This problem came up in the course of thinking about the statistics of the dispersive measurement signal coming from a superconducting qubit. Such qubits have finite excited state lifetimes, typically characterized by an exponential decay parameter $T_1$, which is the mean survival time of the excited state.


Consider a physical system with two states $G$ and $E$. We have a measurement device whose output at any particular time, which we denote $v(t)$, is $+v$ if the system is in $E$ and $-v$ is the system is in state $G$.

The system also has a chance to jump states:

  • If the system is in state $E$, then there is a certain probability per time $T_1$ that it jumps to state $G$.

  • If the system is in $G$, it stays there.

Suppose we observe the system for some total time $T$ and construct the quantity $$x = \int_0^T dt \, v(t) \, .$$ What is the probability distribution of $x(T)$ in the case that the system starts in state $E$?


Consider a case where the system jumps from $E$ to $G$ at time $t$. In this case, we have $$x = \int_0^T v(t) \, dt = \int_0^t v \, dt + \int_t^T -v \, dt = t v + (T - t)(-v) = (2t - T) v \, . \tag{1}$$ Therefore, we need only find the distribution of $t$.

Following the reasoning in this other answer, we can show that the probability density for the system to stay in $E$ until time $t$ and then transition to $G$ at time $t$ is $$P_t(t) = \frac{1}{T_1} e^{-t / T_1} \, .$$ $P_t$ is normalized for $t \in [0, \infty ]$, but our maximum time is $T$, so we have to adjust the normalization. Let's ignore that though, because we can always compute the normalization factor at the end.

The value of $x$ is related to $t$ through a deterministic linear transformation, i.e. Eq. (1). Therefore, we can use standard rules for transforming probability distributions to find the distribution of $x$. The result should be $$P_x(x) = \mathcal{N} \exp \left( -\frac{x/v + T}{2 T_1} \right) \qquad x \in [ -vT, vt ] \tag{2}$$ where $\mathcal{N}$ is the normalization factor.

I don't like that $P_x$ increases as $x$ decreases. Intuitively, if the system starts out in $E$, it should be more likely that the system stays there, causing our measurement device to read $v(t) = +v$ and therefore leading to positive value of $x$.

Furthermore, consider the limit $T_1 \to \infty$. In that limit, the system never switches state, so if we prepare it in $E$ it will stay there forever, leading to $x = vT$ with probability unity.

It seems, therefore, that the expression in Eq. (2) must be incorrect. I think the mechanics of the calculation are correct, so there must be a conceptual mistake somewhere.

  • $\begingroup$ Note to self: the problem is in the assumption that we can account for the range of possible $t$ in $P_t$ entirely by adjusting the normalization factor. That's wrong. In cases when the system doesn't jump (which happens $F = \exp(-T/T_1)$ fraction of the time), the value of $x$ is as if the system jumped at $t=T$. Therefore, $P_t$ needs a an extra delta function term $F \delta(t - T)$. $\endgroup$
    – DanielSank
    Commented Jun 12, 2018 at 6:33

1 Answer 1


The time $t$ at which the system switches from $E$ to $G$ is distributed as $P_t(t) = (1 / T_1) \exp(-t / T_1)$, as noted. But we can't just truncate this distribution at the measurement time $T$, rather for all switching times greater than $T$, we get the maximum measurement result $x = x_\textrm{max} \equiv vT$; this occurs with total probability $P_\textrm{max} \equiv \int_t^\infty P_t(t) dt = \exp(-T/T_1)$. This probability is entirely concentrated on the single outcome $x_\textrm{max}$, so, indeed, if we want to write a probability density function for $x$, it will appear as a delta function: $$ P_x(x) = \frac{1}{2 v T_1} \exp(-t(x) / T_1) + P_\textrm{max}\delta(x-x_\textrm{max}) $$ where $t(x) = (x/v + T)/2$. This indeed matches the expected behavior that as $T_1 \rightarrow \infty$ the distribution becomes $P_x(x) \rightarrow \delta(x-x_\textrm{max})$ and we always get the maximum measurement result.

  • $\begingroup$ There is a small issue with the Px(x) density as currently written: it's only valid for x in [-vT, +vT], e.g. for x=2vT one would expect Px(x) to be zero, but the formula yields a positive density. Another way to think about it is to realize that the delta function term originated by collecting the entire density of [+vT, +inf] and placing it at x=+vT. Thus, Px(x) must be zero on [+vT, +inf]. This also gives us normalization. $\endgroup$ Commented Jun 19, 2018 at 6:15
  • $\begingroup$ @AdamZalcman are you just saying we need to indicate the support of $P_x$? $\endgroup$
    – DanielSank
    Commented Apr 25, 2019 at 22:28

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.