3
$\begingroup$

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?

$\endgroup$
2

1 Answer 1

3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.