Qbit pure vs mixed state space

  • #1
cianfa72
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TL;DR Summary
About the difference between Qbit quantum pure vs mixed states
According to this Wikipedia entry a quantum pure qbit state is a ray in the Hilbert space ##\mathbb H_2## of dimension 2. In other words a qbit pure quantum state is a point in the Hilbert projective line.

Now my question: is an arbitrary vector in ##\mathbb H_2## actually a "mixed" state for the qbit ? Thanks.
 
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  • #2
cianfa72 said:
Now my question: is an arbitrary vector in ##\mathbb H_2## actually a "mixed" state for the qbit ? Thanks.
No, a "mixed" state for the qbit is given by an arbirary density matrix (or density operator) on ##\mathbb H_2##. It is actually quite well behaved and understandable, in many ways.
 
  • #3
gentzen said:
No, a "mixed" state for the qbit is given by an arbirary density matrix (or density operator) on ##\mathbb H_2##. It is actually quite well behaved and understandable, in many ways.
So, what does a state vector in ##\mathbb H_2## represent ?
 
  • #4
cianfa72 said:
So, what does a state vector in ##\mathbb H_2## represent ?
The zero vector represents nothing. A non-zero vector represent a pure state, because it represents a ray in the Hilbert space ##\mathbb H_2##. But I guess you knew that already. So how can we help you?
 
  • #5
gentzen said:
The zero vector represents nothing. A non-zero vector represent a pure state, because it represents a ray in the Hilbert space ##\mathbb H_2##.
Sorry, we said that a qbit pure state is a point in Hilbert projective line. So basically do all ##\mathbb H_2## non-null vectors on the same ray represent the same qbit pure state ?
 
  • #6
cianfa72 said:
So basically do all ##\mathbb H_2## non-null vectors on the same ray represent the same qbit pure state ?
Yes.
 
  • #7
cianfa72 said:
all ##\mathbb H_2## non-null vectors on the same ray represent the same qbit pure state
Is the above a general statement ? In other words for any quantum system described by a separable Hilbert space (with a finite or countable basis), are the pure states points in the corresponding projective spaces (i.e. the set of rays) ?
 
  • #8
cianfa72 said:
Is the above a general statement ? In other words for any quantum system described by a separable Hilbert space (with a finite or countable basis), are the pure states points in the corresponding projective spaces (i.e. the set of rays) ?
Yes.
 
  • #9
Ok, suppose we have a quantum system and an hermitian operator on it. Then there exists a basis of orthonormal eigenvectors (even though a single eigenvalue may have geometric molteplicity greater than one). Now any Hilbert's space vector in a ray can be given as linear combination of the above basis eigenvectors.

As far as I know, a system's measurement represented by an hermitian operator basically projects/collaps the system's representative vector on one of the basis's orthonormal vectors.

Therefore a measurement of the qbit pure state projects vectors belonging to a ##\mathbb H_2## ray on one of the two basis orthonormal vectors ##|0 \rangle ## or ##|1\rangle##.

If the above is correct, then we can claim that a pure state is actually an "equivalent class" of Hilbert's space vectors on the same ray.
 
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  • #10
Sorry, I've not a clear understanding about the difference between pure and mixed states. A mixed qbit state is given by a density matrix on ##\mathbb H_2## while a pure qbit state is given by the "equivalence class" of vectors on ##\mathbb H_2##'s rays.

Why this difference ? Why is an arbitrary qbit state not given by an arbitrary vector of ##\mathbb H_2## ?
 
  • #11
cianfa72 said:
Why this difference ? Why is an arbitrary qbit state not given by an arbitrary vector of ##\mathbb H_2##?
Mixed states have nonzero von Neumann entropy, and e.g. a pure state ##|\psi\rangle\langle\psi|## where ##|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)## will not give the same expectations and variances for all observables as a mixed state ##\frac{1}{2}(|0\rangle\langle0| + |1\rangle\langle1|)##. Pure states are only a subset of all possible states.

