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The density operator $\rho$ of a mixed 2-qubit system has $4^2-1=15$ degrees of freedom. We can require Tr[$\rho^2$] $ =1$ so that the system is in a pure state. Now we have 14 degrees of freedom.

If we describe the system with a complex 4-dimensional vector we have with the norm restriction $2 \times4-1=7$ degrees of freedom. The state vector even shows a redundant degree of freedom because $|\psi\rangle$ is the same state as $e^{i\phi}|\psi\rangle$.

So what part of my counting is wrong, or what does it mean, when the density operator and the state vector are supposed to describe the same thing but the density operator has twice the degrees of freedom, although it is the vector state which has an obvious redundant degree of freedom?

bloods
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    A huge gap is your deducting only 1 degree of freedom for a pure state. You actually have $\rho^2=\rho$, a far more restrictive condition. Try an example. – Cosmas Zachos Oct 25 '19 at 21:50
  • @CosmasZachos Indeed, there are only two numbers satisfying $x^2=x$, so this implies that there must be one non-zero eigenvalue, which equals one. -- Maybe you should write this as an answer? (It is equivalent to the positivity condition which I use in my answer, which is of course the canonical one, but in your formulation, parameter counting is much more feasible.) – Norbert Schuch Oct 25 '19 at 22:37
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    @Norbert Schuch . Not sure I'm there: the problem is probably ill-posed for NxN matrices with N>2. For N=2, tr$\rho^2=1$ and $\rho^2=\rho$ deduct 1 d.o.f. , but this never happens again for N>2 for the idempotence eqn. But ρ has N(N-1) d.o.f. (the junk U diagonalizing tfmations to the standard form), while the d.o.f. of his tensor product goes linearly in N (therefore needs a restricted unitary tfmation to diagonalize: too messy to embed in U). That's why I recommended the OP checks N=3 ; he clearly was inspired by the fortuitous working for N=2. – Cosmas Zachos Oct 26 '19 at 13:48
  • @CosmasZachos I didn't check your argument, but it *must* turn out right: If $\rho=\rho^\dagger$, we have that $\rho=\rho^2$ implies $\rho=\rho\rho^\dagger\ge0$, and clearly, positivity, trace one, and $\mathrm{tr},\rho^2=1$ are equivalent to purity. --- Note that with the conditions in the question, also the $N=2$ case shouldn't work out, since $p_0+p_1=1$ and $p_0^2+p_1^2=1$ has two solutions. (Of course, you could argue that the "dimensional" parameter counting still works, but this still points to the issue that in such cases pure parameter counting is not the right thing to do.) – Norbert Schuch Oct 26 '19 at 13:53
  • You mean $\rho^2=\rho$ for purity. For N=2, it works, since tracelessness deducts 1, idempotence 1, so 4-2=2 d.o.f. For the tensor product matrix, we have 2(2) minus normalization, and minus over-all cancelling phase, so 4-2=2. I gather the OP observed that and attempted to generalize to N. – Cosmas Zachos Oct 26 '19 at 13:58
  • @CosmasZachos No: I mean that the two conditions $\mathrm{tr}\rho =1$, and $\rho^2=\rho$, together with $\rho=\rho^\dagger$, are sufficient in *any* dimension to guarantee that you have a pure state. (Since they yield $\mathrm{tr},\rho^2=1$ and $\rho\ge0$, which together with $\mathrm{tr},\rho=1$ is sufficient!) – Norbert Schuch Oct 26 '19 at 15:32
  • Sigh... of course they do: the OP knows that. That is not part of his problem! I would correct his 2N-1 to 2N-2 , but I don't wish to interfere with his original intent. We all understand his miscounting, but I don't see any proposal of a proper counting of deductions, his problem. He knows the right answer! (almost). – Cosmas Zachos Oct 26 '19 at 15:34
  • @CosmasZachos I think the right answer is: "Counting doesn't work that easily with non-linear constraints." Also, I'm not sure the OP knows that: They never mention anything like positivity anywhere in their counting, and it *is* essential. – Norbert Schuch Oct 26 '19 at 17:11

3 Answers3

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For one thing, you are neglecting positivity.

The other - and in some sense, more central - problem is that you pretend that non-linear constraints such as positivity and the non-linear $\mathrm{tr}\,\rho^2=1$ are independent constraints which remove one variable each.

However, neither non-linear constraints nor positivity are so simple. Quadratic equations (in a single variable) can e.g. have different numbers of solutions. Positivity, on the other hand, is a condition on all eigenvalues, so it adds more than one constraint, similar to a matrix-valued equation $A\vec x=\vec b$ (but not the same!)

Essentially, what you have is conditions on the eigenvalues of the density matrix. If you don't insist on positivity, the eigenvalues $p_n$ have to satisfy $\sum p_n=1$, and $\sum p_n^2=1$. These equations have many solutions if the $p_n$s can be negative as well. So you see that positivity is essential to get the unique solution $p_1=1$ and the other $p_n=0$. At the same time, you can also see that positivity does not just reduce the number of parameters by one, since if all but one $p_n$ are zero, we don't need to specify 3 out of 4 eigenvectors, and thus, many parameters suddenly disappear.

