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22.5.3: iii. Exercise Solutions

  • Page ID
    84793
  • Q1

    a. Evaluate the z-componenet of \(\mu_{fi}\):
    \begin{align} \mu_{fi} &=& & \langle 2p_z |erCos\theta |1s\rangle\text{, where }\psi_{1s} = \dfrac{1}{\sqrt{\pi}}\left( \dfrac{Z}{a_0} \right)^{\dfrac{3}{2}}e^{\dfrac{-Zr}{a_0}}\text{, and } \psi_{2p_z} = \dfrac{1}{4\sqrt{2\pi}}\left( \dfrac{Z}{a_0} \right)^{\dfrac{5}{2}}r Cos\theta e^{\dfrac{-Zr}{2a_0}} \\ \mu_{fi} &=& & \dfrac{1}{4\sqrt{2\pi}}\left( \dfrac{Z}{a_0} \right)^{\dfrac{5}{2}}\dfrac{1}{\sqrt{\pi}}\left( \dfrac{Z}{a_0}\right)^{\dfrac{3}{2}}\langle r Cos\theta e^{\dfrac{-Zr}{2a_0}}| erCos \theta |e^{\dfrac{-Zr}{a_0}}\rangle \\ &=& & \dfrac{e}{4\pi\sqrt{2}} \left( \dfrac{Z}{a_0} \right)^4 \int\limits_0^{\infty} r^2dr \int\limits_0^{\pi}Sin \theta d\theta \int\limits_0^{2\pi}d\varphi \left( r^2e^{\dfrac{-Zr}{2a_0}}e^{\dfrac{-Zr}{a_0}} \right) Cos^2\theta \\ &=& & \dfrac{e}{4\pi\sqrt{2}}2\pi \left( \dfrac{Z}{a_0} \right)^4\int\limits_0^{\infty} \left( r^4 e^{\dfrac{-3Zr}{2a_0}} \right) dr \int\limits^\pi_0 Sin\theta Cos^2\theta d\theta \end{align}

    Using integral equation 4 to integrate over r and equation 17 to integrate over \(\theta\) we obtain:

    \begin{align} &=& & \dfrac{e}{4\pi\sqrt{2}}2\pi \left( \dfrac{Z}{a_0} \right)^4 \dfrac{4\!}{\left( \dfrac{3Z}{2a_0} \right)^5}\left( \dfrac{-1}{3}\right) Cos^3\theta \bigg|^\pi_0 \\ &=& & \dfrac{e}{4\pi\sqrt{2}}2\pi \left( \dfrac{Z}{a_0} \right)^4 \dfrac{2^5a_0^54\!}{3^5Z^5} \left( \dfrac{-1}{3}\right) ((-1)^3 - (1)^3) \\ &=& & \dfrac{e}{\sqrt{2}}\dfrac{2^8a_0}{3^5Z} = \dfrac{ea_0}{Z}\dfrac{2^8}{\sqrt{2}3^5} = 0.7499 \dfrac{ea_0}{Z} \end{align}

    b. Examine the symmetry of the integrands for \(\langle 2p_z|ex|1s\rangle \text{ and } \langle 2p_z|ey|1s\rangle \).

    Function Symmetry
    \(2p_z\) - 1
    x + 1
    1s + 1
    y + 1

    Under this operation the integrand of \(\langle 2p_z |ex|1s\rangle\) is (-1)(1)(1) = -1 (it is antisymmetric) and hence \(\langle 2p_z |ex|1s\rangle = 0\).
    Similarly, under this operation the integrand of \(\langle 2p_z |ey| 1s\rangle\) is (-1)(1)(1) = -1 (it is antisymmetric) and hence \(\langle 2p_z |ey|1s \rangle = 0\).

