I Qubit two-state quantum system

cianfa72
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About the Hilbert space representing the state space of two-state quantum system
Very simple question: for a two-state quantum system (qubit), is its state space a two-dimensional Hilbert space over the complex field ##\mathbb C## isomorphic to ##\mathbb C^2## or is it ##\mathbb C^2## itself ?
 
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We are often a bit sloppy with the notation ##\mathbb{C}^n## could mean the n-dimensional complex vector space with no specific inner product or to the Hilbert space with a given Hermitian inner product (since they are all equivalent). Similarly with ##\mathbb{R}^n##. The author should give some context to distinguish their use of the notation.

At a more nuanced level, we often speak of algebraic objects like SO(3) or ##\mathbb{C}^2## as if there is a singular one (as a categorical object). Other times we mean a specific presentation and other examples are merely isomorphic. In the end its not too big an issue so far as I've seen. I've not yet run across a situation where the ambiguity led to an actualizable error.
 
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jambaugh said:
We are often a bit sloppy with the notation ##\mathbb{C}^n## could mean the n-dimensional complex vector space with no specific inner product or to the Hilbert space with a given Hermitian inner product (since they are all equivalent).
You mean ##\mathbb C^n## could mean just the "naked" set of complex number pairs without any further structure. Or the Hilbert space one gets taking a specific Hermitian inner product within it (the standard one).

However I believe it is better to understand the qubit's state space as an abstract 2-dimensional Hilbert state space over the field ##\mathbb C ## isomorphic to ##\mathbb C^2## endowed with its standard Hermitian inner product structure that turns it into a 2-dimensional Hilbert space over ##\mathbb C##.
 
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cianfa72 said:
However I believe it is better to understand the qubit's state space as an abstract 2-dimensional Hilbert state space over the field ##\mathbb C ##
It's not just "better to understand"--this is the qubit's state space.

cianfa72 said:
isomorphic to ##\mathbb C^2## endowed with its standard Hermitian inner product structure that turns it into a 2-dimensional Hilbert space over ##\mathbb C##.
Not just "isomorphic to"--this is the 2-dimensional Hilbert space over the field ##\mathbb{C}##.
 
PeterDonis said:
It's not just "better to understand"--this is the qubit's state space.
Ok, you draw a distinction between the qubit's state space and the 2-dimensional Hilbert space over ##\mathbb C##.
PeterDonis said:
Not just "isomorphic to"--this is the 2-dimensional Hilbert space over the field ##\mathbb{C}##.
I'm not sure that the 2-dimensional Hilbert space over ##\mathbb C## is exactly the set ##\mathbb C^2## endowed with the standard Hermitian inner product structure. This is, let me say, just the "prototype" of it.
 
cianfa72 said:
Ok, you draw a distinction between the qubit's state space and the 2-dimensional Hilbert space over ##\mathbb C##.
Um, no, I'm saying they're the same.

cianfa72 said:
I'm not sure that the 2-dimensional Hilbert space over ##\mathbb C## is exactly the set ##\mathbb C^2## endowed with the standard Hermitian inner product structure. This is, let me say, just the "prototype" of it.
I have no idea what you mean by this.
 
PeterDonis said:
Um, no, I'm saying they're the same.
Ok.

PeterDonis said:
I have no idea what you mean by this.
I mean the following: what is for instance the Euclidean space of dimension 3 ? Is it the set ##\mathbb R^3## as affine space endowed with the Euclidean inner product structure, or is it any abstract set of "points" with an affine structure and an Euclidean inner product?
 
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cianfa72 said:
I mean the following: what is for instance the Euclidean space of dimension 3 ? Is it the set ##\mathbb R^3## as affine space endowed with the Euclidean inner product structure
Um, yes? What else would it be?

cianfa72 said:
or is it any abstract set of "points" with an affine structure and an Euclidean inner product?
What does "any abstract set of points" even mean?

I think you're getting yourself tangled up over words instead of looking at the concepts.
 
PeterDonis said:
What does "any abstract set of points" even mean?
In this specific context it is the set of qubit's quantum states. We assign it a vector space structure over ##\mathbb C## and an Hermitian inner product turning it into a 2-dimensional Hilbert space over ##\mathbb C## (let's call it ##\mathbb H^2##). The latter is then isomorphic to ##\mathbb C^2## with its standard vector space structure over ##\mathbb C## and the standard Hermitian inner product (the isomorphism isn't canonical since requires to pick a basis in both).

