This problem could be done more simply through the application of linear algebra. You want to prove that

$$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$

The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that

$$\begin{aligned}
\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= \langle \psi_1 - \psi_2 | \psi_1 \rangle + \langle \psi_1 - \psi_2 | \psi_2 \rangle \\
&= \langle \psi_1 | \psi_1 \rangle - \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle - \langle \psi_2 | \psi_2 \rangle
\end{aligned}
$$

Because $\psi_1$ and $\psi_2$ are orthogonal and normalized, you know $\langle \psi_i | \psi_j \rangle = \delta_{i j}$. Substituting, the above expression evaluates to $1 - 0 + 0 - 1 = 0$, demonstrating that the two vectors are indeed orthogonal.

Your approach - using the integrals - was also valid, and fundamentally similar to mine here. However, by noting that the relation you used ($\langle \psi_1 | \psi_2 \rangle = \int_{-\infty}^{\infty} \! \psi_1^* \psi_2 \, \mathrm{d}x$) satisfied the definition of an inner product, the integrals can be omitted.

Honestly, the argument you're making here is a mess - the question is based on bad premises. So let me show you how to do it properly, and hopefully that will resolve your confusion.

You're right that in order for a wavefunction to be normalized, it must satisfy

$$\int_\text{all space} P(x)\mathrm{d}x
= \int_\text{all space} \psi^*(x)\psi(x)\mathrm{d}x = 1\tag{1}$$

But this statement:

On top of that, any $\psi$ can also be expressed as $\psi\psi^∗$

is not correct. Given a function $\psi(x)$, you can write $\psi(x)\psi^*(x)$, but that's a different function.

Anyway, given that your wavefunction can be written

$$\psi(x) = a\phi_1(x) + b\phi_2(x)$$

then you just plug that into the normalization condition (1) and get

$$\int_0^L \bigl(a^* \phi_1^*(x) + b^* \phi_2^*(x)\bigr)\bigl(a \phi_1(x) + b \phi_2(x)\bigr)\mathrm{d}x = 1$$

which expands to

$$\begin{multline}
a^*a \int_0^L \phi_1^*(x)\phi_1(x)\mathrm{d}x
+ a^*b \int_0^L \phi_1^*(x)\phi_2(x)\mathrm{d}x \\
+ b^*a \int_0^L \phi_2^*(x)\phi_1(x)\mathrm{d}x
+ b^*b \int_0^L \phi_2^*(x)\phi_2(x)\mathrm{d}x
= 1
\end{multline}\tag{2}$$

Now you can use the identity

$$\int_0^L\phi_1^*(x)\phi_2(x)\mathrm{d}x = \int_0^L\phi_2^*(x)\phi_1(x)\mathrm{d}x = 0$$

which follows from the fact that $\phi_1$ and $\phi_2$ are *orthogonal* functions (it's not enough that they are eigenfunctions of an operator, they have to be orthogonal), and the identity

$$\int_0^L\phi_1^*(x)\phi_1(x)\mathrm{d}x = \int_0^L\phi_2^*(x)\phi_2(x)\mathrm{d}x = 1$$

which simply reflects the fact that $\phi_1$ and $\phi_2$ are *normalized*. (Check for yourself that this is the same as the normalization condition, equation (1).) With these two identities, equation (2) reduces to

$$\lvert a\rvert^2 + \lvert b\rvert^2 = 1$$

The conceptual question I had was that if we have the probability squared here, is it that or the square root of that probability that is your normalization constant?

That all depends, how do you define your normalization constant? It depends on what you're normalizing and how exactly you express it as a function. However you do it, the end requirement for normalization is just that $\lvert a\rvert^2 + \lvert b\rvert^2 = 1$.

As far as using the specific sinusoidal form for the $\phi_i$, you can do that in this case, because you're given enough information to figure out that the eigenfunctions are in fact sinusoidal. But you don't really need to know that they are sinusoidal for the preceding argument to work; all you need to know is that the $\phi_i$s are orthonormal.

## Best Answer

No, orthogonal wavefunctions are in general not orthogonal on every subinterval. This is only true when the

supportof the wavefunctions (the regions in which they are nonzero) don't overlap. Energy eigenstates, for example, generally have overlapping support and are only orthogonal on whatever full interval they're defined on.To see why wavefunctions with overlapping support are not orthogonal on every subinterval, consider the region of overlap. Both wavefunctions are nonzero everywhere in the region (by definition), which means they are both nonzero

somewherein the region. I choose my integration interval to be just that point, and I get a nonzero value.