    c. \begin{align} \tau_R &=& & \dfrac{3\hbar^4 c^3}{4\left( \dfrac{3}{8}\left( \dfrac{e^2}{a_0} \right) Z^2 \right)^3 \left( \left( \dfrac{ea_0}{Z} \right) \dfrac{2^8}{\sqrt{2}3^5} \right)^2}, \\ &=& & \dfrac{3\hbar^4c^3}{4\dfrac{3^3}{8^3}\left( \dfrac{e^6}{a_0^3}\right) Z^6 \left( \dfrac{e^2a_0^2}{Z^2} \right) \dfrac{2^{16}}{(2)3^{10}} } \\ &=& & \dfrac{\hbar^4 c^3 3^8 a_0}{e^8Z^42^8} \\ \text{Inserting } e^2 &=& & \dfrac{\hbar^2}{m_ea_0} \text{ we obtain }: \\ \tau_R &=& & \dfrac{\hbar^4 c^3 3^8 a_0 m_e^4 a_0^4}{\hbar^8 Z^4 2^8} = \dfrac{3^8}{2^8}\dfrac{c^3 a_0^5 m_e^4}{\hbar^4Z^4} \\ &=& &25.6289 \dfrac{c^3 a_0^5 m_e^4}{\hbar^4Z^4} \\ &=& & 256,289 \left( \dfrac{1}{Z^4}\right) x \dfrac{(2.998x10^{10}\text{cm sec}^{-1})^3 (0.529177x10^{-8}\text{cm})^5 (9.109x10^{-28}\text{ g})^4}{(1.0546x10^{-27} \text{ g cm}^2 \text{ sec}^{-1})^4} \\ &=& & 1.595x10^{-9} \text{ sec }x\left( \dfrac{1}{Z^4} \right) \end{align}
    So, for example:

    Atom \(\tau_R\)
    H 1.595 ns
    He\(^+\) 99.7 ps
    Li\(^{+2}\) 19.7 ps
    Be\(^{+3}\) 6.23 ps
    Ne\(^{+9}\) 159 fs


    Q2

    a. \( H = H_0 + \lambda H'(t), H'(t) = V\theta (t), H_0 \varphi_k = E_k\varphi_k, \omega_k = \dfrac{E_k}{\hbar}, i\hbar\dfrac{\partial \psi}{\partial t} = H\psi\)
    let \(\psi (r,t) = i\hbar \sum\limits_j c_j(t)\varphi_j e^{-i\omega_jt} \text{ and insert into the above expression: }\)
    \begin{align} & & & i\hbar\sum\limits_j \left[ \dot{c}_j - i\omega_j c_j \right] e^{-i\omega_j t} \varphi_j = i\hbar\sum\limits c_j (t) e^{-i\omega_j t}(H_0 + \lambda H'(t))\varphi_j \\ & & & \sum\limits_j \left[ i\hbar dot{c}_j + E_jc_j - c_jE_j - c_j\lambda H' \right] e^{-i\omega_j t}\varphi_j = 0 \\ & & & \sum\limits_j \left[ i\hbar\dot{c}_j \langle m|j\rangle - c_j\lambda\langle m|H'|j\rangle \right] e^{-i\omega_j t} = 0 \\ & & & i\hbar\dot{c}_m e^{-i\omega_mt} = \sum\limits_j c_j\lambda H'_{mj}e^{-i\omega_jt} \\ So, & & & \\ & & & \dot{c}_m = \dfrac{1}{i\hbar}\sum\limits_j c_j\lambda H'_{mj}e^{-i(\omega_{jm})} \end{align}

    Going back a few equations and multiplying from the left by \(\varphi_k\) instead of \(\varphi_m\) we obtain:
    \begin{align} & & & \sum\limits_j \left[ i\hbar \dot{c}_j \langle k|j\rangle - c_j\lambda \langle k|H'|j\rangle \right] e^{-i\omega_jt} = 0 \\ & & & i\hbar\dot{c}_ke^{-i\omega_kt} = \sum\limits_jc_j\lambda H'_{kj}e^{-i\omega_jt} \\ So, & & & \\ & & & \dot{c}_k = \dfrac{1}{i\hbar}\sum\limits_j c_j\lambda H'_{kj} e^{-i\omega_j t} \\ \text{Now, let: } & & &\\ & & &c_m = c_m^{(0)} + c_m^{(1)}\lambda + c_m^{(2)}\lambda^2 + ... \\ & & & c_k = c_k^{(0)} + c_k^{(1)}\lambda + c_k^{(2)}\lambda^2 + ... \\ \text{and substituting into above we obtain: } & & & \\ & & & \dot{c}_m^{(0)} + \dot{c}_m^{(1)}\lambda + \dot{c}_m^{(2)}\lambda^2 + ... = \dfrac{1}{i\hbar}\sum\limits_j \left[ c_j^{(0)} + c_j^{(1)}\lambda + c_j^{(2)}\lambda^2 + ... \right] \lambda H'_{mj} e^{-i(\omega_{jm})t} \end{align}


