Griffith's Quantum Mechanics example 2.1 states that:

Suppose a particle starts out in a linear combination of just two stationary states: $$ \Psi(x,0) = c_1 \psi_1(x) + c_2\psi_2(x) $$ Find the probability density and describe the motion.

The solution drives the probability density to be :

$ |\Psi(x,t)|^2 = 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]$

Here's my question:

Integrating that probability density for $ x= -\infty \ldots +\infty $ results in a time dependent value as opposed to a constant:

$$ 1 = \int_{-\infty}^{+\infty} |\Psi(x,t)|^2 dx=\int c_1^2\psi_1^2 \;dx + \int c_2^2\psi_2^2 \; dx+ 2c_1c_2\cos[(E_2 - E_1) t/\hbar]\int\psi_1\psi_2\;dx$$

Now all of the integrals are constants, so the end result is a function of time which makes the wave function impossible to normalize with a constant factor. So what is going on here? What am I not getting?


1 Answer 1


In the cross terms that are time-dependent, the integral is $0$ due to the orthogonality of the wave functions (when integrated over $x$).


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