1
$\begingroup$

Let us suppose that we have an Hamiltonian that describes a quantum system. If one would like to know all of the possible values that the energy of the system described by that hamiltonian, one has to solve

$$ \hat{H}\psi_m = E\psi_m $$

Where $ \hat{H} $ is the hamiltonian operator and $ E $ is the eigenvalue for the eigenstate $ \psi_m $. My question is, if I denote $ \Psi $ by

$$ \Psi = \sum_m c_m \psi_m $$

where the sum is on all the eigenstates, do I get the "famous" $\Psi$ that has the property that $ \lvert\Psi\rvert^2$ is the probability to find the particle(s) 0f the system in some position or the other? And do the $ \psi_m $ also have this property?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Yes to both of your questions, if your wavefunctions are expressed in the position basis.

This expansion is useful because to get the time evolution of the system, you just do: $$\Psi(t)=\sum c_n\psi_ne^{-iE_nt/\hbar}$$

Improve this answer
$\endgroup$
3
  • $\begingroup$ Great! And will the square norm of $\Psi$ and any of the $\psi_m$ give the same information with regards to position? $\endgroup$
    – peppece
    Commented May 6, 2022 at 18:10
  • $\begingroup$ Also, could you clarify what you mean by "position basis"? I am diagonalizing H, so wouldn't the $\psi_m$ form the basis of the energy operator basis? $\endgroup$
    – peppece
    Commented May 6, 2022 at 18:13
  • 3
    $\begingroup$ @peppece By position basis, I mean that we may for example express $\psi(x)=Ae^{-x^2/2}$, rather than expressing it in some other basis like $\psi(p)=Ae^{-p^2/2}$. The probabilistic interpretation holds in either case. $\endgroup$
    – DanDan面
    Commented May 6, 2022 at 18:26

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.