# Wavefunction as a combination of two stationary states - how to find those states?

Lets say we have a particle in a infinite square well which has a wavefunction like this ($A$ is some constant and $d$ is the width of the well):

\begin{align} A\left[ \sin \left(\frac{2 \pi x}{d}\right) + \frac{1}{3}\sin \left(\frac{3 \pi x}{d}\right)\right] \end{align}

How do we know from which stationary states this wavefunction is composed of? I would say that it is a combination of 1st and 2nd excited state, but the anwser in the book is 1. and 3. excited state. How do we find this out?

• Your wavefunction looks more like for infinitely deep well, not finite. – Ruslan Aug 6 '13 at 22:51
• Woops it is an infinite well - I fixed it. – 71GA Aug 6 '13 at 23:50

For 1D Schrödinger equation, $n$th stationary state wavefunction has exactly $n-1$ zeroes. Thus, $k$th excited state wavefunction has exactly $k$ zeroes. Now if you look at your wavefunction, it's composed from two stationary states:

Count their zeroes (don't take borders into account — they are because of infinite height of well). You'll see that you're right, these states are 1st and 2nd excited states. Your book most likely has a typo.

In more general case, as asked in comments, you have to deal with your wavefunction by definition of superposition of states. Let your function $\psi$ be as follows: $$\psi= A\left[ \cos \left(\frac{7 \pi x}{d}\right) + \frac{1}{3}\sin \left(\frac{4 \pi x}{d}\right)\right]$$ And your eigenfunctions are: $$\phi_k=B_k\sin\left(\frac{k\pi x}d\right)$$ To find how much of $\phi_k$ is in your $\psi$, you find scalar product: $$\left<\psi|\phi_k\right>=\int_V \psi^*\phi_k dx=\int_0^d AB_k \sin\left(\frac{k\pi x}d\right)\left[\cos\left(\frac{7\pi x}d\right)+\frac13\sin\left(\frac{4\pi x}d\right)\right]dx$$ As $\cos(x)$ is not orthogonal to any $\sin(ax)$ on $[0,d]$, you will have infinitely many non-zero $\left<\psi|\phi_k\right>$:

For comparison let's see what this method gives for original question where there's just a superposition of states 2 & 3:

Here only these 2nd and 3rd states are non-zero.

• Are you sure that the book is wrong? I need to be sure. – 71GA Aug 6 '13 at 23:54
• What would the states be if i had for example \psi=\begin{align} A\left[ \cos \left(\frac{7 \pi x}{d}\right) + \frac{1}{3}\sin \left(\frac{4 \pi x}{d}\right)\right] \end{align}? I put the cosinus inside instead of sinus. – 71GA Aug 7 '13 at 0:12
• @71GA: I think that Ruslan is using the infinite square well defined on $[0,d]$, and you're using the infinite square well defined on $[-d/2,d/2]$. You should specify in your question precisely what you mean by "infinite square well", as otherwise the equation you wrote is impossible interpret. – Dan Aug 7 '13 at 1:16
• I don't think the example allows for cosine states - those would have non-zero probabilities of being near the left edge of the well - which is infinitely high. The minimum Energy State would be the $sin(\frac{\pi x}{d})$ state, that has one belly. That is the unexcited state. Adding bellies ( represented by the number before the $\pi$) means adding excitations. – yippy_yay Aug 7 '13 at 1:21
• @Sebastian Henckel I know that if the problem is on an interval $0<x<d$ the solutions are all like $\sin(n\pi x/d)$ but I wondered if i had a problem like this: $\psi= A\left[ \cos \left(\frac{7 \pi x}{d}\right) + \frac{1}{3}\sin \left(\frac{4 \pi x}{d}\right)\right]$ could i reckognize this as a superposition of states on an interval $-d/2<x<d/2$ where in general solutions are of form $\sin(n\pi x/d)$ for even and $\cos(n\pi x/d)$ for odd. Could i then rightfully say that i have states 7 and 4 on an interval $-d/2<x<d/2$? – 71GA Aug 7 '13 at 7:52

If you want to do this analytically, you need to know what the stationary states of the system are. Once you have them, you can use the fact that the stationary states are both orthonormal (the inner product of any two stationary states $i$ and $j$ is $\delta_{ij}$) and complete (any state can be expressed as a sum of stationary states).

Consider an arbitrary state $\psi$. Because the stationary states are complete, $\psi$ can be written as $$\psi =\sum_i c_i \phi_i$$

where $\phi_i$ are the stationary states, and $c_i$ are constants. To find the $c_i$ for a given $\phi$, we can use the orthogonality relation $$\langle\phi_i|\phi_j\rangle \equiv\int\phi_i^*(x)\phi_j(x)dx = \delta_{ij}$$

With a little bit of algebra, we eventually arrive at

$$c_i=\langle\phi_i|\psi\rangle =\int\phi_i^*(x)\psi(x) dx$$

The state $\psi$ is then composed of all the $\phi_i$ for which the corresponding $c_i$ is not zero.

• Can you add the equations you wrote in terms of integrals so i can understand more. Please allso leave the Dirac notation. – 71GA Aug 7 '13 at 0:19
• @71GA: I added the integrals as requested. – Dan Aug 7 '13 at 1:13
• Thank you very much. Ok so if we scalar multiply the stationary state $\phi_i$ with a wavefunction $\psi$ and integrate over $x$ we should get the constant $c_i$ which is normaly in front of an stationary state? How come? – 71GA Aug 7 '13 at 8:49
• @71GA you get not the same constant which is in front of a stationary state. That $A$ constant in your wavefunction given in question is to normalize the wavefunction. And here $c_i$ is scalar product of two wavefunctions, much like a Fourier coefficient. – Ruslan Aug 7 '13 at 8:55
• @71GA: Just plug the first equation into the third equation. Then use the second equation. – Dan Aug 7 '13 at 18:26