Question

Deduce the condition for minimum uncertainty in${\mathit{S}}_{\mathbf{x}}$ and${\mathit{S}}_{\mathbf{y}}$(that is, equality in the expression role="math" localid="1658378301742" ${\mathit{\sigma }}_{{\mathbf{S}}_{\mathbf{x}}}{\mathit{\sigma }}_{{\mathbf{S}}_{\mathbf{y}}}\mathbf{\ge }\mathbf{(}\mathbf{ħ}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{\left|}<S_z>\mathbf{\right|}$, for a particle of spin 1/2 in the generic state (Equation 4.139). Answer: With no loss of generality we can pick to be real; then the condition for minimum uncertainty is that bis either pure real or else pure imaginary.

Short answer

The condition for minimum uncertainty in $S_x$and $S_y$(that is, equality in the expression ${\sigma }_{{\mathrm{S}}_{\mathrm{x}}}{\sigma }_{{\mathrm{S}}_{\mathrm{y}}}\ge (\mathrm{ħ}/2)|<{\mathrm{S}}_{\mathrm{z}}>|$ , for a particle of spin 1/2 in the generic state is deduced

Step-by-step solution

Definition of spinor

Spinors are complex vector space elements that may be linked to Euclidean space.Spinors, like geometric vectors and more generic tensors, change linearly when the Euclidean space is rotated slightly.

Deduction of the condition for minimum uncertainty inSx andSyfor a particle of spin 1/2 in the generic state

The most generalized spinor is defined as:

$X=\left(\frac{a}{b}\right)$

Normalization of this state gives,

${\left|a\right|}^2+{\left|b\right|}^2=1$

The expectation values of $S_x$, $S_y$, and $S_z$are:

And

Also,

Write a and b in exponential form as:

$$\begin{gathered} a=\left|a\right|e^{i{\varphi }_a} \\ b=\left|b\right|e^{i{\varphi }_b} \end{gathered}$$

Now,

$$\begin{gathered} a{b}^{*}=\left|a\right|\left|b\right|e^{i\left({\varphi }_a-{\varphi }_b\right)} \\ =\left|a\right|\left|b\right|e^{i\theta } \end{gathered}$$

Where $\theta ={\varphi }_a-{\varphi }_b$is the phase difference between a and b.

Now consider,

localid="1658379536890"

And

Now,

localid="1658380148912"

For

$$\begin{gathered} \frac{{ħ}^2}{4}\left(1-4{\left|a\right|}^2{\left|b\right|}^2{\cos }^2\theta \right)\frac{{ħ}^2}{4}\left(1-4{\left|a\right|}^2{\left|b\right|}^2{\sin }^2\theta \right)\frac{{ħ}^2}{4}\left[\frac{{ħ}^2}{4}{\left({\left|a\right|}^2-{\left|b\right|}^2\right)}^2\right] \\ 1-4{\left|a\right|}^2{\left|b\right|}^2\cos \theta -4{\left|a\right|}^2{\left|b\right|}^2\sin \theta +16{\left|a\right|}^4{\left|b\right|}^4{\cos }^2\theta {\sin }^2{\left|a\right|}^4+{\left|b\right|}^4-2{\left|a\right|}^4{\left|b\right|}^4 \\ 1-4{\left|a\right|}^2{\left|b\right|}^2+16{\left|a\right|}^4{\left|b\right|}^4{\cos }^2\theta {\sin }^2\theta ={\left|a\right|}^4+{\left|b\right|}^4-2{\left|a\right|}^4{\left|b\right|}^4 \\ {\left|a\right|}^4+{\left|b\right|}^4=1+16{\left|a\right|}^4{\left|b\right|}^4{\cos }^2\theta {\sin }^2\theta -2{\left|a\right|}^2{\left|b\right|}^2 \\ {\left|a\right|}^4+{\left|b\right|}^4+2{\left|a\right|}^2{\left|b\right|}^2=1+16{\left|a\right|}^4{\left|b\right|}^4{\cos }^2\theta {\sin }^2\theta \\ \left({\left|a\right|}^2+{\left|b\right|}^2\right)=1+16{\left|a\right|}^4{\left|b\right|}^4{\cos }^2\theta {\sin }^2\theta \end{gathered}$$

$$\begin{gathered} \mathrm{However},{\left|\mathrm{a}\right|}^2+{\left|\mathrm{b}\right|}^2=1 \\ 1=1+16{\left|a\right|}^4{\left|b\right|}^{4}co{s}^2\theta si{n}^2\theta \\ {\left|a\right|}^4{\left|b\right|}^{4}co{s}^2\theta si{n}^2\theta =0 \end{gathered}$$

Or

${\left|a\right|}^2{\left|b\right|}^2\cos \theta \sin \theta =0$

If the angle $\theta =0$or $\pi$, then a and b are relatively real.

If the $\theta =\pm \frac{\pi }{2}$, then a and b are relatively imaginary, and a and b being equal to 0 is a trivial case.

Thus, it is proved that .