Derive the probability of spin at arbitrary angle is cos( )

[Sparky_]

TL;DR Summary

From Dr. Leonard Susskind's Stanford Lecture: Quantum Entanglement, Lecture 4, he sets up a "given particle is spin up along n (arbitrary direction) and discusses : what is probability we measure up along another arbitrary m direction

From Dr. Leonard Susskind's Stanford Lecture: Quantum Entanglement, Lecture 4, he sets up a "given particle is spin up along n (arbitrary direction) and discusses : what is probability we measure up along another arbitrary m directionHe does all of the setup, - calculates the eigenvectors and gives the final answer: $\frac{1}{2}(1-\cos(\theta))$

Months and months ago I took a stab at the work and gave up - got really messy
Few days ago I thought (for fun) I would tackle it again, jumping to the end I ended up with:
$$\frac{1}{2}((1+\cos(\theta) ) (1 - m_3^2 -n_3^2 + (m_3n_3)^2)$$
$$= \frac{1}{2} (1+\cos(\theta))(m_3^2-1)(n_3^2-1)$$

The work is many pages of algebra with m's and n's, fortunately a lot canceled and then others grouped to simplify
My current result seems clean but has the terms $(m_3^2-1)(n_3^2-1)$ (with$1-\cos(\theta)$ factored out
I'm hoping you tell me there is a step to cancel or something

The initial matrices before the turn-the-crank work is:

$$\begin{pmatrix}\sqrt{ \frac{1+m_3} {2} } && \frac{1-m_3}{m+}\sqrt{\frac{1+m_3}{2}} \end{pmatrix}\begin{pmatrix}\sqrt{\frac{1+n_3}{2}} \\ \frac{1-n_3}{n-}\sqrt{\frac{1+n_3}{2}}\end{pmatrix}$$Thanks

Sparky_

 

[vanhees71]

The component of the spin in direction of the unit vector $\vec{n}$ is given by
$$\hat{s}_{\vec{n}}=\frac{\hbar}{2} \vec{n} \cdot \vec{\sigma},$$
where $\vec{\sigma}$ denote the three Pauli matrices.

Without loss of generality in your above description you can choose $\vec{n}=\vec{e}_3$. Then with the standard definition of the Pauli matrices the state your spin is prepared in is represented by the vector $(1,0)$.

Now find the eigenvector of $\vec{m} \cdot \vec{\sigma}$ with the eigenvalue $1$. The probability that $\sigma_{\vec{m}}$ takes the value $+\hbar/2$ is then given by the squared modulus of the 1st component of this eigenvector (think about, why?).

 

[Sparky_]

here is the link to the lecture. He (Dr. Susskind) sets up the QM eigenvectors. I am just trying to do the brute force calculation to get $\frac{1}{2} 1+\cos(\theta))$ . He says that this result can come out of "just" the algebra

(You see I must be close) Unfortunately I am 10+ pages of algebra and did have several careless mistakes along the way
here is the link - go to around 32:00 minutes

 

[vanhees71]

Ok, Let's do the calculation. We just take
$$\vec{m}=\begin{pmatrix}\sin \vartheta \\ 0 \\ \cos \vartheta \end{pmatrix}.$$
The matrix we have to diagonalize is
$$\vec{m} \cdot \vec{\sigma}=\begin{pmatrix} \cos \vartheta & \sin \vartheta \\ \sin \vartheta &-\cos \vartheta \end{pmatrix}.$$
To get the eigenvalues we need the characteristic polynomial
$$\mathrm{det} (\vec{m} \cdot \vec{\sigma}-\lambda \hat{1}) = (\cos \vartheta-\lambda)(-\cos \vartheta-\lambda)-\sin^2 \vartheta=\lambda^2-1,$$
i.e., we get the two eigenvalues
$$\lambda_1=+1, \quad \lambda_2=-1.$$
This means the spin component $s_{\vec{m}}$ can take the values $\hbar/2$ and $-\hbar/2$.

