I have come across the term quantum tomography which, according to Wikipedia, is the

[...]process by which a quantum state is reconstructed using measurements on an ensemble of identical quantum states.

In the second section of the page ("What quantum state tomography is used for"), it is also stated that

Quantum tomography is applied on a source of systems, to determine the quantum state of the output of that source. Unlike a measurement on a single system, which determines the system's current state after the measurement (in general, the act of making a measurement alters the quantum state), quantum tomography works to determine the state(s) prior to the measurements.

I have got confused in the bold part. From the first quote I understand that we have a quantum system on which we are doing some operations, and we want to retrieve the quantum state after applying those operations, using measurement. So how exactly quantum tomography works towards "getting the state(s) prior to the measurements"? It seems a contradiction. What am I missing?

  • $\begingroup$ How familiar are you with the notion of density matrix? Please mention that so I/people can answer this accordingly! $\endgroup$
    – FDGod
    Mar 22 at 21:29
  • $\begingroup$ Actually, I'm not quite familiar with it! $\endgroup$
    – aghin
    Mar 22 at 22:52

2 Answers 2


The first quote you give does not imply any other operations on the ensemble than those that are needed to determine its state. (We might of course intend to do that later, e.g. using the ensemble as input for a quantum computer which does all kinds of useful things, but that is not meant here. Here we just want to check the content of the ensemble.)

Still, we cannot avoid influencing the quantum states in the ensemble that we are studying in this way. Even seemingly harmless things like using the states as control qubits in a CNOT gate, for instance, do have influence by creating entanglement between us (or our equipment) and the states of the ensemble that we are studying.

That is what the second quote is about, it reiterates that we were interested in the states inside the ensemble before we influence them, as opposed to the states afterwards. So that is no contradiction with the previous.

Actually (now that we're at it) just before that second quote the wikipedia text talks about a "measurement on a single system, which determines the system's current state after the measurement". But that is of course a false notion. After a measurement you see one branch of the total wave function, you do not know whether this is the only branch, or whether there are now several branches and you are just in one of them.

You may think that this last issue is just philosophy of the "many worlds" view, but it is really at the heart of the matter here: in quantum tomography you do the analysis on the ensemble precisely to discern between the possibilities that would still be open if we do only one measurement!

We can look at the simplest case of single-qubit states. In general, if we avoid density matrices (see discussion in the comments), we have to include the source when we write the possibilities:

$$ |1\rangle_\text{source}\ \big(\ \alpha\ |\hspace{-2pt}\uparrow\rangle_\text{qubit} + \beta\ |\hspace{-2pt}\downarrow\rangle_\text{qubit} \big) \ + |2\rangle_\text{source}\ \big(\ \gamma\ |\hspace{-2pt}\uparrow\rangle_\text{qubit} + \delta\ |\hspace{-2pt}\downarrow\rangle_\text{qubit} \big). $$

This just describes that for this simplest case the source can make two choices $|1\rangle$ and $|2\rangle$ and in each of those cases create a qubit for us with different coefficients for $|\hspace{-2pt}\uparrow\rangle$ and$|\hspace{-2pt}\downarrow\rangle$, so in total there are already four complex coefficients, $\alpha, \beta, \gamma,$ and $\delta$. Even if we ignore the total phase and believe that the normalization is $1$, this would still leave us with 6 real parameters. If you do use a density matrix you will find out that the information about the qubit alone (ignoring the source that it has left) still requires 3 real parameters to describe it.

But in fact those are technicalities. It should be clear that many repeated measurements are needed, not all along the same axis (and also the guarantee that we have identical copies in the ensemble) to have any chance of finding those real or complex numbers. Just one measurement on a qubit gives only an "up" or "down" result, which is surely not enough (even though it does mean that the qubit is in that state after the measurement, at least in your branch of the total state...)

PS: For philosophers, quantum tomography should perhaps demonstrate that the many worlds view is the only one that can describe all aspects of QM. But it is also possible that philosophers use another kind of logic.

