The probability that spin will be in -x direction

[czaroffishies]

Homework Statement

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?

Homework 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} \[ \left( {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right) \]$$

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!

 

[kuruman]

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!

The method is correct.

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

Or something else?

Thanks!

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

 

[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.

 

[kuruman]

It is correct.

 

[paranormal]

I would approach this problem by first understanding the physical meaning behind the spin states and how they are represented mathematically. The spin states in this case are represented as a linear combination of the +z and -z spin states, with the coefficients representing the probability amplitudes for each state.

To find the probability of the spin being in the -x direction, we need to find the probability amplitude for the -x spin state. This can be done by taking the inner product between the -x spin state and the given state \left|\Psi\right\rangle. Taking the transpose of the -x spin state, we get \left( {\begin{array}{cc} 1 \\ -1 \\ \end{array} } \right). Multiplying this with the given state \left|\Psi\right\rangle, we get:

\left( {\begin{array}{cc} 1 & -1 \\ \end{array} } \right) \left( {\begin{array}{cc} \sqrt{2/3} \\ i\sqrt{1/3} \\ \end{array} } \right) = \sqrt{2/3} - i\sqrt{1/3}

This is the probability amplitude for the -x spin state. To find the probability, we need to square this amplitude, which gives us:

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

Therefore, the probability that the spin will be in the -x direction is 1, or 100%. This makes sense because the given state is a superposition of the +z and -z spin states, and the +z and -z spin states are orthogonal to the -x spin state. This means that the -x spin state is the only possible outcome for a measurement of the x-component of the spin.