# Time evolution of eigenstates superposition

If a system is in a state $$\psi$$ which is superposition of, let's say two, energy eigenfunction, namely $$\psi_1$$ and $$\psi_2$$, so that $$\psi(t)=\psi_1e^{-i\omega_1t}+\psi_2e^{-i\omega_2t}$$ (I am omitting normalization constants for simplicity) then $$\left|\psi(t)\right|^2=\left|\psi_1\right|^2+\left|\psi_2\right|^2+\psi_1^* \psi_2 e^{i(\omega_1-\omega_2)t}+\psi_2^* \psi_1 e^{i(-\omega_1+\omega_2)t}$$ right? But isn't it true that since $$\psi_1$$ and $$\psi_2$$ are orthogonal eigenfunctions then $$\psi_1^* \psi_2$$ and $$\psi_2^* \psi_1$$ must be zero and so one would have: $$\left|\psi(t)\right|^2=\left|\psi_1\right|^2+\left|\psi_2\right|^2$$ How does it work? thanks

• The cross terms disappear when integrated over the space but otherwise they do not cancel punctually. – ZeroTheHero Oct 4 '18 at 0:31
• When we say two functions (that are normalizable) $f, g$ are orthogonal, we mean the integral $\int_{-\infty}^{\infty} f^{*}(x)g(x) \; dx$ is zero, not just the mere product $f^{*}(x)g(x)$. – SpiralRain Oct 4 '18 at 0:31
• IIRC, comments are not for answers (even a terse answer is an answer). – Alfred Centauri Oct 4 '18 at 2:22

Cross-terms like $$\psi_1^*(x)\psi_2(x)$$ integrate to $$0$$ but their product is not $$0$$ everywhere. As a result, $$\psi_1^*(x)\psi_2(x) e^{i(\omega_1-\omega_2)t}\ne 0$$ in general.

Maybe two simple examples will illustrate the point:

1. Infinite-well solutions

For infinite-well wavefunction with $$\psi_1(x)\sim\sin(\pi x/L)$$ and $$\psi_2(x)\sim \sin(2\pi x/L)$$. Certainly then $$\psi_1^*(x)\psi_2(x)\ne 0$$ in general although $$\int_0^L dx\,\psi_1^*(x)\psi_2(x)=0$$.

2. Harmonic oscillator solutions

Take $$\psi_0\sim e^{-\lambda x^2/2}$$ and $$\psi_1\sim x e^{-\lambda x^2/2}$$. Then again $$\int_{-\infty}^\infty \psi_0^*(x)\psi_1(x)=0$$ but $$\psi_1(x)^*\psi_0(x)\sim x e^{-\lambda x^2}\ne 0$$ except at $$x=0$$ and $$x=\pm \infty$$.

While it's true that

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

it isn't necessarily true that

$$\psi^*_1(x)\psi_2(x) = 0$$

Why? Because we can insert the identity

$$1 = \int\,\mathrm{d}x\,|x\rangle\langle x|$$

in the first equation to get

$$\int\,\mathrm{d}x\,\langle \psi_1|x\rangle\langle x|\psi_2\rangle = \int\,\mathrm{d}x\,\psi^*_1(x)\psi_2(x) = 0$$

But the fact that the integral is zero doesn't imply that the integrand is zero, correct?

• I was almost finished composing this answer when my computer locked up. I see that ZTH has already answered in the meantime. The draft was still available so I've posted it anyhow. – Alfred Centauri Oct 4 '18 at 2:59

Your notation is ambiguous so it's not quite clear what you're actually asking. If by $$\psi_1$$ you mean the position basis-wavefunction $$\psi_1(x)$$, then as the other answers point out, only the integrals of the cross-terms with respect to $$x$$ vanish, not the pointwise products. If by $$\psi_1$$ you mean the basis-independent ket $$|\psi_1\rangle$$, and by $$\psi_1^*$$ the corresponding bra $$\langle \psi_1|$$, then your last equation is correct as written (with no need for any integration), and simply says that the magnitude of $$|\psi\rangle$$ is preserved over time, so if it's correctly normalized at one time then it remains correctly normalized at all other times.