1) A method for evaluating the energy of helium states

To understand why doubly excited helium is not allowed in nature, we should show, to my eyes, that if we try to build this state we'll find a positive energy, so that at least one electron is pushed to the continuum. I wonder if exists a reasonable approximate way to see that (we don't have exact solution of helium problem). The charming method used in problem 5.11 of Griffiths works well in the ground state, giving -75 eV instead of -79 eV (you can do the calculus and check by seeing pag 92 of Amnov Yariv's Dover book, this method gives a first ionization value of 20,4 eV, in eV numerically equal to $-13,6 \cdot 4 + \frac{e}{4 \pi \epsilon_0} \cdot \frac{5}{4a}$), so it looks well suited to be used in excited state too. The method consists in considering the wave function as the product $\psi_1 \psi_2$ of two wave function of hydrogenoid atom (if $\psi_1 = \psi_2$ we obtain a symmetric wave function, and this choice should be accompanied by considerations about spin, but let's go on), and then in evaluating the energy of the system by calculating (the first term is the sum of the energy of two $Z=2$ hydrogenoid atoms, nothing but $\left< T \right> + \left< U \right> $ but ignoring the presence of the other electron, the second one is the positive interaction energy of the two electrons) $$ 4 ( E_{n_1} + E_{n_2} ) + \frac{e^2}{4 \pi \epsilon_0} \left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right> $$ So mutual repulsion of the electrons is ignored in calculating the wave function (we work with approximate wave functions), but we take it into account in evaluating the energy of the system. It sounds rough but it works well for the ground state so it looks reasonable try to use it in excited states.

2) Calculus of the energy of the helium first double excited state

I'm sorry this part is long, if you are not interested in calculus you can jump to part 3 (the last one, with question): in part 2 i tried to write all in a way that everyone can check using pen and paper (as for me, luckily I exploited the software Maxima).

