the allowed energies. I solved this numerically, and found that the probability was maximized for $$n \ = \ 2$$. We know that:

$$E_n \ = \ \frac{n^2 \pi^2 \hbar^2}{2ma^2}$$

So we can simply plug in our value of $$n$$ to get the most probable energy:

$$E_2 \ = \ \frac{2 \pi^2 \hbar^2}{ma^2}$$