A quantum cloud server like from IBM typically returns a string of 0's and 1's such as 0010.

An observable such as $Z_1 \otimes Z_2 ... \otimes Z_{n}$ has two eigenvalues +1 or -1. How do I relate this to the 0010 output? Note that there is no one-to-one correspondence (it is many-to-one).

I recall here that eigenvalues of an operator are the possible results of the measurement.

I started reasoning as follows. The observable in the case of a quantum program (a Projective measurement, see Sec 2.2.5 of Mike and Ike) is given by:

$P = \sum_m m P_m$

where $m$ is an integer between $0$ to $2^n-1$ and $P_m$ is the corresponding operator such as |$0011><0011|$.

Now, if I think about the register as doing a measurement for the following observable (a Hermitian matrix):

$ Z = Z_0 \otimes Z_1 ... \otimes Z_{n-1}$

then the eigenvalues for such an operator are just two in number: $+1$ and $-1$ and hence does not retain the dimensionality information.

Thus I ended up with the conclusion that there is no link: P and Z are two entirely different observables whose eigenvalues cannot be linked (conceptually or mathematically). In particular, P is the right way of mathematically thinking about the register output of a quantum program and not Z.

Am I thinking right? I have a nagging feeling I am missing something here and it might be possible to relate these two pictures.

Many thanks for your time.

(p.s There is another question that preceded this question which can be found here: How are measurements done exactly on IBM cloud computers?)

  • $\begingroup$ Are you sure that the measurement you're doing is $Z_0\otimes Z_1\otimes\ldots\otimes Z_{n-1}$ as compared to $n$ separate measurements $Z_0$, $Z_1$, ..., $Z_{n-1}$? $\endgroup$
    – DaftWullie
    Commented Oct 28, 2022 at 8:16
  • $\begingroup$ Thank you DaftWullie. This small comment clarifies a tremendous lot! I believe the program output is being measured separately. But should I not be doing it together (the tensor version)? Because if suppose the output of the circuit is m, then it is obtained by the action of the projector $P_m$ which is a tensor product (as per the notations in my question above). p.s. I do not understand the physical realization of measuring "together" but that is my other question in a separate thread (as mentioned in my query above). Thank you. $\endgroup$
    – Sam
    Commented Oct 28, 2022 at 8:44

2 Answers 2


The single measurement $Z_0 \otimes Z_1 \otimes \dots \otimes Z_{n-1}$ limits the amount of information you can extract about the state because as you've realized, that only gives you 1 bit's worth of information. As DaftWullie mentioned, you can extract more information with the $N$ measurements of $Z_0 (\otimes I_1 \otimes \dots \otimes I_{n-1}), Z_1, \dots Z_{n-1}$.

You're able to take these multiple measurements because the qubits are spatially separated so you can read out their states individually. For example, the state of spin defect qubits in solids and trapped neutral atom qubits are generally read-out by pumping each qubit with a particular laser and then measuring the amount of photo-luminescence that's produced (as either $|0\rangle$ or $|1\rangle\}$ is a brighter state in such a protocol)

  • $\begingroup$ Thank you so much Chris. I was thrown off by the following identity/: $Z_1 Z_2 = (Z_1 \otimes I_2) (I_1 \otimes Z_2)$ The LHS if I measure will give me +1 or -1. "The RHS represents sequential measurements in time." (Incorrect) This last statement I think is where I made the error. The RHS is the same as LHS. What we are doing in fact is different. We measure $Z_1 \otimes I_2$ then write down the result.; then measure $I_1 \otimes Z_2$ and write down the result. I wonder if there is a mathematical way of writing this measurement procedure that captures the "pause & write" operation. $\endgroup$
    – Sam
    Commented Oct 29, 2022 at 5:24
  • $\begingroup$ Sure. Using the POVM formalism you gave earlier, you could write it mathematically as $P = \sum_{m_0,\dots,m_{n-1}} (m_0\dots m_{n-1}) P_{m_0}\otimes\dots \otimes P_{m_{n-1}}$, where $(m_0\dots m_{n-1})$ is the measured output in binary $\endgroup$
    – Chris E
    Commented Oct 29, 2022 at 7:51
  • $\begingroup$ Thank you Chris! But how does this capture "pause and write"? It still uses the same tensor form analogous to $Z_0 \otimes \cdots \otimes Z_n $. In this case, it simply becomes $ |m_0><m_0| \otimes \cdots \otimes|m_{n-1}><m_{n-1}| $. Could you elaborate a bit more as it is still not obvious to me? Thank you. $\endgroup$
    – Sam
    Commented Oct 30, 2022 at 7:13
  • $\begingroup$ Ah, good question that I didn't explain super well. Let's stick with projective measurements for simplicity. For the "pause and write" set, we can break up our measurement set for each qubit. Explicitly, the $j$th qubit has $\{I_0\otimes I_1 \otimes \dots \otimes P_j \otimes \dots \otimes I_{n-1}\}$ where $P_j$ is $|0\rangle \langle 0|_j$ for $Z$ eigenvalue $+1$ or $|1\rangle \langle 1|_j$ for $Z$ eigenvalue $-1$ . Taken all together, these give the set of projectors of the form $P_{m_0} \otimes \dots \otimes P_{m_{n-1}}$. $\endgroup$
    – Chris E
    Commented Oct 31, 2022 at 4:41
  • $\begingroup$ For the total set where $Z_0 \otimes \dots Z_{n-1}$ are measured together, there are only two projectors: the sum of all of the previous projectors with an even number of $|1\rangle \langle 1|_j$s for a $+1$ eigenvalue, and the sum of those with an odd number of $|1\rangle \langle 1|_j$s for a $-1$ eigenvalue. The difference between these sets is that since you get don't get information from every qubit in the second case, you don't fully collapse the superposition of the wavefunction like in the first "pause and write" case. Does that make sense? $\endgroup$
    – Chris E
    Commented Oct 31, 2022 at 4:54

I want to answer my question regarding the link between $\sum mP_m$ and measuring $Z_0 \otimes \cdots Z_{n-1}$. The link can be understood from Exercise 10.3 in in Chapter 10 of Mike and Ike. Using a three qubit example, there we show that measuring $Z_1 \otimes Z_2$ followed by $Z_2 \otimes Z_3$ is equivalent to measuring the four projectors $P_0, P_1, P_2$ and $P_3$.


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