91影视

A function's Taylor series is an infinite sum of terms expressed in terms of the function's derivatives at a single point. For most ordinary functions, the function and the sum of its Taylor series are identical around this point.

(a)

A function of Taylor series expansion about is given by ,

$$\begin{gathered} f\left(x+x_0\right)=\underset{n=0}{\overset{鈭�}{鈭�}}\frac{1}{n!}{x}_0^n{\left(\frac{d}{dx}\right)}^{n}f\left(x\right) \\ f(蠒+蠄)=\underset{n=0}{\overset{鈭�}{鈭�}}\frac{1}{n!}{蠄}^n{\left(\frac{d}{d蠒}\right)}^{n}f\left(蠒\right) \end{gathered}$$

Using, $L_z=\frac{\mathrm{鈩�}}{i}\frac{d}{d蠒}$

And $\frac{d}{d蠒}=\frac{i{L}_z}{\mathrm{鈩�}}$

$\begin{array}{rcl}f(蠒+蠄)& =& \underset{n=0}{\overset{鈭�}{鈭�}}\frac{1}{n!}{蠄}^n{\left(\frac{i{L}_z}{\mathrm{鈩�}}\right)}^{n}f\left(蠒\right)\\ & =& \underset{n=0}{\overset{鈭�}{鈭�}}\frac{1}{n!}{\left(\frac{i{L}_z蠄}{\mathrm{鈩�}}\right)}^{n}f\left(蠒\right)\\ & & \end{array}$

Known that $e^x=\underset{n=0}{\overset{鈭�}{鈭�}}\frac{x^n}{n!}$

$鈭�f(蠒+蠄)=e^{\left(\frac{u渭蠄}{h}\right)}f\left(蠒\right)$

(b)

If M is a matrix, such that$M^2=1$, then

$\begin{array}{rcl}{e}^{iM蠒}& =& 1+iM蠒+\frac{{\left(iM蠒\right)}^2}{2!}+\frac{{\left(iM蠒\right)}^3}{3!}+鈥�\\ & =& 1+iM蠒-\frac{M^2{蠒}^2}{2!}-\frac{i{M}^3{蠒}^3}{3!}+鈥�\\ & =& 1+iM蠒-\frac{1}{2}{蠒}^2-\frac{iM{蠒}^3}{3!}+鈥�.\left[鈭�M^2=1\right]\\ & =& \left(1-\frac{1}{2}{蠒}^2+\frac{1}{4!}{蠒}^4-鈥�.\right)+iM\left(蠒-\frac{{蠒}^3}{3!}+\frac{{蠒}^5}{5!}+鈥�.\right)\\ & =& \cos 蠒+iM\sin 蠒\\ & & \end{array}$

represents the rotation through an angle $蠒$Rotation,

$\begin{array}{rcl}R& =& e^{i{蟽}_x\frac{蟺}{2}}\\ & =& \cos \left(\frac{蟺}{2}\right)+i{蟽}_x\sin \left(\frac{蟺}{2}\right)\\ & =& i{蟽}_x\\ & =& i\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\\ & & \end{array}$

Such that,

$\begin{array}{rcl}R{蠂}_{+}& =& i\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ & =& i\left[\begin{array}{c}0\\ i\end{array}\right]\\ & =& i{蠂}_{-}\end{array}$

Therefore, the matrix converts the spin-up into a spin-down with a factor i .

(c)

Here,

$\begin{array}{rcl}R& =& e^{i{0}_y\frac{蟺}{4}}\\ & =& \cos \frac{蟺}{4}+i{蟽}_y\sin \frac{蟺}{4}\\ & =& \frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}{蟽}_y\\ & =& \frac{1}{\sqrt{2}}\left[\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+i\left[\begin{array}{cc}0& -i\\ i& 0\end{array}\right]\right]\\ & =& \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\end{array}$

And

$\begin{array}{rcl}R{蠂}_{+}& =& \frac{1}{\sqrt{2}}\left[\begin{array}{cc}1& 1\\ -1& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ & =& \frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ -1\end{array}\right]\\ & =& \frac{1}{\sqrt{2}}\left({蠂}_{+}-{蠂}_{-}\right)\\ & =& {{蠂}_{-}}^{\left(x\right)}\\ & & \end{array}$

The spin-up along z has become spin-down along$x^{\text{'}}$.

(d)

Here,

$\begin{array}{rcl}R& =& e^{i{蟽}_{,}\frac{2蟺}{2}}\\ & =& e^{i{蟽}_z蟺}\\ & =& \cos 蟺+i{蟽}_z\sin 蟺\\ & =& -1\\ & & \end{array}$

As a result, rotating the spinor$360$degrees alters the sign of the spinor.

It doesn't matter, though, because the sign of $蠂$is arbitrary.

(e)

Let鈥檚 consider that

$\begin{array}{rcl}(蟽路\hat{n}{)}^2& =& {\left({蟽}_x{n}_x+{蟽}_y{n}_y+{蟽}_z{n}_z\right)}^2\\ & =& \left({蟽}_x{n}_x+{蟽}_y{n}_y+{蟽}_z{n}_z\right)\left({蟽}_x{n}_x+{蟽}_y{n}_y+{蟽}_z{n}_z\right)\\ & =& {蟽}_x^2{n}_x^2+{蟽}_y^2{n}_y^2+{蟽}_z^2{n}_z^2+n_x{n}_y\left({蟽}_x{蟽}_y+{蟽}_y{蟽}_x\right)+n_y{n}_z\left({蟽}_y{蟽}_z+{蟽}_z{蟽}_y\right)+n_z{n}_x\left({蟽}_x{蟽}_z+{蟽}_z{蟽}_x\right)\\ & & \end{array}$

However, since${蟽}_x^2={蟽}_y^2={蟽}_z^2=1$

Also,${蟽}_x$,${蟽}_y$ and ${蟽}_z$are anti-commuting with each other

$\begin{array}{rcl}{蟽}_x{蟽}_y+{蟽}_y{蟽}_x& =& {蟽}_y{蟽}_z+{蟽}_z{蟽}_y\\ & =& {蟽}_z{蟽}_x+{蟽}_x{蟽}_z=0\\ (蟽路\hat{n}{)}^2& =& n_x^2+n_y^2+n_z^2\\ & =& {\hat{n}}^2\\ & =& 1\\ (蟽路\hat{n})& =& 1\\ 鈭�e^{\frac{(蟽路n)蠒}{2}}& =& e^{\frac{i蠒}{2}}\\ & =& \cos \frac{蠒}{2}+i(蟽路\hat{n})\sin \frac{蠒}{2}\\ & & \end{array}$

Thus, it is proved that $e^{i(\mathbf{蟽}路\hat{n})蠁/2}=\cos (蠁/2)+i(\hat{n}路\mathbf{蟽})\sin (蠁/2)$.