[edit] - Made state examples more consistent
 
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  • #13
So, as far as I can tell, vectors in n-dimensional Hilbert space (possibly infinite-dimensional but separable - i.e. with a countable basis) represent just quantum system's pure states (actually points in the corresponding projective space built from it).

System's mixed states, instead, are just statistical ensemble of pure states described by a density operator/matrix. Of course even a pure state can be described by a particular density operator.

See also Quantum machine learning.
 
  • #14
Morbert said:
e.g. a pure state ##|\psi\rangle\langle\psi|## where ##|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)## will not give the same expectations for all observables as a mixed state ##\frac{1}{2}(|0\rangle\langle0| + |1\rangle\langle1|)##. Pure states are only a subset of all possible states.
Sorry, why do you call ##|\psi\rangle\langle\psi|## a pure state ? As stated before a pure state is a normalized vector ##|\psi\rangle## in system's Hilbert space (i.e. a point in the corresponding projective space), while here you claim it is a density operator/matrix.
 
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  • #15
cianfa72 said:
System's mixed states, instead, are just statistical ensemble of pure states described by a density operator/matrix. Of course even a pure state can be described by a particular density operator.

See also Quantum machine learning.
Your link is good enough for learning important basics about density matrices.
It could have been more explicit about the significance of Lemma 3 that the density matrix allows to evaluate arbitrary expectation values. The significance is the following: Different statistical ensembles of pure states can lead to the same density matrix. But Lemma 3 shows that no measurement (of the considered type) would be able to distinguish those different statistical ensembles. This is because the probability generating function (discrete case) or the characteristic function (continuous case) allow to completely determine a probability distribution from its expectation values.

cianfa72 said:
Sorry, why do you call ##|\psi\rangle\langle\psi|## a pure state ? As stated before a pure state is a normalized vector ##|\psi\rangle## in system's Hilbert space (i.e. a point in the corresponding projective space), while here you claim it is a density operator/matrix.
Because the density matrix corresponding to a pure state is just as good as the pure state itself. It tried to explain why above.
 
  • #16
cianfa72 said:
Sorry, why do you call ##|\psi\rangle\langle\psi|## a pure state ? As stated before a pure state is a normalized vector ##|\psi\rangle## in system's Hilbert space (i.e. a point in the corresponding projective space), while here you claim it is a density operator/matrix.
From the message I linked
A. Neumaier said:
In standard quantum physics (i.e., that not using the ##C^*## algebra terminology), a state is a density operator (i.e., a Hermitian positive semidefinite trace 1 operator) ##\rho## on a Hilbert space ##H##, and it is pure when it has rank one and mixed otherwise. In the pure case (only) it can be represented by a state vector ##\psi## as ##\rho = \psi\psi^*##, uniquely up to a phase. Hence state vectors always represent pure states.
Both pure and mixed states are density operators/matrices. A density operator that is a projector onto a subspace of dimensionality 1 (and hence associated with an element ##\psi## of Hilbert space) is a pure state.
 
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  • #17
Morbert said:
Both pure and mixed states are density operators/matrices. A density operator that is a projector onto a subspace of dimensionality 1 (and hence associated with an element ##\psi## of Hilbert space) is a pure state.
Ah ok so, from a formal point of view, a quantum system's state is a density/operator (and not a vector in the Hilbert space associated to the quantum system).

Then if a density operator is actually a projector (i.e. it is associated with a vector ##|\psi\rangle## in Hilbert space) then it will be a pure state.
 
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  • #18
cianfa72 said:
Ah ok so, from a formal point of view, a quantum system's state is a density/operator (and not a vector in the Hilbert space associated to the quantum system).

Then if a density operator is actually a projector (i.e. it is associated with a vector ##|\psi\rangle## in Hilbert space) then it will be a pure state.
Yes, with the caveat about your "i.e." A projector does not have to be rank 1, and hence does not have to be associated with a vector in Hilbert space. A mixed state can be a weighted sum of projectors of varying ranks, but a pure state is a projector of rank 1.
 
