In the book Introduction to Quantum Mechanics (by David Griffith) there is an Example 2.1:

Suppose a particle starts out in a linear combination of just two stationary states:

$$\Psi(x,0)~=~c_1\Psi_1(x)+c_2\Psi_2(x) $$

(Two keep it simple I´ll assume that the costants $c_n$ and the states $\Psi_n$ are real.) What is the wave function $\Psi(x,t)$ at subsequent times? Find the probability density, and describe its motion.

Solution: The first part is easy:


where $E_1$ and $E_2$ are the energies associated with $\psi_1$ and $\psi_2$. It follows that:

$$|\Psi(x,t)|^2~=~ (c_1\psi_1e^{iE_1/\hbar}+c_2\psi_2e^{iE_2/\hbar})(c_1\psi_1e^{-iE_1/\hbar}+c_2\psi_2e^{-iE_2/\hbar}) \\=~c_1^2\psi_1^2+c_2^2\psi_2^2+2c_1c_2\psi_1\psi_2cos[(E_2-E_1)t/\hbar]$$

Evidently the probability density oscillates sinusoidally. at an angular frequency $(E_2-E_1)/\hbar$; this is certainly not a stationary state. But notice that it took a linear combination of states (with different energies) to produce motion.

My problem is: The cosinus function equals zero at certain times. So, when I normalize the funkction at this time: $$\int_ {-\infty}^\infty c_1^2\psi_1^2+c_2^2\psi_2^2\,\text{d}x~=~1$$

On the other hand at certain times the cosinus function equals one. And then the integral $$\int_ {-\infty}^\infty c_1^2\psi_1^2+c_2^2\psi_2^2\,\text{d}x~+~\int_ {-\infty}^\infty2c_1c_2\psi_1\psi_2\,\text{d}x$$ should be bigger than 1. That seems impossible, so where is my mistake?

I am sorry when the question is unclear in some parts. I´m not used to speak english on science topics.


2 Answers 2


As eigenfunctions of the Hamiltonian, the wavefunctions $\psi_i$ are orthogonal to each other w.r.t. the standard scalar product, i.e. $$ \int \psi_1(x)\psi_2(x)\mathrm{d}x = 0$$ so the summand you are worried about vanishes at all times regardless of the value of the cosine.

  • $\begingroup$ Alright, thank you. But does that mean than, that the time has no effect at all? That would seem weird too, would it not ? My thought is, that $\int_a^b\psi_1(x)\psi_2(x)\text{d}x$ can never be below zero. But that also meens, that it can never be above zero? So part with the cosinus function and therefore the time seems to have no physical meaning at all? $\endgroup$ Commented Mar 31, 2016 at 16:01
  • $\begingroup$ @BenitoMcLanbeck: 1. That integral can easily be below zero. 2. The integral is not zero when you don't integrate over the entire real line - time has an effect on the probability of finding the particle in particular regions - it just can't change the overall probability to find the particle at all. $\endgroup$
    – ACuriousMind
    Commented Mar 31, 2016 at 16:16
  • $\begingroup$ Yes, I found the mistak in my thinking. Thanks a lot! $\endgroup$ Commented Mar 31, 2016 at 16:41

The trick here is that if your states are orthogonal, which they should if they are eigenstates, then integration over space will give you zero no matter what will be the value of cos at that moment.


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