Quantom Mechanics Chp8, 11

QMHW 8, 11

8.3

[
\psi(x) =\begin{cases} A(x+a/2) & x\in [-a/2 , 0]\
-A(x-a/2) & x\in [0, a/2]\
0 & \text{elsewhere}
\end{cases}\qquad A = \sqrt\frac{12}{a^3}
]
[\frac{d^2\psi}{dx^2} = A \delta(x +a/2) - 2A \delta(x) + A \delta(x-a/2)]

[\begin{aligned}\braket H &= \braket{-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V}\
&=-\frac{A\hbar^2}{2m} [\psi(-a/2) - 2\psi(0) + \psi(a/2)] -\alpha\psi(0)\
&= \frac{6\hbar^2}{ma^2} - \alpha\frac{3}{a}
\end{aligned}]
let the derivative be zero for $\braket H$ to $a$
[ \frac{\partial \braket H}{\partial a} = 0\implies a = \frac{4\hbar^2}{m\alpha}]
Hense [\braket H = -\frac{3m\alpha^2}{8\hbar^2}]

8.6

a

[\begin{aligned}\braket H &= \braket{H^0 + H’}\
&=\braket{\psi | H^0 + H’ | \psi}\
&\leq \braket{\psi^0 | H^0 +H’|\psi^0}\
&= \braket{\psi^0 | H^0 | \psi^0} + \braket{\psi^0 | H’ | \psi^0}\
&= E^0 + E^1
\end{aligned}]

b

7.15:

[E_{gs}^{(2)} = \sum_{m\neq gs} \frac{|\braket{\psi_m^0|H’|\psi_{gs}^0}|^2}{E_{gs}^0 - E_{m}^0} ]
since $E_{gs}^0 - E_{m}^0<0$, we have$$ E_{gs}^{(2)}< 0$$

8.17

a

[\psi(x) = A \cos(\pi x/a)]
[\begin{aligned}\braket H &= \braket{\psi | H | \psi}\
&= \int_{-a/2}^{a/2} \frac{\hbar^2A^2}{2m}\frac{\pi^2}{a^2} \cos^2\left(\frac{\pi x}{a}\right) + \frac 1 2 m\omega^2 x^2A^2 \cos^2\left(\frac{\pi x}{a}\right) dx\
&= 2 \int_0^{a/2} \frac{\hbar^2A^2}{2m}\frac{\pi^2}{a^2} \cos^2\left(\frac{\pi x}{a}\right) + \frac 1 2 m\omega^2 x^2A^2 \cos^2\left(\frac{\pi x}{a}\right) dx\
x =at:&\
&= 2\int_0^{1/2} \left[\frac{\hbar^2A^2}{2m}\frac{\pi^2}{a^2} \cos^2(\pi t) + \frac 1 2 m\omega^2 (at)^2A^2 \cos^2(\pi t) \right]adt\
\end{aligned}]

[
\begin{aligned}
\frac{\partial \braket H}{\partial a} &= \frac{\partial}{\partial a}2\int_0^{1/2} \left[\frac{\hbar^2A^2}{2m} \frac{\pi^2}{a^2}\cos^2(\pi t) + \frac 1 2 m\omega^2 (at)^2A^2 \cos^2(\pi t) \right]adt\
&= 2\int_0^{1/2}\left[-\frac{\hbar^2A^2}{2m} \frac{\pi^2}{a^2}\cos^2(\pi t) + \frac 3 2 m\omega^2 (at)^2A^2 \cos^2(\pi t) \right]dt\
&=- \frac{\hbar^2A^2}{m} \frac{\pi^2}{a^2} \int_0^{1/2} \cos^2(\pi t)dt + 3m\omega^2 a^2A^2 \int_0^{1/2}t^2\cos^2(\pi t )dt\
&= - \frac{\hbar^2A^2}{4m} \frac{\pi^2}{a^2} + 3m\omega^2 a^2A^2\left[ \frac{1}{48} - \frac{1}{8\pi^2}\right]\
&=0\
\end{aligned}]

[a = \left[\frac{4\pi^4\hbar^2}{m^2\omega^2(\pi^2-6)}\right]^{1/4}]

[\braket H = \frac{\hbar^2A^2\pi^2}{4ma^2} + A^2 m \omega^2 a^2\left[ \frac{1}{48} - \frac{1}{8\pi^2}\right] ]

