$\newcommand{\ket}[1]{\left|#1 \right\rangle}$ $\newcommand{\bra}[1]{\left\langle #1 \right|}$

For a physical process in a mesoscopic/macroscopic system, how exactly can one deduce the time that process takes?

In the microscopic world, quantum systems transition from one state $\ket{\psi (t_i)}$ to another $\ket{\psi(t_f)}$ according to the time evolution operator $T(t)=\exp(-iHt/\hbar)$. Therefore, the time for any process to occur, $\Delta t = t_f-t_i$, only depends on the Hamiltonian $H$. Moreover, $\Delta t$ is not determined, only transition rates/amplitudes $\bra{\psi(t_f)} \ket{\psi(t_i)}$ are given by the theory - enabling us to make probabilistic predictions.

When we enlarge the system to mesoscopic or even macroscopic size, further complication is introduced as we no longer can keep track of all microstates: the description of the system employs thermodynamics/statistical mechanics with probability theory and average quantities such as half-life $t_{1/2}$ (in the case of decays).

On the whole, it would thus seem that time intervals are not sharply determined. Yet in the "real/classical" world we measure time with clocks and the same processes always seem to take the exact same time (chemical reactions, mechanical systems, etc.).

Where does that precise time come from? Are those sharp intervals only understandable in some thermodynamic limit?

  • 1
    $\begingroup$ "But quantum processes are intrinsically random..." no, they aren't. $\endgroup$
    – DanielSank
    Commented Feb 8, 2015 at 18:34
  • $\begingroup$ Well the time evolution certainly is deterministic, but because of measurement/decoherence a stochastic element is introduced into our description. Let us please not get into an argument about different interpretations of QM - that is not the point of my question. $\endgroup$ Commented Feb 8, 2015 at 18:41
  • $\begingroup$ The "duration" of a physical process needs to be looked at for every individual process to find a satisfying definition for duration. As such, this question is too broad. $\endgroup$
    – ACuriousMind
    Commented Feb 8, 2015 at 18:45
  • $\begingroup$ I would define duration as the time it takes from one point/vector in the phase/state space to another. That should apply to all physical systems. $\endgroup$ Commented Feb 8, 2015 at 18:48
  • $\begingroup$ I wasn't trying to start an "interpretations" discussion. I don't those discussions are scientific. It's just that if you start out assuming quantum processes are random you'll confuse yourself. You mentioned decoherence/measurement. The answer to your original question lies in understanding the dynamics of those processes. They don't just happen magically. Decoherence arises when a system couples to some other degrees of freedom which you aren't keeping track of. The speed of the process is basically just the coupling strength between the system and those other degrees of freedom. $\endgroup$
    – DanielSank
    Commented Feb 8, 2015 at 20:10

1 Answer 1


Where does that precise time come from? Is it just the energy / Hamiltonian $H$ that determines it?


$\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1 |}$ Suppose I have an atom which can be in one of two states $\ket{0}$ or $\ket{1}$. Suppose that atom is coupled to another atom via an interaction Hamiltonian like this

$$H_I = g(\sigma_+ \sigma_- + \sigma_- \sigma_+) \,.$$

In the basis $\{ \ket{00}, \ket{01}, \ket{10}, \ket{11}\}$ the matrix representation of $H_I$ is

$$H_I = g \left( \begin{array}{cc} 0&0&0&0 \\ 0&0&1&0 \\ 0&1&0&0 \\ 0&0&0&0 \end{array} \right) $$

which is just $\sigma_x$ on the $\ket{10}, \ket{01}$ subspace. The time evolution on that subspace is therefore

$$ U(t) = \exp\left[ -i \frac{tg}{\hbar} \sigma_x \right] = \cos(\omega t) \mathbf{I} - i \sin(\omega t) \sigma_x $$ where $\omega \equiv g/\hbar$ and $\mathbf{I}$ denotes the identity operator.

Now suppose that I actually only observe the first atom. In other words, the second atom is part of the "environment". In this case the information I have on my atom is represented as a density matrix. To compute that density matrix, I form the density matrix of the entire system and then trace over the atom I don't have under my control. Let's do that.

