Orthogonality of summed wave functions

Question

Problem. I know that the two wave functions $\Psi_1$ and $\Psi_2$ are all normalized and orthogonal. I now want to prove that this implies that $\Psi_3=\Psi_1+\Psi_2$ is orthogonal to $\Psi_4=\Psi_1-\Psi_2$.

My naive solution. From the premises, we know that $$\int_{-\infty}^\infty \Psi_1^*\Psi_1 dx=\int_{-\infty}^\infty \Psi_2^*\Psi_2 dx=1$$ and $$\int_{-\infty}^\infty \Psi_1^*\Psi_2 dx=\int_{-\infty}^\infty \Psi_2^*\Psi_1 dx=0$$

We also have $(z_1+z_2)^*=z_1^*+z_2^*$

$$\int_{-\infty}^\infty \Psi_3^*\Psi_4 dx = \int_{-\infty}^\infty (\Psi_1+\Psi_2)^*(\Psi_1-\Psi_2)dx \\ =\int_{-\infty}^\infty(\Psi_1^*+\Psi_2^*)(\Psi_1-\Psi_2)dx\\ =\int_{-\infty}^\infty(\Psi_1^*\Psi_1-\Psi_1^*\Psi_2+\Psi_2^*\Psi_1-\Psi_2^*\Psi_2)dx\\ =1-0+0-1=0\,,$$

which is equivalent with what we wanted to prove. Is this a legitimate proof? Is there any simpler way to do this? I am afraid I still haven't grasped how wave functions behave mathematically, so I may have missed somethings very obvious here.

Edit: The solution manual somehow uses normalization factors for $\Psi_3$ and $\Psi_4$. How are these factors when you don't actually know the exact functions? And how does this relate to the concept of orthogonality?

Answer 1

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.