I Understanding Spin States in 2D Vector Spaces

[robphippen]

TL;DR

Does this quote from 'Quantum Mechanics: The Theoretical Minimum' make it self-evident that a spin state can be represented by a 2d vector space?

There is a passage in this book where I don't follow the logic;

In this short quotation from 'Quantum Mechanics: The Theoretical Minimum' by Leonard Susskind and Art Friedman

• \mathcal{A} represents the apparatus that is performing the measurement
• the apparatus can be oriented (in principle) in any direction
• the item being measured is a single spin

Let’s begin by labeling the possible spin states along the three coordinate axes.

If \mathcal{A} is oriented along the z axis, the two possible states that can be prepared correspond to \sigma_{z} = ± 1. Let’s call them up and down and denote them by ket-vectors | u > and | d >. Thus, when the apparatus is oriented along the z axis and registers + 1, the state | u > has been prepared.

On the other hand, if the apparatus is oriented along the x-axis and registers − 1, the state | l > has been prepared. We’ll call it left.

If \mathcal{A} is along the y axis, it can prepare the states | i > and | o > (in and out).

You get the idea. The idea that there are no hidden variables has a very simple mathematical representation: the space of states for a single spin has only two dimensions. This point deserves emphasis:
All possible spin states can be represented in a two-dimensional vector space.”

The book goes on to make it crystal clear that this final statement, that a spin can be represented via a 2d vector space, is true and I totally follow those arguments. So I am not asking if the statement is true.

However, the wording of the quote above seems to imply that the argument above in itself is enough to establish that a spin can be represented via a two-dimensional vector space. That is not obvious to me from this section alone. Can anyone enlighten me?

 

[vanhees71]

I think that's the didactical problem with this "bra-kets first" approach. If you want to restrict yourself to the most simple 2D Hibert space problem, you have to introduce spin 1/2 out of the blue sky. There is not way to make this plausible without just stating the fact or to develop quantum theory along a more conventional heuristical basis until it comes to an understanding that angular momentum is described by an algebra following from representation theory of the rotation group SO(3) or rather its ray representations, which substitutes SO(3) with its covering group SU(2) and thus the possibility of half-integer spins, which have no classical analogon (while orbital angular momentum has, but there you have only integer-spin representations).

On the other hand it doesn't matter too much, which 2D Hilbert space system you consider, and you can just take the electromagnetic field, which has two polarization degrees of freedom (the natural choice of a basis are the helicity eigenstates or the left- and right-circular polarization states). Then you can argue that on a quantum level em. waves of a given frequency/wavelength can be absorbed and emitted only in discrete portions of energy and momentum $\hbar \omega$ and $\hbar \vec{k}$, and then you only consider the polarization states of photons of these kind. This is a bit more "intuitive" if you have some preknowledge about classical wave optics and polarization states of em. waves.

 

[PeroK]

However, the wording of the quote above seems to imply that the argument above in itself is enough to establish that a spin can be represented via a two-dimensional vector space. That is not obvious to me from this section alone. Can anyone enlighten me?

I agree, I don't think it's obvious. There's a mathematical point about the 2D complex vector space that it admits three sets of orthogonal bases with the required mutual relationship:

Basis vectors in basis "z" are an "equal" linear combination of basis vectors "x" and likewise of basis vectors "y".

And the same for basis vectors in bases "x" and "y".

These can be constructed simply as, for example:
$$\ket {z+} = [1, 0], \ \ \ket{z-} = [0, 1]$$$$\ket {x+} = \frac 1 {\sqrt 2}[1, 1], \ \ \ket{x-} = \frac 1 {\sqrt 2}[-1, 1]$$$$\ket {y+} = \frac 1 {\sqrt 2}[1, i], \ \ \ket{y-} = \frac 1 {\sqrt 2}[-1, i]$$You can check for yourself that these three bases have the stated mutual properties.

And, critically, this then represents the structure of the superpositions we see experimentally in terms of electron spin:

Spin-up in the z-direction is an equal superposition of spin-up and spin-down in the x-direction, and in the y-direction. The same for spin-up and down in the x and y directions.

 

[PeterDonis]

Spin-up in the z-direction is an equal superposition of spin-up and spin-down in the x-direction, and in the y-direction.

Not the way you've written them. For the statement you make here to be true--more precisely, for the $z$ direction spins to look the same in the $x$ basis as the $x$ spins look in the $z$ basis, and similarly for $z$ and $y$--the minus signs in $\ket{x-}$ and $\ket{y-}$ need to be on the second terms, not the first.

 

[PeroK]

Not the way you've written them. For the statement you make here to be true--more precisely, for the $z$ direction spins to look the same in the $x$ basis as the $x$ spins look in the $z$ basis, and similarly for $z$ and $y$--the minus signs in $\ket{x-}$ and $\ket{y-}$ need to be on the second terms, not the first.

I don't think you can get all the coefficients the same. Expressing $\ket{y}$ states in terms of $\ket{x}$ states introduces different coefficients, such as $\frac{ 1 \pm i}{\sqrt 2}$.

In any case, I'm using Sakurai's convention. See equations 1.1.9a and 1.1.9b (page 9) from Modern QM (Revised Edition).

 

[PeterDonis]

Expressing $\ket{y}$ states in terms of $\ket{x}$ states

I was only talking about expressing the $z$ states in the $x$ or $y$ basis. With the convention I described, the $z$ states in the $x$ basis look the same as the $x$ states in the $z$ basis, and similarly for $z$ and $y$.

I'm using Sakurai's convention.

Ok. I'm more familiar with the other convention, the one I described, but of course it is a convention and can go either way.

 

[robphippen]

I agree, I don't think it's obvious. There's a mathematical point about the 2D complex vector space that it admits three sets of orthogonal bases with the required mutual relationship:

Basis vectors in basis "z" are an "equal" linear combination of basis vectors "x" and likewise of basis vectors "y".

And the same for basis vectors in bases "x" and "y".

These can be constructed simply as, for example:
$$\ket {z+} = [1, 0], \ \ \ket{z-} = [0, 1]$$$$\ket {x+} = \frac 1 {\sqrt 2}[1, 1], \ \ \ket{x-} = \frac 1 {\sqrt 2}[-1, 1]$$$$\ket {y+} = \frac 1 {\sqrt 2}[1, i], \ \ \ket{y-} = \frac 1 {\sqrt 2}[-1, i]$$You can check for yourself that these three bases have the stated mutual properties.

And, critically, this then represents the structure of the superpositions we see experimentally in terms of electron spin:

Spin-up in the z-direction is an equal superposition of spin-up and spin-down in the x-direction, and in the y-direction. The same for spin-up and down in the x and y directions.

Many thanks, clearly and succinctly stated. Also thanks for the reassurance that I haven’t missed a blindingly obvious direct inference from the text.