Suppose say that I have a quantum state $\vert\psi\rangle$ at time $t = 0$, which is now evolved by a hamiltonian $H$

$$e^{-iHt}\vert\psi\rangle$$. I can ask the question, how much of initial state is still available in the evolved state, this is called survival probability.

$$P(t) = \vert\langle \psi\vert e^{-iHt}\vert\psi\rangle\vert^2$$. Is there a circuit implementation for this survival probability? Any idea how to implement this or source to any link which discusses about this circuit implementation is appreciable.

  • 1
    $\begingroup$ One may approximate the time evolution operator $e^{-iHt}$ using a quantum circuit, however, further assumptions on $H$ are usually needed. The keyword is "Hamiltonian simulation". If you're interested in a specific (type of) Hamiltonian, I recommend that you specialize your question such that a concrete answer is possible. Otherwise, I suggest that you read up a bit on the literature. arxiv.org/abs/2101.07808 contains a decent overview over the available methods (but there is no review paper AFAIK) $\endgroup$ Commented Dec 6, 2022 at 9:50
  • $\begingroup$ @MarkusHeinrich I was actually thinking of Trotterizing the evolution part (assuming the evolution is Trotterizable) and do a swap test between initial state and evolved state, performing a proper measurement on ancilla qubit, i can find the mod square of overlap between these states. But I don't know whether this will work. Just wondering, is there a more efficient way to find this probability. I am trying to simulate Rabi oscillations using quantum circuit, lets say. $\endgroup$ Commented Dec 6, 2022 at 10:01
  • $\begingroup$ Is $\psi$ unknown or why do you want to do a swap test? You would also need two copies for each run. If $\psi$ is known and a "simple" state, I'd rather rotate the basis measurement accordingly. Anyway, without more details, you may not get a useful answer. $\endgroup$ Commented Dec 6, 2022 at 12:04

1 Answer 1


We can calculate this survival probability $P(t) = |\langle \psi | e^{-i \mathcal{H}t} | \psi \rangle|^2$ in two ways. The first corresponds to the so-called Hadamard test, as shown in the figure below,

enter image description here

where $H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ is the Hadamard gate and $|\psi \rangle$ is some $n$-qubit state. The probability of measuring the ancillary qubit in state $|0\rangle$ is $\frac{1 + |\langle \psi | e^{-i \mathcal{H}t} | \psi \rangle|^2}{2} \equiv \frac{1 + P(t)}{2}$.

Alternatively, we can avoid the use of an ancillary qubit and the corresponding control of the time evolution provided that we have access to the circuit $U$ such that $U |0\rangle = |\psi \rangle$. In such case, we can simply use the following ancilla-free circuit

enter image description here

By measuring all $n$ qubits, the probability of measuring all qubits in state $|0 \rangle$ (i.e., retrieving the fiducial state) is $|\langle 0 | U^{\dagger} e^{-i \mathcal{H} t} U | 0 \rangle|^2 = | \langle \psi | e^{-i \mathcal{H} t} | \psi \rangle |^2 \equiv P(t)$.

Of course, the remaining question is how to implement the time-evolution operator $e^{-i \mathcal{H}t}$, which is the Hamiltonian Simulation problem mentioned by @MarkusHeinrich. The simplest option is Trotterization, but more sophisticated approaches such as truncated Taylor series with linear combination of unitaries method or qubitization are also available.

  • $\begingroup$ Yeah, I already figured out the way and added them in comments. Thanks for the answer. $\endgroup$ Commented Dec 6, 2022 at 18:28

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