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# QS 11: Plane wave = Beam of particles?

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Let us consider the momentum eigenstates $\kk{\vv{p}\beta}$ of a system, where $\vv{p}$ are momentum eigenvalues and $\beta$ a set of discrete eigenvalues needed to make a complete set. We normalize them as
\begin{eqnarray*} \bb{\vv{p}\beta}\kk{\vv{p'}\beta'}=\ddd{\vv{p}-\vv{p'}}\dd{\beta\beta'}\end{eqnarray*}
Let $\phi$ be a state which has more or less well defined momentum $\vv{k}$ and a fixed value of $\beta=\beta_0$. This means that the wave functions $f(\vv{p})$ defined by $\bb{\vv{p}\beta}\kk{\phi}=f(\vv{p})\dd{\beta\beta_0}$ has a very sharp peak at $\vv{p}=\vv{k}$. As $\phi$ is a state vector of unit norm $(\phi,\phi)=1$,
\begin{eqnarray*} \intp |f(\vv{p})|^2 &=&1,\qquad (p_0=\sqrt{\vv{p}^2+m^2})\mbox{ relativistic}\\
\int {\rm d}^3p |f(\vv{p})|^2 &=&1,\qquad  \mbox{non-relativistic particles}.\end{eqnarray*}
Now consider a very large number $N$ of copies of the system all prepared in the identical state $\phi$ but then spatially translated form the place of their preparation to positions $\vv{r}_i,i=1,\dots,N$, the points being distributed with a uniform density $\rho$ in space in a large volume.  These systems then are in states $\phi_i$ whose momentum wave-functions are given by
\begin{eqnarray*}\bb{\vv{p}\beta}\kk{\phi_i}=f(\vv{p})\dd{\beta\beta_0}
\exp(i\vv{r}_i.\vv{p}/\hbar)\end{eqnarray*}
Suppose we want to make measurements on an observable $B$ on all these systems.  The expectation value of $B$ in these states will be  (we do it for relativistic particles)
\begin{eqnarray*} \frac{1}{N}\sum_i\bb{\phi_i}|B\kk{\phi_i} &=&
\frac{1}{N}\sum_i\intp\intpp\exp[i(\vv{p-p'}).\vv{r}_i/\hbar] \times\\
&&f(\vv{p'})^*f(\vv{p}) \bb{\vv{p'}\beta_0}|B\kk{\vv{p}\beta_0} \end{eqnarray*}
If the points $\vv{r}_i$ are closely spaced we can replace the sum over points $\vv{r}_i$ by $\rho\intx$. Carrying out the integration over $\vv{r}$ we get
\begin{eqnarray*} \langle B\rangle=\frac{\rho(2\pi\hbar)^3}{N}\intp|f(\vv{p})|^2
\bb{\vv{p}\beta_0}|B\kk{\vv{p}\beta_0} \end{eqnarray*}
If the observable $B$ is such that it {\em does not} have a peak at $\vv{p}=\vv{k}$ then the matrix element can be pulled out at value $\vv{k}$ because that is where the main contribution comes from, and the expectation value, to a good approximation, is
\begin{eqnarray*} \langle B\rangle=\frac{\rho(2\pi\hbar)^3}{N}
\bb{\vv{k}\beta_0}|B\kk{\vv{k}\beta_0} \intp|f(\vv{p})|^2 \end{eqnarray*}
the last factor is equal to one {\em and the dependence on the momentum profile of the state $\phi$ vanishes completely!}
(The same calculation holds for the non-relativistic momentum wave functions.)

Therefore all that matters is that the momentum is close to $\vv{k}$ and that $B$ does not have a very sharply peaked  expectation value in momentum basis at $\vv{p}=\vv{k}$.  Thus we get the fundamental result :

{\em The expectation value of a quantity $B$ obtained by measuring on $N$ copies of the system in a state with momentum very close to $\vv{k}$ (and other eigenvalues equal to $\beta_0$),these copies being distributed in space with uniform density $\rho$ in a volume $V$ is}
\begin{eqnarray}\langle B\rangle
=\frac{\rho(2\pi\hbar)^3}{N}\bb{\vv{k}\beta_0}|B\kk{\vv{k}\beta_0}
=\frac{(2\pi\hbar)^3}{V}\bb{\vv{k}\beta_0}|B\kk{\vv{k}\beta_0}.\end{eqnarray}
The transition rate (formulas (\ref{transition1}) and (\ref{transition2})) for a beam of particles with sharp momentum $\vv{k}$ is therefore
\begin{eqnarray}n_\xi=\frac{2\rho}{\hbar}(2\pi\hbar)^3\bb{\vv{k}\beta_0}|B\kk{\vv{k}\beta_0}.
\label{transition3}\end{eqnarray}

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