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

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    $\begingroup$ The cross terms disappear when integrated over the space but otherwise they do not cancel punctually. $\endgroup$ – ZeroTheHero Oct 4 '18 at 0:31
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    $\begingroup$ 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 $f^{*}(x)g(x)$. $\endgroup$ – SpiralRain Oct 4 '18 at 0:31
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    $\begingroup$ IIRC, comments are not for answers (even a terse answer is an answer). $\endgroup$ – 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?

  • $\begingroup$ 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. $\endgroup$ – 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.


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