Can a free particle have a definite energy in quantum mechanics?

[dRic2]

It's been a long time since my last exam on QM, so now I'm struggling with some basic concept that clearly I didn't understand very well.

1) The Sch. Eq for a free particle is $-\frac {\hbar}{2m} \frac {\partial ^2 \psi}{\partial x^2} = E \psi$ and the solutions are plane waves of the form $\psi(x) = Ae^{1kx} + Be^{-ikx}$. This functions can not be normalized thus they do not represent a physical phenomenon, but if I superimpose all of them with an integral on $k$ I get the "true" solution (the wave packet). This implies that a free particle with definite energy does not exist (only superposition of states with different energies can exist). This bugs me a lot. For example, think about an atom hit by an ionizing radiation: at some point an electron will be kicked out of the shell and now, if I wait some time, I have a free electron (so a free particle) and what about its energy? It should be defined by the law of conservation of energy...

2) I'm reading some lecture notes about scattering. Why does everyone take the incoming particle to be described by the state $\psi_i = e^{i \mathbf k \cdot \mathbf r}$ if it is not normalizable ? It seems to me they all assume the particle to be inside a box of length $L$ and forget about about the normalization constant. But why ?

 

[bhobba]

I think you need to see the Rigged Hilbert Space formulation - it gets rid of a lot of these type of issues. See attached file. Its tough going and does not prove the central generalized Eigenvalue Theorem. I have a proof if you are interested - also attached.

As a warm up I suggest a book I believe should be in any applied mathematicians/physicists library and of course taught in such courses (it would be a good assignment):
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Thanks
Bill

 

[vanhees71]

If you read really good lecture notes or book, they'd tell you that in scattering theory the plane-wave solutions usually presented are handwaving.

Usually you do perturbation theory to derive S-matrix elements. Using plane waves for the in- and out-states you get matrix elements
$$S_{fi} \propto \delta^{(4)}(p_{\text{in}}-p_{\text{out}}).$$
Though I use relativistic notation here, the same holds true for the non-relativistic case either.

Now formally, the S-matrix element must be squared to yield transition probabilities, but you can't calculate $|S_{fi}|^2$, because it simply doesn't make sense to square a $\delta$-distribution!

There are two ways out: The first one is pragmatic and also regularizes a lot of mathematical trouble too (like the issue with Haag's theorem in relativistic QFT): Just think of your lab as a huge but finite box. To have still traveling-wave energy eigensolutions impose periodic boundary conditions. Then the momentum spectrum gets discrete, and the corresponding eigenfunctions are normalizable. You have $\vec{p} \in \frac{2 \pi}{L} \mathbb{Z}^3$ if you choose a cube of length $L$ as this "quantization volume". The eigenstates are $$u_{\vec{p}}=\frac{1}{\sqrt{L^3}} \exp(\mathrm{i} \vec{p} \cdot \vec{x},$$
and now you have
$$\langle u_{\vec{p}}|u_{\vec{p}'}\rangle = \int_{V} \mathrm{d}^3 \vec{x} u_{\vec{p}}^*(\vec{x}) u_{\vec{p}'}(\vec{x})=\delta_{\vec{p},\vec{p}'},$$
i.e., you get a Kronecker $\delta$ for the discrete momenta and not a $\delta$ distribution as in the infinite-volume case, which you can take at the end of the calculation. The upshot is that if you write
$$S_{fi}=\delta_{ji}-\mathrm{i} (2 \pi)^4 M_{fi} \delta^{(4)}(p_{\text{in}}-p_{\text{out}}),$$
the squared matrix element (leaving out the no-scattering piece) to be taken in cross-section calculations is
$$|T_{fi}|^2 (2 \pi)^4 \delta^{(4)}(p_{\text{in}}-p_{\text{out}}).$$
The more physical solution is to indeed use wave packets for the in- and out-states. Then nowhere squares of $\delta$ distributions occur, and again at the end of the calculation you can take the limit of wave packets with sharply defined momenta. This is carefully discussed in

M. Peskin, D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley Publ. Comp., Reading,
Massachusetts (1995).

For non-relativistic QM you find a very good discussion of scattering theory in

A. Messiah, Quantum Mechanics, Dover Publications, New York (1999).

For the free non-relativistic particles, it's more convenient to simply use momentum eigenstates, which are also energy eigenstates. The solution of the time-dependent Schrödinger equation thus can be written as a Fourier integral (I use natural units with $\hbar=1$):
$$\psi(t,x)=\langle x|\exp(-\mathrm{i} \hat{H} t)|\psi_0 \rangle = \int_{\mathbb{R}} \mathrm{d} p \langle x|\rangle p|\exp(-\mathrm{i} \hat{H} t)|\psi_0 \rangle = \int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} \exp\left (\mathrm{i} p x -\frac{\mathrm{i} p^2 t}{2m} \right) \tilde{\psi}_0(p),$$
where
$$\tilde{\psi}_0(p)=\langle p|\psi_0 \rangle=\int_{\mathbb{R}} \mathrm{d} x \langle p|\langle x \rangle \langle x|\psi_0 \rangle = \int_{\mathbb{R}} \mathrm{d} x \frac{1}{\sqrt{2 \pi}} \exp(-\mathrm{i} p x) \psi_0(x)$$
is the initial wave function (square integrable!).

 

[dRic2]

If you read really good lecture notes or book, they'd tell you that in scattering theory the plane-wave solutions usually presented are handwaving.

Using plane waves for the in- and out-states

The more physical solution is to indeed use wave packets for the in- and out-states.

So basically it would make more sense to use the wave packet, but to avoid heavy calculations, at an introductory level, the authors prefer to choose a simple plane waves as basis for the initial and final states?

I'll check for sure all the references that you (@vanhees71 @bhobba) gave me, but at my current level it will take some time... so you might see me back here in a while.

I have just one more question: does the wave packet actually represent a free particle ? Thinking about the ionized electron that I mentioned in my first example, is it correct to describe it by a wave packet ? Now it seems to me that the wave packet is a powerful tool, but lacks a complete correspondence with reality. I may have a wrong idea though.

 

[vanhees71]

The wave function of an electron in non-relativistic QM (which cannot be ionized, because it's a fundamental charged particle already, so that there's nothing left to ionize) has a clear meaning: It's modulus squared is the probability distribution of its position as a function of time, no more no less. If you can isolate the electron enough from its environment, i.e., no other charges or fields around, it will propagate according to the free Schrödinger equation. Thus free wave packets have the usual "physical reality" as any quantum state has: It's a probabilistic description of the to be expected outcomes of measurements on the accordingly prepared electron.

 

[dRic2]

Thank you. I'm very rusty, need to study this topics again (possibly better! )