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In tomography, we can use Pauli operators to estimate the qubit state, and by performing a substantial number of measurements one can estimate their expectation values. Define the estimates as $\bar\sigma_i$.

The expectation values of Pauli matrices are given by:

$$\mathrm{Tr} \left( \rho \,\sigma _{i} \right) =\left \langle \sigma _{i} \right \rangle \tag{1}$$

and we can write down the estimate of a single qubit state as:

$$\tilde \rho =\frac{1}{2}\left ( \mathbb{I} + \sum\limits_{i=1}^3 \bar{\sigma} _{i}\,\sigma _{i} \right) \quad. \tag{2}$$

The writer of the stated source in section 2.4.3, mentions that this technique might lead to states that are not physical. How might this technique entail such a consequence?

Here is the link to the reference.

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2 Answers 2

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TL;DR: The point is that the experimentally obtained mean values $\bar\sigma_j$ are in general not equal to the theoretical expectation values $\langle \sigma_j\rangle_\rho$ and hence your estimate of $\rho$ is in general not equal to it; in particular, it can happen that it is not a density operator at all.


Answer: Let's say we identically prepare $3N$ qubits each in a state described by $\rho$ and on each such system we measure one of the observable corresponding to $\sigma_1,\sigma_2,\sigma_3$; suppose further that each of these observables is measured $N$ times. From these measurements we can extract the average values

$$\bar \sigma_j(N) = \frac{1}{N}\sum\limits_{n=1}^N \sigma_j(n) \quad , \tag{1}$$

where $\sigma_j(n)$ denotes the $n$-th outcome of the measurement of $\sigma_j$, for $j=1,2,3$. The relation to the expectation value, determined by $\rho$, is given by

$$\langle \sigma_j\rangle_\rho:=\mathrm{Tr}\rho \,\sigma_j = \lim\limits_{N\to\infty} \bar\sigma_j(N) \tag{2} \quad .$$

As you can see, for a finite $N$, which is inherent to every experiment, we can only estimate the expected value by the average, i.e. for sufficiently large $N$ we can hope that $\bar\sigma_j(N) \approx \langle \sigma_j\rangle_\rho$.

Now the important point is that the operator $\tilde\rho$, estimated from the average values, might fail to be a density matrix. It is easy to show that it is a density matrix if and only if $$\sum\limits_{j=1}^3\bar\sigma_j(N)^2 \leq 1 \tag 3 \quad .$$

If, however, $\tilde \rho$ happens to be a density matrix but $\tilde \rho \neq \rho$, then there exists at least one observable corresponding to a hermitian operator $A$ such that $\mathrm{Tr}\rho A\neq \mathrm{Tr}\tilde \rho A$, cf. here. So the estimated density matrix $\tilde \rho$ yields wrong predictions. Whether or not this is experimentally verifiable or important depends on the exact situation, tho.

Finally, note that $\rho=\tilde\rho$ if and only if $\bar\sigma_j = \langle \sigma_j\rangle_\rho$, which holds in general only for $N\to \infty$, as can be seen by equation $(2)$.

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  • $\begingroup$ Do you mean that the obtained density matrix might not meet the features of a density matrix or not? or you mean, the density matrix is not equal to the real one. $\endgroup$
    – physicino
    Jan 3, 2023 at 20:44
  • $\begingroup$ because using non-physical state is a little bit controversial. $\endgroup$
    – physicino
    Jan 3, 2023 at 20:46
  • $\begingroup$ I got it completely. Thank you for your consideration. $\endgroup$
    – physicino
    Jan 3, 2023 at 21:03
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For a density matrix of the form $\rho =\frac{1}{2}\left ( \mathbb{I} + \sum \bar{\sigma _{i}}\sigma _{i} \right)$ to be physical the vector $(\bar{\sigma _{x}},\bar{\sigma _{y}},\bar{\sigma _{z}})/2$ needs to have norm smaller or equal to 1/2. This can be seen by first diagonalizing this matrix, and then noticing that if the norm of this vector is bigger than 1/2, then you will have negative eigenvalues (and therefore negative probabilities).

But how could a state tomography give such results? For example, consider the state $\rho = \mathbb{I}/2$, which is a valid state. Assume you only perform 3 measurements (which would be a horrible tomography), one in each direction, and all of them give $+1$ as a result. This is plausible, and has a probability $1/8$, since for each direction you have 50% chance to get $+1$. At this stage your state estimation is given by the vector $(\bar{\sigma _{x}},\bar{\sigma _{y}},\bar{\sigma _{z}})/2= (1,1,1)/2$, and this state would be unphysical, as the vector has norm bigger than half. You can verify that the resulting density matrix will have negative eigenvalues! Of course if you measure in all three directions infinitely many times, all the averages will converge to the correct ones, and then your resulting state will be physical, since the norm of $||(\bar{\sigma _{x}},\bar{\sigma _{y}},\bar{\sigma _{z}})/2|| \leq 1/2$ is imposed by physics.

Practically there is a non-zero probability that the tomography will give a non-physical state, but in general for a large amount of measurements the state will be close to a valid one, meaning if there is a negative probability it will be very small.

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  • $\begingroup$ OP seems to be confused by the wording of "physical" or "non-physical", which indeed is a bit confusing. So you might want to elaborate what a physical density matrix is. I guess the best is to avoid such phrasings at all. $\endgroup$ Jan 3, 2023 at 21:02
  • $\begingroup$ Wow. I understand it now. $\endgroup$
    – physicino
    Jan 3, 2023 at 21:04
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    $\begingroup$ @TobiasFünke Agreed. That's why I wanted to have an explicit mention for the negative eigenvalues. The word "physical" for density matrix is redundant imo, as density matrix already assumes it's positive and trace 1. This is the fault of the lecture notes imo, using a handwavy language such as nonphysical density matrices $\endgroup$
    – peep
    Jan 3, 2023 at 21:06
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    $\begingroup$ @peep The last paragraph is not entirely true. If you have a pure state, there will be a 50% percent probability, regardless of the number of samples taken, that you get an unphysical state. (The pure state is at the surface of the Bloch sphere, and a random whatever small error will have a 50% chance of taking you out of the sphere.) $\endgroup$ Jan 4, 2023 at 23:29
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    $\begingroup$ But what is relevant is whether the state is unphysical. The fact that it is "only slightly unphysical" doesn't make this any better: It is still unphysical. Which is indeed a real challenge people in tomography need to address, since ideally they want to prepare pure states, and then the question is what the best physical guess is, given a result $\bar\sigma$, which is unphysical with 50% probability. $\endgroup$ Jan 6, 2023 at 22:36

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