91影视

The Hamiltonian operator's eigen functions k are stationary states of the quantum mechanical system, each with its own energy . They reflect the system's allowed energy states and may be confined by boundary requirements.

(a)

We have an electron in a spin state given by:

$\mathcal{x}=A\left(\begin{array}{c}1-2i\\ 2\end{array}\right)$ 鈥︹�� (1)

To find the constant we set:

$$\begin{gathered} {\left|\mathcal{x}\right|}^2=1 \\ {\left|A\right|}^2\left(5+4\right)=1 \\ 9{\left|A\right|}^2=1 \\ A=\frac{1}{3} \end{gathered}$$

Therefore, the constant by normalizing $\mathcal{x}isA=\frac{1}{3}$.

(b)

To find $S_z$we need to write $\mathcal{x}$as a linear combination of the eigenspinors of $S_z$as:

$\mathcal{x}=a\left(\begin{array}{c}1\\ 0\end{array}\right)+b\left(\begin{array}{c}0\\ 1\end{array}\right)$

Compare with equation (1) we get:

$\mathcal{x}=\frac{1-2i}{3}\left(\begin{array}{c}1\\ 0\end{array}\right)+\frac{2}{3}\left(\begin{array}{c}0\\ 1\end{array}\right)$

The possible values of localid="1656332066764" $S_{z}are\sout{h}/2and-\sout{h}/2$with probabilities of ${\left|\left(1-2i\right)/3\right|}^2=5/9$and 4/9 that is: $\frac{\sout{h}}{2}$has a probability of $\frac{5}{9}$, and $\frac{\sout{h}}{2}$ with probability $\frac{4}{9}$.

The expectation value of $S_z$is:

We know the following values,

$$\begin{gathered} {\mathcal{x}}^T=\frac{1}{9}\left(1+2i2\right) \\ {S}_z=\frac{\sout{h}}{2}\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \\ \mathcal{x}=\left(\begin{array}{c}1-2i\\ 2\end{array}\right) \end{gathered}$$

The expectation value will be,

Therefore, the expectation value of.

(c)

The operator S_x is given by,

$S_x=\frac{\sout{h}}{2}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

Find the eigenvalues of this operator as:

$$\begin{gathered} \left|\begin{array}{cc}-位& -\frac{\sout{h}}{2}\\ \frac{\sout{h}}{2}& -位\end{array}\right|=0 \\ {位}^2-\frac{{\sout{h}}^2}{4}=0 \\ 位=卤\frac{\sout{h}}{2} \end{gathered}$$

Solving the equation to find the eigenspinors:

$\frac{\sout{h}}{2}\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}鈭�\\ 尾\end{array}\right)=卤\frac{\sout{h}}{2}\left(\begin{array}{c}鈭�\\ 尾\end{array}\right)$

The eigenspinors are,

$$\begin{gathered} {\mathcal{x}}_{+}^{\left(x\right)}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right) \\ {\mathcal{x}}_{+}^{\left(x\right)}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\end{array}\right) \end{gathered}$$ 鈥︹�� (2)

Write the eigenspinors of $S_x$as:

$\mathcal{x}=\frac{\sqrt{2}}{3}\left[a\left(\begin{array}{c}1\\ 1\end{array}\right)+b\left(\begin{array}{c}1\\ -1\end{array}\right)\right]$

To find the constant a and b . We set this equation equal to equation ( I )as:

$$\begin{gathered} a+b=1-2 \\ a-b=2 \\ a=\frac{3}{2}-i \\ b=-\frac{1}{2}-i \end{gathered}$$

The eigenspinor will be:

$\mathcal{x}=\frac{\sqrt{2}}{3}\left[\left(\frac{3}{2}-i\right)\left(\begin{array}{c}1\\ 1\end{array}\right)+\left(-\frac{1}{2}-i\right)\left(\begin{array}{c}1\\ -1\end{array}\right)\right]$

According to this equation the possible values are $\sout{h}/2$with probability ${\left|\left(\sqrt{2}/3\right)\left(3/2-i\right)\right|}^2=13/18$and $-\sout{h}/2$with probability 5/18.

That is: $\frac{\sout{h}}{2}$, with probability $\frac{13}{18}$; $-\frac{\sout{h}}{2}$, with probability$\frac{5}{18}$

The expectation value of $S_x$is:

We know the following values,

$$\begin{gathered} {\mathcal{x}}^T=\frac{1}{9}\left(1+2i2\right) \\ {S}_x=\frac{\sout{h}}{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right) \\ \mathcal{x}=\left(\begin{array}{c}1-2i\\ 2\end{array}\right) \end{gathered}$$

The expectation value will be,

Therefore, the expectation value of

(d)

In this part we follow the same method in part to find the eigenvalues and their probabilities,

$\mathcal{x}=\frac{\sqrt{2}}{3}\left[\left(\frac{1}{2}-2\mathrm{i}\right)\left(\begin{array}{c}1\\ \mathrm{i}\end{array}\right)+\frac{1}{2}\left(\begin{array}{c}1\\ -\mathrm{i}\end{array}\right)\right]$

The possible values are $\sout{h}/2$with probability ${\left|\sqrt{2}/3\left(1/2-2\mathrm{i}\right)\right|}^2=17/18$and $-\sout{h}/2$with probability 1/18

That is: $\frac{\sout{h}}{2}$, with probability $\frac{17}{18}$; $-\frac{\sout{h}}{2}$, with probability$\frac{1}{18}$.

The expectation value of$S_{y}is:$

Therefore, the expectation value of.