# Second Measurement of Spin

1. Sep 25, 2012

### gametheory

So in my quantum class we learned that if you measure spin in one direction and get h/2 and then in another direction that it will be (plus or minus)h/2 as well. I was wondering how you would know the probability of it being the positive value vs the negative value. It's a function of the angle between the two directions, right?

2. Sep 25, 2012

I'm just a beginner myself, so wait for other replies. I've found that if spin is measured along a given axis of a spin½ particle, then the probability (p) that spin then measured along another axis will have the same sign (+ or -) is:

p = cos(α)/2 + 50%,

where α is the angle the new axis makes with the original axis.

Good luck!

Last edited: Sep 25, 2012
3. Sep 25, 2012

### gametheory

That looks like it works. Do you know where this comes from? I figured it involves the pauli matrices in each of two orthogonal directions acting on the possible wavefunctions, but I wasn't sure how to arrange that into an equation.

4. Sep 25, 2012

### dipstik

you wanna look at stern-gerlach experiment i think. should be some good reading.

5. Sep 25, 2012

### Bill_K

Under a finite rotation, angular momentum states transform into each other with the aid of a unitary matrix. Naturally this matrix is called the rotation matrix, and written d(j)mm'(β) = (j m'|exp(iβ/ħ Jy)|j m) where β is the angle. For j = 1/2, d(j)mm'(β) = $$\left(\begin{array}{cc}cos(β/2) &sin(β/2)\\- sin(β/2)&cos(β/2)\end{array}\right)$$ Thus, if you have a state with Jz = +1/2 and describe it in terms of the spin states quantized along an axis inclined at an angle β to the z axis, the probability amplitude of finding Jβ = +1/2 is cos(β/2) and the probability amplitude of finding Jβ = -1/2 is sin(β/2).

6. Sep 25, 2012

I'm afraid I made mine up from a table of some spin correlations.

7. Sep 25, 2012

### DrChinese

I think that:

Cos^2(theta/2) = Cos(theta)/2 + .5

Where theta = {0, 45, 60, 90 degrees} that looks pretty good.

8. Sep 27, 2012

Interesting! My daughter's old precalculus book lists the "Double Angle Formulas":

cos = 2cos2θ -1 = 1 - 2sin2θ = cos2θ - sin2θ, also

sin = 2sinθcosθ

This raised a follow-up question, if I may. (I'll start a new thread if needed.)
Over what range of angles is the quantum spin correlation formula considered complete?

The table I had listed five spin correlations from 0° to 180°. Classical trigonometry addresses 360° in its unit circle. But it seems that quantum spin is associated with a 720° rotation, at least for fermions.

Last edited: Sep 27, 2012