It is well known that a quantum computer is reversible. This means that it is possible to derive an input quantum state $|\psi_0\rangle$ from an output $|\psi_1\rangle$ of an algorithm described by a unitary matrix $U$ simply by applying transpose conjugate to $U$, i.e.

\begin{equation} |\psi_0\rangle = U^\dagger|\psi_1\rangle \end{equation}

In article Arrow of Time and its Reversal on IBM Quantum Computer an algorithm for a time reversal and going back to an input data $|\psi_0\rangle$ is proposed. Steps of the algorithm are following:

  1. Apply a forward time unitary evolution $U_\mathrm{nbit}|\psi_0\rangle = |\psi_1\rangle$
  2. Apply an operator $U_\psi$ to change $|\psi_1\rangle$ to $|\psi_1^*\rangle$, where the new state $|\psi_1^*\rangle$ is complex conjugate to $|\psi_1\rangle$
  3. Apply an operator $U_R$ to get "time-reversed" state $|R\psi_1\rangle$
  4. Finally, apply again $U_\mathrm{nbit}$ to obtain the input state $|\psi_0\rangle$

According to the paper, the algorithm described above simulates reversal of the time arrow. Or in other words, it simulates a random quantum fluctuation causing a time reversal.

Clearly, when the algorithm is run on a quantum computer, it returns back to initial state but without application of an inverse to each algorithm step. The algorithm simply goes forward.

My questions are these:

  1. Why it is not possible to say that an application of $U^\dagger$ on output of algorithm $U$ is reversal of time arrow in general case?
  2. It is true that above described algorithm returns a quantum computer to an initial state but it seems that the algorithm simply goes forward. So where I can see the a reversal of time arrow?
  3. The authors of the articles have found out that when a number of qubit involved in the time reversal algorithm is increasing, the effect of time reversal diminishes:

    • How is it possible to reverse time for few qubits and concurently to preserve flowing of time in forward direction for another qubits?

    • Does this mean that time flows differently for different qubits?

    • When do the qubits return to commnon time frame to be possible to use them in another calculation?

  • $\begingroup$ The arrow of time is a statistical concept (a law of large numbers really). It's not a meaningful notion at the level of qubits. See Emilio Pisanty's answer here. 2 and 3 are non-questions in the sense that they're based on faulty premises and misunderstandings of the notion of the time arrow as a thermodynamic concept. $\endgroup$ Jan 4, 2020 at 7:46

1 Answer 1


Of course if we have unitary evolution $$|\psi_1\rangle = U|\psi_0\rangle$$ then $$|\psi_0\rangle = U^\dagger|\psi_1\rangle$$

I did not read the paper, but evidently the authors do something different, based on the following: the Schrödinger equation $$i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi$$ changes its form if we substitute $t\rightarrow -t$ to complex conjugate: $$-i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi$$ and its solution is also complex conjugate.

So time reversal is antiunitary operator $U_RK$ where $U_R$ is unitary and $K$ is complex conjugation.

  • $\begingroup$ Thanks. However, could you (or anybody else) please answer my other questions (no. 2 and 3)? $\endgroup$ Dec 27, 2019 at 20:50

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