Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

It says in Griffith's chapter 2.1, that:

$$\tag{2.14} \Psi(x,t)~=~\sum_{n=1}^{\infty}c_n\,\psi_n(x) e^{(-iE_n t/\hbar)}$$ It so happens that every solution to the (time-dependent) Schrodinger equation can be written in this form [...].

By "solution" does he mean solutions on separable form

$\tag{2.1} \Psi(x,t)~=~\psi(x)f(t),$

which he stated to begin with?

`

share|improve this question
    
2.14 is pretty much equivalent to saying that every vector can be written as a linear combination of a set of basis vectors. –  Joren Jun 7 '13 at 13:46
    
@Joren: It looks similar, but it is not the same. When you represent an exact eigenfunction $\psi_k$ of the total Hamiltonian $\hat{H}$ as a linear combination of the eigenstates of a non perturbed Hamiltonian $\hat{H}_0$ like $\psi_k=\sum c_{kn}\psi_n^{(0)}$, in experiment you will never find any state with energy $E_n^{(0)}$. –  Vladimir Kalitvianski Jun 7 '13 at 21:42

2 Answers 2

up vote 6 down vote accepted

No, by a solution he means any solution, i.e. a function $\Psi(x,t)$ that satisfies Schrödinger's equation and that will typically not be writable in the $a(x)b(t)$ separated form. The claim that any solution may be written in the summation form you quoted is the claim that the solutions in the separated form are "sufficient" because the most general solution may be written as their superposition.

There isn't really any ambiguity in the text.

share|improve this answer
1  
Fair enough thanks:) –  user25504 Jun 7 '13 at 11:04
    
@user25504: A general solution of this type does not have a certain energy. In a measurement process one observes different eigenstates with probabilities $|c_n|^2$, not one eigenstate. Certain and conserved is the mean energy of the state $\bar{E}=\langle \Psi|\hat{H}|\Psi\rangle$ (if the Hamiltonian is time-independent, of course). –  Vladimir Kalitvianski Jun 7 '13 at 12:30

That wave function is a series

$$\Psi(x,t)=\sum_{n}c_{n}\psi_{n}(x)e^{-iE_{n}t/\hbar} $$

every term (of index, say $j$) looks like a separable solution

$$\Psi_{j}(x,t)=\psi_{j}(x)e^{-iE_{j}t/\hbar}=\psi(x)f(t)$$

and because the equation is linear, if $\Psi_{j}$ is a solution of the time dependent Schrödinger equation, then

$$\Psi (x,t)=\sum_{n}c_{n}\Psi_{n}(x,t) $$

is the most general solution.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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