# Why does measurement in computational basis result in classic probability?

I am working on a problem related to finding the limits on the joint probability distributions/correlations of three or more quantum systems who share entangled states, after measurement.

I have been told that measurement in the computational basis is equivalent to having a classical probability distribution and is in fact not quantum at all. But I don't understand why this is the case. What other basis do I need to consider? Or if I need to have a general representation of measurement in an arbitrary basis what should that be like?

• I find the assertion about the computational basis to be surprising. There is no reason to prefer one basis over another in this sense. Perhaps using an entangled basis gives interesting results Commented Nov 25, 2021 at 14:57
• Not sure I really understand the question. What do you mean by the distribution is not quantum? Commented Nov 25, 2021 at 15:25
• That was part of my question too!
– Pegi
Commented Nov 26, 2021 at 13:02

The ideal probability of obtaining some measurement result $$i$$ (with associated measurement operator $$M_i$$) from some state $$\rho$$ is $$Tr(M_i \rho) = p_i$$. For the computational basis, $$M_0 = |0\rangle\langle 0|$$ and $$M_1=|1\rangle\langle 1|$$.

If you can only measure in the computational basis, consider trying to tell whether you have the state $$\rho_0 = |+\rangle\langle+|$$ or $$\rho_1 = \frac{1}{2} I$$. Let $$p_{ij}$$ denote the probability of outcome $$i$$ in state $$j$$. It is easy to show that $$p_{00} = p_{01} = p_{10} = p_{11} = \frac{1}{2}$$, which is to say that $$\rho_1$$ and $$\rho_2$$ are functionally equivalent as far as you can tell. If you cannot distinguish a superposition (quantum) from a mixed state (classical), then your quantum states are not any more useful to you than classical states that happen to have the same outcome probabilities w.r.t. your measurement system.

Instead, if you can measure in the $$x$$-basis (so $$M_0 = |+\rangle\langle +|$$ and $$M_1 = |-\rangle\langle -|$$), you would find that $$p_{00} = 1, p_{10} = 0$$ and $$p_{01} = p_{11} = \frac{1}{2}$$. By measuring in other bases, we can properly distinguish superpositions from mixed states. Another way to think about it is that quantum states are points lying on/in the Bloch sphere. If we want to distinguish one point from another, we need to know all three coordinates, which correspond to the expectation values of the operators $$Z,X,Y$$, each of which define a two-element POVM.

The measurements we use when dealing with the Bloch sphere are typically projector-valued measurements, called PVMs. You can think of a PVM as picking some pure state on the surface of the Bloch sphere and defining it to be "outcome 0", and the state antiparallel to it will be "outcome 1". PVMs are a special case of positive operator-valued measures (aka POVMs) which are even more general, since they need only refer to a set of positive semidefinite matrices that sum to the identity. Here are a few previous answers that might be useful as well:

• So is that why only using the computational basis measurement doesn't give us a quantum correlation?
– Pegi
Commented Nov 26, 2021 at 13:03
• "Quantum correlation" could mean a couple of different things, but to me the phrase "measurement in the computational basis is equivalent to having a classical probability distribution" is probably referring to the fact that the main diagonal of a density matrix defines a classical probability distribution over the computational basis (or any chosen basis). It's of course a bit contrived to exclude the off-diagonal coherences, because in reality that also suggests the inability to perform quantum gates at all, in which case you might as well be using a classical device. Commented Nov 26, 2021 at 13:25
• In this case it means the joint probability distribution.
– Pegi
Commented Nov 28, 2021 at 17:22
• Then I would say another reason is that one of the interesting properties of entanglement (a kind of quantum correlation) is that a maximally entangled state is entangled regardless of the choice of basis, which would also not be possible to observe if you restrict yourself only to the computational basis. Commented Nov 28, 2021 at 17:28
• Thank you for your explanation. Can you think of a resource for studying this issue (about different bases and entanglement) more extensively?
– Pegi
Commented Nov 28, 2021 at 17:30