# How can Spin-up z and spin-down z span the space?

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##\def \sqx{\frac{1}{\sqrt{2}}}##
If Szu represents spin-up in the z-direction and Szd represents spin-down in the z-direction then the vector which represents spin-up in the x-direction is given by the superposition of the z states $$|S_{zu}\rangle =\begin{bmatrix}1 \\ 0\end{bmatrix}, |S_{zd}\rangle =\begin{bmatrix}0 \\1\end{bmatrix}$$ and

$$|S_{xu}\rangle =\begin{pmatrix} \sqx \\ \sqx\end{pmatrix} = \sqx \left[ |S_{zu}\rangle + |S_{zd}\rangle\right]$$

How can the two z-vectors span the entire space? How can the spin in the x-direction be a linear combination of the spin vectors in only the z-direction?

Last edited:
edguy99

## Answers and Replies

ShayanJ
Gold Member
These are not vectors in the 3 dimensional space we live in. These are quantum states, vectors in a Hilbert space. In this case the Hilbert space is two dimensional and any two orthonormal vectors can serve as a basis for it. So you can either use Szu and Szd or Sxu and Sxd, etc.

Nugatory
The two dimensions complex spinor can be projected to the three dimensional space through the Pauli matrices, σx, σy and σz.

if you have the spinor |S> and you want the probability to find a spin up (or down) for the x direction you just have to do: |Sx> = σx|S>.

If the amplitudes are properly normalized ||Sx>|2 = 1

These are not vectors in the 3 dimensional space we live in. These are quantum states, vectors in a Hilbert space. In this case the Hilbert space is two dimensional and any two orthonormal vectors can serve as a basis for it. So you can either use Szu and Szd or Sxu and Sxd, etc.
Three dimensional euclidean space is a Hilbert Space so just saying the vectors are in a Hilbert Space is not really illuminating. Saying it is a two dimensional Hilbert space is a little more illuminating. What 2-dimensional Hilbert Space is it? What are the dimensions? And why could't you use the more familiar 3-dimensional Euclidean space?

Last edited:
ShayanJ
Gold Member
Three dimensional euclidean space is a Hilbert Space so just saying the vectors are in a Hilbert Space is not really illuminating.
Fair point. So let's say its a different Hilbert space than the 3 dimensional Euclidean space.
What 2-dimensional Hilbert Space is it? What are the dimensions? And why could't you use the more familiar 3-dimensional Euclidean space?
Its a two dimensional complex Hilbert space with the inner product that you know for different spin states.
The dimensions are + and - in any direction you want.
Because we're not talking about vectors in 3-dimensional Euclidean space,we're talking about the spin state of a spin-1/2 particle. A spin-1/2 particle has 2 orthogonal spin states so we should use a 2-dimensional Hilbert space for describing it. Also we never use the Euclidean space as the Hilbert space for the state for any degrees of freedom of any quantum system, simply because northwest is not a quantum state, but pointing towards northwest is! Its the same reason the Euclidean space is not the Hilbert space for the translational degrees of freedom.

edguy99
edguy99
Gold Member
I like to visualize an axis of spin on a bloch sphere. When viewed from the front, an axis of spin that is up (ie. above the equator), may be on the left side of vertical or the right side of vertical. If that same axis of spin is viewed from the top, the ones on the left side of the vertical will be spin down (from the viewpoint of the top) and the ones on the right side of the vertical will be spin up.

The axis of spin of a spin up particle, has a 50% chance of being up when viewed from the top and a 50% chance of being down when viewed from the top.

stevendaryl
Staff Emeritus
It might be helpful to look at the explicit relationship between vectors and spinors.

Let $\hat{A}$ be a unit vector in some direction, and let $\psi$ be the two-component spinor corresponding to the spin state of a particle that is spin-up in direction $\hat{A}$, and let $\chi$ be the spinor for spin-down relative to $\hat{A}$. These relationships can be summarized in terms of the Pauli spin matrices $\sigma_x, \sigma_y, \sigma_z$ by:

$(\hat{A} \cdot \vec{\sigma}) \psi = \psi$
$(\hat{A} \cdot \vec{\sigma}) \chi = - \chi$

It's convenient to rewrite this in terms of projection operator:

$P_{\hat{A}} = \frac{1}{2} (1 + \hat{A} \cdot \vec{\sigma})$

In terms of $P_{\hat{A}}$, we can write:

$P_{\hat{A}} \psi= \psi$
$P_{\hat{A}} \chi = 0$

So $P_{\hat{A}}$ "projects out" the component of a spinor that is spin-up relative to $\hat{A}$.

