Determine the direction of a spin state given the state

In summary: We can write the wave function as follows:$$|\psi>=\frac{e^{-i\frac{\pi}{12}}}{\sqrt{3}}|+>+e^{i\frac{\pi}{4}}\frac{e^{-i\frac{5\pi}{12}}}{\sqrt{3}}|->$$Then$$\cos(\frac{\theta}{2})=e^{-i\frac{\pi}{12}}/\sqrt{3}$$$$\phi = -\frac{\pi}{12}$$In summary, the conversation discusses the calculation of the quantum state in terms of phase factors and the use of the Bloch sphere representation. It is determined that the method used
  • #1
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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
 
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  • #2
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
 
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  • #3
Gaussian97 said:
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!
 
  • #4
Here's an idea: $$\tan \frac \theta 2 = \frac{|1 - i\sqrt 3|}{|1+i|} = \sqrt 2$$
 
  • #5
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.
 
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  • #6
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
$$
\cos(\frac{\theta}{2})=\sqrt{\frac{1}{3}}
$$
$$
\phi=\frac{13\pi}{12}
$$
 
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  • #7
A neat suggestion.
 

What is a spin state?

A spin state refers to the quantum mechanical property of a particle, such as an electron, that describes its intrinsic angular momentum. It can have two possible values: spin up or spin down.

How is the direction of a spin state determined?

The direction of a spin state is determined by the orientation of the particle's spin relative to a chosen axis. This can be measured using techniques such as magnetic resonance imaging or electron spin resonance.

What are the possible values for a spin state?

The possible values for a spin state are spin up (+1/2) and spin down (-1/2). These values are based on the spin quantum number, which is a fundamental property of particles.

Can a particle have a spin state other than spin up or spin down?

No, a particle can only have a spin state of spin up or spin down. This is a fundamental property of particles and cannot be changed.

How does the spin state of a particle affect its behavior?

The spin state of a particle can affect its behavior in various ways, such as determining its magnetic properties and how it interacts with other particles. It is also a key factor in many quantum phenomena and technologies.

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