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Firstly, I know this is a physics problem, but this forum is so much more active, and I'm sure someone here could help me. The problem is the following:

Given a spin-1/2 particle, and the generic state:

$$ |\psi\rangle = e^{i\phi}cos(\theta)|z_+\rangle + > e^{-i\phi}\sin(\theta)|z_{-}\rangle $$

find $\Delta S_x$, $\Delta S_y$ and $\langle S_z \rangle$ to show $\Delta S_x \Delta S_y \ge \frac{\hbar}{2} |\langle S_z \rangle|$

I think this should be pretty straightforward. Given I know the uncertainty of an opperator is given by

$$ \Delta A =\sqrt{\langle A^2\rangle - \langle A \rangle^2 } $$

I also know that

\begin{align*} S_z= \frac{\hbar}{2}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad S_x= \frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\qquad S_y= \frac{\hbar}{2}\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}\\ \implies S_z^2=S_x^2=S_z^2= \frac{\hbar^2}{4}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align*}

So after a bunch of calculations I get:

\begin{align*} \langle S_z\rangle&=\frac{\hbar}{2}\cos(2\theta)\\ \langle S_x\rangle&=\hbar\sin(\theta)\cos(\theta)\cos(2\phi), \qquad \langle S_x^2\rangle=\frac{\hbar^2}{4}\\ \langle S_y\rangle&=-\hbar\sin(\theta)\cos(\theta)\sin(2\phi), \qquad \langle S_y^2\rangle=\frac{\hbar^2}{4} \end{align*}

Finaly:

\begin{align*} \Delta S_x = \sqrt{\frac{\hbar^2}{4} -(\hbar \sin(\theta)\cos(\theta)\cos(2\phi))^2}\\ \Delta S_y = \sqrt{\frac{\hbar^2}{4} -(-\hbar \sin(\theta)\cos(\theta)\sin(2\phi))^2}\\ \frac{\hbar}{2}|\langle S_z\rangle|= \frac{\hbar^2}{4}|\cos(2\theta)| \end{align*}

Somehow I simply can't prove what is asked from this. Either it is a simple step I am not seeing, or I have some calculation mistake. I would appreciate any help

H44S
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1 Answers1

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All you have left is to show that $$\sqrt{\frac{\hbar^2}{4} -(\hbar \sin(\theta)\cos(\theta)\cos(2\phi))^2}\sqrt{\frac{\hbar^2}{4} -(\hbar \sin(\theta)\cos(\theta)\sin(2\phi))^2} \ge \frac{\hbar^2}{4}|\cos(2\theta)|$$

You can multiply by $\frac{4}{\hbar^2}$ to make it $$\sqrt{1 -(2\sin(\theta)\cos(\theta)\cos(2\phi))^2}\sqrt{1 - (2\sin(\theta)\cos(\theta)\sin(2\phi))^2} \ge |\cos(2\theta)|$$

Or equivalently, since $2\sin(\theta) \cos(\theta) = \sin(2\theta)$, $$\sqrt{1 -(\sin(2\theta)\cos(2\phi))^2}\sqrt{1 - (\sin(2\theta)\sin(2\phi))^2} \ge |\cos(2\theta)|$$

Square both sides, and use that $|\cos(2\theta)| = \sqrt{1-\sin^2(2\theta)}$ to make it $$\left(1-\left(\sin\left(2\theta\right)\cos\left(2\phi\right)\right)^{2}\right)\left(1-\left(\sin\left(2\theta\right)\sin\left(2\phi\right)\right)^{2}\right) \ge 1-\sin^{2}\left(2\theta\right)$$

The LHS simplifies to (after using that $\sin^2(2\phi) + \cos^2(2\phi) = 1$) $$1-\sin^{2}(2\theta)+\left(\sin(2\theta)\sin(2\phi)\right)^{2}\left(\sin(2\theta)\cos(2\phi)\right)^{2}$$

which must be greater than or equal to $1-\sin^2(2\theta)$, since the last term is always nonnegative.

Varun Vejalla
  • 8,873