The wave function of hydrogen in the state $n=2$ & $l=0$ (I'll consider only the s orbital) is $$ \psi = \frac{1}{2 \sqrt{2 \pi a^3}} \left( 1 - \frac{r}{2a} \right) \exp \left( - \frac{r}{2a} \right) $$ In hydrogenoid atoms we have to replace $e^2 \rightarrow Ze^2$ so here we have to replace $a \rightarrow \frac{a}{2}$, finding $$ \psi = \frac{1}{\sqrt{\pi a^3}} \left( 1 - \frac{r}{a} \right) \exp \left( - \frac{r}{a} \right) $$ By supposing that the whole wave function is simply the product of two wave function like that, we have $$ \Psi = \frac{1}{\pi a^3} \left( 1 - \frac{r_1}{a} \right) \left( 1 - \frac{r_2}{a} \right) \exp \left( - \frac{r_1 + r_2}{a} \right) $$ where $r_1$ and $r_2$ are the distance from the origin (the nucleus) of particle 1 and 2 respectively. Now observe that here $\Psi$ and $\hat{Q}=\frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|}$ are real numbers, so we can write $\left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right>$ in this way $$ \int \int \frac{|\Psi|^2}{|\mathbf{r}_1 - \mathbf{r}_2|} d^3 \mathbf{r}_2 d^3 \mathbf{r}_1 $$ In the internal integral, particle 1 coordinates are constants, so we are allowed to orientate the coordinates so that $\theta_1 = 0$ (in the external integral $\theta_1$ will vary, but this is not important, at the moment we are interested in solving the internal integral). Starting from spherical coordinates $\mathbf{r}_i = r_i (\sin \theta_i \cos \phi_i , \sin \theta_i \sin \phi_i, \cos \theta_i )$ it is so easy to obtain that if $\theta_1= 0$ we have $|\mathbf{r}_1 - \mathbf{r}_2| = \sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos \theta_2} $. Considering that the integration of the polar angle of the two particle will give $(2\pi)^2$ we can write $\left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right> $ in this way (in our case $|\Psi|^2$ has spherical symmetry so it isn't function of azimuthal angles and we can carry it out of the inner integral) $$ 4\pi^2 \int_0^\infty \int_0^\pi \int_0^\infty |\Psi|^2 \int_0^\pi \frac{1}{\sqrt{r_1^2 + r_2^2 - 2 r_1 r_2 \cos \theta_2} } \sin \theta_2 d \theta_2 r_2^2 dr_2 \sin \theta_1 d \theta_1 r_1^2 d r_1 $$ Now, the primitive of $\frac{\sin x}{\sqrt{a+b \cos x}}$ is $-\frac{2}{b} \sqrt{a+b \cos x}$, so $$ \int_0^\pi \frac{\sin x}{\sqrt{a+b \cos x}} dx = \frac{2}{b} \left( \sqrt{a+b} - \sqrt{a-b} \right) $$ In our case $a=r_1^2 + r_2^2$ and $b=-2 r_1 r_2$ so, considering that $\sqrt{x^2} \equiv |x|$ and that $r_1 , r_2 > 0$, we have that the internal of the big integrals written above is $\frac{r_1 + r_2 - |r_1 - r_2|}{r_1 r_2}$: so is $\frac{2}{r_1}$ if $r_2 < r_1$ and $\frac{2}{r_2}$ if $r_2 > r_1$. Therefore the internal integral (the one in $\theta_2$) bring us simple functions, but it forces us to break the external integral in two. So we get (let's insert the square of the $\Psi$ written above) $$ \left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right> = \frac{8}{a^6} \int_0^\infty \left( 1 - \frac{r_1}{a} \right)^2 r_1^2 \exp \left( -\frac{2 r_1}{a} \right) \int_0^\pi \sin \theta_1 $$ $$ \left[ \frac{1}{r_1} \int_0^{r_1} \left( 1 - \frac{r_2}{a} \right)^2 r_2^2 \exp \left( -\frac{2 r_2}{a} \right) dr_2 \right. + $$ $$ \left. \int_{r_1}^\infty \left( 1 - \frac{r_2}{a} \right)^2 r_2 \exp \left( -\frac{2 r_2}{a} \right) dr_2 \right] d \theta_1 d r_1 $$ Now observe that $$ \int \left( 1 - \frac{x}{a} \right)^2 x \exp \left( - \frac{2x}{a} \right) dx = - \frac{4x^3 - 2 ax^2+2a^2x+a^3}{8a} \exp \left( - \frac{2x}{a} \right) +c $$ $$ \int \left( 1 - \frac{x}{a} \right)^2 x^2 \exp \left( - \frac{2x}{a} \right) dx = - \frac{2x^4+2a^2x^2+2a^3x+a^4}{4a} \exp \left( - \frac{2x}{a} \right) +c $$ Exploiting wich we can show that (if $a,b>0$) $$ \frac{1}{b} \int_0^b \left( 1 - \frac{x}{a} \right)^2 x^2 \exp \left( - \frac{2x}{a} \right) dx + \int_b^\infty \left( 1 - \frac{x}{a} \right)^2 x \exp \left( - \frac{2x}{a} \right) dx = $$ $$ \frac{a^3}{4b} - \frac{ 2 b^3 + 2 a b^2 + 3 a^2 b + 2 a^3}{8b} \exp \left( - \frac{2b}{a} \right) $$ This can in turn be exploited to resolve the integrals in $r_2$ written above. The integral in $\theta_1$ simply involve a doubling so we get that $\left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right>$ is (in our case $b$ is $r_1$) $$ \frac{16}{a^6} \int_0^\infty \left( 1 - \frac{r_1}{a} \right)^2 r_1^2 \exp \left( -\frac{2 r_1}{a} \right) \left[ \frac{a^3}{4{r_1}} - \frac{ 2 r_1^3 + 2 a r_1^2 + 3 a^2 r_1 + 2 a^3}{8 r_1} \exp \left( - \frac{2 r_1}{a} \right) \right] d r_1 $$ Now we "only" have to solve this ugly integral to obtain the information we are looking for. The primitive of $$ x^2 {{\left( 1-\frac{x}{a}\right) }^{2}} {\exp \left( {-\frac{2x}{a}} \right) } \left[ \frac{a^3}{4x}-\frac{\left( 2x^3+2a x^2+3a^2x+2a^3\right) }{8x} {\exp \left( {-\frac{2x}{a}} \right) } \right] $$ is $$ \frac{256x^6+128a x^5+288a^2 x^4+32a^3 x^3-104a^4 x^2+204a^5x+51a^6}{4096a} \exp \left( {-\frac{4x}{a}} \right) $$ $$ -\frac{4a^2 x^3-2a^3 x^2+2a^4x+a^5}{32} {\exp \left( {-\frac{2x}{a}} \right) } $$ Knowing this we can immediately find that $\left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right> = \frac{77}{256 a}$ (where $a$ is as always the Bohr radius). So the interaction energy $\frac{e^2}{4 \pi \epsilon_0} \left< \frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} \right> $ is about 8 eV: much smaller that I expected. Indeed, if we consider only the interaction of the electrons with nucleus, the energy of the system is $$ E = -4 \cdot 13,6 \left( \frac{1}{n_1^2} + \frac{1}{n_2^2} \right) $$ If $n_1=n_2=2$ we have $E=-2\cdot 13,6$ eV$\approx -27$ eV. Summing the interaction energy we find that the total energy is still decidely negative (-20 eV, roughly speaking): the state is bound.

3) Final comment and question

I don't understand this result: the method works pretty well with the ground state, giving -75 eV instead of the true value -79 eV. So I'm surprised that it doesn't works at all in the first double excited state: this one is not observed, so I expected a positive energy (or at least a negative energy near zero, presumably pushed to a positive value by a more accurate analysis, with effects of nucleus finite mass, relativity, spins, and other effects I still don't know). This strongly negative result suggests that the good result in the ground state is due to a coincidence. I wrong?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.