Is a quantum wave function always the continuous sum of energy eignstates?

Question

To illustrate, I will take the step potential situation in one dimension.

We have a particle of mass $m$ coming from $-\infty$. The potential is $V(x)=0$ if $x<0$ and $V_0$ if $x>0$.

The usual way to treat the situation is to look for energy eigenstates (eg. take $E=2V_0$) and solve the time-independent Schrodinger equation.

A solution has the form $$\Psi_w(x)= \begin{cases} A_1e^{ik_1x} + B_1e^{-ik_1} \text{ if } x<0\\[3ex] A_2e^{-ik_2x} \text{ if } x> 0 \end{cases}$$

with $k_1$ and $k_2$ depending on $E$ and $B_1, A_2$ related to $A_1$ by continuity.

(see wiki)

This solution can't be normalized. Usually, the trick to give a physical sense to this kind of solution (free particle) is to say a non-stationnary solution is the integral (a fourier transform) of these eigenstates.

If I understand correctly, that means if I can express the initial state as $\Psi(x,0)=\int_{-\infty}^{+\infty}A(w)\Psi_w(x)dw$

then I can compute the evolution as $\Psi(x,t)=\int_{-\infty}^{+\infty}A(w)\Psi_w(x)e^{iwt}dw$

The wikipedia page about the Schrodinger equation suggests it is always possible to express the intial state as an integral of this form (it says it is a type of "spectral theorem", but is it? In my case, it does not seem obvious at all: what kind of energy repartition should I imagine here for an incoming wavepacket ? For a free particle, it seems right because we can use a fourier transform. But is it the case here too ?

Answer 1

First of all, it is important to know the system you're dealing with. There are two important questions here:

Q1: What are the possible energies?

In the example above you speak about a 'free' particle, this means that it can have all possible energies (a continuous spectrum). Note that this is not always the case (in most cases not).

Q2: How do I represent the state of a particle?

It is one of the axioms of the theory of quantum mechanics that the possible physical measurable quantities of a physical observable (represented as an operator; think of this as the energy operator, momentum operator, etc.) are eigenvalues of that operator. In this specific example we're dealing with the 'energy operator', also known as the hamiltonian $\hat{H}$. When our system has a certain energy (we don't necessarily know the value), it thus needs to be in a superposition of such energy eigenstates. When these eigenstates are discrete we use a sum, when continuous we use an integral. However, in a lot of textbooks one uses a summation to represents a sum over discrete indices and an integral over the continuous indices. So $\sum$ can be a $\sum$ or $\int$ depending on the conditions.

It may help to give a specific example, consider a system that has two possible energies (like the spin of an electron that is placed in a magnetic field, however this is not important for the following): $E_1$ and $E_2$. We denote the corresponding states (think of this as your wavefunctions) as $|E_1>$ and $|E_2>$. We know the particle has a certain energy, so it is in a lineair superposition of the possible energy eigenstates: $|\Psi> = c_1|E_1> + c_2|E_2> = \sum_{i=1}^2 c_i|E_i>$. We see that in this example, there is no integration but a summation over the discrete indices.

To conclude, yes, in your example of the free particle the energies are continuous so that one needs to integrate over the possible energy eigenstates, but this need not always be the case, as illustrated in my example above.

EDIT:

An additional remark that wasn't mentioned above is that you can indeed write every state $|\Psi>$ of the system as a lineair superposition of the energy eigenstates $|E_\omega>$, in this case an integration (as the energies form a continuous spectrum).