Why is the following property regarding the general solution of the Schr. Eq. valid only for the free-particle case?

In my quantum mechanics lecture notes(see picture at the bottom), they say the plane wave basis $$\{\phi_k(x)=\frac{1}{(2\pi)^{3/2}}\exp(ik\cdot x)\}$$ is so general that any $$\psi \in L^2(\mathbb{R}^3)$$ can be expressed in that basis, as equation $$(34)$$ shows and then they write in equation $$(35)$$ but highlighting that this is for the special case of the S.E. for the free-particle.

My question is, why is this expresion not valid in general? I don't think I need a free particle to do it, because the S.E always has the time-dependent part solved by $$e^{-\frac{iEt}{\hbar}}$$ and we only deal with the eigenvalue equation for the hamiltonian to get the space-dependent part $$\psi(x)$$ and costruct the whole solution as $$\psi(x,t)=\psi(x)e^{-\frac{iEt}{\hbar}}$$ In the case of the free partiche the solution of the eigenvalue equation of the hamiltonian is $$\psi(x)=\phi_k(x) = e^{ik.x}$$. If the particle was not free the complete solution would still be $$\psi(x,t)=\psi(x)e^{-\frac{iEt}{\hbar}}$$ with $$\psi(x)$$ the solution of the corresponding eigenvalue problem and the equation (35) should be true in general

Can someone clarify this ?

There is nothing contradictory here. There are a few points of clarification which should resolve your confusion. I will ignore the issues of normalization (i.e., factors of $$2\pi$$) and will work in one-dimension. However, the underlying point should be clear.

• You can always express any wave-function as \begin{align} \psi(x)=\int dk\ a_k\exp(ikx) \end{align} because the momentum eigenfunctions form a complete basis. Or, given that the wave-function at any time can still be expressed as a linear combination of the momentum eigenfunctions, we can write \begin{align} \psi(x,t)=\int dk\ a_k(t)\exp(ikx) \end{align} Notice that since the wave-function potentially changes with time, the linear combination of the momentum eigenfunctions that produces the wave-function correspondingly changes with time. Thus, the time-dependence of the coefficients $$a_k(t)$$.
• One can also always write any wave-function as \begin{align} \psi(x)=\int dn\ a_n\phi_n(x) \end{align} where $$\phi_n(x)$$ are the energy eigenstates. The reason that we can do this is again the same, since the Hamiltonian is a Hermitian operator, we know that the energy eigenstates form a complete basis. But, most importantly, we know that the time-evolution operator is a linear operator. And we also know that a wave-function that is an energy eigenfunction will only pick up a phase, namely, $$\exp(-iE_nt)$$. Thus, the time-evolution of the wave-function that is the linear combination that we wrote above will be given by \begin{align} \psi(x,t)=\int dn\ a_n\exp(-iE_nt)\phi_n(x) \end{align}
• Now, in the special case of the free particle, we know that the energy eigenfunctions are momentum eigenfunctions, i.e., $$\phi_n=\exp(inx)$$. Thus, we can write the time-dependent wavefunction as \begin{align} \psi(x,t)=\int dn\ a_n\exp(-iE_nt+inx) \end{align}

In other words, in the case of the free-particle, we know that $$a_k(t)=a_k\exp(-iE_kt)$$. However, in a generic case, we do not know the form of $$a_k(t)$$. Thus, we can only go so far as saying $$\psi(x,t)=\int dk\ a_k(t)\exp(ikx)$$.

• Thanks for the reply, but I still don't see why $a_k(t)=a_k\exp(-iE_kt)$ is not always valid if the particle is not free. If the particle is not free the hamiltonian has a potential part, but that doesn't change the validity of the separation of variables method used when deriving the time independent S.E., which shows that the solution is always $(space function)*exp(-iE_kt)$ – mathlover Jan 13 at 17:58
• where the space function is always the eigenfunction solution of the eigenvalue equation – mathlover Jan 13 at 18:04
• @mathlover Look at the second bullet point, in particular, the last expression in the second bullet point. You can always write $a_n(t)=a_n\exp(-iE_nt)$. But the $n$ will not be enumerating the momentum eigenstates unless the particle is free. – Dvij D.C. Jan 13 at 18:11
• what would it be enumerating then? Maybe the problem is that I fail to see that the statement "And we also know that a wave-function that is an energy eigenfunction will only pick up a phase, namely, exp(−iEnt)" does not happen for a non-free particle. Since I think it happens in general. – mathlover Jan 13 at 18:17
• @mathlover Yes, obviously, that is a general statement. I thought I made it clear. It holds true for all energy eigenstates regardless of the fact as to whether the Hamiltonian is a free particle Hamiltonian or not. $n$ would be enumerating the eigenfunctions of the relevant Hamiltonian, whatever they might happen to be. They'd be enumerating momentum eigenfunctions only in the case of the free particle because only in that case would the momentum eigenfunctions also be the energy eigenfunctions. Does that make sense? – Dvij D.C. Jan 13 at 18:20