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

Question

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 ?

Answer 1

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 $\psi(x,t)=\int dk\ a_k(t)\exp(ikx)$.