3
$\begingroup$

I was working through Griffiths' Introduction to Quantum Mechanics, specifically the part about the 1D infinite square well potential (situated between $x = 0$ and $x = a$). To my understanding, this allows for multiple wave functions, each associated with a discrete level of energy:

$$\Psi_n(x, t) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)\,e^{-i\frac{E_n}{\hbar}t}$$

where:

$$E_n = \frac{n^2\pi^2\hbar^2}{2a^2m}.$$

This is where it starts to get confusing to me. Does this mean that only one of these wave functions describes the particle state? Or is it that a general solution can be obtained by combining all the above possible wave functions to get the following one:

$$\Psi(x, t) = \sqrt{\frac{2}{a}}\sum_{n=1}^{+\infty}C_n\, \sin\left(\frac{n\pi}{a}x\right)\,e^{-i\frac{E_n}{\hbar}t}$$

This is how the textbook says the general solution is determined, but what's confusing me, in this case, are the $C_n$'s (the equation was taken directly from the book). How did they get here, even though they were not present in the first equations? And how are we to determine them?

$\endgroup$
1
  • $\begingroup$ Hint (to see what the $C_n$ are): Calculate $(\Psi_n,\Psi)$, i.e. the scalar product between $\Psi_n(x,t)$ and $\Psi(x,t)$. $\endgroup$
    – Jakob
    Oct 3 at 14:35
4
$\begingroup$

I have now understood that the general solution is not simply a sum of the stationary states, but rather a LINEAR COMBINATION of the stationary states, thus the presence of the $C_n$'s.

I also understood that the way to compute the $C_n$'s is to use the orthogonality of the stationary states to get:

$$C_n = \int_0^a\Psi(x,0)\psi_n^*(x)dx$$

$\endgroup$
1
  • $\begingroup$ Indeed, though it should be stated it is because the Schrödinger equation is a linear PDE. $\endgroup$
    – Triatticus
    Oct 3 at 23:04
2
$\begingroup$

Those wavefunctions (note: not "wave equations") are the energy eigenstates of the infinite square well. A general solution can be written as a linear combination of the eigenstates, where the coefficients of that linear combination are determined by the initial conditions set up by the physical situation. Any particle state described by a single energy eigenstate will evolve trivially in time (that is, it's probability distribution will not change in time, it merely picks up a complex phase). By contrast, a particle state describe by a linear combination will evolve in time: its probability distribution will appear to change due to beating between the eigenstates make up the linear combination that describes the physical state, each of which evolves with a different characteristic frequency.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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