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Chemistry LibreTexts

Solutions 12

  • Page ID
    52479
  • Q1

    The homework question assigned We are given two states:

    \[ | v =0 \rangle^{g} = \dfrac{1}{\pi^{1/4}} e ^{-q_1^2/2} = \dfrac{1}{\pi^{1/4}} \exp\left[-\dfrac{k}{2\hbar \omega} (R-R_{eq})^2\right] = | g \rangle\]

    and

    \[ | v =0 \rangle^{e} = \dfrac{1}{\pi^{1/4}} e ^{-q_2^2/2} = \dfrac{1}{\pi^{1/4}} \exp\left[-\dfrac{k}{2\hbar \omega} (R-R_{eq}-\delta)^2\right] = | e \rangle\]

    and required to evaluate the overlap integral: \(\langle g | e \rangle \)

    To do this, we'll need to evaluate Gaussian integrals of the form \(\int_{-\infty}^{\infty} \exp(-\alpha x^2) dx = (\dfrac{\pi}{\alpha})^{1/2} \) and 

    \[ \int_{-\infty}^{\infty} \exp(-\alpha x^2 + \beta x) dx = \exp( \dfrac{\beta^2}{4 \alpha}) (\dfrac{\pi}{\alpha})^{1/2} \]

    Just like our previous problems involving wavefunctions, we must normalize our wavefunctions if they aren't already:

    \[\langle g | g \rangle  =  \dfrac{1}{\pi^ {1/2}} \int_{-\infty}^{\infty} \exp\left[-\dfrac{k}{\hbar \omega} (R-R_{eq})^2\right] dR  \]

    As a Gaussian integral where

    • \(x = R- R_{eq} \),
    • \( dx = dR \), and
    • \( \alpha = (\dfrac{k}{\hbar \omega})^{1/2} \)

    we have

    \[\langle g | g \rangle   = \dfrac{1}{\pi^{1/2}} (\dfrac{\pi \hbar \omega}{k})^{1/2} = (\dfrac{\hbar \omega}{k})^{1/2} \]

    making our normalization constant

    \[C = (\dfrac{k}{\hbar \omega})^{1/4} \]

    Our new ground state vector is therefore:

    \[ | g \rangle  = ( \dfrac{k} {\pi \hbar \omega})^ {\dfrac{1}{4}} \exp\left[-\dfrac{k}{2 \hbar \omega} (R-R_{eq})^2 \right] \]

    We can perform the same procedure on the excited state:

    \[\langle e | e \rangle = \dfrac{1}{\pi^ {1/2}} \int_{-\infty}^{\infty} \exp\left[-\dfrac{k}{\hbar \omega} (R-R_{eq}-\delta)^2\right] dR  \]

    This is a similar Gaussian integral  \(x = R- R_{eq} -\delta\), \( dx = dR \)  and \( \alpha = (\dfrac{k}{\hbar \omega})^{1/2} \). The integral gives the same result as the ground state. This makes sense since if we look closely at the wave functions, the only difference between them is the constants which subtract from \(R\). The normalized excited state is:

    \[ | e \rangle  = ( \dfrac{k} {\pi \hbar \omega})^ {\dfrac{1}{4}} \exp\left[-\dfrac{k}{2 \hbar \omega} (R-R_{eq}-\delta)^2 \right] \]

    We are almost ready to evaluate the overlap integral. The integral will result in a product of exponentials which leads to a sum of square terms. Playing around with a sum of squares, let us keep in mind:

    \[ a^2 + (a + b)^2 = 2a^2 + 2ab + b^2 \]

    Proceeding to evaluate the overlap integral, we have:

    \[ \langle e | g \rangle =  (\dfrac{k} {\pi \hbar \omega})^{1/2}\int_{-\infty}^{\infty}  \exp\left[-\dfrac{k}{2 \hbar \omega} (R-R_{eq})^2 \right] \exp \left[-\dfrac{k}{2 \hbar \omega} (R-R_{eq}-\delta)^2 \right]  dR\]

    \[= ( \dfrac{k} {\pi \hbar \omega})^{1/2}\int_{-\infty}^{\infty}  \exp\left[-\dfrac{k}{2 \hbar \omega} \left[ (R-R_{eq})^2 + (R-R_{eq}-\delta)^2  \right] \right ] dR\]

