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This question follows the "Probabilistic Simulation of Quantum Circuits with the Transformer" paper by Carrasquilla et al. In the Formalism section on page 2 the authors state that probability of getting a measurement vector $a$, the $P(a) = Tr[M^{(a)}\varrho]$ is equal to the product of probabilities $P(a_1)P(a_2)...P(a_N)$ if $|\psi\rangle = \prod_{i}\otimes|\psi_i\rangle$ . Here, $N$ is the number of subsystems and $a$ is a measurement result (for qubits, $a_i \in \{0,1\}$, if I understand it correctly) and $P(a_i)=Tr[M^ {(a_i)}|\psi_i\rangle\langle\psi_i|]$. From these considerations, the authors proceed to the equation (1), showing how to estimate $\varrho$ from the measurement statistics:

$$ \varrho = \sum_{a,a^\prime} P(a^\prime)T^{−1}_{a,a\prime}M^{(a)} $$ where $$ T_{a,a\prime} = Tr[M^{(a)}M^{(a\prime)}] $$

My questions are the following:

  1. Could you please explain the math behind the $\varrho$ formula? What do authors call Trace in the $T$ formula? Is it a partial trace? What space do we trace-out?
  2. Is it still valid if $|\psi\rangle$ is not a product, i.e. $\neq \prod_{i}\otimes|\psi_i\rangle$ ? How do authors expect to deal with entangled states?
  3. A question regarding the measurement operator $M$. Does it make sense to talk about "entangled" multi-qubit measurements which are not tensor products of single-qubit measurements $M^{(a)}$ ?
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  • $\begingroup$ Thanks to Carlo Beenakker's answer and the author's previous paper, I've realized that $T_{a,a^\prime}$ denotes the component $[a,a^\prime]$ of the matrix. – Grwlf 9 mins ago $\endgroup$
    – Grwlf
    Aug 30, 2022 at 11:35

1 Answer 1

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To build intuition for the reconstruction formula of the density matrix (known as quantum state tomography), it helps to consider the case that the operators $M^{(a)}$ are projection operators, $M^{(a)}=|\Psi^{(a)}\rangle\langle\Psi^{(a)}|$. Here $a=(a_1,a_2,\ldots a_N)$ is a string of single-qubit measurement outcomes, and $\Psi^{(a)}=\psi_{a_1}\otimes\psi_{a_2}\otimes\cdots\otimes\psi_{a_N}$ is the corresponding product state. We assume that the $\Psi^{(a)}$'s form an orthonormal set spanning the Hilbert space.

In this case the overlap matrix $T$ is the unit matrix and the formula for $\rho$ is the usual representation of a mixed state as a convex combination of pure states, $$\rho=\sum_{a}P(a)M^{(a)}=\sum_aP(a)|\Psi^{(a)}\rangle\langle\Psi^{(a)}|.$$ In the more general case the overlap matrix $T$ accounts for non-orthogonal $M^{(a)}$'s.

Turning now to the specific questions:

  1. The trace in the definition of the overlap matrix is a full trace over the Hilbert space.
  2. The formula for $\rho$ applies to any state, pure or mixed, entangled or not. The choice for a product state $\Psi^{(a)}$ is only the choice for a particular basis.
  3. The set of measurement operators $M^{(a)}$ need not be constructed as a product of single-qubit measurements. The formula holds for any set of positive semi-definite operators that sum to the identity.
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