Essence: You can't say that every equation removes one parameter if the equations are not linear, or if they are matrix-valued.

Norbert Schuch
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  • Norbert, actually I disagree on this, positivity does not change the dimension. – lcv Oct 26 '19 at 15:40
  • @lcv You are right, not directly. But in combination with other constraints, like $\mathrm{tr}, \rho^2=1$ and $\mathrm{tr},\rho=1$, it does so heavily: Just count how the number of parameters changes when you in addition impose positivity! So it is really the intersection of linear/quadratic constraints with the positive cone. This type of effect is something which will not happen with linear constraints (there is a clear counting), so (part of) the point is really the positivity imposed in combination with those other constraints. But admittedly, there is no unique way of ... – Norbert Schuch Oct 26 '19 at 17:09
  • ... attributing this effect. – Norbert Schuch Oct 26 '19 at 17:09
  • @NorbertSchuch a very similar point was made in this answer to essentially the same question – glS Oct 26 '19 at 19:24
  • Yes this is probably when sets of constraints become incompatible – lcv Oct 26 '19 at 19:55
  • a more geometric argument could be to work in the Bloch repr. If $\rho=(1+r\cdot\sigma)/N$ then positivity imposes $r\le1/|\min\lambda_i|$ with the min over the eigenvalues of $\hat r\cdot\sigma$ (see https://physics.stackexchange.com/a/425101/58382). Then $\text{tr}\rho^2=1$ imposes $|r|=\sqrt{N-1}$. For both to be satisfied we need to directions such that $\hat r\cdot\sigma$ has a minimum eigenvalue such that $1/|\lambda|$ is on the hypersphere of radius $\sqrt{N-1}$. This doesn't easily lead to the correct dimension, but nicely shows the geometrical complexity of the constraint I think – glS May 27 '20 at 19:57
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Essentially the reason why you see the discrepancy in your calculation is that purity imposes not one constraint but of order of $N^2$.

Let me list some relevant spaces together with their respective dimensions.

  • Set of complex $N\times N$ matrices. $N^2$ complex dimension, $2 N^2$ real dimension.

  • Subset of Hermitian matrices. These matrices are made of $N$ real numbers on the diagonal and $(N^2-N)/2$ complex number off the diagonal. The total dimension is $N^2$ real dimension.

  • Subset of positive matrices. This is also a positive cone. This condition does not change the dimension of the set. Hence this set has still dimension $N^2$ (real).

  • Subset of density matrices. This is the intersection of the positive cone with the hyperplane $\mathrm{tr}(\rho) =1$. Note that this is only one real equation. Hence the dimension of this set is $N^2-1$. A common representation is the generalized Bloch form which makes this explicit.

  • Subset of pure states. Here is where the complication happens. This is due to the fact that purity imposes several constraint, since it is the requirement that $\rho^2=\rho$. It can be shown that this is equivalent to the fact that $\rho = |\psi\rangle \langle \psi |$. In this latter form it is much easier to compute its dimension and one sees that this set has dimension $2 N -2$. Note that this is only a subset of the $N^2 -2$ dimensional boundary of the set of general density matrices.

lcv
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  • But you are just computing the d.o.f. of a pure state density matrix through the OP's "2nd calculation" (including the irrelevancy of the over-all phase), so the d.o.f. of a tensor product matrix. The OP evidently wants to see how this known number comes out of incremental pruning off the d.o.f. of an arbitrary pure state ρ independently of the tensor product reduction. You are tweaking and rearranging the problem. – Cosmas Zachos Oct 26 '19 at 15:31
  • Absolutely. The reason why I don't do it is that it is not entirely trivial. As you saw yourself (I think) if you simply count naively it doesn't work. But it can be done, I'll post the calculation if I get the time. – lcv Oct 26 '19 at 15:38
  • As I've recommended to others, N=3 should be sufficient, since N=2 is virtually trivial. Again, to abstract his problem: Without exploiting the equivalence to a tensor product matrix of complex vectors, how do you show an idempotent, hermitian, trace 1 matrix has 2N-2 d.o.f ? – Cosmas Zachos Oct 26 '19 at 15:43
  • Actually I think that the case $N=3$ is essentially as hard as the general case. – lcv Oct 27 '19 at 14:03
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For a pure state $|\psi\rangle$, the corresponding density operator is the projector

$$\hat\rho=|\psi\rangle\langle\psi|.\tag1$$

The trace property is a necessary, but not a sufficient condition for a density operator to represent a pure state. That's why imposing this condition doesn't reduce enough the number of degrees of freedom for an arbitrary operator, even in a finite-dimensional Hilbert space.

The easiest way to see that, for a finite-dimensional Hilbert space, $\hat\rho$ has exactly the same number of degrees of freedom (up to the phase factor you've already mentioned) as $|\psi\rangle$, is the definition $(1)$. From it you can infer that, if you take a matrix representation of $\hat\rho$, all the rows in this matrix will be linearly dependent (its rank is 1).

Ruslan
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