# Orthogonality of summed wave functions

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?

• Yes, you do need normalization factors so that $\int |\Psi_3|^2 dx = 1$, but your proof as it stands is correct. You directly calculated their inner product and found it to be zero, hence orthogonal vectors. – webb Apr 2 '14 at 16:42
• Braket notation might be simpler, but yours is good enough. – jinawee Apr 2 '14 at 16:49
• I don't understand why I would need $\int |\Psi_3|^2 dx=1$. And how would the bracket notation help? – Oskar Henriksson Apr 2 '14 at 16:55
• You will probably get to Dirac/braket notation later in your course. But it would help by doing away with the integrals in this problem. – BMS Apr 2 '14 at 16:56
• @PoetryInMotion: The underlying assumption is that the symbols "plus ($+$)" and "minus ($-$)" that you used to express $\Psi_3$ and $\Psi_4$ in terms of the "orthonormal basis" states $\Psi_1$ and $\Psi_2$ do in fact represent the corresponding arithmetic operations between the complex number values of inner products. Consequently: $\langle \Psi_1 - \Psi_2 | \Psi_1 + \Psi_2 \rangle \text{=(means the same as)=} \langle \Psi_1 | \Psi_1 \rangle + \langle \Psi_1 | \Psi_2 \rangle - \langle \Psi_2 | \Psi_1 \rangle - \langle \Psi_2 | \Psi_2 \rangle$ which can be readily evaluated further (being $0$) – user12262 Apr 2 '14 at 17:37

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.

• Ah, that makes sense! However, I still don't understand what the normalization factors have to do with the question. Both $\Psi_3$ and $\Psi_4$ happens have the normalization factor $1/\sqrt{2}$, but can't have anything to do with their orthogonality, can it? – Oskar Henriksson Apr 2 '14 at 19:59
• Yeah, I'm not sure why you'd need the normalization factors here. It's clear that if $\langle \psi_1 | \psi_2 \rangle = 0$, then $\langle k \psi_1 | \psi_2 \rangle = k \langle \psi_1 | \psi_2 \rangle = 0$. – Shivam Sarodia Apr 2 '14 at 22:48
• @PoetryInMotion Can you explain the context in which the solution manual used the normalization factors? – Shivam Sarodia Apr 3 '14 at 0:11
• Oh, sorry, I missed that I was supposed to both investigate the orthogonality of $\Psi_3$ and $\Psi_4$ and normalize $\Psi_3$ and $\Psi_4$. The explains why the solution manual disused normalization constants. – Oskar Henriksson Apr 3 '14 at 21:29
• All right. Since you correctly found the normalization factor above, I take it you don't need help with that part. – Shivam Sarodia Apr 3 '14 at 23:18