Consider that the hydrogen atom is placed in a constant electric field $F$ along the negative $z$-direction.

Hamiltonian $$H=H_0+H_1$$where $$H_0=-\frac{1}{2}\bigtriangledown^2-\frac{1}{r}\qquad,\qquad H_1=-Fr\cos\theta\qquad (\text{in a.u.})$$ $H_0~$ is the unperturbed Hamiltonian and $~H_1~$ is a small perturbation.

The second order correction to the ground state energy of Hydrogen is given by $$E^{(2)}_{1s}=\sum_{m, ~(m~\ne ~n)}\frac{|\langle\phi_m^{(0)}|H_1|\phi_n^{(0)}\rangle|^2}{E_n^{(0)}-E_m^{(0)}}\tag1$$

Let there exists an operator $~A~$ such that $$H_1|\phi_n^{(0)}\rangle=(AH_0-H_0A)|\phi_n^{(0)}\rangle\tag2$$

Using $(2)$, from $(1)$ we have $$E^{(2)}_{1s}=\langle\phi_n^{(0)}|H_1A|\phi_n^{(0)}\rangle\tag3$$

Thus we see that determination of the second order correction depends on finding the operator $~A~$ defined by equation $(2)$. It can be shown that the operator $~A~$ depends on the coordinates only, and, in spheriacal polar coordinates, it is given by $$A=-F\left(\frac{r}{2}+1\right)r\cos \theta$$

From here my problem starts. I have no idea how to find this operator $~A~$.

Can any one provide me the full calculation in simple language ?

Thank you in advance.

Edit: If you have another way to find the second order correction to the ground state energy of Hydrogen (Stark effect), then you are also welcome. But please give the full explanation about each and every step.

  • $\begingroup$ There’s no point in showing my work now. Your key confusion, I believe, was in not understanding that the solution for $A$ was only valid for the ground state. And, as I stated, you had a sign mistake, so the proof you were asking for was impossible. $\endgroup$
    – G. Smith
    Commented Aug 26, 2019 at 16:12
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Commented Aug 26, 2019 at 16:14

1 Answer 1


Let's use the convention for the wavefunction of the hydrogen atom as in Wikipedia. We're interested in the second-order correction to the energy of the ground state due to an electric field. In appropriate units the ground state Hamiltonian is $H_0 = -\frac{1}{2}\nabla^2 -\frac{1}{r}$. If we add an electric field in the $z$ direction, the Hamiltonian becomes $H=H_0+H_1$ with $H_1 = Fr\cos{\theta}$. One can choose $F$ positive or negative depending on whether the field is in the positive or negative $z$ direction. The sign choice we make differs from OP's unnumbered equation so that in the end we get $A$ as in the OP's question; if we chose $H_1$ as in the OP's question, then the sign in $A$ would be flipped. I will follow Aruldhas section 9.5.

