# The probability that spin will be in -x direction!

1. Jul 19, 2010

### czaroffishies

1. The problem statement, all variables and given/known data

A spin 1/2 particle is in the state $$\left| \Psi \right\rangle$$ = $$\sqrt{2/3}\left|\uparrow\right\rangle + i\sqrt{1/3}\left|\downarrow\right\rangle$$

A measurement is made of the x-component of the spin. What is the probability that the spin will be in the -x direction?

2. Relevant equations

Spin states are represented as linear combinations of the spin in the +z direction and -z direction, since these spins form an orthonormal basis set:
http://en.wikipedia.org/wiki/Spin-½#Mathematical_description

In this case, spin in -x direction is represented by

$$\sqrt{1/2} $\left( {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right)$$$

3. The attempt at a solution

P = $$\left|\left\langle\leftarrow\left|\Psi\right\rangle\left|^{2}$$
= the transpose of the $$\leftarrow$$ matrix, times the $$\Psi$$ matrix, squared.

When calculating this straightforwardly, I will end up with a complex probability because of the i term in the $$\Psi$$ matrix. That doesn't make sense!

So, do I just take this complex number and find its magnitude in the complex plane, and then square that?

Or something else?

Thanks!

2. Jul 19, 2010

### kuruman

The method is correct.
To find the probability you multiply the projection by its complex conjugate. Did you do that or did you just square it?

3. Jul 19, 2010

### czaroffishies

That is a fantastic idea!

P = $$\left(\sqrt{1/2}*\sqrt{2/3}-i\sqrt{1/2}*\sqrt{1/3}\right)*\left(\sqrt{1/2}*\sqrt{2/3}+i\sqrt{1/2}*\sqrt{1/3}\right)$$

= 1/2

Is this correct?

Actually, this is the same answer I got when I used the magnitude-in-the-complex-plane method. Hrm. Coincidence, or no... I will look into it.

4. Jul 19, 2010

### kuruman

It is correct.