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$\newcommand{\vec}[1]{\mathbf{#1}} \newcommand{\dd}{\mathrm{d}}$I'm reading Landau and Lifshitz' book on non-relativistic quantum mechanics and I have some doubts about a passage in the chapter about elastic scattering. I have the French edition of 1966 so I cannot quote precisely, but it should be in §125, from around equation (125.10).

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?

yellon
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1 Answers1

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  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.)

  2. On the other hand, in scattering theory, the initial state is not normalized. In this case there is no absolute notion of probability. Instead the scattering cross-section is by definition normalized relative to the flux of the incident beam. L&L's normalization of the initial wave function to the unit current density is designed to fulfill this definition. It is not arbitrary.

References:

[L&L] L.D. Landau & E.M. Lifshitz, QM, Vol. 3, 3rd ed, 1981; $\S$126.

Qmechanic
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