# Are energy eigenstates in infinte well orthogonal?

According to my limited understanding of this subject, all eigenfunctions of operators are orthogonal, i.e. $\psi_1\times \psi_2 = 0$

However, as I learned about infinte wells, the eigenfunction of the energy operator is $\psi_n = C_nsin\frac{n\pi x}{L}$,but

$\psi_1$ x $\psi_2=sin\frac{\pi x}{L}sin\frac{2\pi x}{L}$ which is not equal to $0$ for most values of $x$.

Is there something wrong about my understanding of orthogonality? I would appreciate any clarifications on the subject.

Orthogonality means:

$$\int \psi_i^* \psi_j = \delta^i_j$$

Not $\psi_i \times \psi_j = 0$ everywhere.

• Apologies, but could I get a further clarification? – Krykt Nov 12 '17 at 10:36
• I understand that the $\delta_j^i$ means that the sum is $0$ when $i$ $\ne j$, which seems to be the case for me as $\psi_1$ is a not the same as $\psi_2$. So going by $\int \psi_1^*\psi_2=\delta_i^j$, is the sum not $0$? Thanks!! – Krykt Nov 12 '17 at 10:42
• @Krykt: no, the function $\psi_1^* \psi_2$ doesn't have to be zero everywhere. It can some regions where it is negative and some where it is positive. We only require that when integrated the negative and positive regions cancel out to give zero. – John Rennie Nov 12 '17 at 10:44
• Ah, yes. That was exactly what I needed to know. Thanks, really! – Krykt Nov 12 '17 at 11:34

Let's go back to ordinary vectors for a moment. If I have some $\vec a \in \mathbb R^n$ and $\vec b \in \mathbb R^n$, then I can define the inner product for this vector space as,

$$\langle a, b \rangle = \vec a \cdot \vec b = \sum_{i=1}^n a_i b_i = |\vec a| |\vec b|\cos\theta$$

where $\theta$ is the angle between the vectors. What does orthogonality mean? It means these two vectors would be perpendicular to each other and so $\theta = \pi/2$ implying that the inner product must vanish, i.e. $\langle a, b\rangle = 0$ for $\vec a$ and $\vec b$ orthogonal.

When we talk about functions, we define a different version of the inner product, namely,

$$\langle f, g \rangle = \int f^{*}g \, \mathrm dx .$$

So, following on from the vector example, we say $f$ and $g$ are orthogonal if $\langle f, g \rangle = 0$. For your set of functions $\psi_i$ this translates to, $\langle \psi_i, \psi_j\rangle = \delta_{ij}$.

This is always zero except when $i = j$, which makes sense since the same function cannot be orthogonal to itself, hence the appearance of the Kronecker delta. Note that in some cases there may need to be some weight function in the definition of the inner product.