I QM Qubit state space representation by Projective Hilbert space

[cianfa72]

TL;DR Summary

About the representation of Qubit state space set by mean of projective space built over the Hilbert space of dimension 2

I'd like to discuss some aspects of quantum systems' state space from a mathematical perspective.

Take for a instance a qubit, e.g. a two-state quantum system and consider the set of its pure states. This set as such is a "concrete" set, namely the "bag" containing all the qubit's pure states. Such a set, however, doesn't carry a vector space structure much less an Hilbert space structure.

Nevertheless, quantum physics employs the abstract Hilbert space $\mathcal H$ of dimension 2 to model such a concrete set. Namely we take the set of rays of the abstract space $\mathcal H$ and map/define a one-to-one/onto correspondence (bijective map) between the set of qubit's pure states and set of $\mathcal H$ rays (namely the elements of the projective Hilbert space $P(\mathcal H)$ built over $\mathcal H$).

Up to this point no "concrete" or specific "instances" of the abstract Hilbert space of dimension 2 are involved (note that such abstract Hilbert space of dimension 2 is unique by its very definition).

Next fixing a basis in $\mathcal H$ one can define a straightforward isomorphism with $\mathbb C^2$ which is a concrete instance/realization of the abstract Hilbert space $\mathcal H$. Let's call $\ket{\uparrow}$ and $\ket{\downarrow}$ such basis elements in $\mathcal H$.

From the above it follows that $\ket{\uparrow}$ or $\ket{\downarrow}$ are not qubit states themselves, any of them actually represents a qubit state. Formally what is really going on is taking the ray the vector $\ket{\uparrow}$ or $\ket{\downarrow}$ belongs to in order to select/pick the unique corresponding element within $P(\mathcal H)$. Such an element maps back (bijectively) to an element of the qubit state space's set.

What do you think, does it actually make sense ?

 

[SergejMaterov]

The chosen basis provides a convenient coordinating, but does not change the geometry of the state space itself

 

[cianfa72]

The chosen basis provides a convenient coordinating, but does not change the geometry of the state space itself

Yes, of course. My doubt concerns symbols like $\ket{\uparrow}$ and $\ket{\uparrow}$. They aren't qubit's pure states, they actually represent them. In other words any of them is an element of the abstract Hilbert space of dimension 2 such that the ray it defines actually represents a qubit's pure state.

 

[SergejMaterov]

The vectors ∣0⟩ and ∣1⟩ are not cubit states in themselves, but only their representatives. The only physical content is the rays, that is, the points of the project space.

 

[cianfa72]

The vectors ∣0⟩ and ∣1⟩ are not cubit states in themselves, but only their representatives. The only physical content is the rays, that is, the points of the project space.

I'd better say: the physical content is in the rays which represent qubit's pure states, however themselves (the rays) are not.

Morever $\ket{0}$ or $\ket{1}$ like $\ket{\uparrow}$ or $\ket{\downarrow}$ can be understood as elements of the 2D abstract Hilbert space or as elements of $\mathbb C^2$.

 

[SergejMaterov]

$$H \rightarrow \mathbb{C}^2,\quad a|0\rangle + b|1\rangle \mapsto \begin{pmatrix}a \\b\end{pmatrix}$$
This is already the realization of an abstract space, and not the physical state itself.
The physical state of a qubit is a class of vectors {λ(a,b)} = point ([a:b]) in $\mathbb{CP}^1$

 

[cianfa72]

$$H \rightarrow \mathbb{C}^2,\quad a|0\rangle + b|1\rangle \mapsto \begin{pmatrix}a \\b\end{pmatrix}$$
This is already the realization of an abstract space, and not the physical state itself.

Yes, picking a basis $\{ \ket{0} , \ket{1} \}$ in the abstract Hilbert space $\mathcal H$ of dimension 2 defines the above isomorphism. $\begin{pmatrix}a \\b\end{pmatrix}$ is the mapped vector into $\mathbb{C}^2$ (the complex numbers pair coincides incidentally with the mapped vector's coordinates when chosing the canonical basis of $\mathbb{C}^2$).

The physical state of a qubit is a class of vectors {λ(a,b)} = point ([a:b]) in $\mathbb{CP}^1$

As before, I'd say point([a:b]) $\in \mathbb{CP}^1$ isn't a qubit's pure state itself, it represents it.

