# How is it possible that a superposition of z+ and z- can ever equal x-?

1. Apr 8, 2012

### skynelson

Suddenly I am at a loss with something I used to think I understood!

From Consistent Quantum theory, Griffiths, pg 51:

Our basis is |z+>, |z->

I can write |w+> = +cos(α/2)exp(-iθ/2) |z+> + sin(α/2)exp(iθ/2) |z->

In this case, if I choose α = π/2 and θ = π, then this |w+> points in the -x direction (or so he says):

|w+> = (1/√2)(-i) |z+> + (1/√2)(i) |z-> = (i/√2) (|z-> - |z+>)

Can somebody remind me how this superposition of states can correspond to an orthogonal direction, x?

The image in my head is of the z axis, and how no two vectors purely along the z axis can ever add up to a vector along the x axis. I am sure that I am missing the main point here of QM, Hilbert spaces, orthogonal states, and non-commuting operators.

Thanks!

2. Apr 8, 2012

### Matterwave

The states |z+> and |z-> means that when you measure the z-component of the spin, you will get +1 or -1 with certainty. It doesn't mean that the spin points entirely along the z-axis. You can still take a measurement of the x-component of spin of a particle in either |z+> or |z->. You will get +1 with 50% probability and -1 with 50% probability. This is why you shouldn't think of the spin as a definite vector in space.

3. Apr 8, 2012

### martinbn

This vector is an eigenvector for the operator that represents the observable "spin along the x-axis".

4. Apr 8, 2012

### genericusrnme

Is this the same griffiths that did the abomination that is 'introduction to quantum mechanics'?

5. Apr 8, 2012

### Chopin

Like Matterwave said, you can't actually think of a spinor as a definite vector in space. It shares some features in common with it, but the numerical components of a spinor don't have an easy mapping to directions the way that a vector does.

The important thing about spinors is that there are transformations on them which recreate the behavior of 3-dimensional rotations. To understand this, think about a spinor just as a 2-component object, without trying to attach any geometric meaning to it at all. Like any 2-component object, transformations on it are represented by 2x2 matrices.

Write a 3-dimensional rotation of angle $\theta$ around an axis $e$ as $R(e\theta)$. For any rotation, we can then define a matrix $D(R(e\theta))$, such that $D(R(e\theta))D(R(f\phi)) = D(R(e\theta)R(f\phi))$. This means that applying multiple matrices in succession to a spinor is the same as applying the single matrix which corresponds to the resultant rotation--the matrices compose with each other in the same way that rotations do. Therefore, we have defined a set of matrices which is isomorphic to 3-dimensional rotations, on a 2-dimensional object.

The components of the spinor bear no direct resemblance to 3-dimensional space, but now we know that, given any spinor, it makes sense to talk about rotating it in a specific direction, just by applying the appropriate matrix. So in your case, if you start with the state (1,0) and arbitrarily decide that that means "downward", and apply a 90-degree Y-rotation to it, then it makes sense to say that it now points "rightward", etc., because we know that the transformations will follow the right pattern when composed on top of each other.

The generic term for this process is called representation theory, and in this case we say that the matrices $D(R(e\theta))$ form a 2-dimensional representation of the 3-dimensional rotation group, which is known as SO(3). There's also a 3-dimensional representation of SO(3): the plain old set of rotation matrices we all know and love. Furthermore, it can be shown that we can construct a similar set of matrices in any dimension--this is actually done when dealing with higher-order angular momentum/spin combinations in QM.

Last edited: Apr 8, 2012
6. Jun 15, 2012

### skynelson

Thank you for the clear thought provoking replies.

7. Jun 15, 2012

### Delta Kilo

[STRIKE]In Soviet Russia [/STRIKE]in spinor space |z+> and |z-> are orthogonal but |z+> and |x+> are not!