  • #19
Morbert said:
A projector does not have to be rank 1
But because the trace of a projector is equal to its rank, a density matrix which happens to be a projector does have rank 1. (The trace of a density matrix is 1.)
 
  • #20
Morbert said:
Yes, with the caveat about your "i.e." A projector does not have to be rank 1, and hence does not have to be associated with a vector in Hilbert space. A mixed state can be a weighted sum of projectors of varying ranks, but a pure state is a projector of rank 1.
Ah ok, a projector ##\rho## of rank 1 (a pure state) is always of the form ##\rho = |\psi\rangle \langle \psi|## for a vector ##|\psi\rangle## in the system's Hilbert space.
 
  • #21
gentzen said:
But because the trace of a projector is equal to its rank, a density matrix which happens to be a projector does have rank 1. (The trace of a density matrix is 1.)
Yeah. I wasn't sure if the "i.e." was pertaining to "density operator" or to "projector"
 
  • #22
cianfa72 said:
Ah ok, a projector ##\rho## of rank 1 (a pure state) is always of the form ##\rho = |\psi\rangle \langle \psi|## for a vector ##|\psi\rangle## in the system's Hilbert space.
Yes, and as @gentzen mentioned: A projector that is of rank > 1 cannot by itself be a state, mixed or pure. If a state is a projector it must be rank 1, and hence pure.
 
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  • #23
Morbert said:
Yes, and as @gentzen mentioned: A projector that is of rank > 1 cannot by itself be a state, mixed or pure. If a state is a projector it must be rank 1, and hence pure.
Ok, therefore a density matrix representing a mixed state is not a projector.
Basically a pure state is a dyadic tensor product of a vector times itself.
 
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  • #24
Sorry, is the notation ##|\psi\rangle \langle \psi|## actually a shorthand for ##|\psi\rangle \otimes |\psi\rangle## ?
 
  • #25
cianfa72 said:
Sorry, is the notation ##|\psi\rangle \langle \psi|## actually a shorthand for ##|\psi\rangle \otimes |\psi\rangle## ?
No, it is more that if ##\ket{\psi}## would correspond to a vector ##v##, then ##\bra{\psi}## corresponds to the dual vector ##v^T## (or rather ##v^H## i.e. transposed and complex conjugated). So ##\ket{\psi}\bra{\psi}## then corresponds to the matrix ##v v^H##.
 
  • #26
gentzen said:
No, it is more that if ##\ket{\psi}## would correspond to a vector ##v##, then ##\bra{\psi}## corresponds to the dual vector ##v^T## (or rather ##v^H## i.e. transposed and complex conjugated). So ##\ket{\psi}\bra{\psi}## then corresponds to the matrix ##v v^H##.
Sorry, then ##\ket{\psi}\bra{\psi}## should be actually the (dyadic) product tensor of the ket ##\ket{\psi}## times the dual-vector (bra) ##\bra{\psi}## ?

Edit: it is a dyad according to Wikipedia (tensor of order two and rank 1).
 
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  • #27
cianfa72 said:
Ah ok so, from a formal point of view, a quantum system's state is a density/operator
If you're using the density operator formalism, yes.

cianfa72 said:
(and not a vector in the Hilbert space associated to the quantum system).
A density operator corresponding to a pure state will have a corresponding Hilbert space vector (or ray if we want to ignore normalization). A density operator corresponding to a mixed state will not.
 
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  • #28
PeterDonis said:
A density operator corresponding to a pure state will have a corresponding Hilbert space vector (or ray if we want to ignore normalization).
And from what said in previous posts, a density operator corresponding to a pure state will be in the form ##\ket{\psi}\bra{\psi}## for some normalized vector ##\ket{\psi}## in system's Hilbert space.
 