    \begin{align} \text{first order: } & \\ & \dot{c}_m^{(0)} = 0 \Rightarrow c_m^{(0)} = 1 \\ \text{second order: } & \dot{c}_m^{(1)} = \dfrac{1}{i\hbar}\sum\limits_jc_j^{(0)} H'_{mj} e^{-i(\omega_{jm})t} \\ (n + 1)^{st} \text{ order: } & \\ & \dot{c}_m^{(n)} = \dfrac{1}{i\hbar}\sum\limits_j c_j^{(n-1)}H'_{mj} e^{-i(\omega_{jm})t} \\ \text{Similarly: } & \\ \text{first order: } & \\ & \dot{c}_k^{(0)} = 0 \Rightarrow c_{k\ne m}^{(0)} = 0 \\ \text{second order: } & \\ & \dot{c}_k^{(1)} = \dfrac{1}{i\hbar}\sum\limits_j c_j^{(0)}H'_{kj} e^{-i(\omega_{jk})} \\ \text{(n + 1)}^{st}\text{ order: } & \\ & \dot{c}_k^{(n)} = \dfrac{1}{i\hbar} \sum\limits_j c_j^{(n-1)} H'_{kj} e^{-i(\omega_{jk})t} \\ \text{So, } & \\ & \dot{c}_m^{(1)} = \dfrac{1}{i\hbar} c_m^{(0)}H'_{mm} e^{-i(\omega_{mm})t} = \dfrac{1}{i\hbar}H'_{mm} \\& c_m^{(1)}(t) = \dfrac{1}{i\hbar}\int\limits_0^t dt' V_{mm} = \dfrac{V_{mm}t}{i\hbar} \\ \text{ and similarly, } & \\ & \dot{c}_k^{(1)} = \dfrac{1}{i\hbar}c_m^{(0)} H'_{km} e^{-i(\omega_{mk})t} = \dfrac{1}{i\hbar}H'_{km}e^{-i(\omega_{mk})t} \\ & c_k^{(1)}(t) = \dfrac{1}{i\hbar}V_{km} \int\limits_0^t dt' e^{-i(\omega_{mk})t'} = \dfrac{V_{km}}{\hbar\omega_{mk}}\left[ e^{-i(\omega_{mk})t} - 1 \right] \\ & \dot{c}_m^{(2)} = \dfrac{1}{i\hbar}\sum\limits_j c_j^{(1)}H'_{mj} e^{-i(\omega_{jm})t} \\ & \dot{c}_m^{(2)} = \sum\limits_{j\neq m} \dfrac{1}{i\hbar}\dfrac{V_{jm}}{\hbar\omega_{mj}} \left[ e^{-i(\omega_{mj})t} - 1 \right] H'_{mj} e^{-i(\omega_{jm})t} + \dfrac{1}{i\hbar}\dfrac{V_{mm}t}{i\hbar} H'_{mm} \\ & c_m^{(2)} = \sum\limits_{j\neq m} \dfrac{1}{i\hbar}\dfrac{V_{jm}V_{mj}}{\hbar\omega_{mj}}\int\limits_0^t dt' e^{-i(\omega_{jm})t'} \left[ e^{-i(\omega_{mj})t'} - 1 \right] - \dfrac{V_{mm}V_{mm}}{\hbar^2} \int\limits_0^t t'dt' \\ &= \sum\limits_{j\ne m} \dfrac{V_{jm}V_{mj}}{i\hbar^2\omega_{mj}} \int\limits_o^t dt' \left[ 1 - e^{-i(\omega_{jm})t'} \right] - \dfrac{|V_{mm}|^2}{\hbar^2}\dfrac{t^2}{2} \\ &= \sum\limits_{j\ne m} \dfrac{V_{jm}V_{mj}}{i\hbar^2\omega_{mj}} \left( t - \dfrac{e^{-i(\omega_{jm})t}-1}{-i\omega_{jm}} \right) - \dfrac{|V_{mm}|^2}{\hbar^2}\dfrac{t^2}{2} \\ &= \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{mj}}{\hbar^2\omega_{mj}^2}\left( e^{-i(\omega_{jm})t} - 1 \right) + \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{mj}}{i\hbar^2 \omega_{mj}}t - \dfrac{|V_{mm}|^2 t^2}{2\hbar^2} \\ \text{Similarly, } & \\ \dot{c}_k^{(2)} &= \dfrac{1}{i\hbar}\sum\limits_j c_j^{(1)} H'_{kj} e^{-i(\omega_{jk})t} \\ &= \sum\limits_{j\ne m} \dfrac{1}{i\hbar} \dfrac{V_{jm}}{\hbar \omega_{mj}}\left[ e^{-i(\omega_{mj})t} - 1 \right] H'_{kj} e^{-i(\omega_{jk})t} + \dfrac{1}{i\hbar}\dfrac{V_{mm}t}{i\hbar} H'_{km} e^{-i(\omega_{mk})} \\ c_k^{(2)}(t) &= \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{kj}}{i\hbar^2\omega_{mj}} \int\limits_0^t dt' e^{-i(\omega_{jk})t'} \left[ e^{-i(\omega_{mj})t'} - 1 \right] - \dfrac{V_{mm}V_{km}}{\hbar^2} \int\limits_0^t t' dt' e^{-i(\omega_{mk})t'} \\ &= \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{kj}}{i\hbar^2\omega_{mj}} \left( \dfrac{e^{-i(\omega_{mj} + \omega_{jm})t} - 