    Summing the exponents can be done by noticing the algebraic form of \( a^2 + (a + b)^2 = 2a^2 + 2ab + b^2 \), where in our case we have \(a = R-R_{eq}\) and \( b = -\delta\) 

    \( (R-R_{eq})^2 + (R-R_{eq}-\delta)^2  = 2(R-R_{eq})^2 - 2 \delta (R-R_{eq}) + \delta^2 \)

    With this we simplify the overlap integral to: 

    \[=  ( \dfrac{k} {\pi \hbar \omega})^{1/2}\int_{-\infty}^{\infty}  \exp \left[-\dfrac{k}{2 \hbar \omega} \left[  2(R-R_{eq})^2 - 2 \delta (R-R_{eq}) + \delta^2 \right] \right ] dR \]

    \[=  ( \dfrac{k} {\pi \hbar \omega})^{1/2}\int_{-\infty}^{\infty}  \exp \left[  -\dfrac{k}{ \hbar \omega}  (R-R_{eq})^2 + \dfrac{k \delta}{ \hbar \omega} (R-R_{eq}) -\dfrac{k}{2 \hbar \omega} \delta^2 \right] dR \]

    \[= ( \dfrac{k} {\pi \hbar \omega})^{1/2} \exp(-\dfrac{k  \delta^2 }{2 \hbar \omega})  \int_{-\infty}^{\infty}  \exp \left[  -\dfrac{k}{ \hbar \omega}  (R-R_{eq})^2 + \dfrac{k \delta}{ \hbar \omega} (R-R_{eq})\right] dR \]

    This has the form: 

    \[ \int_{-\infty}^{\infty} \exp(-\alpha x^2 + \beta x) dx = \exp( \dfrac{\beta^2}{4 \alpha}) (\dfrac{\pi}{\alpha})^{1/2} \]

    In our case

    • \( \alpha = \dfrac{k}{\hbar \omega} \)
    • \(\beta = \dfrac{k \delta}{\hbar \omega} \)    

    so

    \[ \dfrac{\beta^2}{4 \alpha} = (\dfrac{k \delta}{\hbar \omega} )^2 (\dfrac{\hbar \omega}{4k}) = (\dfrac{k \delta^2}{4 \hbar \omega} ) \]

    and

    \[ (\dfrac{\pi}{\alpha})^{1/2} = (\dfrac{\pi \hbar \omega}{k})^{1/2} \]

    Therefore we have:

    a, b )

    \[\langle g | e \rangle  = ( \dfrac{k} {\pi \hbar \omega})^{\frac{1}{2}}  \exp(-\dfrac{k  \delta ^2 }{2 \hbar \omega})  \exp(\dfrac{k \delta ^2}{4 \hbar \omega} ) (\dfrac{\pi \hbar \omega}{k} )^{\frac{1}{2}} = \exp(-\dfrac{k \delta^2}{4 \hbar \omega} ) \]

    c). From the behavior of a negative exponential, it can only evaluate to one if \(\delta = 0\), which is the situation of overlap with the same state.

    d). \(S\) has no dependence on the value of \(R\) but only on the relative positions which are expressed in \(\delta\).

    Q2

    This question requires us to evaluate the expression, \(\exp(-\dfrac{k \delta^2}{4 \hbar \omega} ) \), for different values of \(\delta\), given \(\dfrac{k}{\hbar \omega} =200 \; Å^{-2}\)

    We effectively evaluate  \(\exp( - 50 \delta^2 ) \) :

    \(\delta \) ( Å)  \(\exp( - 50 \delta^2 ) \)
    0 1
    0.05 0.8825
    0.1 0.6065
    0.5 \(3.727 \times 10^{-6}\)