  • #29
cianfa72 said:
And from what said in previous posts, a density operator corresponding to a pure state will be in the form ##\ket{\psi}\bra{\psi}## for some normalized vector ##\ket{\psi}## in system's Hilbert space.
You just said the same thing I said in different words.
 
  • #30
I'm still confused about the meaning of a writing like this: $$\ket{0}$$
Is it an actual vector in qbit's 2-dimensional Hilbert space or is it the "column vector" of the coordinates/components in a given Hilbert space basis ?
 
  • #31
cianfa72 said:
I'm still confused about the meaning of a writing like this: $$\ket{0}$$
Is it an actual vector in qbit's 2-dimensional Hilbert space
Yes, it is.

cianfa72 said:
or is it the "column vector" of the coordinates/components in a given Hilbert space basis ?
If you want a density matrix, then it is easiest to see it as a column vector in a specific basis.

Otherwise, you should imagine a density operator. This is not difficult either, it is just a different picture. In that case, you "imagine" how it applies/operates on a given vector. For example, if you apply the operator ##\ket{\psi}\bra{\psi}## to the vector ##\ket{x}##, you get ##\ket{\psi}\bra{\psi}\ \ket{x}=\ket{\psi}\ (\langle\psi|x\rangle)=(\langle\psi|x\rangle)\ \ket{\psi}##.
 
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  • #32
gentzen said:
For example, if you apply the operator ##\ket{\psi}\bra{\psi}## to the vector ##\ket{x}##, you get ##\ket{\psi}\bra{\psi}\ \ket{x}=\ket{\psi}\ (\langle\psi|x\rangle)=(\langle\psi|x\rangle)\ \ket{\psi}##.
Ah yes, as I said before in this thread ##\ket{\psi}\bra{\psi}## should be actually a dyadic tensor product.
 
  • #33
cianfa72 said:
Sorry, then ##\ket{\psi}\bra{\psi}## should be actually the (dyadic) product tensor of the ket ##\ket{\psi}## times the dual-vector (bra) ##\bra{\psi}## ?

Edit: it is a dyad according to Wikipedia (tensor of order two and rank 1).
If the "dyad picture" works nice for you, then you can think like that. However, the typical contexts where the "dyad picture" arises is in formulas from classical electrodynamics that are written with explicit scalar products (and vector products). In this context, the "dyad picture" is a convenient way to generalize the behavior of scalar products, such that a "tensor like" value can be put into a scalar product, and you get out a vector, instead of a scalar.
 
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  • #34
cianfa72 said:
Is it an actual vector in qbit's 2-dimensional Hilbert space or is it the "column vector" of the coordinates/components in a given Hilbert space basis ?
You seem to be confused about what the thing you are calling a "column vector" actually is. It is a representation of the vector itself in a particular basis. The vector itself is an element of a vector space, which is an abstract space that exists and has certain properties independent of any particular choice of representation/basis. If you choose a basis, then you can write the vector as a column vector in that basis.

A ket like ##\ket{0}## can represent either the abstract vector itself, or the column vector in a specific basis. It depends on how you want to view the label ##0## that you put on the ket. Is that label referring to the particular abstract vector that you have in mind? Or is it referring to the column vector that represents that abstract vector in the particular basis in which that vector is an eigenvector whose eigenvalue corresponds to the label ##0##? It could be either; the only person who knows which is the person who wrote down the ket, which in this case is you.
 
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  • #35
PeterDonis said:
Or is it referring to the column vector that represents that abstract vector in the particular basis in which that vector is an eigenvector whose eigenvalue corresponds to the label ##0##?
A vector is an eigenvector for an operator regardless of the picked basis. Therefore I believe what you meant is that if ##0## was understood as the label for an eigenvalue, then ket ##\ket{0}## would be the "column vector" representing the associated eigenvector in the basis in which an element is that eigenvector. That means it could be for example in the form $$\ket{0}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$
 
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