# How do you visualize the spin of electron in your head?

1. Aug 14, 2010

### orienst

How do you visualize the spin of electron in your head? I can’t imagine such a vector which has the same projection in any direction and at any time. Is it an understandable object in three dimension?

Last edited: Aug 14, 2010
2. Aug 16, 2010

### Demystifier

It cannot be imagined. This is actually about contextuality of QM. Electron does not HAVE the same projection in any direction at any time. Instead, by changing the kind of measurement you perform on it, its properties change in time. So, if you measure spin in the z direction, it will have the value 1/2 in THAT direction, but not in other directions.

If you want to know what is its spin when you DON'T measure it at all, then you should know that standard QM does not give an answer to this question. Such a question can only be answered within non-standard hidden-variable interpretations of QM, such as the Bohmian interpretation.

3. Aug 17, 2010

### orienst

Thanks, Demystifier !Perhaps I didn’t express clearly. When I say “a vector a which has the same projection in any direction and at any time”, I mean I’ve made a measurement in a random direction at some time. Certainly I will get $$\hbar$$/2 or -$$\hbar$$/2. So I say that the electron spin has projection of the same magnitude in any direction.

4. Aug 17, 2010

### Demystifier

That is fine, but I don't see why it is difficult to visualize it.

But it is misleading to say that. Let me illustrate it by a common-sense example.

Suppose that I want to determine where I am. So first I go to check if I am in China. The only way to check it is to go to China to see if I am there. And if I do that, I find out that I really am in China. But then I do the same for America, and find out that I am in America too. In fact, wherever I try to find out if I am there, I find that I am.

And yet, I CANNOT say that I am everywhere. All I can say is that I am wherever I try to observe if I am there. The type of observation I chose to perform influences the result of that observation. This is called - contextuality.

5. Aug 17, 2010

### orienst

6. Aug 17, 2010

### Demystifier

If you understood the point of my story, you should not have any problems to visualize such a vector. Apparently, you still have such problems, which means that you have not understood the point of my story.

7. Aug 20, 2010

### Freshtictac

i imagine the cloud... though it is a fixed point in time and space it moves fast enough to create a cloud out of one particle.

8. Aug 20, 2010

### TheAlkemist

I like this explanation! Thanks!

9. Aug 21, 2010

### orienst

Thanks! I think what you mean is that the measurement collapse the spin state. But what does the spin vector like before your measurement?

10. Aug 21, 2010

### Tomsk

A neat way of visualizing spin is with the Bloch Sphere. In fact this lets you visualize any two state quantum system, such as the ground and excited state of an atom, photon polarization and so on.

The state space can be visualized as a unit sphere in three dimensions (i.e. the surface of a unit ball, the interior of the ball is important for mixed states). The quantum state $|\psi\rangle$ is described in terms of the density matrix $\rho = |\psi\rangle \langle\psi |$. This has the advantage of removing the arbitrary global phase. The density matrix can be written in terms of a unit vector s called the Bloch Vector like this: $\rho = \frac{1}{2}(\mathbb{I}+s\cdot \sigma)$ where I is the 2x2 identity matrix and sigma means a vector of Pauli matrices. So the state is entirely encoded in s, which is really easy to visualize, since it's just a unit vector in normal 3D space!

The Bloch vector is really useful for calculating expectation values. The expectation value of the spin in a direction n is just $\langle S_n \rangle = \frac{\hbar}{2}n \cdot s$ (n is a unit vector). Also, the Bloch vector after a measurement along n is $\pm n$ with probability $P_\pm = \frac{1}{2}(1 \pm n \cdot s)$ (+ means spin up, - means spin down). In fact any hermitian operator A that acts on a spin-1/2 can be written as a scalar coefficient of the identity matrix A_S and a vector A_V like this: $A = A_S \mathbb{I} + A_V \cdot \sigma$. It turns out that the expectation value is given by $\langle A \rangle = A_S + A_V \cdot s$.

So you can describe a single spin-1/2 system in terms of a 3-component vector, which is easy to visualize. This doesn't generalize to higher spins, or multiple spin-1/2 systems very nicely though (although the two problems are related - I think you need multiple Bloch vectors, but the number is not obvious, for N spin-1/2 systems you need 2^N-1 or something). I hope that helps a bit. Sometimes spinors are easier to work with though!

11. Aug 21, 2010

### Naty1

I like post #4 as one way of thinking about it.

Another way is to consider the observables associated with quantum spin....half integer spin [in addition to integer spin] is hard to "picture" for example, angular momentum, precession, Pauli Exclusion Principle, degress of freedom, quanta, all affect submicroscope particle behavior and are aspects of "spin"...

So I'd suggest you consider "observable" effects...

A related viewpoint is Noether's Theorem: http://en.wikipedia.org/wiki/Noether's_theorem
Spin is a type of angular momentum.

In the same way that visualizing a spinning macroscopic top, for example, only reveals a very few of it's physical properties, "visualizing" subatomic spins, if it could be done, most likely would only reveal superficial physical characteristics.

For example, spin leads directly to the Pauli exclusion principle
(http://en.wikipedia.org/wiki/Pauli_exclusion_principle) but I doubt even if you could "visualize" spin itself you'd realize all it's implications.

Last edited: Aug 22, 2010
12. Aug 23, 2010

### fala

the hamiltonian of a system must be invariant by a small rotation around an arbitary axis in 3d.this gives the generators namely angular momentum. in 4d gives angular momentum and spin.
you can study neother theorm in QFT.

13. Aug 23, 2010

### Demystifier

As I said in #2, standard QM does not offer an answer to this question. It can only be answered within some hidden-variable interpretations of QM.

To be specific, let me explain a Bohmian view (which by no means is the only possible view). A particle does not have a spin at all. And you never really measure spin. In the Stern-Gerlach apparatus (which is supposed to measure spin) all what you really measure is the final position of the particle traveling through a magnetic field. This position may be up or down, which you (if you are not a bohmian) may interpret as being caused by interaction between magnetic field and magnetic moment of the particle caused by the particle spin. But that is only an interpretation, you never observe spin as such. All you really observe are particle positions. And according to the Bohmian interpretation, this is the only property that a particle really possesses - a position at each time - even if you don't measure it.

Last edited: Aug 23, 2010
14. Nov 5, 2011

### fala

spin is an intrinsic angular momentum of Dirac field. you can derive it by investigation of invariance of Dirac Lagrangian under a infinitesimal variation of spinors. spinors are imaginary entities.you can not define spin as function of space-time(it may be possible in string theory??)

15. Nov 5, 2011

### juanrga

Spin in QM is not given by a vector, therefore it cannot be visualized as such. If you want a 3D visualization think of it as a kind of cone with height $S_z$ and module $S$,

http://t1.gstatic.com/images?q=tbn:ANd9GcSezTVXJy-R7VTSaShcjf-7TkyNDlYbnbKi7p0AQbpUesZKq9wPUA

although this is only a very rough view.

Last edited: Nov 5, 2011