Sorry to insist, I believe it goes like the Euclidean space $E^2$ vs $\mathbb R^2$ endowed with the standard Euclidean structure. The former is the abstract Euclidean space while the latter is the concrete $\mathbb R^2$ set equipped with the standard Euclidean inner product. "Points" in the former are just mathematical objects, what is relevant is only the structure given by the Euclidean structure. "Points" in the latter are real number pairs instead.

 

[SergejMaterov]

Exactly right. You’re drawing the perfect analogy. Point [a:b] in $\mathbb C\mathrm P^1$ isn’t literally the qubit state, but represents the unique physical pure state (ray) in the abstract $\mathcal H$. All the physics lives in the projective geometry; picking $\mathbb C^2$ is only a convenient coordinate chart.

 

[cianfa72]

Point [a:b] in $\mathbb C\mathrm P^1$ isn’t literally the qubit state, but represents the unique physical pure state (ray) in the abstract $\mathcal H$. All the physics lives in the projective geometry; picking $\mathbb C^2$ is only a convenient coordinate chart.

Ok, so your point is as follows: take the set of qubit's pure states (this is a concrete set, namely the set of qubit's physical pure states). We can map each of them uniquely to a ray in the abstract Hilbert space $\mathcal H$ of dimension 2. Picking a basis in the abstract $\mathcal H$ defines a (basis-dependent) isomorphism to $\mathbb C^2$.

A ray in the abstract $\mathcal H$ maps back uniquely to a qubit's pure state. A point in the "concrete" $\mathbb C\mathrm P^1$ represents a ray in $\mathcal H$ which maps back uniquely into the qubit's pure states set.

 

[SergejMaterov]

Exactly. All the physics lives on the projective space $\mathbb C\mathrm P^1$ (or Bloch sphere); $\mathbb C^2$ is just a convenient, basis‑dependent coordinate chart.

 

[cianfa72]

take the set of qubit's pure states (this is a concrete set, namely the set of qubit's physical pure states). We can map each of them uniquely to a ray in the abstract Hilbert space $\mathcal H$ of dimension 2. Picking a basis in the abstract $\mathcal H$ defines a (basis-dependent) isomorphism to $\mathbb C^2$.

Just to finalize: consider a physical experiment consisting of measuring the spin of 1/2 spin particle along the $z$-axis. We can choose to map the spin-up along $z$-axis state to the ray in the abstract $\mathcal H$ given/identified by the vector labeled $\ket{\uparrow} \in \mathcal H$ and the spin-down along $z$-axis state to the ray given by the vector labeled $\ket{\downarrow} \in \mathcal H$. The only requirement to be fulfilled is that they must be chosen as orthogonal in $\mathcal H$.

 

[gentzen]

Take for a instance a qubit, e.g. a two-state quantum system and consider the set of its pure states. This set as such is a "concrete" set, namely the "bag" containing all the qubit's pure states. Such a set, however, doesn't carry a vector space structure much less an Hilbert space structure.

I am sure you know what you want, but you might miss some physics, if you ignore mixed states:

Exactly. All the physics lives on the projective space $\mathbb C\mathrm P^1$ (or Bloch sphere); $\mathbb C^2$ is just a convenient, basis‑dependent coordinate chart.

I admit that capturing physics ignored by pure states, like temperature, is challenging. More challenging than just allowing mixed states. But the density matrix representation of mixed states at least gives a nice understanding why pure states are so special.

 

[cianfa72]

But the density matrix representation of mixed states at least gives a nice understanding why pure states are so special.

Yes, I know quantum systems' states (pure & mixed) can be represented by density operators acting on abstract Hilbert spaces of relevant dimension (e.g. dimension 2 for qubit). Note that such density operators are represented by a density matrix when a basis is fixed/picked in the abstract $\mathcal H$ by virtue of the basis-dependent isomorphism $\mathcal H \to \mathbb C^2$.

 

[cianfa72]

Suppose to perform a Stern-Gerlach experiment to measure the spin of a spin-1/2 particle by orienting the equipment along the z-axis. In the abstract $\mathcal H_2$ model this physical measurement is represented by the Hermitian operator $S_z$. It acts on elements/vectors in $\mathcal H_2$, even though the actual spin state is represented by a ray.