1}{-i\omega_{mk}} - \dfrac{e^{-i(\omega_{jk})t} - 1}{-\omega_{jk}} \right) - \dfrac{V_{mm}V_{km}}{\hbar^2} \left[ e^{-i(\omega_{mk})t'} \left( \dfrac{t'}{-i\omega_{mk}} - \dfrac{1}{-(i\omega_{mk})^2} \right) \right]^t_0 \\ &= \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{kj}}{\hbar^2\omega_{mj}} \left( \dfrac{e^{-i(\omega_{mk})t} - 1}{\omega_{mk}} - \dfrac{e^{-i(\omega_{jk})t} - 1}{\omega_{jk}}\right) + \dfrac{V_{mm}V_{km}}{\hbar^2\omega_{mk}} \left[ e^{-i(\omega_{mk})t'} \left( \dfrac{t'}{i} - \dfrac{1}{\omega_{mk}} \right) \right]^t_0 \\ &= \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{kj}}{E_m - E_j} \left( \dfrac{e^{-i(\omega_{mk})t} - 1}{E_m - E_k} - \dfrac{e^{-i(\omega_{jk})t}}{E_j - E_k} \right) + \dfrac{V_{mm}V_{km}}{\hbar (E_m - E_k)} \left[ e^{-i(\omega_{mk})t} \left( \dfrac{t}{i} - \dfrac{1}{\omega_{mk}}\right) + \dfrac{1}{\omega_{mk}} \right] \end{align}
    So, the overall amplitudes \(c_m \text{, and } c_k\), to second order are:
    \begin{align} c_m(t) &= 1 + \dfrac{V_{mm}t}{i\hbar} + \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{mj}}{i\hbar (E_m - E_j)}t + \sum\limits_{j\ne m}'\dfrac{V_{jm}V_{mj}}{\hbar^2 (E_m - E_j )^2}( e^{-i(\omega_{mk})t} - 1) - \dfrac{|V_{mm}|^2t^2}{2\hbar^2} \\ c_k(t) &= \dfrac{V_{km}}{(E_m - E_k)}\left[ e^{-i(\omega_{mk})t} - 1 \right] + \dfrac{V_{mm}V_{km}}{(E_m - E_k)^2}\left[ 1 - e^{-i(\omega_{mk})t} \right] + \dfrac{V_{mm}V_{km}}{(E_m - E_k)}\dfrac{t}{\hbar i}e^{-i(\omega_{mk})t} + \sum\limits_{j\ne m}'\dfrac{V_{jm}V_{kj}}{E_m - E_j}\left( \dfrac{e^{-i(\omega_{mk})t} - 1}{E_m - E_k} - \dfrac{e^{-i(\omega_{jk})t} - 1}{E_j - E_k} \right)\end{align}
    b. The perturbation equations still hold:
    \begin{align} \dot{c}_m^{(n)} &= \dfrac{1}{i\hbar}\sum\limits_j c_j^{(n-1)}H_{mj}' e^{-i(\omega_{jm})t}\text{ ; } \dot{c}_k^{(n)} = \dfrac{1}{i\hbar} \sum\limits_j c_j^{(n-1)} H_{kj}' e^{-i(\omega_{jk})t} \\ \text{So, }c_m^{(0)} &= 1 \text{ and } c_k^{(0)} = 0 \\ \dot{c}_m^{(1)} &= \dfrac{1}{i\hbar} H_{mm}' \\ c_m^{(1)} &= \dfrac{1}{i\hbar}V_{mm} \int\limits_{-\infty}^t dt' e^{-i(\omega )t'} = \dfrac{V_{km}}{i\hbar (-i\omega_{mk} + \eta )} \left[ e^{-i(\omega_{mk} + \eta)} \right] \\ &= \dfrac{V_{km}}{E_m - E_k + i\hbar\eta} \left[ e^{-i(\omega_{mk} + \eta )t} \right] \\ \dot{c}_m^{(2)} &= \sum\limits_{j\ne m}' \dfrac{1}{i\hbar}\dfrac{V_{jm}}{E_m - E_j + i\hbar\eta}e^{-i(\omega_{mj} + \eta )t}V_{mj}e^{\eta t}e^{-i(\omega_{jm} )t} + \dfrac{1}{i\hbar}\dfrac{V_{mm}e^{\eta t}}{i\hbar \eta}V_{mm} e^{\eta t} \\ c_m^{(2)} &= \sum\limits_{j\ne m}' \dfrac{1}{i\hbar}\dfrac{V_{jm}V_{mj}}{E_m - E_j + i\hbar\eta}\int\limits_{-\infty}^t e^{2\eta t'} dt' - \dfrac{|V_{mm}|^2}{2\hbar^2\eta^2}\int\limits_{-\infty}^t e^{2\eta t'}dt' &= \sum\limits_{j\ne m}' \dfrac{V_{jm}V_{mj}}{i\hbar 2\eta (E_m - E_j + i\hbar\eta )}e^{2\eta t} - \dfrac{|V_{mm}|^2}{2\hbar^2\eta^2}e^{2\eta t} \\ \dot{c}_k^{(2)} &= \sum\limits_{j\ne m}' \dfrac{1}{i\hbar}\dfrac{V_{jm}}{E_m - E_j + i\hbar\eta}e^{-i(\omega_{mj} + \eta )t}H_{kj}' e^{-i(\omega_{jk})t} \end{align}