For the eigenvector to the eigenvalue +1 we must have
$$\vec{m} \cdot \vec{\sigma} \begin{pmatrix} a\\ b \end{pmatrix}=\begin{pmatrix}\cos \vartheta a + \sin \vartheta b \\ \sin \vartheta a -\cos \vartheta b\end{pmatrix}=\begin{pmatrix} a\\ b \end{pmatrix}.$$
From the first component we get
$$b=\frac{1-\cos \vartheta}{\sin \vartheta}a.$$
Setting $a=N \sin \vartheta$ we get the eigenvector
$$|m_{\vec{m}}=+\hbar/2 \rangle=N \begin{pmatrix}\sin \vartheta \\ 1-\cos \vartheta \end{pmatrix}.$$
The normalization facto must be chosen such that this vector has norm 1,
$$|N|^2 [(1-\cos \vartheta)^2 + \sin \vartheta^2]=|N|^2 2(1-\cos \vartheta) =1.$$
Up to an irrelevant phase factor thus
$$N=\frac{1}{\sqrt{2(1-\cos \vartheta)}}.$$
So we get
$$ |m_{\vec{m}}=+\hbar/2 \rangle=\frac{1}{\sqrt{2(1-\cos \vartheta}} \begin{pmatrix} \sin \vartheta \\ 1-\cos \vartheta \end{pmatrix}.$$
So the probability is
$$P=|\langle m_z=\hbar/2|m_{\vec{m}}=\hbar/2 \rangle|^2=\frac{\sin^2 \vartheta}{2 (1-\cos \vartheta)}.$$
With the formulae
$$\sin \vartheta=2 \sin (\vartheta/2) \cos(\vartheta/2), \cos \vartheta=\cos^2(\vartheta/2)-\sin^2(\vartheta/2)=1-2 \sin^2 \vartheta \; \Rightarrow \; 1-\cos \vartheta=2\sin^2 (\vartheta/2)$$
you finally get
$$P=\cos^2(\vartheta/2)=\frac{1}{2}(1+\cos \vartheta).$$
That's not what you say, but of course for $\vartheta=0$ we must get $P=1$. So this seems to be the correct answer.

 

[Sparky_]

I don't think we are on the same page, I am trying to see if I am close to the $\frac{1}{2}(1-\cos(\theta))$ result
I feel like I am close and don't want to give up again
From the lecture I did the inner product (squared) and worked through a lot of mess

My question is not on setting up the inner product of the eigenvectors, it is given Dr. Suskkind's inner product (squared) from lecture what remaining step do I need to finish / end up with $\frac{1}{2}(1-\cos(\theta))$

Thanks

 

[Sparky_]

the approach I am chasing is with n = $n_1 +\n_2 + n_3$ and m = $m_1 + m_2+m_3$
the angle between is $\theta$
$n- = n_1 - i*n_2 and n+ = n_1 + i*n_2$ likewise for m- and m+

$$\begin{pmatrix}\sqrt{ \frac{1+m_3} {2} } && \frac{1-m_3}{m+}\sqrt{\frac{1+m_3}{2}} \end{pmatrix}\begin{pmatrix}\sqrt{\frac{1+n_3}{2}} \\ \frac{1-n_3}{n-}\sqrt{\frac{1+n_3}{2}}\end{pmatrix}$$

squaring and matrix multiplication:

$$\sqrt{\frac{1+m_3+n_3+m_3n_3}{4}}(\frac{m_+n_- + 1 - m_3 - n_3 + m_3n_3}{m_+n_-}) x \sqrt{\frac{1+m_3+n_3+m_3n_3}{4}}(\frac{m_-n_+ + 1 - m_3 - n_3 + m_3n_3}{m_-n_+})$$

dot dot dot... lots of m's and n's and errors and going backwards to find where I got off ... i get

$$\frac{1}{4}(2+2m_1n1 + 2m_2n_2 + 2m33n_3 - 2m_3^2 -2n_3^2 + 2(m_3n_3)^2 -2m_3^2(m_3n_3)-2n_3^2(m_3n_3) + 2(m_3n_3)^3 - 2(m_1n_1m_3^2 - 2m_1n_1n_3^2 + 2m_1n_1(m_3n_3)^2-2m_2n_2m_3^2 -2m_2n_2n_3^2 + 2m_2n_2(m_3n_3)^2)$$

$$\frac{1}{2}( 1+\cos(\theta) -m_3^2 ( 1+\cos(\theta)) -n_3^2 ( 1+\cos(\theta) )+(m_3n_3)^2( 1+\cos(\theta)))$$

where I am hoping I am a step or so away (or I am there and just don't see it) to the result:
$$\frac{1}{2}(1+\cos(\theta))$$

 

[PeroK]

$$\frac{1}{2}( 1+\cos(\theta) -m_3^2 ( 1+\cos(\theta)) -n_3^2 ( 1+\cos(\theta) )+(m_3n_3)^2( 1+\cos(\theta)))$$where I am hoping I am a step or so away (or I am there and just don't see it) to the result:
$$\frac{1}{2}(1+\cos(\theta))$$

If those two are equal, then $m_3^2n_3^2 - m_3^2 - n_3^2 = 0$, which can't be true for all choices of $m_3$ and $n_3$. So, you must have made a mistake.