As far as I can tell, the standard approach here is to use two orthonormal vectors in $\mathcal H_2$ labeled as $\ket{\uparrow}$ and $\ket{\downarrow}$ to represent the spin-up along z-axis vs spin-down along z-axis physical states, respectively, as results of the above experimental setup (note that any pair of orthonormal vectors in $\mathcal H_2$ is suitable). Logically the next step is define the basis-dependent isomorphism to map such abstract $\mathcal H_2$ vectors to their representative elements in $\mathbb C^2$. The standard approach is map $\ket{\uparrow}$ to $\ket{0}$ and $\ket{\downarrow}$ to $\ket{1}$. In using the latter symbols the following definition is understood: $$\ket{0} \equiv
\begin{bmatrix}1 \\0\end{bmatrix}, \ket{1} \equiv \begin{bmatrix}0 \\1\end{bmatrix}$$
Then, by using such mappings, the $S_z$ operator will get the 2x2 matrix representation we all know, right ?

 

[SergejMaterov]

Yes, right, that’s exactly how you get the familiar 2×2 matrix.

1. The mapping is basis-dependent: a different orthonormal choice (or a different ordering or phase convention for the basis vectors) gives a different matrix representation — but physically equivalent (related by a unitary change of basis).

2. Because physical states are rays (global phase irrelevant), you may multiply each basis vector by a phase without changing physics; that only changes the matrix by the corresponding similarity transform if you change both basis vectors consistently.

3. Hermiticity of the operator in H becomes Hermiticity of the matrix: A†=A.

 

[cianfa72]

Yes, right, that’s exactly how you get the familiar 2×2 matrix.

1. The mapping is basis-dependent: a different orthonormal choice (or a different ordering or phase convention for the basis vectors) gives a different matrix representation — but physically equivalent (related by a unitary change of basis).

You mean pick a different orthonormal basis in $\mathcal H_2$ mapping they elements to vectors in $\mathbb C^2$ as prescribed by the "old" basis-dependent isomorphism (i.e. by the isomorphism mapping the previously picked basis's vectors $\ket{\uparrow}, \ket{\downarrow}$ into $\begin{bmatrix}1 \\0\end{bmatrix}, \begin{bmatrix}0 \\1\end{bmatrix}$ respectively).

The $S_z$ matrix representation obtained this way will be related to the "familiar" one by a unitary change of basis.

 

[SergejMaterov]

$$Sz=\frac{ℏ}{2}σ_z$$ ,where $σ_z$ is the Pauli z-matrix.

 

[cianfa72]

$$Sz=\frac{ℏ}{2}σ_z$$ ,where $σ_z$ is the Pauli z-matrix.

Yes, the above is the "familiar" 2 x 2 matrix representation for the $S_z$ operator (involving the Pauli z-matrix). By changing the basis vectors used in $\mathcal H_2$ -- without changing the basis-dependent isomorphism $\phi: \mathcal H_2 \rightarrow \mathbb C^2$ -- results in a new $S_z$ matrix representative related to the above (the familiar one) by a unitary change of basis in $\mathcal H_2$.

 

[SergejMaterov]

If $v_1, v_2$ are chosen to be the eigenvectors of $S_z$ with eigenvalues +ℏ/2,−ℏ/2 and you define the basis-dependent isomorphism φ: $H_2→C^2$ by $$φ(v_1)=(1,0)^T,φ(v_2)=(0,1)^T$$ then $S_z$ is represented by the familiar matrix
$$S_z=\frac{ℏ}{2}\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} = \frac{ℏ}{2}σ_z$$
Your further remark is correct provided the new basis $w_1,w_2$ are also eigenvectors of $S_z$ (they can differ from $v_1,v_2$ by phases). If you choose a different orthonormal basis that is not composed of $S_z$-eigenvectors, then the matrix representation changes by a unitary similarity transform. Concretely,
if $w_j=\sum_i v_i U_{ij}$ with unitary $U$, the matrix of $S_z$ in the $w$-basis is
$$S_z^{(w)}=U^†S_z^{(v)}U$$, so you generally get a different 2×2 matrix (unitarily equivalent to the diagonal one).
So: yes, if you always map the eigenbasis of $S_z$ to the computational basis you get the same diagonal matrix; if you pick a different orthonormal basis but also insist on mapping that basis to (1,0),(0,1), then the matrix will only stay the same in the special case that the new basis consists of the same eigenvectors (up to phases). Physically nothing changes — the operator is basis-independent; its matrix representation depends on your chosen isomorphism.

Example: take

$$w_1=\frac{v_1+v_2}{\sqrt2}, w_2=\frac{v_1-v_2}{\sqrt2}$$.
Mapping $w_1↦(1,0)^T,  w_2↦(0,1)^T$ gives

$$S_z^{(w)}=\frac{ℏ}{2}\sigma_x,$$
which is the same operator written in a different basis (not a contradiction — just a basis change).