    c. In part a. the \(c^{(2)}(t)\) grow linearly with time ( for \(V_{mm}\) = 0) while in part b. they remain finite for \(\eta \langle 0\). The result in par a. is due to the sudden turning on of the field.
    d.
    \begin{align} |c_k(t)|^2 &= \bigg| \sum\limits_j \dfrac{V_{jm}V_{kj}e^{-i(\omega_{mk} + 2\eta )t}}{(E_m - E_j + i\hbar\eta )(E_m - E_k + 2i\hbar\eta )} \bigg|^2 \\ &= \sum\limits_{jj'} \dfrac{V_{kj}V_{kj'}V_{jm}V_{j'm}e^{4\eta t}}{\left[ (E_m-E_j)(E_m-E_{j'}) + i\hbar\eta (E_j-E_{j'}) + \hbar^2\eta^2 \right] ((E_m - E_k)^2 + 4\hbar^2\eta^2)} \\ \text{ Now, look at the limit as } \eta \rightarrow 0^+:& \\ & \dfrac{d}{dt} |c_k(t)|^2 \ne 0 \text{ when } E_m = E_k \\ & lim_{\eta \to 0^+} \dfrac{4\eta}{((E_m - E_k)^2 + 4\hbar^2\eta^2)}\alpha \delta (E_m - E_k) \\ \text{So, the final result is the 2}^{nd}\text{ order golden rule expression:}& \\ & \dfrac{d}{dt}|c_k(t)|^2 \dfrac{2\pi}{\hbar}\delta (E_m - E_k) \text{lim}_{\eta\to 0^+} \bigg| \dfrac{V_{jm}V_{kj}}{(E_j - E_m - i\hbar\eta )} \bigg|^2 \end{align}

