Second exam

[next] [prev] [prev-tail] [tail] [up]

3.2 Second exam

3.2.1 Questions
3.2.2 my solution to second problem (post exam)

3.2.1 Questions

3.2.1.1 First question
3.2.1.2 Second question

3.2.1.1 First question

PDF

3.2.1.2 Second question

PDF

3.2.2 my solution to second problem (post exam)

$H = \frac{e ℏ B}{2 m} (\begin{matrix} 0 & - i \\ i & 0 \end{matrix})$ The eigenvalues can be found to be $E_{+} = \frac{e ℏ B}{2 m} = ℏ λ, E_{-} = - \frac{e ℏ B}{2 m} = - ℏ λ$ where $λ = \frac{e B}{2 m}$ . The associated normalized eigenvectors are $\begin{aligned} v_{1} & = \frac{1}{\sqrt{2}} (\begin{array}{c} 1 \\ i \end{array}) \\ v_{2} & = \frac{1}{\sqrt{2}} (\begin{array}{c} i \\ 1 \end{array}) \end{aligned}$

To ﬁnd the spin state for $t > 0$ , we need to solve the Schrodinger equation $i ℏ \frac{\partial}{\partial t} | ψ ⟩ = H | ψ ⟩$ In the basis of $H$ the above becomes $\begin{aligned} i ℏ (\begin{array}{c} x_{1}^{'} (t) \\ x_{2}^{'} (t) \end{array}) & = (\begin{array}{c} E_{+} & 0 \\ 0 & E_{-} \end{array}) (\begin{array}{c} x_{1} (t) \\ x_{2} (t) \end{array}) \\ = (\begin{array}{c} ℏ λ & 0 \\ 0 & - ℏ λ \end{array}) (\begin{array}{c} x_{1} (t) \\ x_{2} (t) \end{array}) \end{aligned}$

Hence $\begin{aligned} i ℏ x_{1}^{'} (t) & = ℏ λ x_{1} (t) \\ i ℏ x_{2}^{'} (t) & = - ℏ λ x_{2} (t) \end{aligned}$

$\begin{aligned} x_{1}^{'} (t) & = - i λ x_{1} (t) \\ x_{2}^{'} (t) & = i λ x_{2} (t) \end{aligned}$

The solution is $\begin{aligned} x_{1} (t) & = x_{1} (0) e^{- i λ t} \\ x_{2} (t) & = x_{2} (0) e^{i λ t} \end{aligned}$

In the original basis, this becomes $\begin{aligned} (\begin{array}{c} X_{1} (t) \\ X_{2} (t) \end{array}) & = x_{1} (t) {\vec{v}}_{1} + x_{2} (t) {\vec{v}}_{2} \\ (1) & = \frac{1}{\sqrt{2}} x_{1} (0) e^{- i λ t} (\begin{array}{c} 1 \\ i \end{array}) + \frac{1}{\sqrt{2}} x_{2} (0) e^{i λ t} (\begin{array}{c} i \\ 1 \end{array}) \end{aligned}$

What is left is to determine $x_{1} (0), x_{2} (0)$ . We are told that at $t = 0$ , $(\begin{matrix} X_{1} (0) \\ X_{2} (0) \end{matrix}) = (\begin{matrix} 1 \\ 0 \end{matrix})$ since $(\begin{matrix} 1 \\ 0 \end{matrix})$ is the eigenstate associated with $\frac{ℏ}{2}$ of $S_{z}$ . Therefore the above becomes (at $t = 0$ ) $\begin{aligned} (\begin{array}{c} 1 \\ 0 \end{array}) & = \frac{1}{\sqrt{2}} x_{1} (0) (\begin{array}{c} 1 \\ i \end{array}) + \frac{1}{\sqrt{2}} x_{2} (0) (\begin{array}{c} i \\ 1 \end{array}) \\ = \frac{1}{\sqrt{2}} (\begin{array}{c} 1 & i \\ i & 1 \end{array}) (\begin{array}{c} x_{1} (0) \\ x_{2} (0) \end{array}) \end{aligned}$

Hence $\begin{aligned} (\begin{array}{c} x_{1} (0) \\ x_{2} (0) \end{array}) & = \sqrt{2} {(\begin{array}{c} 1 & i \\ i & 1 \end{array})}^{- 1} (\begin{array}{c} 1 \\ 0 \end{array}) \\ = \sqrt{2} \frac{(\begin{array}{c} 1 & - i \\ - i & 1 \end{array}) (\begin{array}{c} 1 \\ 0 \end{array})}{1 - (i^{2})} \\ = \frac{\sqrt{2}}{2} (\begin{array}{c} 1 & - i \\ - i & 1 \end{array}) (\begin{array}{c} 1 \\ 0 \end{array}) \\ = \frac{\sqrt{2}}{2} (\begin{array}{c} 1 \\ - i \end{array}) \\ = \frac{1}{\sqrt{2}} (\begin{array}{c} 1 \\ - i \end{array}) \end{aligned}$

Therefore $\begin{aligned} x_{1} (0) & = \frac{1}{\sqrt{2}} \\ x_{2} (0) & = \frac{- i}{\sqrt{2}} \end{aligned}$

Hence the solution (1) becomes $\begin{aligned} (\begin{array}{c} X_{1} (t) \\ X_{2} (t) \end{array}) & = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} e^{- i λ t} (\begin{array}{c} 1 \\ i \end{array}) + \frac{1}{\sqrt{2}} \frac{- i}{\sqrt{2}} e^{i λ t} (\begin{array}{c} i \\ 1 \end{array}) \\ = \frac{1}{2} e^{- i λ t} (\begin{array}{c} 1 \\ i \end{array}) - \frac{i}{2} e^{i λ t} (\begin{array}{c} i \\ 1 \end{array}) \end{aligned}$

Therefore $\begin{aligned} X_{1} (t) & = \frac{1}{2} e^{- i λ t} - \frac{i^{2}}{2} e^{i λ t} \\ X_{2} (t) & = i \frac{1}{2} e^{- i λ t} - \frac{i}{2} e^{i λ t} \end{aligned}$

Or $\begin{aligned} X_{1} (t) & = \frac{1}{2} e^{- i λ t} + \frac{1}{2} e^{i λ t} \\ X_{2} (t) & = - \frac{1}{2 i} e^{- i λ t} + \frac{1}{2 i} e^{i λ t} \end{aligned}$

Or $\begin{aligned} X_{1} (t) & = \cos (λ t) \\ X_{2} (t) & = \sin (λ t) \end{aligned}$

Or $\begin{aligned} X_{1} (t) & = \cos (\frac{e B}{2 m} t) \\ X_{2} (t) & = \sin (\frac{e B}{2 m} t) \end{aligned}$

Hence $| ψ ⟩ = (\begin{matrix} \cos (\frac{e B}{2 m} t) \\ \sin (\frac{e B}{2 m} t) \end{matrix})$

[next] [prev] [prev-tail] [front] [up]