Here's a cool fact about $P_{\hat{A}}$ and $\psi$:

$\psi \psi^\dagger = P_{\hat{A}}$

where $\dagger$ means the complex-conjugate of the transpose. So there is a strange sense in which $\psi = \sqrt{P_{\hat{A}}}$ (sort of).

If you use the Pauli spin matrices, then you can find the components of $\hat{A}$ in terms of the components of $\psi$. Letting $\psi = \left( \begin{array} \\ a \\ b \end{array} \right)$, then

$A_z = |a|^2 - |b|^2 = 2 |a|^2 - 1$ (the spinor is normalized so that $|a|^2 + |b|^2 = 1$
$A_x = a^* b + a b^*$
$A_y = i (a^* b - b^* a)$

You can invert these to find $a$ and $b$ in terms of $\hat{A}$. This is easier if you represent $\hat{A}$ in spherical coordinates:

$A_z = cos(\theta), A_x = sin(\theta) cos(\phi), A_y = sin(\theta) sin(\phi)$

Then in terms of $\theta$ and $\phi$,

$a = cos(\theta/2) e^{-i \phi/2}$
$b = sin(\theta/2) e^{+i \phi/2}$

The relationship is two-to-one, in the sense that if you rotate $\hat{A}$ through $2\pi$ about any axis, you get back to the same vector, but the corresponding spinor $\psi$ changes sign. This can be seen easily from the above relationship if you choose the z-axis. Then rotating corresponds to changing $\phi$, so if you rotate by $2 \pi$, that means changing the phase of $a$ by $e^{-i (2\pi)/2} = -1$ and changing the phase of $b$ by $e^{+i (2\pi)/2} = -1$

So as far as how a two-component spinor can represent arbitrary directions in 3-D space, it sort of makes sense: They both are characterized (up to a normalization and overall phase) by the pair of real numbers $\theta$ and $\phi$.

edguy99 and ShayanJ
How can the two z-vectors span the entire space?
When the outcome of a measurement of a particular observable can take on only one of two values, we represent the underlying state as a vector in C2 (and the distinct possible measurements as orthogonal vectors, and thus a basis).

How can the spin in the x-direction be a linear combination of the spin vectors in only the z-direction?
You can discover, experimentally, that x-direction spin and y-direction spin are non-commuting. This suggests they share a state space. And indeed, one discovers vectors (and corresponding projectors) in C2 that model the results of measuring spin in the three physical axes.

First of all, thanks to everyone for the really great responses. I do appreciate them. I have already learned so much by investigating each of the answers which I am continuing to do.

I have many questions, but the simplest is regarding this response...

When you say above "that x-direction spin and y-direction spin are non-commuting. This suggests they share a state space.". What do you mean by they "share a state space? and why does not commuting suggest that?

IANAP. Hopefully others will correct me for any imprecisions.

By "state space" we mean the Hilbert space used to represent a state vector. If the measurement of observable A and observable B commute, it means that we can simultaneously get well-defined values for each of them. In that case, the state space for observable A and observable B can be considered separately. They're two unrelated vector spaces (and if we want to represent the pair of observables, we use the tensor product of those spaces). This "allows" them to evolve independently. (I put "allow" in quotes because obviously the math does not make anything happen, but it's a good way of thinking about things.)

If they cannot be measured simultaneously, it is because there is a state vector in some space that encodes both properties in some way. When observable A is measured, this projects the state vector onto one of the eigenstates of A (i.e., one of the two orthogonal vectors representing distinct measurements). This can in general change the distribution of outcomes for B (whose probabilities can be calculated by projecting the state vector onto B's eigenstates).

Maybe this can be put more simply. I can give you the space C2, and three different basis sets {x+, x-}, {y+, y-}, {z+, z-}, and when you take the projection of x+ onto y- (for example) you will get a vector whose length will help you predict how likely an electron in x+ will be measured in y- (as confirmed by experiment). This suggests that the model I gave you is indeed a good model for electron spin.