We're interested in the shift in the ground state, that is the $n=1, l=0$ (and therefore $m=0$) state of the hydrogen atom. We write $\phi^{(0)}_n$ for the $n$-th $l=0$ eigenstate of $H_0$. By direct computation, the first order energy shift vanishes. The second order energy shift is given by $$E^{(2)}_{1}=\sum_{m, m\neq 1}\frac{|\langle\phi_m^{(0)}|H_1|\phi_1^{(0)}\rangle|^2}{E_1^{(0)}-E_m^{(0)}}$$ This should be compared with OP's eq. $(1)$. However this can be written in a different form which can be more suited to our uses. The first order correction to the ground state wavefunction is $$\phi^{(1)}_1= \sum_{m, m\neq 1}\frac{\langle\phi_m^{(0)}|H_1|\phi_1^{(0)}\rangle}{E_1^{(0)}-E_m^{(0)}}\phi_m^{(0)}$$ so that the corrected wavefunction (to first order) would be $\phi^{(0)}_1+\phi^{(1)}_1$. Therefore one can also write the second order energy shift as $$E^{(2)}_{1}=\langle\phi_1^{(0)}|H_1|\phi_1^{(1)}\rangle$$ This formula is not very useful as it is. But now suppose that we could find an operator $A$ such that $$A|\phi_1^{(0)}\rangle = |\phi_1^{(1)}\rangle$$ Then we get $E^{(2)}_{1}=\langle\phi_1^{(0)}|H_1 A|\phi_1^{(0)}\rangle$, which is substantially simpler than evaluating the full infinite sum, at least if $A$ is simple enough. To determine $A$ we can't simply use its definition, because we would need to compute $\phi_1^{(1)}$ and that's as bad as computing the infinite sum. So we take a different route. The eigenvalue equation $H \phi = E \phi$ can be written to first order (not zeroth order) as $$H_1\phi_1^{(0)} +H_0\phi_1^{(1)} =E_1^{(1)} \phi_1^{(0)} +E_1^{(0)}\phi_1^{(1)}$$ But the first order energy shift $E_1^{(1)}$ vanishes, so using $A$ and $H_0$ we get $$(AH_0-H_0A)\phi_1^{(0)} = H_1\phi_1^{(0)}\tag{2'}$$ This should be compared with OP's eq. $(2)$, the main difference being that the relation must only be satisfied for the $n=1$ state (and in fact, as stated, OP's eq. $(2)$ is not true for $n\neq 1$). Note that if $A$ is a solution of $(2')$ and $B$ is any operator which commutes with the hamiltonian, then $A+B$ will also be a solution of $(2')$. Now as a simplifying assumption, suppose that $A$ is a scalar function (i.e. it contains no derivatives). After all, we don't care about a specific solution, any solution will do. Since $A$ is a scalar, it commutes with the $1/r$ part of $H_0$. So expanding $(2')$ using the chain rule we get $$\phi_1^{(0)}\nabla^2 A +2 \nabla A \cdot \nabla \phi_1^{(0)}=2H_1 \phi_1^{(0)}$$ In OP's choice of units and with the Wikipedia convention, $\phi_1^{(0)} = C e^{-r/n}$ where $C$ is some constant. So in fact we get $$\nabla^2 A - 2\frac{\partial A}{\partial r}=2H_1$$ As for how to solve this, it is usually very hard to solve PDEs and in many cases most of it is guesswork. The only $\theta$ dependence in $H_1$ is through $\cos{\theta}$ which is equal to $P_1(\cos{\theta})$, where $P_1$ is a Legendre polynomial. These are eigenfunctions of the laplacian, so we can guess $A = B(r) \cos{\theta}$ and in fact the $\theta$ dependence vanishes, and we get $$\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial B}{\partial r}\right)-2\frac{1}{r^2}B-2\frac{\partial B}{\partial r}=2Fr$$ or expanding $$\frac{\partial^2 B}{\partial r^2}+2\left(\frac{1}{r}-1\right)\frac{\partial B}{\partial r}-2\frac{1}{r^2}B=2Fr$$ which is an equation for $B$ only. We can then look for a series solution $B(r)=F \sum_l a_l r^l$, which gives $A$ as in the OP's question. Alternatively, we see that the coefficients of the equation are singular at $r=0$. Therefore expanding for small $r$ we get the equation $$\frac{\partial^2 B}{\partial r^2}+2\frac{1}{r}\frac{\partial B}{\partial r}-2\frac{1}{r^2}B=0$$ This is an equidimensional equation which has a solution of the form $r^k$. Substituting and solving the quadratic we get $k=1$ or $k=-2$. We choose $k=1$, which is non singular, so in fact $B\approx r$ near zero. It is therefore useful to write $B(r) = r F G(r)$ (we put in $F$ to get rid of it). Substituting we get a simpler equation $$\frac{\partial^2 G}{\partial r^2}+(4-2r)\frac{\partial G}{\partial r}-2G=2r$$ This you can plug into WolframAlpha.They give the solution in a complicated form, but some things cancel out and you can choose the constants to get rid of the nasty part, getting the particular solution $G = -(r/2+1)$.

  • $\begingroup$ You were faster ಥ_ಥ $\endgroup$
    – Olexot
    Commented Aug 26, 2019 at 9:53
  • $\begingroup$ @Olexot I know the feeling! :) $\endgroup$
    – John Donne
    Commented Aug 26, 2019 at 9:55
  • 1
    $\begingroup$ Very thorough explanation! By the way, it seems there is no sign mistake. If you choose the field along the z-axis, which means that you choose $H_1=-Fr\cos\theta$, you'll obtain minus sign in $A$ operator. If you choose the opposite field direction as you did, there will be plus sign $\endgroup$
    – Olexot
    Commented Aug 26, 2019 at 10:08
  • 1
    $\begingroup$ Thanks for your time. @John Donne ~ Let me give some time to cross check the whole matter. Again thank you vary much. $\endgroup$
    – nmasanta
    Commented Aug 26, 2019 at 10:25
  • 1
    $\begingroup$ @Olexot I checked the calculation again and I think the sign does need to be flipped. If you look at $(2')$, it's clear that if you flip the sign of $H_1$ you'll also need to flip the sign of $A$. So we disagree on whether a $+$ or a $-$ in $H_1$ will give rise to a $-$ in $A$. As a check that it's the $+$ that gives rise to the $-$, Aruldhas gets the same sign as in this answer. Moreover, the correction needs to be negative (as is clear from the second order energy shift formula), so $A$ and $H_1$ must have opposite sign $\endgroup$
    – John Donne
    Commented Aug 26, 2019 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.