Quantom Mechanics Week 1

QM Week 1

1.9

$$\Psi (x,t) = A e^{-a[(mx^2/\hbar)+ it]}$$

a

$$\int |\Psi|^2 dx = A^2\sqrt{\frac{\pi \hbar}{2am}}=1$$

$A = \left ({\frac{2am}{\pi \hbar }}\right)^{\frac 1 4}$

b

$$i \hbar \frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V\Psi$$

$$\frac{\partial \Psi}{\partial t} = -ia \Psi\qquad \frac{\partial^2 \Psi}{\partial x^2} = \left[-\frac{2am}{\hbar} + \left(\frac{2amx}{\hbar}\right)^2\right] \Psi$$

$$\therefore V(x) = 2a^2 m x^2$$

c

1. Since $x |\Psi|^2$ is an odd function:

$$<x> = \int_{-\infty}^\infty x |\Psi|^2 dx =0$$

$$\begin{aligned}<x^2> &= \int_{-\infty}^\infty \Psi^* x^2 \Psi dx\\&=\int_{-\infty}^\infty A^2 x^2 \exp(-2ax^2/\hbar)dx\\&=\frac{\hbar}{4am}\end{aligned}$$

3. $$<p> = m\frac{d<x>}{dt} = 0$$

4. $$<p^2> = \int_{-\infty}^\infty \Psi^* (-i\hbar\frac{\partial}{\partial x})^2\Psi dx=am\hbar$$

5. $$\sigma_x^2 = <x^2> - <x>^2 =\frac{\hbar}{4am}$$

   $$\sigma_p^2 = <p^2> - <p>^2 = am\hbar$$

   $$\sigma_p\sigma_x = \frac{\hbar}2\geq\frac{\hbar}2$$

Which is consistent with the uncertainty principle.

1.14

a

$$P_{ab} = \int_{a}^b \Phi^* \Phi dx$$

$$\frac{dP_{ab}}{dt} = \int_{a}^b \frac{\partial}{\partial t}\Phi^* \Phi dx = \frac{i\hbar}{2m}\left(\Psi^* \frac{\partial \Psi}{\partial x}- \frac{\partial \Psi^*}{\partial x}\Psi\right)\Bigg|_a^b = J(a,t) - J(b,t)$$

$J$ describes the probability at point $x$ time $t$ per unit time. It has the unit: $s^{-1}$

b

$$J(x,t) = \frac{i\hbar}{2m}\left(-\Psi \frac{2amx}{\hbar}\Psi^*+\Psi^* \frac{2amx}{\hbar}\Psi\right)=0$$

1.15

$$\begin{aligned}\frac{d}{dt}\int_{-\infty}^\infty \Psi_1^*\Psi_2dx & = \int_{-\infty}^\infty \frac{\partial\Psi_1^*}{\partial t}\Psi_2+ \Psi_1^* \frac{\partial \Psi_2}{\partial t}dx\\ &=\int_{-\infty}^\infty\left(-\frac{i\hbar}{2m}\frac{\partial^2\Psi_1^*}{\partial x^2}+\frac{iV}\hbar \Psi_1^*\right)\Psi_2 + \Psi_1\left(\frac{i\hbar}{2m}\frac{\partial^2\Psi_2}{\partial x^2}- \frac{iV}\hbar\Psi_2\right)dx\\ &=\int_{-\infty}^\infty \frac{i\hbar}{2m}\left(-\frac{\partial^2\Psi_1^*}{\partial x^2}\Psi_2+\Psi_1^*\frac{\partial^2\Psi_2}{\partial x^2}\right)dx\\ &=\frac{i\hbar}{2m}\left(-\frac{\partial\Psi_1^*}{\partial x}\Psi_2+\Psi_1^*\frac{\partial\Psi_2}{\partial x}\right)\Bigg |_{-\infty}^\infty\\ &=0 \end{aligned}$$

1.16

a

$$\int |\Psi|^2 dx = \int_{-a}^a A^2(a^2-x^2)^2dx =1$$

$$A = \sqrt{\frac{15}{16a^5}}$$

b

$$<x> = \int_{-a}^a xA^2(a^2-x^2)^2dx=0$$

c

$$\begin{aligned}<p> &= \int_{-a}^a \Psi^* (-i\hbar \frac{\partial }{\partial x})\Psi dx\\ &=\int_{-a}^a 2i\hbar A^2 x(a^2-x^2)dx\\&=0 \end{aligned}$$

d

$$<x^2> = \int_{-a}^a x^2A^2(a^2-x^2)^2dx=\frac{a^2}{7}$$

e

$$\begin{aligned}<p^2> &= \int_{-a}^a A^2(a^2-x^2)(-i\hbar \frac{\partial }{\partial x})^2(a^2-x^2)dx\\ &=\int_{-a}^a 2A^2\hbar^2(a^2-x^2)dx\\ &=\frac{5\hbar^2}{2a^2} \end{aligned}$$

f

$$\sigma_x = \sqrt{<x^2> - <x>^2 }= \sqrt\frac{a^2}{7}$$

g

$$\sigma_p = \sqrt{<p^2>-<p>^2} = \sqrt\frac{5\hbar^2}{2a^2}$$

h

$$\sigma_p\sigma_x = \sqrt{\frac 5 {14}}\hbar\geq \frac \hbar 2$$

1.17

a

$$\begin{aligned} \frac{dP}{dt} &= \int \frac{\partial \Psi^*}{\partial x}\Psi+\Psi^* \frac{\partial \Psi}{\partial x}dx\\ &=\int \left(\frac{-\hbar i}{2m}\frac{\partial^2 \Psi^*}{\partial x^2} + \frac{V_0i\Psi^*}{\hbar} - \frac{\Gamma \Psi^*}{\hbar}\right)\Psi + \Psi^* \left(\frac{\hbar i}{2m}\frac{\partial^2 \Psi}{\partial x^2} - \frac{V_0i\Psi}{\hbar} - \frac{\Gamma i \Psi}{\hbar}\right)dx\\ &=\int \frac{i\hbar}{2m}\left(-\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \Psi^* \frac{\partial^2 \Psi}{\partial x^2}\right) - \frac{2\Gamma}\hbar \Psi^*\Psi dx\\ &=\int \frac{i\hbar}{2m}\left(-\frac{\partial^2 \Psi^*}{\partial x^2}\Psi + \Psi^* \frac{\partial^2 \Psi}{\partial x^2}\right)dx - \frac{2\Gamma}\hbar \int\Psi^*\Psi dx\\ &=- \frac{2\Gamma}\hbar P\end{aligned}$$

b

Solve the ODE of (a):

$$P(t) = c \exp(-\frac{2\Gamma}\hbar t)$$

assume that $c=1$, the lifespan $\tau$ is: $$\tau =\frac\hbar{2\Gamma}$$