# Time Dependence of Sum of Two Stationary States

Here's a question that has been nagging at me:

Suppose I've got a sum of two stationary states from the particle in the box:

$$\Psi(x,t)=c_1\psi_1(x)e^{-iE_1t/\hbar}+c_2\psi_2(x)e^{-iE_2t/\hbar}$$

The probability density is $$\Psi^*(x,t)\Psi(x,t)$$

Which works out to $$\lvert{c_1}\rvert\psi_1^*(x)\psi_1(x)+\lvert{c_2}\rvert\psi_2^*(x)\psi_2(x)+c^*_1c_2\psi_1^*(x)\psi_2(x)\cdot e^{i(E_1-E_2)t/\hbar}+c^*_2c_1\psi_2^*(x)\psi_1(x)\cdot e^{i(E_2-E_1)t/\hbar}$$

The book I'm reading points out that this a time-dependent equation. But I'm confused, because I thought the stationary states were all perpendicular to each other when you integrate over the spatial domain. Doesn't that mean that the time-dependent terms all zero out when you try to compute any sort of average value of something like position or momentum? If so, isn't the time dependence here sort of irrelevant?

The sum of two stationary states with different energies is not an eigenstate of $$H$$ and thus not stationary, as your example perfectly illustrates.

If the energies were identical, you could factor a common time exponential, and this would disappear when computing the probability density. When the energies are NOT identical, there is no such common exponential in time and a time dependence proportional to $$t \Delta E$$ emerges in the probability density, exactly as shown by your example.

In particular spatial averages of observables will be time dependent, oscillating in your example at frequency $$\omega= \Delta E/\hbar$$.

Note that orthogonality results from integrating over space and indeed if you do this the time-dependence will go but that’s not useful if you are computing an average position which would involve an integral of the type \begin{align} \langle x\rangle = \int dx\, x \left(|c_1|^2\psi_1^2(x)+|c_2|^2 \psi_2^2(x)+ 4 c_1c_2 \psi_1(x)\psi_2(x) \cos(\omega t)\right) \end{align} when $$c_1$$ and $$c_2$$ are real, and with $$\psi_1(x)$$ and $$\psi_2(x)$$ also real. The last term in $$x\psi_1(x)\psi_2(x)\cos(\omega t)$$ is not guaranteed to be $$0$$ by orthogonality. This is in line with the observation that the probability density is time-dependent for combinations of states with different energies.

The same conclusion holds if once considers complex coefficients and wavefunctions.

• But see my question. Aren't the wave functions orthonormal? If yes, don't they integrate to 0 with respect to x? If they integrate to zero, isn't 0 times any function of time zero? And then, don't the time dependent terms get zeroed out? Commented Nov 22, 2020 at 21:36
• Got it. Thank you! Commented Nov 22, 2020 at 21:46
• The wavefunctions are indeed orthonormal which means that $\int \psi_1^*(x) \psi_2(x) dx = 0$, but it doesn't tell you anything when you multipy by an arbitrary function under the integral, for instance $\int x \psi_1^*(x) \psi_2(x) dx \neq 0$ in general. Commented Nov 22, 2020 at 21:53