Consider the quantum mechanical system of an atom with Hamiltonian $\hat{H}_0$ and assume that we know the solutions of the eigenvalue problem $\hat{H}_0|n\rangle=E_n|n\rangle$ for $n=0,1,2,\ldots$. Let now $\Psi_n(x)$ denote the position space vector corresponding to $|n\rangle$ where we are restricting ourselves to one dimension of position.

Assume now that the atom sits at $x=0$ and that a charged particle passes the atom with velocity $v$ such that the perturbing potential $V(x)=V_0e^{-|x-x_0(t)/\lambda|}$ is introduced where $\lambda\in\mathbb{R}$ is the reach of the potential $V(x)$ and $x_0(t):=vt$.

Assignment: If, in the distant past for $t\to-\infty$, the system is in the ground state $|0\rangle$, compute the probability to find the atom in a state $|n\rangle, n>0$ for $t\to\infty$ (Compute the relevant integral piecewise by considering the two cases $x<x_0(t)$ and $x\ge x_0(t)$ and use that $\Psi_n(x)$ is mostly located around the atom's position in order to expand the phase factor in the integral around $x=0$). What conditions do $\Psi_n(t)$ have to satisfy such that the transition probability is positive?

My try: If we denote by $V_i(t)$ the perturbing potential in the interaction picture, we get: $V_i(t)=V_0\cdot\left( e^{i\hat{H_0}t/\hbar} \circ e^{-|x-x_0(t)|/\lambda} \circ e^{-i\hat{H}_0t/\hbar} \right)$ where $\circ$ denotes composition.

We now plug this into the formula for the transmission probability

$P(|0\rangle\to|n\rangle)=\left| \frac{1}{\hbar}\int_{-\infty}^{\infty}dt' \langle n|V_i(t')|0\rangle \right|^2=\frac{1}{\hbar^2}\left|\int_{-\infty}^{\infty} dt \,e^{i(E_n-E_0)t/\hbar}\cdot \langle n|V_0e^{-|x-x_0(t)|/\lambda}|0\rangle\right|^2$

and by distinguishing between by $x<x_0(t)$ and $x\ge x_0(t)$ and switching to position space, this is equal to

$\frac{1}{\hbar^2}\cdot \left| \int_{-\infty}^{\infty}dt \, e^{i(E_n-E_0)t\hbar}\cdot V_0\cdot\left( \langle\Psi_n(x)|e^{(x_0(t)-x)/\lambda}\cdot\mathbb{1}_{x\ge x_0(t)}|\Psi_0(x)\rangle + \langle\Psi_n(x)|e^{(x-x_0(t))/\lambda}\cdot\mathbb{1}_{x< x_0(t)}|\Psi_0(x)\rangle \right) \right|^{\,2}$

How do I continue? In particular, since the integral is wrt. time and not position, how do I use the two cases $x<x_0$ and $x\ge x_0$ and make a Taylor expansion?

  • $\begingroup$ Do you have any information regarding $H_0$ or the functional form of $\psi_0(x)$ ? $\endgroup$
    – Hans Wurst
    Jul 11, 2021 at 12:00
  • $\begingroup$ Your substitution $|n\rangle\to|\Psi_n(x)\rangle$ is misleading; recall $|n\rangle=\int dx |x\rangle\langle x|n\rangle=\int dx |x\rangle \Psi_n(x)$ and you will see that position is very relevant. This may lead you to the overlap formula $\langle n|V|0\rangle=\int dx \Psi_n^*(x) V(x) \Psi_0(x)$... $\endgroup$ Jul 11, 2021 at 18:12
  • $\begingroup$ @HansWurst No, no information on those is given but the exact form of $H_0$ shouldn't be important anyway since the $\Psi_n(x)$ are eigenstates of the Hamiltonian. $\endgroup$
    – test123
    Jul 12, 2021 at 7:07

1 Answer 1


The relevant integral is $$\langle n|V|0\rangle=\int_{-\infty}^\infty dx \Psi_n^*(x) V(x)\Psi_0(x)=\int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x)e^{-|x-x_0(t)|}.$$ If we use the variable $\omega_n=(E_n-E_0)/\hbar$, the probability becomes \begin{aligned} P&=\frac{1}{\hbar^2}\left|\int_{-\infty}^\infty dt e^{i\omega_n t}\int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x)e^{-|x-x_0(t)|}\right|^2\\ &=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) \int_{-\infty}^\infty dt e^{i\omega_n t}e^{-|x-x_0(t)|}\right|^2\\ &=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) \left(e^{-x}\int_{-\infty}^{\lambda x/v} dt e^{i\omega_n t}e^{v t/\lambda}+ e^x\int_{\lambda x/v}^{\infty} dt e^{i\omega_n t}e^{-v t/\lambda} \right)\right|^2\\ &=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) e^{i\lambda \omega_n x/v}\left(\frac{\lambda}{v+ i\omega_n\lambda}+ \frac{\lambda}{v- i\omega_n\lambda} \right)\right|^2\\ &=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) e^{i\lambda \omega_n x/v}\frac{2\lambda v}{v^2+\omega_n^2\lambda^2}\right|^2\\ &\vdots \end{aligned} This should get you on your way!

  • $\begingroup$ Thanks a lot but when I now try using a Taylor expansion for $\Psi_n^{*}(x)$ and $\Psi_0(x)$ around $x=0$ up to some order $k$, I obtain a linear combination of integrals of the form $\int_{-\infty}^{\infty}dx \, x\cdot e^{i\lambda\omega x/v}$ which aren't even finite. Maybe something else was meant by the particle being "located around $x=0$"? $\endgroup$
    – test123
    Jul 12, 2021 at 7:05
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    $\begingroup$ The question says to Taylor expand the phase factor; nothing is said about expanding the wavefunctions $\endgroup$ Jul 12, 2021 at 13:28
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    $\begingroup$ You might have missed squaring some constants in the denominator but no big deal. In bra-ket form, the overlap is simply $\langle n|x|0\rangle$ - no need to state the the states are functions of $x$ unless you're explicitly using wavefunctions. Further simplification requires knowledge of the eigenstatest themselves (eg in a quantum harmonic oscillator the operator $x$ only connects adjacent eigenstates...) $\endgroup$ Jul 12, 2021 at 15:23
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    $\begingroup$ That's standard wavefunction stuff. The idea is that $\hat{V}$ is an operator that has spatial eigenstates $\hat{V}|x\rangle=V(x)|x\rangle$. Then we can use identities $\int dx |x\rangle\langle x|$ and $\langle x|x^\prime\rangle=\delta(x-x^\prime)$ all over the place to find $\langle n|\hat{V}|0\rangle=\int dx\int dx^\prime \langle n|x\rangle \langle x|V(x^\prime)|x^\prime\rangle\langle x^\prime| 0\rangle=\int dx \Psi_n^*(x) V(x)\Psi_0(x)$ $\endgroup$ Jul 13, 2021 at 13:57
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    $\begingroup$ Then the time dependences can be reinstated after switching pictures, or you can simply use $\langle n|e^{i H_0t/\hbar}=\langle n|e^{i E_nt/\hbar}$ and $e^{-i H_0t/\hbar}|0\rangle=e^{-i E_0t/\hbar}|0\rangle$... $\endgroup$ Jul 13, 2021 at 13:59

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