There is no need to consider that the wave function is not $c^2$ continuous at the edges since $\psi(x) = 0$ at the edges. The delta functions are cancelled: $\delta(x - a/2)\cos(\pi x/a) = 0$

h = 1.054571817646156e-34
m = 1.67262192369e-27
k = 1
w = math.sqrt(k / m)
pi = math.pi

a = (4 * pi**4 * h**2 / (m**2 * w**2 * (pi**2 - 6)))**(1/4)
A2 = 2 / a
H = A2 * (h**2 * pi**2 / (4 * m * a**2) + m * w**2 * a**2 * (1/48 - 1 / (8 * pi**2)))

print(H)           # 修正后的能量期望值
print(h * w / 2)   # 基态能量 ℏω/2

1.0511203283622713e-11
1.2892800808307563e-21

差 10个数量级这能对吗哈哈😄

b

11.5

a

According to 11.14 and 11.15
[
\dot c_a = -\frac{i}{\hbar}[c_a H’{aa} + c_b H’{ab}e^{-i\omega_0 t}],\quad
\dot c_b = -\frac{i}{\hbar}[c_b H’{bb} + c_a H’{ba}e^{i\omega_0 t}]
]
Using the pertubation theory, assume
[c_a(0) = 1,; c_b(0) = 0]

• Zero Order
  [c_a^{(0)} = 1 ,; c_b^{(0)} = 0]

• First Order
  [\frac{dc_a^{(1)} }{dt} = -\frac i \hbar H’{aa} c_a^{(0)}\implies c_a^{(1)}(t) = 1-\frac i \hbar \int_0^t H’{aa}(t’) dt’]

  [\frac{dc_b^{(1)} }{dt} = -\frac i \hbar \left[ H’{ba}e^{i\omega_0 t}\right]\implies c_b^{(1)} (t) = -\frac i \hbar \int_0^t H’{ba} (t’) e^{i\omega_0 t’}dt’ ]

[|c_a^{(1)} (t)|^2 + |c_b^{(1)} (t)|^2 = ? ]

b

Check for $\dot d_a$, the same applies to $\dot d_b$

[\begin{aligned}\frac{\mathrm{d}d_a}{\mathrm{d}t} &= \frac{\mathrm{d} } {dt} e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ } c_a\
&=c_a\frac{d}{dt} e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ } + e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ } \dot c_a\
&=c_a \frac{d}{dt} e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ } -\frac{i}{\hbar}[c_a H’{aa} + c_b H’{ab}e^{-i\omega_0 t}] e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ }\
& = -\frac{i}{\hbar}c_b H’{ab}e^{-i\omega_0 t} e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ }\
& = -\frac i \hbar c_b e^{i\phi} H’{ab} e^{-i\omega_0 t}e^{\frac i \hbar \int_0^t H’{bb} (t’) dt’ } \
& = -\frac i \hbar c_b e^{i\phi} H’{ab} e^{-i\omega_0 t}d_b \
\end{aligned}]

c

• Zero Order
  [d_a^{(0)}(t) =1,; d_b^{(0)}(t) = 0]
• First Order
  [\dot d_a^{(1)} = 0 \implies d_a^{(1)}(t) = 1]
  [\dot d_b^{(1)} = -\frac i \hbar e^{-i\phi} H’{ba} e^{i\omega_0 t}e^{\frac i \hbar \int_0^t H’{aa} (t’) dt’ } = -\frac i \hbar H’{ba} e^{i\omega_0 t}e^{\frac i \hbar \int_0^t H’{bb} (t’) dt’ }]
  [d_b^{(1)}(t) = 1-\frac i \hbar \int_0^t H’{ba}(t’‘) e^{i\omega_0 t’‘}e^{\frac i \hbar \int_0^{t’‘} H’{bb} (t’) dt’ } dt’']

Notice that $d_b^{(1)}(t)$ involves $H'_{ba}$ and $H'_{bb}$ compared to where $c_a^{(1)}(t)$ only involves $H'_{aa}$ and $c_b^{(1)}(t)$ only involves $H'_{ba}$.

11.15

a

[L^2 = L_x^2 + L_y^2 + L_z^2]
[\begin{aligned}[L^2, z] &= L^2z - zL^2\
&=(L_x^2 + L_y^2 + L_z^2)z - z( L_x^2 + L_y^2 + L_z^2)\
&=
\end{aligned}]

11.24

a

[\Psi (t) = \sum c_n (t) \psi_n e^{-iE_n t/\hbar}]
According to the Scrodinger equation:
[\hat H \Psi = i\hbar \frac{d\Psi}{dt}\qquad \text{where } \hat H = \hat H^0 + \hat H’(t) ]
we have:

