I have a quick question. Particle in 1D box with length $L$ and wave equation $$ \psi(x,t=0)=Ax^3(L-x) $$ If I want to express time evolution do I just add time factor at end like this? $$ \psi(x,t)=x^3(L-x)e^{-iEt/\hbar} $$ Or am I suppose to use superposition of wave functions describing particle in a box? Then the time dependence would be $$ \psi(x,t)=\sum_{n=1}^\infty c_n\psi_n(x)e^{-iE_nt/\hbar}, $$ where $$ \psi_n(x)=\sqrt{\frac{2}{L}}\sin(n \pi x/L) $$ $$ c_n=\sqrt{\frac{2}{L}}A\int_0^L\sin(n\pi x/L)x^3(L-x). $$

  • 1
    $\begingroup$ To answer your own question, check if your second equation satisfies the TDSE. $\endgroup$
    – G. Smith
    Nov 23, 2020 at 18:21

1 Answer 1


For time-independent Hamiltonian, We know that $$|\Psi(t)\rangle=U(t)|\Psi(0)\rangle=e^{-i Ht/\hbar}|\Psi(0)\rangle$$

I'm using $|n\rangle$ to show the energy eigenbasis. $$|\Psi(0)\rangle=\sum_n|n\rangle\langle n|\Psi(0)\rangle=\sum_n\int |n\rangle\langle n|x\rangle\langle x|\Psi(0)\rangle dx $$ $$=\sum_n\left(\int \phi_n(x)\psi(x,0)dx \right)|n\rangle$$ $$|\Psi(t)\rangle=\sum_n\left(\int \phi_n(x)\psi(x,0)dx \right)e^{-i Ht/\hbar}|n\rangle=\sum_n\left(\int \phi_n(x)\psi(x,0)dx \right)e^{-i E_nt/\hbar}|n\rangle$$

$$\psi(x,t) = \langle x|\Psi(t)\rangle=\sum_n\left(\int \phi_n(x)\psi(x,0)dx \right)e^{-i E_nt/\hbar}\phi_n$$

where $$\phi_n=\langle x|n\rangle=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$ That's cleared what you should use.


Your Answer

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

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