    Q3

    For the sudden perturbation case:
    \begin{align} |c_m(t)|^2 &= 1 + \sum\limits_j' \dfrac{V_{jm}V_{mj}}{(E_m - E_j)^2}\left[ e^{-i(\omega_{jm})t} - 1 + e^{i(\omega_{jm})t} - 1 \right] + O(V^3) \\ |c_m(t)|^2 &= 1 + \sum\limits_j' \dfrac{V_{jm}V_{mj}}{(E_m - E_j)^2} \left[ e^{-i(\omega_{jm}t)} + e^{i(\omega_{jm})t} - 2\right] + O(V^3) \\ |c_k(t)|^2 &= \dfrac{V_{km}V_{mk}}{(E_m - E_k)^2}\left[ -e^{-i(\omega_{mk})t} - e^{i(\omega_{mk})t} + 2 \right] + O(V^3) \\ 1 - \sum\limits_{k\ne m}' |c_k(t)|^2 &= 1 - \sum\limits_k' \dfrac{V_{km}V_{mk}}{(E_m - E_k)^2}\left[ -e^{-i(\omega_{mk})t} - e^{i(\omega_{mk})t} + 2 \right] + O(V^3) \\ &= 1 + \sum\limits_k' \dfrac{V_{km}V_{mk}}{(E_m - E_k)^2}\left[ e^{-i(\omega_{mk})t} + e^{i(\omega_{mk})t} - 2 \right] + O(V^3) \end{align}
    \(\therefore\) to order \( V^2, |c_m(t)|^2 = 1 - \sum\limits_k' |c_k(t)|^2\), with no assumptions made regarding \(V_{mm}.\)
    For the adiabatic perturbation case:
    \begin{align} |c_m(t)|^2 &= 1 + \sum\limits_{j\ne m}' \left[ \dfrac{V_{jm}V_{mj}e^{2\eta t}}{i\hbar 2\eta (E_m - E_j + i\hbar\eta )} + \dfrac{V_{jm}V_{mj}e^{2\eta t}}{-i\hbar 2\eta (E_m - E_j + i\hbar\eta )} \right] + O(V^3) \\ &= 1 + \sum\limits_{j\ne m}' \dfrac{1}{i\hbar 2\eta}\left[ \dfrac{1}{(E_m - E_j + i\hbar\eta)} - \dfrac{1}{( E_m - E_j - i\hbar\eta )} \right] V_{jm}V_{mj} e^{2\eta t} + O(V^3) \\ &= 1 + \sum\limits_{j\ne m}' \dfrac{1}{i\hbar 2\eta}\left[ \dfrac{-2i\hbar\eta}{(E_m - E_j)^2 + \hbar^2\eta^2} \right] V_{jm}V_{mj}e^{2\eta t} + O(V^3) \\ &= 1 - \sum\limits_{j\ne m}' \left[ \dfrac{V_{jm}V_{mj}e^{2\eta t}}{(E_m - E_j)^2 + \hbar^2\eta^2} \right] + O(V^3) \\ |c_k(t)|^2 &= \dfrac{V_{km}V_{mk}}{(E_m - E_k)^2 + \hbar^2\eta^2}e^{2\eta t} + O(V^3) \end{align}
    \(\therefore \text{ to order } V^2, |c_m(t)|^2 = 1 - \sum\limits_{k}' |c_k(t)|^2\), with no assumptions made regarding \(V_{mm}\) for this case as well.