Question

(a) For a function$f\left(\varphi \right)$that can be expanded in a Taylor series, show that $f(\varphi +\phi )=e^{i{L}_z\phi /\hslash }f\left(\varphi \right)$ (where is an arbitrary angle). For this reason, $L_z/\hslash$ is called the generator of rotations about the Z-axis. Hint: Use Equation $4.129$ , and refer Problem $\mathbf{3}.\mathbf{39}$.More generally, $\mathbf{L}·\hat{n}/\hslash$ is the generator of rotations about the direction $\hat{n}$, in the sense that $exp(i\mathbf{L}·\hat{n}\phi /\hslash )$effects a rotation through angle$\phi$ (in the right-hand sense) about the axis $\hat{n}$ . In the case of spin, the generator of rotations is $\mathbf{S}·\hat{n}/\hslash$. In particular, for spin $1/2$ ${\chi }^{\text{'}}=e^{i(\mathbf{\sigma }·\hat{n})\phi /2}\chi$tells us how spinors rotate.

(b) Construct the $(2\times 2)$matrix representing rotation by ${180}^{\circ }$about the X-axis, and show that it converts "spin up" $\left({\chi }_{+}\right)$into "spin down"$\left({\chi }_{-}\right)$ , as you would expect.

(c) Construct the matrix representing rotation by ${90}^{\circ }$about the Y-axis, and check what it does to

${\chi }_{+}$

(d) Construct the matrix representing rotation by ${360}^{\circ }$about the -Zaxis, If the answer is not quite what you expected, discuss its implications.

(e) Show that$e^{i(\mathbf{\sigma }·\hat{n})\phi /2}=cos(\phi /2)+i(\hat{n}·\mathbf{\sigma })sin(\phi /2)$

Short answer

a) Spinor rotate in$e^{\left(\frac{u\mu \psi }{h}\right)}f\left(\varphi \right)$.

b) The matrix converts the spin-up into a spin-down with a factor i .

c) The spin-up alonghas become spin-down along x .

d) Rotating the spinor$360$degrees alters the sign of the spinor.

e) It is proved that $e^{i(\mathbf{\sigma }·\hat{n})\phi /2}=\cos (\phi /2)+i(\hat{n}·\mathbf{\sigma })\sin (\phi /2)$.

Step-by-step solution

Definition ofTaylor series.

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.

The rotation of spiner.

(a)

A function of Taylor series expansion about is given by ,

$$\begin{gathered} f\left(x+x_0\right)=\sum _{n=0}^{\infty }\frac{1}{n!}{x}_0^n{\left(\frac{d}{dx}\right)}^{n}f\left(x\right) \\ f(\varphi +\psi )=\sum _{n=0}^{\infty }\frac{1}{n!}{\psi }^n{\left(\frac{d}{d\varphi }\right)}^{n}f\left(\varphi \right) \end{gathered}$$

Using, $L_z=\frac{\hslash }{i}\frac{d}{d\varphi }$

And $\frac{d}{d\varphi }=\frac{i{L}_z}{\hslash }$

$\begin{array}{rcl}f(\varphi +\psi )& =& \sum _{n=0}^{\infty }\frac{1}{n!}{\psi }^n{\left(\frac{i{L}_z}{\hslash }\right)}^{n}f\left(\varphi \right)\\ & =& \sum _{n=0}^{\infty }\frac{1}{n!}{\left(\frac{i{L}_z\psi }{\hslash }\right)}^{n}f\left(\varphi \right)\\ & & \end{array}$

Known that $e^x=\sum _{n=0}^{\infty }\frac{x^n}{n!}$

$\therefore f(\varphi +\psi )=e^{\left(\frac{u\mu \psi }{h}\right)}f\left(\varphi \right)$

Construct the (2×2) matrix representing rotation by  180∘ about the x axis, and conversion of "spin up" χ+  into "spin down" χ-.

(b)

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

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

represents the rotation through an angle $\varphi$Rotation,

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

Such that,

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

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

Construct the matrix representing rotation by  90∘about the  Y-axis, and check what it does to χ+ . 

(c)

Here,

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

And

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

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

Step 5:Construct the matrix representing rotation by  360∘about the  Z -axis, If the answer is not quite what you expected.

(d)

Here,

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

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

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

Step 6:Show that  ei(σ·n^)φ/2=cos(φ/2)+i(n^·σ)sin(φ/2) 

(e)

Let’s consider that

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

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

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

$\begin{array}{rcl}{\sigma }_x{\sigma }_y+{\sigma }_y{\sigma }_x& =& {\sigma }_y{\sigma }_z+{\sigma }_z{\sigma }_y\\ & =& {\sigma }_z{\sigma }_x+{\sigma }_x{\sigma }_z=0\\ (\sigma ·\hat{n}{)}^2& =& n_x^2+n_y^2+n_z^2\\ & =& {\hat{n}}^2\\ & =& 1\\ (\sigma ·\hat{n})& =& 1\\ \therefore e^{\frac{(\sigma ·n)\varphi }{2}}& =& e^{\frac{i\varphi }{2}}\\ & =& \cos \frac{\varphi }{2}+i(\sigma ·\hat{n})\sin \frac{\varphi }{2}\\ & & \end{array}$

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