# Arbitrary normalisation of a free particle wave function


While studying the rate of transitions in the continuous spectrum (dealing with free particles of given momenta) due to some potential $U$, it is written that «we normalise the outgoing wave function, with momentum $\vec p'$, as the Dirac delta in the momentum space \begin{equation*} \psi_{\vec{p}'}(\vec{x})=\frac1{(2\pi\hbar)^{3/2}}e^{\frac{i}{\hbar}\vec{p}'\cdot\vec{x}} \end{equation*} and the incoming wave function to unit current density \begin{equation*} \psi_{\vec{p}}(\vec{x})=\sqrt{\frac{m}{p}}e^{\frac{i}{\hbar}\vec{p}\cdot\vec{x}} \end{equation*} therefore the probability given by Fermi's golden rule \begin{equation*} \dd w_{\vec{p}\vec{p}'}=\frac{2\pi}{\hbar}\bigl\lvert \langle\vec{p}'|U|\vec{p}\rangle\bigr\rvert^2\delta\bigl(E(\vec{p})-E(\vec{p}')\bigr)\,\dd\nu \end{equation*} represents the differential cross section of the scattering process».

Here $m$ is the mass of the particle, $p=\lVert\vec{p}\rVert$ and $\dd\nu$ represents an "interval of states", in this case $\dd p_x\dd p_y\dd p_z$.

Now, my question: can the normalisation factor of a free particle be arbitrary? My feeling is that the authors did it "because it works" and because it gives the desired result, but probably I just don't know something that happens behind the curtains of this derivation. I get that free particle wave functions cannot be normalised anyway in $\mathbb{R}^3$, but does this mean that I can multiply them by whatever (constant scalar factor) I want?

When the equation for the golden rule for transitions between continuous spectrum states was introduced (§43 in my edition), the authors in fact wrote that $\dd w$ cannot be considered as a transition rate, since it doesn't even have the correct units (I guess that depends on how you "count the states": I could have used e.g. $\dd\nu=\dd p_x\dd p_y\dd p_z/\hbar^3$ as well).

How do I resolve all this arbitrariness?

• Something is very fishy about the normalization to unit current density; it doesn't even have the right units. Definitely wavefunctions should always be normalized to 1 (this is what normalization to Dirac delta does). My gut is telling me that the "normalization to unit current density" should actually be happening in Fermi's golden rule, not at the level of the wavefunction. Perhaps it truly should be in this "interval of states" (or as wikipedia calls it density of final states). I do want to point out though that free particle wavefunctions do have proper normalization (eg Dirac) – Aaron Nov 5 '16 at 4:06
• If I had to choose how to normalise the incoming wave function, I would choose too the Dirac delta in momentum space, like the outgoing wave, since the property of the particle after all is having a definite momentum. Also, does the choice of such density of states (in terms of the momentum coordinates) follow from the normalisation in the momentum space? It seems likely to me since (final) momentum is the only "variable" in Fermi's rule. – yellon Nov 5 '16 at 13:07
• Normalization is always determined by your definition of inner product on the Hilbert space. Also, the density of states is determined by the definition of inner product. So in an indirect way, the normalization should determine your density of states. Given this, what I said before doesn't seem to be true. The only other place these extra factors could legitimately come from then is the differential cross section. Unfortunately, I don't have a copy of the book, so I don't know how this is defined for them. – Aaron Nov 5 '16 at 17:52
• I never thought about it in this way... So would it go something like this: I represent the states in the "momentum picture" (which is natural since they are free particles) that is as functions in $L^2(\mathbb{R}^3)$ with the measure $\mathrm{d}p_x\mathrm{d}p_y\mathrm{d}p_z$, thus free particles states are represented by (and normalised as) Dirac deltas $\delta(\mathbf{p}-\mathbf{p}')$ and the density of states is the chosen integration measure? (I'm not sure if "measure" is a correct terminology from a mathematical point of view, but the meaning should be clear) – yellon Nov 5 '16 at 18:56
• Roughly speaking, that's correct. You can think about a quantum system on a finite size $L^3$ system and taking the $L \rightarrow \infty$ limit, and the connections are more clear since you don't have to worry about distributions (at least more clear to me). – Aaron Nov 5 '16 at 21:56

1. On one hand, Fermi's golden rule states that transition probability rate is $$\frac{dP}{dt}~=~\frac{2\pi}{\hbar} | \langle f | V| i\rangle|^2 \rho_f. \tag{1}$$ It is assumed that the initial state $|i\rangle$ is normalized. The final states $|f\rangle$ do not have to be normalized. (The latter can be seen by putting the system in a potential box with volume $L_xL_yL_z$, and take the limit $L_xL_yL_z\to \infty$. The normalization of $|f\rangle$ and $\rho_f$ would scale in a way so that the formula (1) remains invariant.)
[L&L] L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S$126.