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In the paper Maximum Likelihood, Minimum Effort, given an orthonormal Hermitian operator basis $\{\sigma_i\}_{i=1}^{d^2}$ of $d \times d$ matrices and a set of measured values $m_{ij}$ corresponding to the $j$th measurement of the expectation value $\sigma_i$ applied to the true state $\rho_0$, the authors give the definition of a matrix $\mu$ as $$ \mu = \frac{1}{d}\sum_i m_i\sigma_i, $$ where $m_i = \sum_{j=1}^n \frac{m_{ij}}{n}$ and $n$ is the number of measurements. After defining this matrix, further optimization is done to approximate the density matrix of the true state.

I'm trying to implement this procedure for a single qubit as follows.

  1. To get $m_i$, I am first measuring the true state in the $X$, $Y$, $Z$ bases. With the counts for each circuit, I'm running the following code:
m = 0
try:
    m += count['0'] # measurement of |0> has eigenvalue of +1
except:
    pass
try:
    m -= count['1'] # measurement of |1> has eigenvalue of -1
except:
    pass
m /= shots
  1. Using the respective $m_i$, I'm calculating the matrix $\mu$ according to the formula as (where ms is a list with the three values calculated in step 1):
mu = (ms[0] * Z + ms[1] * X + ms[2] * Y) / 3.

However, after inputing the resulting matrix through the Fast algorithm for Subproblem 1 described later on in the paper, I'm not getting the expected results as I always get all the eigenvalues $\lambda_i$ set to $0$. I don't think there is any problem in my implementation of the Fast algorithm for Subproblem 1 as I tested it with the values given in Figure 1 of the same paper and I got the expected results.

Therefore, I suspect there is something wrong in my calculation of the matrix $\mu$. Is there something I'm missing or interpreted incorrectly?

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A quick thing before answering to the actual question: your first code snippet can be replaced by

m = (count.get('0', 0) - count.get('1', 0)) / shots

Now, if we follow the formulas you wrote, in $2$ dimensions (which is $1$ qubit), you should have $2\times 2 = 4$ projectors. So you are missing one projector. In fact, you are missing the identity. The set of matrices you should consider is given in the equation between equations 4 and 5 in the paper you linked:

$$ \left\{ \sigma_0, \sigma_1, \sigma_2, \sigma_3 \right\} = \left\{ I, X, Y, Z \right\}. $$

In your point 2., in the formula, the dimension $d$ is $2$. Adding the identity and fixing this typo gives:

mu = (I + ms[0] * Z + ms[1] * X + ms[2] * Y) / 2

Note that the factor in front of the identity is always $1$.

PS: when I learned about state tomography, I found this PDF to be quite useful to start.

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  • $\begingroup$ Thanks for the help! The code now works as expected. My misunderstanding was that I did considered to include the identity in mu but didn't figure out what should be the factor, thanks! And just one note, I think you meant to change 3 for 2 in the last codeblock (at least that way it works for me). $\endgroup$
    – epelaaez
    Aug 10 at 13:00
  • 1
    $\begingroup$ Right, I just fixed the issue, I copy-pasted your code and forgot to change the 3 to 2, thank you! $\endgroup$ Aug 10 at 13:03

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