By the way, from a physical viewpoint, what is relevant are actually the elements of the projective space ##\mathbf P(\mathbb H^2)## represented by the Riemann or the Bloch sphere.

Although we had a thread some time ago about it, I ask: consider an "instance" of Bloch sphere, do points on it represent superposition of qubit's spin about a single given axis in (physical) 3D space ?
 
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  • #10
cianfa72 said:
In this specific context it is the set of qubit's quantum states.
Ok, but that's not an "abstract set of points." It already has a particular structure.

cianfa72 said:
We assign it a vector space structure over ##\mathbb C## and an Hermitian inner product turning it into a 2-dimensional Hilbert space over ##\mathbb C## (let's call it ##\mathbb H^2##).
No, we don't "assign" this--we discover that the set of a qubit's quantum states has this structure by looking at how qubits behave physically in experiments. (With one caveat, which you give later on--see below.)

cianfa72 said:
The latter is then isomorphic to ##\mathbb C^2## with its standard vector space structure over ##\mathbb C## and the standard Hermitian inner product (the isomorphism isn't canonical since requires to pick a basis in both).
Um, no, it isn't "isomorphic"--it's the same thing. I don't understand why you think these are somehow two different mathematical objects that are "isomorphic". They're just one mathematical object.

cianfa72 said:
from a physical viewpoint, what is relevant are actually the elements of the projective space ##\mathbf P(\mathbb H^2)## represented by the Riemann or the Bloch sphere.
Yes, that's correct.
 
  • #11
cianfa72 said:
onsider an "instance" of Bloch sphere
What does this mean?

cianfa72 said:
do points on it represent superposition of qubit's spin about a single given axis in (physical) 3D space ?
What do you think?
 
  • #12
PeterDonis said:
Um, no, it isn't "isomorphic"--it's the same thing. I don't understand why you think these are somehow two different mathematical objects that are "isomorphic". They're just one mathematical object.
I'd say they are actually two "instances" of the same mathematical object.

PeterDonis said:
What do you think?
Yes, fixed an axis in the physical world and a spin measurement apparatus aligned along it, the Bloch sphere North and South poles represent "spin-up" and "spin-down" along that axis respectively. Every other point on the sphere represents a superposition of the above two states, and not the spin along a different axis in space.
 
  • #13
cianfa72 said:
I'd say they are actually two "instances" of the same mathematical object.
You keep throwing these words around without explaining what they mean.

cianfa72 said:
Yes, fixed an axis in the physical world and a spin measurement apparatus aligned along it, the Bloch sphere North and South poles represent "spin-up" and "spin-down" along that axis respectively.
If you choose to interpret it that way, yes.

cianfa72 said:
Every other point on the sphere represents a superposition of the above two states
Yes.

cianfa72 said:
and not the spin along a different axis in space.
No. Every point on the Bloch sphere represents an eigenstate of qubit spin along some axis.

In other words, a qubit's spin about some axis is also a superposition of spin states about some other axis. They're not two different things only one of which is true. They're two different ways of viewing the same quantum state.
 
  • #14
PeterDonis said:
No. Every point on the Bloch sphere represents an eigenstate of qubit spin along some axis.

In other words, a qubit's spin about some axis is also a superposition of spin states about some other axis. They're not two different things only one of which is true. They're two different ways of viewing the same quantum state.
Ah ok, so a pair of antipodal points on the Bloch sphere represent the "spin-up" and "spin-down" states respectively along some spatial axis (along which a spin measurement apparatus alike Stern-Gerlach might be aligned).
 
  • #15
cianfa72 said:
a pair of antipodal points on the Bloch sphere represent the "spin-up" and "spin-down" states respectively along some spatial axis
Yes.
 
  • #16
All 2 dimensional complex vector spaces are isomorphic so I am not sure that the question is important.
 
  • #17
jbergman said:
All 2 dimensional complex vector spaces are isomorphic so I am not sure that the question is important.
Exactly, although they are isomorphic that doesn't mean they are actually the same.
 
