I am currently trying to gain a fuller understanding of the meaning of various spin states and their relation to the direction of measurement by a Stern-Gerlach device. I came across two spin-${1 \over 2}$ states, the first of which given by:

$$\left|\psi_1\right\rangle={1 \over 2} \left|+z\right\rangle+{i\sqrt3 \over 2}\left|-z\right\rangle $$

I know that if a spin-${1 \over 2}$ particle is prepared spin up along an axis specified by:


Then its spin state is given by:

$$\left|+n\right\rangle=\cos{{\theta \over 2}}\left|+z\right\rangle+e^{i\phi}\sin{{\theta \over 2}}\left|-z\right\rangle$$

By inspection, then, I determined that for $\left|\psi_1\right\rangle$ possible values for $\theta$ and $\phi$ are $\theta = {2\pi \over 3}$ and $\phi={\pi \over 2}$. Knowing these angles I then determined the direction in which $\left|\psi_1\right\rangle$ is spin up. This made sense.

The second state I encountered, however, is given by:

$$\left|\psi_2\right\rangle={-i \over 2}\left|+z\right\rangle+{\sqrt3 \over 2}\left|-z\right\rangle$$

According to the expression I gave above for $\left|+n\right\rangle$, however, in order for the amplitude for $\left|+z\right\rangle$ to be complex, the argument passed to the cosine function must be complex. In turn, this means $\theta$ must be complex. This doesn't make sense to me, because my understanding is that $\hat{\boldsymbol{n}}$ is a vector in ordinary 3-dimentional space.

This lead me to believe I am determining $\theta$ incorrectly. If I am making a mistake in determining the values for $\theta$ and $\phi$, I am curious to know what it is. If my method is correct, then why am I coming up with a complex angle for the spin direction of $\left|\psi_2\right\rangle$?

  • $\begingroup$ The expression for the $|+n\rangle$ state also includes a global phase (which is usually not mentioned explicitly). You could assume a global phase of $\pi/2$ for $|\psi_2\rangle$, i.e. multiply it with $i$ and find $|\psi_2\rangle \equiv |\psi_1\rangle$ $\endgroup$ – jayann Jan 15 '15 at 19:18

Recall that two ket vectors which differ by a global phase represent identical physical states. In this case, $|\psi_2\rangle = i|\psi_1\rangle$, and the states are therefore equivalent. One can always use this global phase freedom to make the coefficient of $\lvert+z\rangle$ real, and then a parametrisation in terms of the angles $\theta$ and $\phi$ becomes straightforward.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.