- #1

PhysicsKush

- 29

- 4

- Homework Statement
- Verify that ##\Delta S_{x} = \sqrt{\langle S_{x}^{2} \rangle - \langle S_{x}\rangle^{2}} = 0## for the state ##\ket{+x}##.

- Relevant Equations
- $$\Delta S_{x} = \sqrt{\langle S_{x}^{2} \rangle - \langle S_{x}\rangle^{2}} = 0$$

$$ \ket{+x} = \alpha \ket{+z} + \beta \ket{-z}$$

$$ \langle S_{x} \rangle = |\alpha|^{2} \left( \frac{\hbar}{2}\right) + |\beta|^{2} \left( \frac{\hbar}{2}\right)$$

By definition , ##\ket{+x} = \alpha \ket{+z} + \beta \ket{-z}.## Therefore we proceed ,

\begin{align*}

\langle S_{x} \rangle &= \lvert \alpha \rvert^{2} \left(\frac{\hbar}{2}\right) + \lvert \beta\rvert^{2} \left(-\frac{\hbar}{2}\right) = (\alpha^{2} - \beta^{2})\left(\frac{\hbar}{2}\right).\\

\left(\langle S_x \rangle \right)^{2} &= \left(\alpha^{2} - \beta^{2} \right)^{2} \left(\frac{\hbar^{2}}{4}\right). \\

\therefore \Delta S_{x} &= \sqrt{\left(\frac{\hbar^{2}}{4}\right) - \left(\frac{\hbar^{2}}{4}\right)(\alpha^{2} - \beta^{2})^{2}} \\

&= \frac{\hbar}{2}\sqrt{1- \left(\alpha^{2} - \beta^{2}\right)^{2}} \\

&= \frac{\hbar}{2}\sqrt{1-(\alpha^{2} - (1-\alpha^{2}))^{2}} \\

&= \frac{\hbar}{2}\sqrt{1-(4\alpha^{4} - 4\alpha^{2} +1)} \\

&= \frac{\hbar}{2}\sqrt{4\alpha^{2} - 4\alpha^{4}} \\

&= \hbar \alpha \beta \neq 0 ?

\end{align*}

I don't understand what I'm missing or where I'm going wrong. Any indications would be appreciated.

\begin{align*}

\langle S_{x} \rangle &= \lvert \alpha \rvert^{2} \left(\frac{\hbar}{2}\right) + \lvert \beta\rvert^{2} \left(-\frac{\hbar}{2}\right) = (\alpha^{2} - \beta^{2})\left(\frac{\hbar}{2}\right).\\

\left(\langle S_x \rangle \right)^{2} &= \left(\alpha^{2} - \beta^{2} \right)^{2} \left(\frac{\hbar^{2}}{4}\right). \\

\therefore \Delta S_{x} &= \sqrt{\left(\frac{\hbar^{2}}{4}\right) - \left(\frac{\hbar^{2}}{4}\right)(\alpha^{2} - \beta^{2})^{2}} \\

&= \frac{\hbar}{2}\sqrt{1- \left(\alpha^{2} - \beta^{2}\right)^{2}} \\

&= \frac{\hbar}{2}\sqrt{1-(\alpha^{2} - (1-\alpha^{2}))^{2}} \\

&= \frac{\hbar}{2}\sqrt{1-(4\alpha^{4} - 4\alpha^{2} +1)} \\

&= \frac{\hbar}{2}\sqrt{4\alpha^{2} - 4\alpha^{4}} \\

&= \hbar \alpha \beta \neq 0 ?

\end{align*}

I don't understand what I'm missing or where I'm going wrong. Any indications would be appreciated.