  • #18
cianfa72 said:
Exactly, although they are isomorphic that doesn't mean they are actually the same.
Right but isomorphism is sufficient. We don't really care about sameness. I would disagree also with @PeterDonis and say they are only isomorphic and *not* the same, but when talking about these vector spaces most don't even make these distinctions.
 
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  • #19
jbergman said:
All 2 dimensional complex vector spaces
Can you give examples of more than one?
 
  • #20
PeterDonis said:
Can you give examples of more than one?
Complex polynomials of degree 2 or less with complex coefficients is isomorphic to ##\mathbb C^2## as a vector space.
 
  • #21
jbergman said:
Complex polynomials of degree 2 or less with complex coefficients is isomorphic to ##\mathbb C^2## as a vector space.
Is it? It takes three complex constants to define such a polynomial, not two.
 
  • #22
jbergman said:
isomorphism is sufficient. We don't really care about sameness.
I would put this a bit differently. I would say that the abstract space involved is the space of 2-tuples of complex numbers, with an appropriate vector space structure, and there is only one such abstract space, mathematically speaking; but we can interpret this space in different ways. But this might be more a matter of how we describe the math in ordinary language than anything else.
 
  • #23
PeterDonis said:
Is it? It takes three complex constants to define such a polynomial, not two.
You are right. Good catch. Ok, the polynomials of degree 1.
 
  • #24
PeterDonis said:
I would say that the abstract space involved is the space of 2-tuples of complex numbers, with an appropriate vector space structure, and there is only one such abstract space, mathematically speaking; but we can interpret this space in different ways.
No, I don't think so. The abstract space involved is a set of "abstract points" with the appropriate vector space structure over the field ##\mathbb C##. Mathematicians don't know what "points" are (they admit don't know what they're talking about), what is relevant is the assigned structure alone. ##\mathbb C^2## is just an "instance" or "realization/incarnation" of the abstract vector space of dimension 2 over the field ##\mathbb C##.
 
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  • #25
PeterDonis said:
I would put this a bit differently. I would say that the abstract space involved is the space of 2-tuples of complex numbers, with an appropriate vector space structure, and there is only one such abstract space, mathematically speaking; but we can interpret this space in different ways. But this might be more a matter of how we describe the math in ordinary language than anything else.

If we are being pedantic a vector space is just a set with the additional structure that satisfies the axioms of a vector space.

So we just need an abelian group structure on that set and a distributive multiplication of a field.

The polynomials of max degree 1 with complex coefficients satisfy these axioms.

The isomorphism is a linear invertible map from the polynomials to ##\mathbb C^n## or the 2-tuple of complex numbers as you describe it. There are actually multiple possible isomorphisms here as vector spaces.

But again these distinctions aren't very important for physics or even much math.
 
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  • #26
cianfa72 said:
No, I don't think so.
And yet you have given no difference between this...

cianfa72 said:
a set of "abstract points" with the appropriate vector space structure over the field ##\mathbb C##.
...and this:

cianfa72 said:
an "instance" or "realization/incarnation" of the abstract vector space of dimension 2 over the field ##\mathbb C##.
Unless and until you can state some actual difference between an "instance" or "realization" of an "abstract space" and the "abstract space" itself, you have no basis for claiming that they are different.
 
  • #27
jbergman said:
There are actually multiple possible isomorphisms here as vector spaces.
Can you describe two distinct ones?
 
  • #28
PeterDonis said:
Can you describe two distinct ones?
Just pick two different basis pairs (i.e. not the same pair), the isomorphism depends on the basis you pick on both vector spaces.
 
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  • #29
PeterDonis said:
Can you describe two distinct ones?
##f(ax+b)=(a,b)## and ##g(ax+b)=(b,a)## for example. When we think of polynomials there isn't really a natural sense in which coefficient should map to the 1st coordinate or the 2nd. Of course these choices don't matter. We could even choose something like ##h(ax+b) = (3a,b)##. Now if we have an orientation and hermitian inner product on both vector spaces you are more constrained.
 
  • #30
cianfa72 said:
Just pick two different basis pairs (i.e. not the same pair), the isomorphism depends on the basis you pick on both vector spaces.
No, isomorphism between two vector spaces is not basis dependent.
 

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