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Suppose I have a wave-function over a Hilbert-space of (complex) dimension $N$. It has $2 N-2$ real degrees of freedom, after normalization and removing the phase. It seems to me that I can measure these degrees of freedom with $2N-2$ measurements, first by projecting on each of the basis states, and then by making a measurement for each of the relative phases.

Lucien Hardy in his 2001 paper about the "reasonable axioms" (https://arxiv.org/abs/quant-ph/0101012) says it takes instead $N^2 -2$ measurements to completely determine the state, by which he means the density matrix (I am referring to normalized pure states, otherwise it's $N^2$). He calls these "fiducial" measurments.

I understand where the $N^2$ comes from -- it's the number of real entries that you need to specify a generic hermitian matrix over a complex vector space of dimension $N$.

What I don't understand is why do I need $N^2-2$ measurements if I know that the density matrix of a pure state can be written as a tensor product of the wave-function and therefore has only $2N-2$ degrees of freedom? Why do the measurements that I mentioned above (projection on basis plus relative angles) not entirely determine the density matrix? Or do I misunderstand what the "fiducial" measurements are?

glS
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WIMP
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  • Could you point to where in the paper this statement is? – user2723984 May 26 '20 at 10:28
  • It says on page 2 "The number of degrees of freedom, $K$, is defined as the minimum number of probability measurements needed to determine the state, or, more roughly, as the number of real parameters required to specify the state." He then goes on to show that in quantum mechanics $K=N^2$ (states are not normalized and this includes mixed states, hence the difference of 2 to my question).

    The "fiducial" measurements are defined at the beginning of section 6.3: "We will call the probability measurements labeled by $k = 1$ to $K$ used in determining the state the fiducial measurements"

     – WIMP May 26 '20 at 10:33
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    As far as I read the paper, the statement $K=N^2$ is never claimed to be a lower bound for quantum states, but an upper bound, since they include mixed, pure, normalized, and unnormalized states. So I don't think the paper implies that you need $N^2 - 2$ measurements if you know that the state you are looking at is a pure state. – Marius Ladegård Meyer May 26 '20 at 11:23
  • See the top of page 9 for a discussion. – Marius Ladegård Meyer May 26 '20 at 11:42
  • But there's only one degree of freedom difference between mixed and pure states (the requirement that the trace of the square of the density matrix equals one), so I don't see how this helps. – WIMP May 26 '20 at 13:43
  • @WIMP The last statement you make is not true (even though it seems as if): In combination with the positivity of $\rho$, this gives an extremely strong constraint, as you correctly observe! I remember having a discussion about that, probably here (or maybe on quantumcomputing.se), I'm not sure if I will be able to find it (believe it was only in comments, along the same lines, though with more details in a concrete case). – Norbert Schuch May 26 '20 at 16:53
  • (Basically it's like saying you are on a high-dimensional manifold and then you fix some coordinate. Usually this only removes one parameter, but if it is the boundary of the manifold, this can completely constrain you to a point.) – Norbert Schuch May 26 '20 at 16:54
  • This makes no sense, then the dimension of the space just would not be $N^2-2$ which is also in conflict with what Hardy writes. – WIMP May 27 '20 at 04:18
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    about the parameter counting using the "one constraint" of $\operatorname{tr}(\rho^2)$, @NorbertSchuch might be thinking of this or maybe this threads – glS May 27 '20 at 11:39
  • I haven't read the paper you link, but the question makes me think of compressive sensing techniques for state tomography, see e.g. this paper. If you know that the state is has rank $r$ then you only need $O(rN \log^2 N)$ measurement settings. But if you know that the state is pure, then sure $2(N-1)$ measurement settings are enough. I don't quite get why you think otherwise. – glS May 27 '20 at 11:50
  • @glS, thanks this is very helpful. Why I think otherwise? Well have a look at the paper, or at least the quotes above. He writes in the paper that $K$ is "the minimum number of probability measurements needed to determine the state" and then goes on to show $K=N^2$, where $N $is the dimension of the Hilbert-space (the number of entries on the trace of the density matrix). So, well. I don't know how to make sense of this. (As I said above, subtract two to get normalized, pure states only.) – WIMP May 27 '20 at 13:49
  • @WIMP it seems to me that the author is simply referring to generic states, not restricting to pure ones, which are characterised by $N^2-1$ degrees of freedom. You cannot simply subtract a few dimensions to get the pure ones, as per the threads linked above – glS May 27 '20 at 15:56
  • @glS Indeed, it was the latter linked thread I had in mind. Thanks for finding it! – Norbert Schuch May 27 '20 at 19:27

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