We can use time dependant perturbation theory (specifically Fermi's Golden Rule) to calculate the transition rate (probability of transition per unit time) from one energy eigenstate to another. However, it seems to me that a 'transition' from one state to another can only occur if the system is measured and happens to collapse onto the desired final state. If no measurement occurs, the state can never truly transition - it will simply be in a superposition of initial and final states.

What kind of measurement is taking place (what Hermitian operator are we considering)? How often is this measurement taking place? And if this measurement is taking place regularly, why do we never consider its effects on the dynamics, given that each measurement should collapse the state vector in some way? I suppose different experiments will involve measuring different things, but any example would be helpful.


There is no measurement. "Transition" means the possibility of finding the system in a different state than it was prepared in.

You say it yourself; Fermi's Golden Rule calculates the

probability of transition per unit time

  1. Prepare the system in an eigenstate of the unperturbed Hamiltonian $H_0$ $$\Psi(t=0) = a_i \psi_i $$

  2. Now switch on a perturbation $V(t)$. Since the eigenstates of $H_0$ form a basis, you can expand the state as $$ \Psi(t) = \sum_k a_k(t) \psi_k $$

  3. As by the rules of QM, the modulus squared of the coefficients are the probabilities to find the system in the state $\psi_k$ upon measurement.

  4. Fermi's Golden Rule is an approximate expression for the $|a_k|^2$ given that the system started out in the state $\psi_i$ $$ |a_k|^2 \sim |V_{fi}|^2 t $$ and hence the transition rate (change in probability per time) is $\lambda_{i\rightarrow f} \sim |V_{fi}|^2 $

No measurement takes place. The system evolves unitarily under $H_0 +V$ into a superposition. Without the perturbation a measurement would have yielded $\psi_i$ with certainty, but now there is a non-zero chance of finding the system in a different state. That is meant by making a transition.

  • 1
    $\begingroup$ Thanks for the answer. I'm familiar with points 1-4, what I want to understand more is what you mean by 'finding the system in a different state'. Presumably this means some kind of measurement has been performed? What observable is being measured and how often do we measure it? I'd be happy with any reasonably simple experimental example. $\endgroup$
    – Andrew
    Sep 17 '16 at 7:32

In clarification of your comment (which I can't comment to yet): the type of measurement does not matter per se, it could be a flowing electrical current or emission of a photon for example.

I think you misunderstand the term transition. If there is no transition, all and any of your measurements will always yield the initial state. To give you a clearer picture: If you prepared an atom or molecule in an energetically excited state and forbade any transitions, you would always measure the excited state, independently of your type of measurement.

Now, if transitions are allowed, fluorescence could occur, or intersystem crossing (change from singlet state to triplet state), or non-radiative recombination (conversion to heat). Each of these happens with different transition probabilities after a certain time, and the time-evolution can be calculated with Fermi's Golden Rule. Therefore, after a picosecond you could for example get 90% initial state, 9.8% fluorescence, 0.19% ISC and 0.01% heat. After a nanosecond this could change to 10% initial state, 88% fluorescence, 1.5% ISC and 0.5% heat. Of course you would have to measure thousands of excited states to get these numbers, that is the core principle of quantum mechanics. Any single measurement could yield any of them.

To be more specific, you can measure fluorescence with a photo-diode, ISC with magnetic hyperfine-splitting measurements, heat with a thermoelement, etc and (if you calibrate them right) all of these measurements would add up to a 100%.

Of course, as long as you don't measure, the system is still in a superposition of all the states, but the word 'transition' describes how the probabilities of the individual states change over time (chance of initial state goes down while the others go up).

Does that picture help you?


Andrew, you are right: when a probability is used in quantum mechanics, it is ultimately always in the context of a measurement.

After the unitary evolving of the initial state into the timedepending superposition state there is no collapse (at time t), without the further incredient called "measurementdevice".

To make a truly transition into a collapsed final state possible, you have to open the system including the perturbation for this device or environment of the system.

The transitions in the fermirule are only possible for systems that are open for the environment. The environment is the reason for the collapse of the superposition state into a final one with a probability not equal zero.

The perturbation is only a necessary condition for a transition. The (perturbation+environment) is the sufficient condition for the transition. If there is no physical environnment (system and perturbation are alone in the universe) then there is no collapse and no transitionrate.

So, you are right beeing cautious calling the quantity calculated by the fermirule a "transition rate" without a measuring device.

As measuring device you can take a detector that counts the transitions per time.

For details have a look to "decoherence".


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