Suppose we start in state $\ket{10}$, i.e. our atom is excited and the "environment" atom is not. Using the $U(t)$ we computed above, the state that we have after time $t$ is

\begin{align} \ket{\Psi(t)} &= [\cos(\omega t) \mathbf{I} - i \sin(\omega t) \sigma_x]\ket{10}\\ &= \cos(\omega t) \ket{10} - i \sin(\omega t)\ket{01} \, . \end{align}

The density matrix is therefore

\begin{align} \rho(t) &=\ket{\Psi(t)}\bra{\Psi(t)} \\ &= \cos(\omega t)^2 \ket{10}\bra{10} + \sin(\omega t)^2 \ket{01}\bra{01} \\ &+ i\cos(\omega t)\sin(\omega t) \ket{10}\bra{01} - i \cos(\omega t) \sin(\omega t) \ket{01}\bra{10} \,. \end{align}

The information of just the atom we care about is found by doing the trace over the other atom. To notate that I'll write $\bra{x0}\rho\ket{x0}$ to indicate that we're taking a sandwich with respect to only the second atom's state. In other words, $x$ means "leave that part of the state alone.

\begin{align} \rho_\text{our atom}(t) &= \text{Tr}_\text{other atom} \rho(t) \\ &= \bra{x 0} \rho(t) \ket{x 0} + \bra{x 1}\rho(t) \ket{x 1} \\ &= \cos(\omega t)^2 \ket{1}\bra{1} + \sin(\omega t)^2 \ket{0}\bra{0} \, . \end{align}

At $t=0$ our atom's density matrix is $\ket{1}\bra{1}$. This is a pure state with the atom excited. At slightly longer times, by expanding the $\sin$ for small $t$ our atom's density matrix is (not normalized)

$$\ket{1}\bra{1} + (\omega t)^2 \ket{0}\bra{0} \, .$$

This is a mixed state, which means that the atom has a classical probability of having decayed from $\ket{1}$ to $\ket{0}$. The probability of the decay is just

$$p_\text{decay} \propto (\omega t)^2 = \left(\frac{gt}{\hbar}\right)^2 \, .$$

So you can plainly see that the strength of the coupling term in the Hamiltonian dictates how fast the decay happens. Of course, it's still stochastic (what we computed here is a decay probability), but the point is that the probability grows in time faster if the coupling to the environment is larger.

Decay and transition between energy levels happen without any apparent cause - that is what I mean by "intrinsically random".

I hope in looking at the example in this answer you now see that this is perhaps not the right way to look at it. The appearance of transition probabilities comes purely out of the fact that we ignore the environment degrees of freedom. When you do that, the information you're left with for the system under observation takes on a (partially) classical nature with transition probabilities. It doesn't happen "without any apparent cause", it happens because $H_I$ makes it happen!


An astute reader will have noticed that the process analyzed here isn't really a decay in the usual sense. From the $U(t)$ we computed, you can see that after one period $T = 2\pi / \omega$ the excitation will come back from the "environment" into our atom. It's only in the small time limit that the process appears like a decay. In fact, even the time dependence is weird. In what we usually call "decay", the probability of the transition is proportional to time in the small time limit. This leads to the canonical exponential decay processes. However, in our example above it was proportional to the square of the time. The reason for this is that our "environment" in the example is too small. The system has a simple cyclic nature because of the small number of degrees of freedom. In real life where you have environments with a gazillion degrees of freedom what happens is that the excitation kind of spreads out among the environment and never manages to cycle back into the atom (or whatever else) you're looking at. When that happens you wind up with exponential decays.

  • $\begingroup$ Your answer helped me a lot to understand one part of my original question, thank you! I reworded the remaining part of my question because somehow the community thought it was too broad. $\endgroup$ Commented Feb 9, 2015 at 12:45
  • $\begingroup$ @quantumorsch: Does the bonus section of my answer address your question or are you looking for something more? Do you need an explanation about how macroscopic systems average over many quantum events to give the appearance of the macroscopic rate which is always the same? $\endgroup$
    – DanielSank
    Commented Feb 9, 2015 at 17:02
  • $\begingroup$ Yes, I would definitely appreciate that - thank you! The classical decay rate you derived is still a probability and I dont quite know how to get the thermodynamic limit for macroscopic systems (and as a consequence sharp intervals). $\endgroup$ Commented Feb 9, 2015 at 17:44
  • $\begingroup$ @quantumorsch ok, well I voted to re-open the question. $\endgroup$
    – DanielSank
    Commented Feb 9, 2015 at 17:54

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