# 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

Last edited:
• berkeman

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

Homework Helper
Gold Member
2022 Award
Here's an idea: $$\tan \frac \theta 2 = \frac{|1 - i\sqrt 3|}{|1+i|} = \sqrt 2$$

Homework Helper
Gold Member
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.

• Foracle and PeroK
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
$$\cos(\frac{\theta}{2})=\sqrt{\frac{1}{3}}$$
$$\phi=\frac{13\pi}{12}$$

• Foracle, PeroK and kuruman