  • $\begingroup$ @JosBergevoet same as for the answer of glS, what exactly do you mean by identical copies of the ensamble? Anyway, I don't really see how the problem of one measurement could be related to the many worlds view, although I don't know much about that theory so I would not really go that deep. But yeah, I get what your point is about quantum tomography. $\endgroup$
    – aghin
    Mar 23 at 21:24

Imagine you have a measurement apparatus that always prepares some state $\rho$, but you don't know or anyway want to verify what this state $\rho$ is precisely.

Obviously, the only thing you can do to reach this goal is perform some measurements. But being these quantum states, measurements will destroy or anyway modify the state themselves. Nonetheless, what you can do is perform a number of measurements on a number of different copies of the generated $\rho$, to infer what $\rho$ is. Now, what you'll have estimated is not what the state is after the measurement, because the measurements inevitably change it. You do however manage to characterise what the state $\rho$ that is being measured is. In this way, next time you run the experiment you'll know what the generated state is.

As an extremely simple example (not an example of tomography, an example of why estimation in quantum mechanics retrieves what was there before the measurement), imagine your apparatus generated a qubit in the state $\sqrt p|0\rangle+\sqrt{1-p}|1\rangle$. If you perform computational basis measurements on (several copies of) this state, you'll manage to estimate the value of $p$. After each measurement the state is not $\sqrt p|0\rangle+\sqrt{1-p}|1\rangle$ anymore because it collapsed to $|0\rangle$ or $|1\rangle$ (or was modified in some other way, this depends on the measurement apparatus). However, the information about $p$ tells you something about the state that will be generated next time you run the experiment.

  • $\begingroup$ That example would not reveal whether the source creates mixed states, entangled with the source itself, like: $\sqrt{p}\ |1\rangle_\text{source}\ |\hspace{-2pt}\uparrow\rangle_\text{qubit} \ + \sqrt{1-p}\ \ |2\rangle_\text{source}\ |\hspace{-2pt}\downarrow\rangle_\text{qubit}, $ or whether it just creates pure qubit states: $\sqrt{p}\ |1\rangle_\text{source}\ |\hspace{-2pt}\uparrow\rangle_\text{qubit}+\sqrt{1-p}\ \ |1\rangle_\text{source}\ |\hspace{-2pt}\downarrow\rangle_\text{qubit}. $ So additional to the computational basis measurement, we'll need another one... $\endgroup$ Mar 23 at 13:35
  • 1
    $\begingroup$ @JosBergervoet well, of course. I didn't mean it as an example of tomography. It's an example to address the issue of "estimating what was there before the measurement", which is not really specific to tomography per se. Added a note to stress this $\endgroup$
    – glS
    Mar 23 at 17:22
  • $\begingroup$ It is in fact a very useful example (the single qubit state example, I mean!) So in order to inform OP (and other readers) correctly we should stress that even for this simplest example we need: a) a large number of measurements, and b) measurements along a few different axes. (Just trying to be educational... although it's probably already perfectly clear to all readers now.) $\endgroup$ Mar 23 at 18:05
  • $\begingroup$ The example you gave helped a lot with the understanding of the problem I was asking. However, just to be clear: when you say "perform a number of measurements on a number of different copies of the generated $\rho$", it doesn't go against the no-cloning theorem because these states are, in fact, different (so not really copies) and we want to estimate (reconstruct) what $\rho$ is based on the measurement of all the states we got, right? $\endgroup$
    – aghin
    Mar 23 at 21:09
  • $\begingroup$ Yes, that is what we want and for that we need a large number of equal states, or you can say that they are different but equal, or repeated equal states. Whether we actually can create equal states without violating no-cloning, is the same question as whether we can create blanks or ancilla qubits in a desired state repeatedly. Physically, there are ways to put things in an energetically lowest ground state with good accuracy, and subsequently that can then be transformed to any desired state. And this can be repeated. $\endgroup$ Mar 23 at 22:02

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