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
    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. $\endgroup$ Jun 19, 2018 at 6:15
  • $\begingroup$ @AdamZalcman are you just saying we need to indicate the support of $P_x$? $\endgroup$
    – DanielSank
    Apr 25, 2019 at 22:28

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