1
$\begingroup$

Consider Fermi golden rule

$$\Gamma _{{i\rightarrow f}}={\frac {2\pi }{\hbar }}\left|\langle f|H'|i\rangle \right|^{{2}}\rho $$

I don't understand why $\left|\langle f|H'|i\rangle \right|^{{2}}$ is defined as a probability density.

Since it is an integral over the volume of normalization, in my view, it should be a probability itself and not a density.

What kind of probability density is it? In other words, w.r.t. which variable should I integrate it to get the total probability (i.e. 1)?

$\endgroup$
4
  • 3
    $\begingroup$ What does the derivation of the article you are quoting say? $\endgroup$ Commented Jun 28, 2018 at 20:02
  • $\begingroup$ @CosmasZachos Thanks for the reply, it refers to a "transition probability per unit of time" but I'm not sure if it's $\left|\langle f|H'|i\rangle \right|^{{2}}$ or the $\Gamma _{{i\rightarrow f}}$ itself $\endgroup$
    – Sørën
    Commented Jun 28, 2018 at 20:15
  • 3
    $\begingroup$ The $\Gamma_{i\to f}$ is the probability of the transition per unit time. Dimensional analysis is your friend here: check units. $\endgroup$
    – rob
    Commented Jun 28, 2018 at 20:28
  • 1
    $\begingroup$ Related. $\endgroup$ Commented Jun 29, 2018 at 15:36

2 Answers 2

2
$\begingroup$

"Probability density" is nowhere to be found in the WP article you are link-quoting. I'll try to detail @rob 's recommendation that should have sufficed to answer your question, together with the article linked. Set $\hbar=1$ here for simplicity, easily reinstatable by elementary dimensional analysis. I sense the confusion might reside in the bra-ket notation used.

Recall states such as $|i\rangle$, etc are dimensionless... they are not dimensionful like $|x\rangle$ or $\psi(x)=\langle x|\psi\rangle$. So thinking about spacial probabilities and volume integrals is a canard: you never considered space integrals at the level of this treatment (but you may need them to produce the final matrix element). You just do time-dependent perturbation theory w.r.t. time-varying dimensionless coefficients $a(t)$.

In these units, then, the transition amplitude $\langle f|H'|i \rangle $ has dimensions of frequency, so, inverse time. Squared, it produces the decay rate to a single state k, with energy very, very close to that of i: $$\Gamma_{i\rightarrow k }= \frac{\mathrm{d}}{\mathrm{d}t} \left|a_k(t)\right|^2 \sim {2|\langle k| H'|i\rangle |^2} ~ t. $$ Integrated over a small time interval, it will give you the probability of your single state decay $i \rightarrow k$ during that interval. You know the drill of accounting for a collection of i s...

Putting lots of final states into the decay channels, their uneven contributions to the relevant integral will give you $$ \Gamma_{i \rightarrow f}= {2 \pi} \left | \langle f|H'|i \rangle \right |^{2} \rho , $$ where ρ, the density of states, has dimensions of time, since integration over its argument, frequency or energy, must net one (dimensionless). In total, the decay rate $\Gamma_{i \rightarrow f}$ has dimensions of inverse time, or energy (the width of plots versus energy). "Golden" refers to the rather surprising outcome that it is constant.

$\endgroup$
1
$\begingroup$

$\Gamma_{i \to f}$ refers in some sense to the probability per unit time of an initial state $i$, being found in the particular final state, $f$. The probability of $i$ transitioning to $\textit{some}$ state will be (hopefully) $1$, so $|\langle f|H'|i\rangle|^2$ is a "function" of the final state, such that summing (or integrating) over the final states will give you unity.

For example, you can have an incoming plane wave solution scattering off of a localized potential which is $0$ outside some radius (like a hard sphere). If a ping pong ball bounces off a bowling ball, it has a probability density of being directed in any direction $(\theta, \phi)$ (these are the final states - the momentum is usually uniquely determined by the direction for simple cases so the state itself can be thought of as just the direction). Integrating over a region of $\theta$ and $\phi$ values (final states) gives the probability (not the density) of the ping pong going in that direction.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.