Determine the direction of a spin state given the state

[Foracle]

Homework Statement

Given an unnormalized spin state
$\Psi=(1+i)|+>-(1+i\sqrt{3})|->$
which direction does this spin point to?

Relevant Equations

$|n;+> = cos\frac{\theta}{2}|+>+sin\frac{\theta}{2}e^{i\phi}|->$

From the relevant equation above, there is not imaginary part in the |+> state, so I multiplied the state by (1-i). The state is then :
$\Psi=(2)|+>-(1+\sqrt{3})+i(\sqrt{3}-1)|->$
Then I normalize it :
$\Psi=(\frac{1}{\sqrt{3}})|+>-\frac{1}{2\sqrt{3}}(1+\sqrt{3})+i(\sqrt{3}-1)|->$

From the |n;+> equation above, I concluded from the |+> part that :
$cos\frac{\theta}{2}=\frac{1}{\sqrt{3}}$
But when I did the same thing from the |-> part, I got different value from $cos\frac{\theta}{2}$

My guess is that this method of determining direction of spin state is probably wrong.

Edit: Somehow the Latex is not working, I'm trying to figure out how to fix this

 

[Gaussian97]

I think your method is good.
Notice that the coefficient of the - state gives you $\sin{\frac{\theta}{2}}$, not the cosine, maybe that's the problem.
To use LaTeX you need to use the $ or the # symbols (2before and 2 after the expression)
Take a look at the LaTeX Guide

 

[Foracle]

I think your method is good.
Notice that the coefficient of the - state gives you $\sin{\frac{\theta}{2}}$, not the cosine, maybe that's the problem.
To use LaTeX you need to use the $ or the # symbols (2before and 2 after the expression)
Take a look at the LaTeX Guide

Oh yeah, that's right. Thanks man! Can't believe I missed such that .

About the Latex, I have added the # symbols but it seems like its still not working. Do you know what's wrong with it?
Edit: Never mind, the Latex works now. Thanks again!

 

[PeroK]

Here's an idea: $$\tan \frac \theta 2 = \frac{|1 - i\sqrt 3|}{|1+i|} = \sqrt 2$$

 

[kuruman]

Here is another idea. You have found that
$\Psi=(\frac{1}{\sqrt{3}})|+>-\frac{1}{2\sqrt{3}}(1+\sqrt{3})+i(\sqrt{3}-1)|->$ from which you can identify

$\sin\dfrac{\theta}{2}e^{i\phi}=-\dfrac{1}{2\sqrt{3}}(1+\sqrt{3})+i(\sqrt{3}-1).$
Separate real and imaginary parts, $$\sin\frac{\theta}{2}\cos\phi=-\dfrac{1}{2\sqrt{3}}(1+\sqrt{3})~;~~\sin\frac{\theta}{2}\sin\phi=\sqrt{3}-1$$ and get $\tan\phi$ from this.

 

[Fred Wright]

I have the following suggestions. First, write the wave function in terms of phase factors. We have.
$$
(1+i)|+>= \sqrt{2}e^{i\frac{\pi}{4}}|+>
$$
$$
-(1+i\sqrt{3})|->=-2e^{i\frac{\pi}{3}}|->=2e^{i\frac{4\pi}{3}}|->
$$
$$
|\psi>=\sqrt{2}e^{i\frac{\pi}{4}}|+>+2e^{i\frac{4\pi}{3}}|->
$$
Now normalize the w.f. Since
$$
\sqrt{2}^2 + 2^2=6
$$
$$
|\psi>=\frac{1}{\sqrt{6}}(\sqrt{2}e^{i\frac{\pi}{4}}|+>+2e^{i\frac{4\pi}{3}}|->)
$$
$$
=\sqrt{\frac{1}{3}}e^{i\frac{\pi}{4}}|+>+\sqrt{\frac{2}{3}}e^{i\frac{4\pi}{3}}|->
$$
With the Bloch sphere representation we know that the quantum state doesn't change if we multiply the w.f. by any number of unit norm, i.e. $e^{i\zeta}|\psi>=|\psi>$. For our w.f. multiply by $\zeta=-i\frac{\pi}{4}$ to get
$$
\sqrt{\frac{1}{3}}|+>+\sqrt{\frac{2}{3}}e^{i\frac{13\pi}{12}}|->
$$
Thus
$$
\cos(\frac{\theta}{2})=\sqrt{\frac{1}{3}}
$$
$$
\phi=\frac{13\pi}{12}
$$

 

[kuruman]

A neat suggestion.