[\begin{aligned}
&\sum c_n (t) \hat H^0 \psi_n e^{-iE_n t/\hbar} +\sum c_n (t) \hat H’ \psi_n e^{-iE_n t/\hbar}\ &=i \hbar \left[\sum\dot{c_n} \psi_n e^{-iE_nt/\hbar} + \sum c_n \psi_n \left( \frac{-iE_a}{\hbar}\right) e^{-iE_nt/\hbar}\right] \
\implies& \sum c_n (t) \hat H’ \psi_n e^{-iE_n t/\hbar} = i \hbar\sum\dot{c_n} \psi_n e^{-iE_nt/\hbar}
\end{aligned}]
Tame the inner product with $\psi_m$:
[\begin{aligned} \bigg \lang\psi_m \bigg|\sum c_n (t) \hat H’ \psi_n e^{-iE_n t/\hbar} \bigg \rang &= \bigg \lang \psi_m \bigg | i \hbar\sum\dot{c_n} \psi_n e^{-iE_nt/\hbar} \bigg \rang \ \sum c_n e^{-iE_n t /\hbar } H’{mn} & = i\hbar \dot{c_m} e^{-i E_m / \hbar}\
\dot{c_m} & = -\frac{i}{\hbar}\sum c_n e^{i(E_m- E_n) t /\hbar } H’{mn}
\end{aligned}]

b

[c_m^{(0)} = \begin{cases} 1 & m = N \ 0 & m\neq N \end{cases} ]
since
[\dot c_N = -\frac{i}{\hbar}\sum c_n e^{i(E_m- E_n) t /\hbar } H’{mn} = -\frac{i}{\hbar} H’{NN}]

[c_N^{(1)}(t) = -\frac{i}{\hbar}\int_0^t H’{NN}(t’) dt’ + 1]
and for $m\neq N$ :
[\dot c_m = -\frac{i}{\hbar}\sum c_n e^{i(E_m- E_n) t /\hbar } H’{mn} = -\frac{i}{\hbar} e^{i(E_m- E_N) t /\hbar } H’_{mN} ]

[c_m^{(1)} (t) = -\frac{i}{\hbar} \int_0^t e^{i(E_m- E_N) t’ /\hbar } H’_{mN}(t’) dt’]

c

Since $H'_{MN}$ is constant and $H'$ is only applied in a time window of $T$, we have:
[\begin{aligned}c_M(t) &= -\frac{i}{\hbar} H’{MN}\int_0^T e^{i(E_M- E_N) t /\hbar } dt\
&= -\frac{i}{\hbar} H’{MN} \frac{e^{i(E_M- E_N) T /\hbar } - 1}{i(E_M- E_N)/\hbar}\
|c_M(t)|^2 &= |H’{MN}| \frac{2 - 2\cos((E_M- E_N) T /\hbar )}{(E_M - E_N)^2}\
& = 4 |H’{MN}|\frac{\sin^2 [(E_M- E_N) T /2\hbar ]}{(E_M - E_N)^2}
\end{aligned}]

d

[ \begin{aligned}H’{mN }(t) &= \braket{\psi_m | V\cos(\omega t) | \psi_N }\
& = \cos(\omega t) V{mN} \
&\text{where } V_{mN} = \braket{\psi_m | V | \psi_N }\
\end{aligned}]

[\begin{aligned}c_m(t) &= -\frac{i}{\hbar} \int_0^t e^{i(E_m- E_N) t’ /\hbar } H’{mN}(t’) dt’\
& =-\frac{i}{\hbar}\int_0^t e^{i(E_m- E_N) t’ /\hbar } \cos(\omega t’)V{mN}dt’ \
& =-\frac{i}{\hbar}\int_0^t e^{i(E_m- E_N) t’ /\hbar } [ e^{i\omega t’} - i\sin(\omega t’)]V_{mN}dt’\
& =-\frac{i}{\hbar}\int_0^t e^{i(E_m- E_N + \hbar \omega) t’ /\hbar }V_{mN} dt’ + \frac{i}\hbar \int_0^t e^{i(E_m- E_N) t’ /\hbar } i\sin(\omega t’)V_{mN} dt’ \
\end{aligned}]
Consider on the period $[0,T] , \;T = 2\pi/\omega$, if $E_m - E_N = \hbar \omega_0$ :
[\begin{aligned}c_m(t) &= -\frac{i}{\hbar}\int_0^T e^{i(\omega_0+ \omega) t’ }V_{mN} dt’ + \frac{i}\hbar \int_0^T e^{i\omega_0 t’ }i\sin(\omega t’)V_{mN} dt’ \&\text{since the second componet highly oscillates, its close to 0}\
& = -\frac{i}{\hbar}V_{mN}\int_0^T e^{i(\omega_0+ \omega) t’ } dt’
\end{aligned} ]
When $\omega_0 + \omega = 0$, the intergral gives
[c_m(t) = -\frac{i}{\hbar}V_{mN} T]
else:
[c_m(t) = -\frac{i}{\hbar}V_{mN} \frac{e^{i(\omega_0 +\omega)T} -1}{i(\omega_0 +\omega)}]